junior secondary mathematics175_126707893120100225

Prescribed by the National Curriculum and Textbook Board as a Textbook for class VIII from the academic year 1998

JUNIOR SECONDARY MATHEMATICS

(Nimna Madhyamic Ganit) [For class VIII ]

Translated by

Saleh Matin A. S. M. Noman Alam

Edited by

Surujuddin Ahmed

National Curriculum and Textbook Board, Dhaka.

Published by National Curriculum and Textbook Board,

69-70, Motijheel Commercial Area, Dhaka-1000

[ All rights reserved by the Publisher ]

First Edition : 1998 Reprint : 2009

Computer Compose

Ananda Printers

Cover

Md. Abdul Halim

Design

NCTB, Dhaka

For free distribution from academic year 2010 by the Government of Bangladesh

Website Version developed by : MARS Solutions Limited

Preface

Education is the key to development. A progressively improved education system largely determines the pace and the quality of national development. To reflect the hopes and aspirations of the people and the socio-economic and cultural reality in the context of the post independent Bangladesh, new textbooks were introduced in the beginning of the 1980s following the recommendations of the National Curriculum and Textbook Committee.

In 1994, in accordance with the need for change and development, the textbooks of lower secondary, secondary and higher secondary were revised and modified. The textbooks from classes VI to IX were written in 1995. In 2000, almost all the textbooks were rationally evaluated and necessary revision were made. In 2008, the Ministry of Education formed a Task Force for Education. According to the advice and guidance of the Task Force, the cover, spelling and information in the textbooks were updated and corrected.

The study of arithmetic has been limited to class VI and more importance has been given to the study of algebra, so that application and use of mathematics can be made simple and easy. For this reason, the study of algebra has been introduced from class six. It is hoped that students can easily solve mathematical problems through algebraic formulas. It is necessary to develop mathematical skill in the learners so that they can study science in a better manner. Therefore, new techniques and methods has been presented in a simple and concrete way. As a result, learners can themselves solve their problems without depending fully on teachers. Mathematics is a subject where practice is needed. It cannot be memorised. Exercises after each chapter, therefore, includes both traditional practice questions as well as creative questions. This book of Junior Secondary Mathematics for class VIII is the English Version of the original textbook entitled ‘Nimna Madhyamik Ganit’ written in Bangla. We know that curriculum development is a continuous process on which textbooks are written. Any logical and formative suggestions for improvement will be considered with care. On the event of the golden jubilee of the Independence of Bangladesh in 2021, we want to be a part of the ceaseless effort to build a prosperous Bangladesh. In spite of sincere efforts in translation, editing and printing some inadvertent errors and omissions may be found in the book. However, our efforts to make it more refined and impeccable will continue. Any constructive suggestion towards its further improvement will be gratefully considered.

I thank those who have assisted us with their intellect and efforts in the writing, editing and rational evaluation of this book. We hope that the book will be useful for the students for whom it is written.

(Prof. Md. Mostafa Kamaluddin) Chairman National Curriculum and Textbook Board Dhaka

Index

Chapter Content Page

Arithmetic

One Recurring Decimal, Rational and Irrational Numbers 2

Two Percentage, Profit-Loss 37

Three Measurement and Unit 73

Four Statistics 89

Algebra

One Algebraic Formulas and their applications 116

Two Algebraic Fraction 137

Three Linear Equations and their applications 155

Four Simultaneous Linear Equations and their applications. 173

Five Drawing of Graphs and applications 185

Geometry

One Parallelism of Lines, Congruency of Triangles 202

Two Theorems related to Quadrilaterals 218

Three Area 227

Four Problems 239

Answers 251


ARITHMETIC

Chapter -I

Recurring Decimal, Rational and

lrrational Numbers

1.1 Rational and Irrational Numbers

1, 2, 3, 4, --- etc. are natural numbers. The numbers which can be expressed in the

form of a fraction ab (where, a and b are natural numbers), are called rational

numbers. Therefore rational number is a fraction in the form of ab , where a and b

are natural numbers. Each whole number (integer) is a rational number : a = a1 .

The fractions which have equal value are the same rational number of different

type. For example, the fractions 34 , 6

8 , 912 , ------etc. represent the same rational

number.

Again, 12 ,

43 ,

53 ,

61 ,

103 ,

89 , ------etc. are rational numbers. Zero, natural numbers,

proper and improper fractions are rational numbers.

The numbers that cannot be expressed in the form of fractions ab , are all irrational

numbers. For example, 2 , 3 , 5 , 32 , 3

3 , ...... etc are irrational numbers. There are so many irrational numbers, which are not the roots (Square root, cubic root, .......... ) of either natural or rational numbers. All the fractions are rational numbers. Natural numbers are all rational numbers but those are not generally written in the form of fraction. 1.2. Classification of Decimal Fractions

Rational and irrational numbers are all real numbers. Each real number can be expressed in the form of a decimal fraction. There are three types of decimal factions : terminating decimals, recurring decimals and non-terminating decimals. Terminating Decimals : Here the finite number of digits in the right side of a decimal points. For example, 3.45, 7.893, 12.6032, .... etc. are terminating decimals. Recurring Decimals : Here the digits or some of the digits will occur again and again in the right side of the decimal points.

3

For example, 3.333 ......., 8.4567567 ........, 6.4345674567 ......, etc. are recurring decimals. Non-terminating Decimals: Here the digits in the right side of a decimal point never terminate that is the number of digits in the right side of decimal point will not be finite or some of the digits will not occur again and again and digits in the right side will be unlimited. For example. 3.4513681437 ...........,8.26104035 ............, etc. Terminating decimals and recurring decimals are rational numbers and non-terminating decimals are irrational numbers. The value of an irrational number can be determined upto the required number of places after the decimal point. If the numerator and the denominator of a fraction can be expressed in natural numbers, then that fraction is a rational number. Decimal fraction

In class six we have discussed about decimal fractions. Here again those are repeated. In the formation of numbers of base ten is generally used. Here ten's place, hundred's place, ............ etc. occur irrespectively in the left hand side of unit's place and the place value of each digit is ten times the place value of that digit if it occurs immediately in the right side. The digit in right side is less than ten times that of in units place. The digits in hundred's place is ten times that of tens,..........etc. Therefore digits of tens place is one tenth part of digit in hundred's place digits. Units is one tenth of part ten. In the formation of many number digits tens, hundreds,............... etc. are respectively placed in the left side of units place. Similarly tenth, hundredth, thousandths,..............etc. are placed respectively in the right hand side of units place. In order to identify the digit of units place while thousandth, hundredth, tenth, units, tens, hundreds, ...........etc. are all written together, to express a number of point (.) is placed in the right side of units place. This point is called the decimal point. The portion of the number in the left side of decimal point is a whole number (integer) and the right side is a fraction. For this reason number with a decimal point is called decimal fraction. 94.5372 is a decimal fraction. Here 94 is a whole number and .5372 is a fraction. This decimal fraction is read as ninety four point five three seven two. Here 5 is tenth, 3 is hundredth, 7 is thousandth and 2 is ten thousandth. 0.78 is a proper decimal fraction, where the whole number is zero.. Sometimes 0.78 is written as .78. But it is not better to write like this.

4

The value of any number does not change if any number of zeros, without a significant digit is used after the decimal point. Again any number of zeros is used at the left side of a number, then its value does not also change. For example, 12 = 12.000, 734.52 = 734.5200, 25.003 = 25.00300, 12 = 00012 etc. When one, two,.........zeros are placed at the right side of decimal point before a significant digit, then the value will be respectively becomes hundredth, thousandth,........... etc. of its original value. For example, .1 = One tenth of unit. .01 = One hundredth of unit, .001 = One thousandth of unit,..... etc. When decimal point is moved by one, two, three, ..........places to the right side, then the value of decimal increases by 10, 100, 1000 times..........and again if it is moved to the left side by one, two, three, ... places then the value decreases by 10, 100, 1000, ..........times. Therefore when a decimal is multiplied by 10, 100, 1000, ............then the decimal point is to be placed respectively at the right side after one, two,..........digits and when it is divided by 10,100,............ the decimal point is placed respectively before one, two, ............digits at the left side. For example, 17.4 × 10 = 174, 39.478 × 10 = 394.78, 124.6789 ×100 = 12467.89, etc. Again, 78.453 ÷ 10 = 7.8453, 6789.23 ÷ 100 = 67.8923. etc. Those decimals whose values are the same are called equivalent decimal fractions. For example, 12.34, 12.340, 12.3400,...........etc. are equivalent decimal fractions. Again, those decimals which have equal number of digits to the right side of decimal point are called similar decimal fractions. For example, 0.452, 12.467, 8.350, ............etc. are similar decimal fractions. If necessary, by placing required number of zeros at the right side of decimal fractions, two or more decimal fractions can be made similar. Conversion of Simple Fractions into Decimal Fractions To convert a simple fraction into a decimal fraction, the numerator is to be divided by denominator. Beside multiplying denominator by a digit and changing it into 10, 100 or 1000, ..........etc. and then multiplying the numerator by that number, the simple fractions can be converted into decimal fractions. Remark After dividing numerator by denominator if there is any remainder, it will have to be changed into tenth part. At the time of changing into tenth part, the decimal point should be placed in the quotient and zero should be placed at the right side of the remainder. In this way till the process of division is not completed, a zero will have to be placed at the right side of the number that occurs in the remainder etc.

5

and thus the digits of the hundredth, thousandth, ............parts are obtained in the quotient. 1.3 Conversion of Terminating Decimal Fractions into Simple Fractions To convert a digit decimal fraction into a simple fraction, decimal point is first to be to the extreme right side of all digits. Thus the number will be whole number. If there is one digit at the right side of the decimal point i.e. if it is tenth part, then after multiplying by ten, the decimal point will move at the extreme right side. Similarly, if it is hundredth part it is to be multiplied by 100, in case of thousandth part it is to be multiplied by 1000, according the necessity successive multiplication is to be made by 10000, 100000,.......etc. As a result the decimal point can be moved at the extreme right side. But value of the original decimal fraction will remain unchanged when converted into simple fraction if the whole number thus found is divided by the same number by which the decimal fraction is multiplied. This number will be the denominator of the simple fraction and the numerator of the fraction is obtained when the decimal fraction is multiplied by this number, the whole number that in obtained by replacing the decimal point to the extreme right side the decimal point of the decimal fraction.

For example, 112⋅45 ×100 = 11245, 112⋅45 = 112⋅45 × 100100 . After changing it into

lowest terms fraction thus formed is the fraction of 112.45. Again. 112.45 = 112 + tenth of 4 + hundredth of 5

= 112 + 410 +

5100

= 11200 + 40 + 5

100

= 11245100

Remark To convert a terminating decimal fraction into fractions, its denominator is obtained by putting at the right side of 1 the number of zeros equal to the number of digits at the right side of the decimal point and the numerator will be the whole number that is obtained when decimal point is omitted from the decimal fraction. In this way terminating decimal fraction are to be converted into simple fractions. All the terminating decimals are rational numbers.

6

Addition and Subtraction of Terminating Decimal Fractions In the process of addition and subtraction of decimal fractions, those are first made into similar decimal fractions, then the same procedure like the process of addition and subtraction of whole number is followed. Multiplication of Terminating Decimal Fractions Converting decimal fractions into simple fractions, numerator and denominators are respectively multiplied by numerators and denominators and then, is to be changed into decimal fraction, thus the product of decimal fractions is obtained. To multiply decimal fraction of tenth part by the decimal fraction the process of multiplication of whole number is to be applied in case of numbers ignoring decimal point of decimal fraction and then the decimal point is placed in the product in such a way that the product remains as decimal of (tenth × tenth) or decimal of hundredth. To multiply decimal fraction of tenth part by the decimal fraction of hundredth part, the product remains as decimal of (hundredth × tenth) or decimal of thousandth. Similarly multiplication of other decimal fractions are to be made.

For example, 1⋅2 × 3⋅8 = 1210 ×

3810 =

456100 = 4⋅56

∴ 1.2 × 3.8 = 4.56 Again, 3.43 × 1.2 = 4.116 Remark : Here in two numbers there are two digits in the right side of decimal point of the first number and there is one digit in the right side of the decimal point of the second number. Considering the two numbers as whole numbers, multiplication is made and then putting decimal point after (2 + 1) or 3 digits towards the left from the extreme right hand digit of the product, the actual product is found. Rule of Placing Decimal Point in the Product

In the product number of digits in the right side of decimal point = In the multiplier number of digits in the right side of decimal point + In the multiplication the number of digits in the right side of decimal point. Division of Terminating Decimal Fractions : In the process of division of decimal fractions the dividend and the divisor is first changed into simple fractions and then the dividend fraction is divided by divisor fraction and changed into simplest form and then by dividing numerator by denominator, the quotient will be found.

7

To divide 6.48 by 3.6, it is required to divide 648100 by

3610 .

648100 ÷

3610 =

648100 ×

1036 =

95 = 1⋅8

5 ) 9 ( 1⋅8

5

40

40 0

Alternative Method

6⋅48 ÷ 3⋅6 = 6⋅48 × 103⋅6 × 10 =

64⋅836 = 1⋅8

36 ) 64⋅8 ( 1⋅8

36

288

288 0

Remark

To make divisor a whole number the decimal point is required to be towards the right side after some digits then the decimal point in the dividend to be shifted towards then the right side of the same number of digits. Then the quotient is found after dividing numerator by denominator.

Terminating Decimal Numbers

If the simple fraction is converted into decimal fraction, it will be a terminating decimal fraction if and only if denominator of simple fraction has no prime factors other than 2 and 5. In the process of division for converting simple fraction into decimal no remainder will remain at a stage and therefore a definite number of digits after the decimal point in the quotient will exist. Example 1. Express the following fractions into decimal fractions: 627100 ,

245 ,

920 and

4132.

Solution : 627100 = 6.27

8

245 =

24 × 25×2 =

4810 = 4.8

920 =

9 × 520 × 5 =

45100 = 0.45

32) 41(1.28125

32 90 64 260 256

40 32 80 64 160 160

0

∴ 4132 = 1.28125

Ans: 6.27, 4.8, 0.45, 1.28125 Example 2. Express the decimal fractions into simple fractions : 3.57. 12.045. 9.825 and 0.05

Solution: 3.57 = 3.57×100

100 = 357100 = 3

57100

12.045 = 12.045 × 1000

1000 = 120451000 =

2409200 = 12

9200

9.825 = 9.825 × 1000

1000 = 98251000 =

39340 = 9

3340

0.05 = 0.05 × 100

100 = 5

100 = 120

Ans : 357100, 12

9200 , 9

3340 and

120

N.M.G. -3

9

Example 3. Find out the sum of 42.304 + 23.4 + 15.02 + 121.657 + 12

Solution : 42.304

23.400

15.020

121.657

12.000

214.381 (All decimals are made similar and then added) Ans: 2l4.38l Example 4. (a) Subtract 87.7456 from 123.43

(b) Subtract 298.45 from 523.7083

Solution : (a) 123.4300

87.7456

35.6844 (Decimals are made similar and then subtracted)

Ans 35.6844

(b) 523.7083 298.4500 225.2583 (Decimals are made similar and then subtracted) Ans : 225.2583

Example 5. (a) Multiply 43.28 by 22.674 (b) Multiply 123.35 by 2.568

(c) Multiply 3.674 by 0.26

(d) Multiply 0.4325 by 0.16

Solution:

(a) 43.28 × 22.674 = 4328100 ×

226741000 =

98133072100000 = 981.33072

10

Alternative method : 43.28 22.674

17312

302960

2596800

8656000

86560000

981.33072

Ans: 981.33072. (Here there are 2 digits in the right side of decimal point of multiplicand and 3 digits in the right side of decimal point of multiplier. The decimal point is to be placed in the left side of (2 + 3) or 5 digits of the product). (b) 12335 × 2568 = 31676280. ∴ 123.35 × 2.568 = 316.76280. (Here the decimal point is placed at the left side of

(2 + 3) or 5) ∴ 123.35 × 2.568 = 316.7628 (In decimal fractions '0' in the extreme right side is

omitted)

Ans: 316.7628. (c) 3674 × 26 = 95524 ∴ 3.674×0.26 = 0.95524. (Here the decimal point is placed at the left side of (3+2)

or 5 digits)

Ans: 0.95524 (d) 4325 × 16 = 69200 ∴ 0.4325 × 0.16 = 0.069200 [Here the decimal point is to be placed at the left side of (4 + 2) or 6 digits. But the product consists of 5 digits and so the product has been made of 6 digits by placing 0 at the extreme left and then decimal point has been placed.]

Again 0.069200 = 0.0692

∴ 0.4325 × 0.16 = 0.0692.

Ans : 0.0692

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Example 6. (a) Divide 42.64 by 3.2

(b) Divide 1.656 by 0.36 (c) Divide 1209.11 by 4.025

(d) Divide 0.0005 by 0.008

Solution : (a) 42.64 ÷ 3.2 = 42.64

3.2 = 42.64 × 103.2 × 10 =

426.432

32) 426.4(13.325 32 106 96 104 96 80 64 160 160 0

Ans: 13.325

(b) 1.656 ÷ 0.36 = 1.656 0.36 =

1.656 × 100 0.36 × 100 =

165.636

36) 165.6(4.6 144 216 216 0

Ans : 4.6

(c) 1209.11 ÷ 4.025 = 1209.114.025 =

1209.11 × 10004.025 × 1000 =

12091104025

4025)1209110(300.4 12075

16100 16100

0 Ans: 300.4

12

(d) 0.0005 ÷ 0.008 = 0.0005 × 1000 0.008 × 1000 =

0.58

= 0.5 × 108 × 10 = , [

0.0005 × 10000 0.008 × 10000 =

580 ]

80)500(0.0625 or, 480

200 160

400 400

0

8)0.50( 0.0625 48

20 16

40 40 0

Ans : 0.0625 Remark : At present in our country terminating decimals are used for currency, weight, measurement of length and area, etc. Therefore, it is essential to acquire knowledge about addition, subtraction, multiplication and division of terminating decimals. 1.4 Recurring Decimal Fractions

Some times it is found that, when a numerator of a fraction is divided by its denominator to convert it into decimal fraction the process of division is never terminated. In such cases the same number appears again and again or some numbers repeat alternately in the quotient. When in the right side of the decimal point of decimal fractions a digit appears, again and again or more than one digit repeat successively, then these are called recurring decimal fractions. In recurring decimal fractions, the portion which occurs again and again is called recurring part. 3.3333 ------,10.23456456 ------, etc. are recurring decimal fractions. In the first recurring decimal 3 is recurring part and in the second decimal 456 is the recurring part. To indicate presence of the digit (s) of the recurring part a symbol (.) once again recurring point is used upon the digit (s).

For example, 3.333 ------- = 3. ⋅3 and 10.23456456 ....... = 10.23 ⋅45 ⋅6

Remark : When only one digit recurrs, recurring point is placed upon that digit. When more than one digit recurrs, in that case the recurring points are placed upto the first and the last digit.

13

In recurring decimal fractions, if after decimal point there is no other digit except recurring one then it is called pure recurring decimal and in recurring decimal fractions, if there are one digit or more than one digit after decimal point in addition to recurring one, then it is called mixed recurring decimal. For example,

5. ⋅3 is pure recurring decimal and 8.235 ⋅1 ⋅2 is mixed recurring decimal. If there exists prime factors other than 2 and 5 in the denominator of the fraction, the numerator will not fully be divisible by denominator. As the last digit of successive divisions are 1, 2, ........ , 9 and nothing else, so at one stage the same number will repeat in the remainder. The number in the recurring part is always smaller than that of the denominator.

Example 7. Express 13 into decimal fraction.

Solution: 3)1.0( 0.333.............. 9

10 9 10 9 1

Here in the quotient the number after the decimal points is 3 and the remainder is 1. Now if the process of division is continued in each time 3 will occur in the quotient and the remainder will be 1. The process of division will never be completed.

∴ The required decimal fraction = 0.333 ............= 0. ⋅3

Ans : 0. ⋅3


14

Solution: 13) 1.00(0.07692307........ 91__

90 78___ 120 117__ 30 26__ 40 39__

100 91__ 9

∴ The required decimal fraction = 0.07692307..........= 0. ⋅07692 ⋅3

Ans : 0. ⋅07692 ⋅3


Solution: 26) 101 (3.88461538461 78_ 230 208__ 220

208__ 120 104__ 160 156__ 40 26__ 140 130__ 100 78__ 220

208__ 120 104__ 160 ____156__ 40 26__ 14

15

∴ The required decimal fraction = 3.8 ⋅84615 ⋅3

Ans : 3.8 ⋅84615 ⋅3

Conversion of Recurring Decimal into Simple Fraction

Example 10. Express 0. ⋅3 into simple fraction.

Solution : 0. ⋅3 = 0.3333 .................

0. ⋅3 × 10 = 0.333........ × 10 = 3.333 .................

and 0. ⋅3 × 1 = 0.333......... × 1 = 0.333 .................

By subtracting, 0. ⋅3 × 10 − 0. ⋅3 ×1 = 3 or, 0. ⋅3 × (10 − 1) = 3 or, 0. ⋅3 × 9 = 3

Therefore, 0. ⋅3 = 39 =

13

Ans : 13

Example 11. Express 0. ⋅2 ⋅4 into simple fraction.

Solution : 0. ⋅2 ⋅4 = 0.24242424 ..............

Therefore, 0. ⋅2 ⋅4 × 100 = 0.242424 .......... × 100 = 24.2424...............

and 0. ⋅2 ⋅4 × 1 = 0.242424 ........... × 1 = 0.242424 ...............

By subtracting, 0. ⋅2 ⋅4 × 99

∴ 0. ⋅2 ⋅4 = 2499 =

833

Ans. 8

33

Example 12. Express 5.1 ⋅34 ⋅5 into simple fraction.

Solution: 5.1 ⋅34 ⋅5 = 5.1345345345..........

Therefore, 5.1 ⋅34 ⋅5 × 10000 = 5.1345345........... × 10000 = 51345.345...........

and 5.1 ⋅34 ⋅5 × 10 = 5.1345345 .......... × 10 = 51.345.................

16

By subtracting, 5.1 ⋅34 ⋅5 × 9990 = 51294

∴ 5.1 ⋅34 ⋅5 = 512949990 =

85491665 = 5

2241665

Ans : 5 2241665

Example 13 (a). Express 42.34 ⋅7 ⋅8 into simple fraction.

Solution : 42.34 ⋅7 ⋅8 = 42.34787878 ............................

Therefore, 42.34 ⋅7 ⋅8 × 10000 = 42.347878........× 10000 = 423478.7878 .......

and 42.34 ⋅7 ⋅8 × 100 = 42.347878 .......× 100 = 4234.7878...........

By subtracting, 42.34 ⋅7 ⋅8 × 9900 = 419244

∴ 42.34 ⋅7 ⋅8 = 4192449900 =

34937825 = 42

287825

Ans: 42 287825

Explanation : From the examples given above, it appears that, * The recurring decimal has been multiplied by the number formed by putting at

the right side of 1 the number of zeros equal to the number of digit in the right side of decimal point of the recurring decimal.

* The recurring decimal has been multiplied by the number formed by putting at the right side of 1 the number of zeros equal to the number of digits which are nonrecurring after decimal point of the recurring decimal.

* Second product has been subtracted from the first product. By subtracting second product from first product the whole number has been obtained at the right side. Here it is observed that, the number of non recurring part has been subtracted from the number obtained by removing the decimal and recurring points of recurring decimal fraction.

* The result of subtraction that is obtained above which is a whole number has been divided by the number formed by writing the same number of 9 equal to the number of digits of recurring part and number of zeros equal to the number

of digits of nonrecurring part (all 9's at the left and all 0's at the right has been written in the formation of number).

N.M.G. -4

17

* The denominator of the fraction that is obtained consists of the number of 9 which is equal to the number of digits in the recurring part and number of zeros equal to the number of digits in the nonrecurring part, all these zeros are placed in the right side of all 9's. And the numerator of the fraction in the result that is obtained by subtracting the number of the digits formed by omitting the digits of recurring part from the number formed by removing the decimal and recurring points of recurring decimal.

Remark : Any recurring decimal can also be converted into a simple fraction. All recurring decimals ate rational numbers.

Example 13 (b). Express 5.23 .4 5

.7 into simple fraction.

Solution : 5.23 .

4 5.7 = 5.23457457457 ..............

Therefore, 5.23 .

4 5.7 × 100000 = 523457.457457

and 5.23 .4 5

.7 × 100 = 523.457457

By subtracting, 5.23 .

4 5.7 × 99900 = 522934

∴ 5.23 .

4 5.7 =

52293499900 =

26146749950

Ans : 26146749950

Explanation : In the decimal part there are 5 digits, recurring decimal has been multiplied first by 100000 (5 zeros at the right side of 1). As there are two digits at the left side of recurring part in the decimal portion, so the recurring decimal has been multiplied by 100 (two zeros at the right side of 1). Second product has been subtracted from the first product. In one side of the result of subtraction is a whole number and at the other side the result is (100000−100) = 99900 times the value of the given recurring decimal. Dividing both sides by 99900, the required fraction is obtained. Rule of Transfomation of Rucurring Decimals into Simple Fractions

Numerator of the required fraction = The result obtained by subtracting the number formed by omitting the digits of recurring part and decimal point of the given recurring decimal form. The number found by removing decimal point of the given recurring decimal.

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Denominator of the requried fraction = Number formed by putting the number of 9 equal to the number of digits in the recurring part of given recurring decimal at the left and the number of zeros equal to the number of digits in the nonrecurring part. Now the above rule is directly applied to express some recurring decimals into simple fractions.


Solution : 45.2 ⋅34 ⋅6 = 452346 – 452

9990 = 4518949990 =

2259474995 = 45

11724995

Ans : 45 11724995

Example 15. Express 32. ⋅56 ⋅7 into simple fraction.

Solution : 32. ⋅56 ⋅7 = 32567 – 32

999 = 32535999 =

3615111 =

120537 = 32

2137

Ans : 32 2137


Solution : 0.6 ⋅3 ⋅4 = 634 – 6

990 = 628990 =

314495

Ans: 314495

Example. 17. Express 2.63 ⋅7 into simple fraction.

Solution : 2.63 ⋅7 = 2637 – 263

900 = 2374900 =

1187450 = 2

287450

Ans : 2 287450


Solution : 1.53 ⋅2 ⋅6 = 15326 – 153

9900 = 151739900 = 1

52739900

Ans : 1 52739900

19

Similar Recurring Decimals If the number of digits in the nonrecurring part of recurring decimals are equal and also the numbers of digits in the recurring part are equal, then those are called similar recurring decimals. Other recurring decimals are called non-similar

recurring decimals. For example, 0.3 ⋅45 ⋅6 and 12.6 ⋅79 ⋅3 ; 12. ⋅4 ⋅5 and 6. ⋅3 ⋅2; 9.45 ⋅3 and

125.89 ⋅7 are similar recurring decimals.

Again, 0.3 ⋅45 ⋅6 and 7.45 ⋅78 ⋅9 ; 6.43 ⋅5 ⋅7 and 2.89 ⋅34 ⋅5 are non-similar recurring decimals. The Rule of Changing Non-Similar Recurring Decimals into Similar Recurring Decimals:

The value of the recurring decimal is not changed, if the digits of its recurring part are written again and again. For example:

6.45 ⋅3 ⋅7 = 6⋅45 ⋅373 ⋅7 = 6⋅453 ⋅7 ⋅3 = 6⋅4537 ⋅3 ⋅7. Here each one is recurring decimal, 6.45373737....... is non-terminating decimal. It will be seen that each recurring decimal if converted into a simple fraction has the same value.

64537 – 645

9900 = 6453737 – 645

999900 = 645373 – 6453

99000 = 6453737 – 64537

990000

or, 638929900 =

6453092999900 =

63892099000 =

6389200990000

Multiplying by 101 the numerator and denominator of the frist fraction, the numerator and denominator of the second fraction is obtained. To get the necessary numbers of digits of the recurring and nonrecurring parts, the recurring point can be transfered in such a way that the value of the recurring decimal is not changed. In order to make the recurring decimals similar, number of digits in the nonrecurring part of each recurring decimal is to be made equal to the number of digits of nonrecurring part of that recurring decimal in which greatest number of digits in the nonrecurring part exists and the number of digits in the recurring part of each recurring decimal is also to be made equal to the lowest common multiple of the numbers of digits of recurring part of recurring decimals.

Example 19. Convert 5. ⋅6, 7.3 ⋅4 ⋅5 and 10.78 ⋅42 ⋅3 into similar recurring decimals.

20

Solution : The numbers of digits of nonrecurring parts of 5. ⋅6, 7.3 ⋅4 ⋅5 and 10.78 ⋅42 ⋅3 are respectively 0,1 and 2. Here the number of digits in the nonrecurring part

occurs in 10.78 ⋅42 ⋅3 and that number is 2. Therefore to make the recurring decimals similar the number of digits in the nonrecurring part of each recurring decimal is to be made 2. Again the numbers of

digits to recurring part of 5. ⋅6, 7.3 ⋅4 ⋅5 and 10.78 ⋅42 ⋅3 are respectively 1,2 and 3. The lowest common multiple of 1,2 and 3 is 6. So the number of digits in the recurring part of each recurring decimal would be 6 in order to make them similar.

So, 5. ⋅6 = 5.66 ⋅66666 ⋅6, 7.3 ⋅4 ⋅5 = 7.34 ⋅54545 ⋅4,

10.78 ⋅42 ⋅3 = 10.78 ⋅42342 ⋅3.

Ans : 5.66 ⋅66666 ⋅6, 7.34 ⋅54545 ⋅4, 10.78 ⋅42342 ⋅3. Remark : In order to make the terminating fractions similar the required ; number of zeros is placed after the digits at the extreme right of decimal point of each decimal fraction and thus the number of nonrecurring part in each decimal after the decimal point has been made equal. Again in recurring decimals the numbers of digits or nonrecurring part of decimals after the decimal points are equal and the numbers of digits of recurring parts made equal by using the recurring digits. Beginning from any digit after all the digits of nonrecurring part the recurring part can be taken. Addition and Subtraction of Recurring Decimals

In the process of addition or subtraction of recurring decimals, the recurring decimals are to be made similar. Then the process of addition or subtraction as that of terminating decimals are followed. If addition or subtraction of terminating decimals and recurring decimal together are done, then in order to make recurring decimals similar the number of digits of nonrecurring part of each recurring decimals should be equal to the number of digits between the numbers of digits after the decimal points of terminating decimals and that of the nonrecurring parts of recurring decimals. The number of digits of recurring part of each recurring decimal will be equal to L. C. M. as obtain by applying the rule stated earlier and in case of terminating decimals, necessary numbers of zeros are to be used in its recurring parts. Then the same process of addition and subtraction as done in case of terminating decimals are to be applied. The sum or the difference obtained in

21

this way will not be actual one. It should be observed that in the process of addition of similar decimals if any number is to be earring over after adding the digits at the extreme left of the recurring part of the decimals then that number is added to the sum obtained and thus the actual sum is found. In case of subtraction the number to be carried over is to subtract from the difference obtained and thus actual result is found. The sum or difference which is found in this way is the required sum or difference. Remark : (a) The sum or difference of recurring decimals is also a recurring decimal. In this sum or difference of the number of digits in the nonrecurring part will be equal to the number of digits in the nonrecurring part of that recurring decimal which have the highest number of digits in its nonrecurring part. Similarly, the number of digits in the recurring part of the sum or the result of subtraction will be the equal to L.C.M. of the numbers of digits of recurring parts of recurring decimals. If there is any terminating decimal, the number of digits in the nonrecurring part of each recurring decimal will be equal to the highest numbers of digits that occurs either in the nonrecurring parts of recurring decimals or in the terminating decimals after the decimal point. (b) Converting the recurring decimals into simple fractions, addition and subtraction may be done according to the rule as used in case of simple fractions and then the sum or difference is converted into decimal fractions. But this process needs more time.

Example 20. Convert 1.7643, 3. ⋅2 ⋅4 and 2.78 ⋅34 ⋅6 into similar recurring decimals. Solution : In 1.7643 the number of digits in the nonrecurring part means 4 digits after decimal point and here there is no recurring part.

In 3. ⋅2 ⋅4 the number of digits in the recurring and nonrecurring parts are respectively 0 and 2.

In 2.78 ⋅34 ⋅6 the number of digits in the recurring and nonrecurring parts are respectively 2 and 3. The highest number of digits in the nonrecurring parts is 4 and the L.C. M. of the numbers of digits in the recurring parts i. e. 2 and 3 is 6. The numbers of digits in the recurring and nonrecurring parts of each decimal will be respectively 4 and 6.

∴ 1.7643 = 1.7643 ⋅00000 ⋅0; 3. ⋅2 ⋅4 = 3.2424 ⋅24242 ⋅4; 2.78 ⋅34 ⋅6 = 2.7834 ⋅63463 ⋅4.

Ans: 1.7643 ⋅00000 ⋅0, 3.2424 ⋅24242 ⋅4 and 2.7834 ⋅63463 ⋅4.

22

Example 21. Add : 3. ⋅8 ⋅9, 2.1 ⋅7 ⋅8 and 5.89 ⋅79 ⋅8. Solution : Here the number of digits in the nonrecurring part will be 2 and the number of digits in the recurring part will be 6 which is L. C. M. of 2, 2 and 3. At first three recurring decimals are made similar.

3. ⋅8 ⋅9 = 3.89 ⋅89898 ⋅9

2.1 ⋅7 ⋅8 = 2.17 ⋅87878 ⋅7

5.89 ⋅79 ⋅8 = 5.89 ⋅79879 ⋅8 11.97576574 + 2

11.97 ⋅57657 ⋅6

[ 8 + 8 + 7 + 2 = 25. Here 2 is the number to be carried over. 2 of 25 has been added].

∴ The required sum = 11.97 ⋅57657 ⋅6 or, 11.97 ⋅57 ⋅6.

Ans 11.97 ⋅57 ⋅6. Remark : In the sum the number in the recurring part is 576576. But the value is not changed if 576 is taken as the number of recurring part. Note : To make clear the concept of adding 2 at the extreme right side this addition is done in another method:

3. ⋅8 ⋅9 = 3.89 ⋅89898 ⋅9|89

2.1 ⋅7 ⋅8 = 2.17 ⋅87878 ⋅7|87

5.89 ⋅79 ⋅8 = 5.89 ⋅79879 ⋅8|79

11.97 ⋅57657 ⋅6|55

Here the number is extended upto 2 more digits after the completion of recurring part. The additional digits are separated by drawing a vertical line. Then it is added 2 has been carried over from the sum of the digits at the right side of the vertical line and this 2 is added to the sum of the digit at the left side of the vertical line. The digit in the right side of the vertical line is the same as the digit from which recurring point begins. Therefore both the sums are the same.

23

Example 22. Add : 8.9 ⋅47 ⋅8, 2.346 and 4. ⋅7 ⋅1

Solution : To make the decimals similar, the number of digits of nonrecurring parts would be 3 and that of recurring parts would be 6 which is L. C. M. of 3 and 2.

8.9 ⋅47 ⋅8 = 8.947 ⋅84784 ⋅7

2.346 = 2.346 ⋅00000 ⋅0

4. ⋅7 ⋅1 = 4.717 ⋅17171 ⋅7 16.011019564 + 1

16. 011 ⋅01956 ⋅5 [8 + 0 + 1 + 1 =10. Here the digit in the second place on the left is 1 which is to be carried over. Therefore 1 of 10 is added.]

∴ The required sum = 16. 011 ⋅01956 ⋅5.

Ans: 16. 011 ⋅01956 ⋅5.

Example 23. Subtract 5.24⋅67

⋅3 from 8.2

⋅4⋅3.

Solution : Here the number of digits in the nonrecurring part would be 2 and that of recurring part is 6 which is L.C.M of 2 and 3. Now making two decimal numbers similar subtraction is done.

8.2⋅4⋅3 = 8.24

⋅34343

⋅4

5.24⋅67

⋅3 = 5.24

⋅67367

⋅3

2.99669761 [ Subtracting 6 from 3,1 is to be carried over.]

– 1

2.99⋅66976

⋅0

Ans : 2.99⋅66976

⋅0

Remark : If the digit at the beginning place of recurring point in the number from which deduction to be made is smaller than that of the digit in the number which is to be deducted then 1 is to be subtracted from the extreme right hand digit of the result of subtraction.

24

Note : In order to make the conception clear why 1 is subtracted, subtraction is done in another method as shown below :

8.2⋅4⋅3 = 8.24

⋅34343

⋅4|34

5.24⋅67

⋅3 = 5.24

⋅67367

⋅3|67

2.99⋅66976

⋅0|67

Here both the difference are the same.

Example 24. Subtract 16. ⋅43

⋅7 from 24.45

⋅64

⋅5.

Solution : 24.45⋅64

⋅5 = 24.45

⋅64

⋅5

16. ⋅43

⋅7 = 16⋅43

⋅74

⋅3

8⋅01902 [7 is subtracted from 6,1 is to be carried over. ]

– 1

8⋅01⋅90

⋅1

or, 8⋅⋅01

⋅9

Ans: 8⋅⋅01

⋅9.

Note : 24⋅45⋅64

⋅5 = 24⋅45

⋅64

⋅5/64

16⋅⋅43

⋅7 = 16⋅43

⋅74

⋅3/74

8⋅01⋅90

⋅1/90 or, 8⋅

⋅01

⋅9

Example 25.

(a) Subtract 10⋅418 from 13⋅12⋅78

⋅4

(b) Subtract 8⋅2⋅43

⋅8 from 24⋅435

Solution :

(a) 13⋅12⋅78

⋅4 = 13⋅127

⋅84

⋅7

10⋅418 = 10⋅418⋅00

⋅0

2⋅709⋅84

⋅7

Ans : 2⋅709⋅84

⋅7

N.M.G. -5

25

(b) 24⋅435 = 24⋅435⋅00

⋅0

8⋅2⋅43

⋅8 = 8⋅243

⋅84

⋅3

16⋅191157 – 1

16⋅191⋅15

⋅6

Ans : 16⋅191⋅15

⋅6.

Multiplication and Division of Recurring Decimals: Converting recurring decimals into simple fraction and completing the process of their multiplication or division, the simple fraction thus obtained when expressed into a decimal fraction will be the product or quotient of the given recurring decimals. In the process of multiplication or division amongst terminating and recurring decimals the same method is to be applied. But in the case of division, if both the divisor and dividend are recurring decimals and those are made similar, then the process of division can easily be completed. Remark : The direct method similar to addition or subtraction of recurring decimals is not used for multiplication or division.

Example 26. Multiply 4⋅⋅3 by 5⋅

⋅7

Solution : 4⋅⋅3 =

43 – 49 =

399 =

133

5⋅⋅7 =

57 – 59 =

529

∴ 4⋅⋅3 × 5⋅

⋅7 =

133 ×

529 =

67627 = 25⋅

⋅03

⋅7

Ans : 25⋅ ⋅03

⋅7

Example : 27. Multiply 0.2⋅8 by 42⋅

⋅1⋅8

Solution : 0.2⋅8 =

28 – 290 =

2690 =

1345

26

42⋅⋅1⋅8 =

4218 – 4299 =

417699 =

46411

∴ The required product = 1345 ×

46411 =

6032495 = 12⋅1

⋅8⋅5

Ans: 12⋅1⋅8⋅5.

Example 28. Multiply 14⋅⋅6 by l⋅2

⋅1⋅8

Solution : 14⋅⋅6 =

146 – 149 =

1329 =

443

1⋅2⋅1⋅8 =

1218 – 12990 =

1206990 =

6755

∴ The required product = 4443 ×

67555

= 26815 = 17⋅8

⋅6

Ans : 17⋅ 8⋅6

Example 29. 2⋅5 × 4⋅3⋅5 × 1⋅2

⋅3⋅4 = how much?

Solution : 2⋅5 = 2510 =

52

4⋅3⋅5 =

435 – 4390 =

39290

1⋅2⋅3⋅4 =

1234 – 12990 =

1222990 =

611495

∴ The required product = 15 21

× 196392 9018

× 611495 =

196 × 6118910 =

1197568910

= 13.44062---- Ans: 13⋅44062

Example 30. Divide 7⋅⋅3⋅2 by 0⋅2

⋅7

27

Solution : 7⋅⋅3⋅2 =

732 – 799 =

72599

0⋅2⋅7 =

27 – 290 =

2590 =

518

∴ 7⋅⋅3⋅2 ÷ 0⋅2

⋅7 =

72599 ÷

518 =

725145

9911 ×

218 51

= 29011 = 26.

⋅3⋅6

Ans : 26⋅ ⋅3⋅6

Example 31. Divide 2⋅⋅271

⋅8 by 1⋅9

⋅1⋅2

Solution : 2⋅⋅271

⋅8 =

22718 – 29999 =

227169999 , 1⋅9

⋅1⋅2 =

1912 – 19990 =

1893990

∴ 2⋅⋅271

⋅8 ÷ 1⋅9

⋅1⋅2 =

227169999 ÷

1893990 =

1222716

9999101 ×

10990

18931 =

120101 = 1⋅

⋅188

⋅1

Ans: 1⋅⋅188

⋅1.

Example 32. Divide 9⋅45 by 2⋅ 8⋅6⋅3

Solution : 9⋅45 ÷ 2⋅8⋅6⋅3 =

945100 ÷

2863 – 28990 =

945189

100102

× 99099

2835

= 1189 × 9933

2 × 28359455

= 3310 = 3⋅3

Ans: 3⋅3

Example 33. Simplify :

4⋅⋅2 – 3⋅1

⋅4

4⋅⋅2 + 3⋅1

⋅4 ÷

2⋅⋅2

1⋅⋅5 of

3⋅5

2⋅⋅2 +

1⋅⋅60

⋅2

5⋅⋅97

⋅2

28

Solution : Given Expression :

=

42 – 49 –

314 – 3190

42 – 49 +

314 – 3190

÷

22 – 29

15 – 19

of

3510

22 – 29

÷

1602 – 1999

5972 – 5999

=

389 –

28390

389 +

28390

÷ ⎝⎜⎛

⎠⎟⎞10

209

1 ×

19

147

of ⎝⎜⎛

⎠⎟⎞7

3510 ×

920

4 +

16019995967999

=

380 – 28390

380 + 28390

÷ 11071

of 963404

+ ⎝⎜⎜⎛

⎠⎟⎟⎞1601

9991 ×

9991

5967

= ⎝⎜⎜⎛

⎠⎟⎟⎞97

901 ×

1 90663 ÷

94 +

16015967 =

97663 ×

49 +

16015967

= 3885967 +

16015967 =

19891

59673 =

13 = 0⋅

⋅3

Ans : 0⋅⋅3

Remark: The product of recurring decimals, may be a recurring decimal or not.

For example, 1⋅⋅3 × 2⋅

⋅9⋅9 = 4. The quotient of recurring decimals may also be a

recurring decimal or not. 1.5. Non Terminating Decimals

There are so many decimal fractions in which the number of digits after its decimal point is un-limited, again one or more than one digit does not come successively again and again. Such decimal fractions are called non terminating decimal fractions. For example, 5⋅134248513942307................. is a non terminating decimal number. The square root of 2 is a non terminating decimal. Now if we want to find out the square root of 2 then it will be seen that the number of digits after the decimal point never ends.

29

1) 2 (1⋅4142135 ..................... 1

24 100 96

281 400 281

2824 1190011296

28282 60400 56564

282841 383600282841

2828423 100759008485269

28284265 159063100 141421325

17641775

If the above process is continued for ever then it will never ends. The value upto the definite number of decimal places/approximate value upto some decimal places

It is required to find out the value of such non terminating decimals upto definite number of decimal place. Sometimes it is asked to find out the approximate value upto a definite number of decimal places. But the meaning of both is not the same. The value of the decimal, 5.4325893 ------- upto four decimal places will be 5.4325, but the approximate value of the decimal, 5.4325893-------upto four decimal places will be 5.4326. Here the value upto 2 decimal places and the approximate value upto 2 decimal places are the same. This value is 5.43. In this way the approximate terminating decimals can also be found out. Remark : When it is wanted to find out the value upto some decimal places, then the digits that occur in those places are to be written without any alteration of those digits.

03

Example 34. Find out the square root of 13 and write down the approximate value upto 3 decimal places. Solution :

66 396 400

7205 40000 36025

72105 397500360525

3)13(3.605551---------- 9

721105 36975003605525

7211101 9197500 7211101

1986399

∴ The required square root = 3.605551 ............ ∴ The required approximate value upto 3 decimal places = 3.606. Ans: 3⋅605551................, 3⋅606. Remark : The square roots of those numbers which are not square numbers, i.e. which are not 1,4, 9, 16, 25, ------- or 12, 22, 32, 42, 52-------, these square roots are non terminating decimals, i.e. the numbers are irrational. Example 35. Find out the value and approximate value of 4⋅4623845 ............. upto 1, 2, 3, 4 and 5 decimal places. Solution : The value of 4⋅4623845 upto 1 decimal place is 4⋅4 and approximate value " 1 " " " 4⋅5 Value upto 2 decimal place is 4⋅46 and approximate value upto 2 " " " 4⋅46 Value upto 3 decimal place is 4⋅462 and approximate value upto 3 " " " 4⋅462 Value upto 4 decimal place is 4⋅4623 and approximate value upto 4 " " " 4⋅4624 Value upto 5 decimal place is 4⋅46238 and approximate value upto 5 " " " 4⋅46238

31

i Example 36. Which numbers are rational and which are irrational of the following

(a) 3⋅25 (b) 5 (c) 7 (d) 3⋅2⋅4 (e)

1315 (f) 25 (g) 14

(h) 0 (i)

3456

(j) 5

2 (k) 16517 (l)

218

Solution :

(a) 3⋅25 = 325100 , rational (b) 5 =

51 , rational (c) 7 , irrational, since 7 is not a

square number.

(d) 3⋅2⋅4 =

324 – 3290 =

29290 , rational, (e)

1315 , rational (f) 25 = 5, rational.

(g) 14 = 2 × 7 , irrational. (h) 0 = 0

Any number except zero , rational

(i)

3456

= 34 ÷

56 =

34 ×

65 =

910 , rational. (j)

52 = 5 ×

12 , irrational [... The product

is irrational, when the irrational number multiplied by rational number.]

(k) 16517 , rational (1)

218 =

22 9 =

13 , rational.

Remark : It is noted that, multiplication or division of two irrational number may be rational.

EXERCISE 1 1. Express the following fractions into decimal fractions:

(i) 12 (ii)

34 (iii)

85 (iv) 1

316 (v)

710 (vi) 2

720 (vii) 2

932 (viii)

12161280 (ix) 3

87100

2. Express the following decimal fractions into simple fractions: (i) 0⋅3 (ii) 0⋅25 (iii) 1⋅25 (iv) 3⋅47 (v) 2⋅125 (vi) 2⋅025 (vii) 3⋅675 (viii) 10⋅005 (ix) 12⋅2255

32

3. Find out the sum of the following decimal fractions : (i) 0⋅12, 3⋅02, 2⋅8 (ii) 1⋅3, 3⋅12, 5⋅023, 3 (iii) 4⋅23, 6⋅345, 8⋅8792, 6⋅3458 (iv) 0⋅005, 500, 5⋅505 (v) 1 + 1⋅1 + 1⋅111 + 1⋅001101 (vi) 0⋅0507 + 0⋅507 + 5⋅07 + 50⋅007 + 5 (vii) 7 + 0⋅873 + 0⋅01 + 0⋅0837 4. Subtract the following decimal fractions: (i) 5⋅321 – 2⋅109 (ii) 6⋅01 – 4⋅2301 (iii) 12⋅0145 – 8⋅93 (iv) 6⋅00004 – 0⋅678523 (v) 0⋅001 – 0⋅000635 (vi) 20⋅345 – 12⋅4936 5. Simplify: (i) 15⋅32 – 4⋅689 + 1⋅5639 – 4⋅324 + 6⋅342 (ii) 0⋅2 – 3⋅567 + 6⋅32 – 1⋅6324 – 0⋅0034 – 1⋅3025 (iii) 40⋅0001 – 12⋅001 – 5⋅0001 – 16⋅456 (iv) 2⋅6003 – 3⋅423 + 5⋅5 + 3⋅002 – 5⋅732 6. Find out the product : (i) 4⋅32 × 6 (ii) 6⋅75 × 12 (iii) 40⋅05 × 200 (iv) 0⋅8 × 0⋅8 (v) 5⋅2 × 0⋅8 (vi) 40 × 3⋅7 (vii) 7⋅24 × 5⋅6 (viii) 8⋅32 × 0⋅05 (ix) 0⋅006 × 0⋅025 (x) 34⋅025 × 12⋅8 (xi) 0⋅0306 × 24⋅02 (xii) 0⋅5 × 0⋅5 × 0⋅5 (xiii) 1⋅9 × 1⋅01 × 1⋅001 (xiv) 0⋅02 × 0⋅05 × 0⋅04 (xv) 0⋅2 × 0⋅02 × 0⋅002 × 20 7. Find out the quotient: (i) 12⋅25 ÷ 5 (ii) 432.5 ÷ 25 (iii) 1209⋅6 ÷ 112 (iv) 284⋅26 ÷ 233 (v) 2⋅45 ÷ 54⋅635 (vi) 0⋅00281 ÷ 1⋅405 (vii) 0⋅0006 ÷ 0⋅005 (viii) 4 ÷ 0⋅625 (ix) 84⋅375 ÷ 0⋅0375 8. Simplify: i. (3⋅45 + 2⋅6) × 6⋅34 + 2⋅65 ÷ 1⋅25 –10⋅345 ii. (6⋅27 × 0⋅5) − 0⋅5 of 0⋅75 ÷ 8⋅36 iii. 1⋅175 – 0⋅116 of (1⋅75 ÷ 3⋅5) ÷ { 0⋅1 of (120 ÷ 20) – 0⋅55} iv. 2⋅4 ÷ 0⋅5 of 0⋅5 × 3⋅75 + 3⋅76 ÷ 0⋅16 of 5 9. Express into recurring decimal fractions :

(a) 16 (b)

23 (c)

19 (d)

59 (e)

711 (f) 1

56

(g) 329 (h) 4

1112 (i)

115 (j) 3

815 (k)

1718 (l)

6471980

N.M.G. -6

33

10. Express into simple fractions :

(a) 0 ⋅⋅2 (b) 0⋅

⋅3⋅5 (c) 0⋅

⋅45

⋅6 (d) 0⋅1

⋅3 (e) 1⋅4

⋅2 (f) 3⋅7

⋅8

(g) 5⋅2⋅4⋅3 (h) 8.

⋅89

⋅3 (i) 4⋅0

⋅3⋅2 (j) 6⋅2

⋅30

⋅4 (k) 4.67

⋅8

11. Express into similar recurring decimal fractions :

(a) 1.⋅3, 3.

⋅3 ⋅2 (b) 2.

⋅3, 5.2

⋅3⋅5 (c) 7.2

.6, 4.23

.7 (d) 5.

.7, 8.

⋅3 .4, 6.

⋅24

⋅5

(e) 3.2⋅3, 9.23

⋅4⋅8, 1.2

⋅57

⋅6 (f) 12.32, 2.1

⋅9, 4.32

⋅5⋅6

12. Add:

(a) 0⋅⋅2 + 0⋅

⋅3 (b) 0 ⋅

⋅7 + 0 ⋅

⋅8 (c) 0⋅1

⋅2 + 0⋅24

⋅3 (d) 0⋅4

⋅5 + 0⋅13

⋅4

(e) 8 + 2⋅3⋅4 + 7⋅23

⋅8 (f) 2⋅0

⋅5 + 8⋅0

⋅4 + 7⋅018 (g) 2⋅24

.5 + 5⋅

⋅8⋅9 + 0⋅

⋅34

⋅1

(h) 0⋅00⋅6 + 0⋅

⋅9⋅2 + 0⋅0

⋅13

⋅4 (i) 2⋅

⋅7 + 0 ⋅3

⋅9⋅5 + 5⋅12

⋅74

⋅6

13. Subtract:

(a) 3⋅⋅4 – 2⋅1

⋅3 (b) 5.

⋅1⋅2 – 3⋅4

⋅5 (c) 18.

⋅32

⋅5 – 9⋅

⋅5⋅6

(d) 6⋅4⋅3⋅2 – 2⋅75

⋅34

⋅6 (e) 15 – 12⋅3

⋅4⋅5 (f) 8⋅49 – 5⋅3

⋅5⋅6

(g) 19⋅34⋅5 – 13⋅2

⋅34

⋅9 (h) 7⋅1

⋅60

⋅7 – 5⋅23

⋅4 (i) 9⋅

⋅07

⋅6 – 4⋅234

(j) 4⋅.8⋅5 – 1⋅

⋅469

⋅3 (k) 3⋅8

⋅97

⋅2 – 0⋅00

⋅8⋅9

14. Multiply:

(a) 0⋅⋅3 × 0⋅6 (b) 1⋅1

⋅3 × 2⋅

⋅6 (c) 2⋅

.4 × 0⋅

⋅8⋅1 (d) 7⋅

⋅1⋅8 × 4⋅0

⋅3 (e) 0⋅

⋅0⋅9 × 0⋅7

⋅3

(f) 0⋅75 × 1⋅⋅6 (g) 0⋅6

⋅2 × 0⋅

⋅3 (h) 42⋅

⋅1⋅8 × 0⋅2

.8 (i) 1⋅

⋅2 × 1⋅

⋅1⋅2 × 0⋅0

⋅8⋅1

15. Divide :

(a) 0⋅⋅6 ÷ 0⋅

⋅9 (b) 0⋅

⋅3 ÷ 0⋅

⋅6 (c) 2⋅

⋅4 ÷ 0⋅0

⋅4 (d) 0⋅3

⋅5 ÷ 1⋅

⋅7 (e) 2⋅3

⋅7 ÷ 0⋅4

⋅5

(f) 0⋅7⋅3⋅2 ÷ 0⋅02

⋅7 (g) 12⋅1

⋅8⋅5 ÷ 42⋅

⋅1⋅8 (h) 0⋅

⋅3 ÷ 0⋅75 (i) 55⋅

⋅85

⋅1 ÷ 9⋅

⋅6

(j) 0⋅⋅6 ÷ 0⋅

⋅0⋅9 (k) 0⋅

⋅1⋅6 ÷ 1⋅

⋅3 (l) 1⋅

⋅18

⋅5 ÷ 0⋅

⋅2⋅4

34

16. Find out the square root (upto four decimal places) and write down the approximate values of the square roots upto three decimal places.

(a) 12 (b) 1⋅25 (c) 0⋅⋅5 (d) 0⋅

⋅2⋅5 (e) 1⋅3

⋅4

(f) 7⋅03⋅6 (g) 14⋅3

⋅45

⋅6 (h) 7 (i) 3⋅437 (j) 5⋅1

⋅30

⋅2

17. Write down which are rational and irrational of the following numbers :

(a) 0⋅⋅4 (b) 9 (c) 1⋅35 (d) 11 (e)

713 (f)

63 (g)

87

(h) 2748 (i)

2337

(j) 126315 (k) 12⋅0

⋅3⋅9 (l) 5⋅

⋅63

⋅9

18. Simplify:

(a) (0⋅⋅3 of 0⋅8

⋅3) ÷ (0⋅5 × 0⋅

⋅1) + 0⋅3

⋅5 + 0⋅0

⋅8,

(b) 1⋅5 ÷ 0⋅075 × 3⋅25 ÷ 1⋅3 × 3⋅3 ÷ 2⋅25 – 1⋅3.6

(c) [(6⋅27 × 0⋅5) ÷ {(0⋅5 of 0⋅75) × 8⋅36}] ÷ {(0⋅25 of 0⋅1) × (0⋅75 of 21⋅.3) ×

0⋅5}

(d) 2⋅8 of 2⋅ ⋅2 ⋅7

1⋅ ⋅3 ⋅6 +

4⋅ ⋅4 – 2⋅8 ⋅3

1⋅ ⋅3 + 2⋅62 ⋅9 of 8⋅2

Multiple Choice Questions: [Mark (√) on the correct answer] 1. Which one of the following is a rational number?

(a) 3 8 (b) 2

(c) 3 7 (d) 5

4

2. Which one of the following decimal fractions are similar? (a) 4.37, 43.7 (b) 0.530, 0.817 (c) 12.34, 12.346 (d) 0.20, 20

35

3. Which one of the following fractions can be expressed in to terminating decimal fraction ?

(a) 73 (b)

31

(c) 1641 (d)

135

4. Which one of the following is in recurring decimal form of 338 ?

(a) 42.0 & (b) 402.0 &

(c) 0.2.4.0 (d) 42.0 &&

5. If 74.0 & is converted in simple fraction then which one of the following would be the result?

(a) 9047 (b)

9043

(c) 9943 (d)

9947

6. Which one of the following is the correct value of 4 ÷ 0.125? (a) 0.64 (b) 6.4 (c) 3.2 (d) 32 7. (i) The product of two irrational numbers will be irrational. (ii) 0 is a rational number. (iii) All those numbers which are not square number, the square

root of those numbers are irrational. Which one of the following is correct on the basis of the above

informations? (a) i and ii (b) i and iii (c) ii and iii (d) i, ii and iii Answer the questions (8–10) on the basis of the following

informations:

0..6, 0.

.9, 0.

.5 are three recurring decimal fractions and

988 is a

rational fraction.

36

8. If 9.0 & is expressed in simple fraction then which one of the following would be the result?

(a) 1 (b) 910

(c) 101

(d) 91

9. If 6.0 & is divided by 5.0 & then which one of the following would be the result?

(a) 910 (b) 1.2

(c) 2.1 & (d) 38.0 &

10. Which one of the following is the approximate value of 988 upto

three decimal place? (a) 0.283 (b) 0.284 (c) 0.285 (d) 0.286 CREATIVE QUESTIONS 1. 2031.5593.0,40.8 &&&&& and are three recurring decimal fractions. (a) convert 40.8 & into simple fraction. (b) Add the above three fractions after expressing into similar

recurring decimal fractions.

(c) Find the square root of 5.1.30

.2 upto four decimal place and

find the approximate value of the square root upto three decimal places.

2. 12⋅.18

.5, 42⋅

.1.8 and 0⋅2

.8 are three recurring fractions.

(a) Convert 581.12 && into simple fraction. (b) Divide 581.12 && by 81.42 && (c) Find the product of the three numbers.

3. 2.8 of 72.2 && , 1⋅.3.6, 38.24.4 && − , 962.23.1 && + and 8.2 are some fractions.

(a) Convert 2.8 of 72.2 && into simple fraction. (b) Divide 38.24.4 && − by 962.23.1 && + and then multiply the quotient by 8.2. (c) Divide the result of (a) by 63.1 && . Sum this quotient with the

result of (b) and find the square root of the sum upto two

decimal place.

Chapter II Percentage, Profit - Loss

2.1 Percentage

'Percent' means per hundred, i.e. per one hundred. The word, 'percent' is used to determine the quantity per one hundred. 15 percent means 15 per 100.15 percent, 15

100 or, 0.15 have the same meaning. The symbol (%) is used for percent. In short,

15 percent is written as 15%. Percentage is a fraction. Its denominator is 100 and numerator is the number, calculated as percentage. Percentage is used in as calculating interest, gain and loss, in increasing and decreasing rate of population etc. 23 parts of 100 parts of

Tk. 1 can be expressed in any one of forms Tk. 23100 , Tk. ⋅23 or, 23% of Tk. 1.

Some simple fractions and decimal fractions are expressed as percentages: Simple fraction Decimal fraction Hundredth Percentage

14 0⋅25

25100 25%

12 0⋅50

50100 50%

34 0⋅75

75100 75%

32 1⋅50

150100 150%

185 3.60

360100 360%

Any simple or decimal fraction can be expressed as percentage and also percentage can be expressed either in simple or decimal fraction.

Some percentages are expressed in simple and decimal fractions. Percentage Hundredth Decimal fraction Simple fraction

20% 20100 0⋅20

15

66% 66100 0⋅66

3350

375% 375100 3.75

154 or, 3

34

38

Remark : (a) As improper fractions are greater than 1 so percentages can be greater than

100%. (b) 100% expresses whole of the quantity. 200% or, 300% express twice or thrice

of the whole of the quantity respectively. . Expression of Simple Fractions as Percentages

Example 1 : Express 35 and

78 as percentages.

Solution : 35 =

3 × 10020

51 × 100 = 60100 = 60%

78 =

7 × 10025

82 × 100 = 175

2 × 100 =

1752

100 = 1752 % or, 87

12 %

Ans : 60% and 8712 %

Example 2 : Express 5 2125 and 3

916 as percentages.

Solution :

5 2125 =

14625 =

146 × 1004

251 × 100 = 584100 = 584%

3 916 =

5716 =

57 × 10025

164 × 100 = 1425

4 × 100 =

14254

100 = 1425

4 % or, 356 14 %

Ans : 584% and 35614 %

Expression of Decimal Fractions as Percentages Example 3. Express 0⋅55 and 4⋅36 as percentages.

Solution : 0⋅55 = 0.55 × 100

100 = 55100 = 55%

4.36 = 4.36 × 100

100 = 436100 = 436%

Ans : 55% and 436%

39

Example 4. Express 2⋅678 and 6.0345 as percentages.

Solution : 2⋅678 = 2.678 × 100

100 = 267.8100 = 267⋅8%

6⋅0345 = 6⋅0345 × 100

100 = 603⋅45%

Ans : 267⋅8% and 603⋅45%

Expression of Percentages in Simple Fractions and Decimal Fractions

Example 5. Express 64% and 22416 % in simple fractions.

Solution : 64% = 6416

10025 =

1625

22416 % =

13456

100 = 2691345

6 × 10020 =

269120 = 2

29120

Ans : 1625 and 2

29120

Example 6. Express 2413 % and 256⋅6% in simple fractions.

Solution : 2413 % =

733 % =

733

100 = 73

3 × 100 = 73300

256.6% = 256610 % =

256610 % =

256610100 =

1283 2566105 × 100 =

1283500 = 2

283500

Ans : 73300 and 2

283500

Example 7. Express 56% and 126% in decimal fractions.

Solution : 56% = 56100 = 0.56.

126% = 126100 = 1⋅26.

Ans : 0⋅56 and 1⋅26.

40

Examples 8. Express 12318 % and 56⋅7% in decimal fractions.

Solution : 12318 % =

9858 % =

9858

100 = 123⋅125

100 = 1⋅23125

56⋅7% = 56⋅7100 = 0⋅567

Ans: 1⋅23125 and 0⋅567. Percentage of Numbers

Example 9. (a) What is 5% of 100? (b) What is 8% of 150? (c) What is 22% of 450? (d) What is 25% of 680?

Solution : (a) 5% = 5

100

Therefore, 5% of 100 = 1100 × 5

1001 = 5.

(b) 8% = 8

100

∴ 8% of 150 = 630150 ×

28100204

1

= 12

(c) 22% = 22100

∴ 22% of 450 = 450 × 22100 = 99

(d) 25% = 25100

∴ 25% of 680 = 680 × 25100 = 170

Ans: (a) 5 (b)12 (c) 99 (d) 170.

N.M.G. -7

41

One Number as a Percentage of Another Number

Example 10. (a) What percentage is 5 of 25? (b) What percentage is 24 of 40? (c) What percentage is 37 of 45?

Solution : (a) 5 is 525 of 25.

When the fraction 5

25 is expressed as percentage, it becomes

525 = (

5251

× 1004)% = 20%

Ans : 20%

(b) 24 is 2440 of 40.

When the fraction 2440 is expressed as percentage, it becomes

2440 = (

246

40101

× 10010)% = 60%

Ans : 60% | ■

(c) 37 is 3745 of 45

When the fraction 3745 is expressed as percentage, it becomes

3745 =

⎝⎜⎜⎛

⎠⎟⎟⎞37

459 × 10020 % =

7409 % = 82

29 %

Ans : 8229 %

Number from the Percentage

Example 11. Find the number whose 60% is 48.

42

Solution : 60% = 48

∴ 1% = 4860

∴ 100% = 48124

601551

× 100 20= 80.

Ans : 80. Example 12. Find the number whose 75% is 90. Solution : 75% = 90

∴ 1% = 9075

∴ 100% = 9030

7531

× 4100 = 120

Ans : 120 To Express the Ratio as Percentage

If one of two quantities of the same kind is expressed as a fraction of the other quantity, this fraction is said to be the ratio of their sizes provided that both the quantities are in the same unit.

For example, the ratio of Tk. 2 and Tk. 3 is written as 2 t 3 = 23

The symbol ÔtÕ is the mathematical symbol of ratio. Percentage is also a fraction, whose numerator is the required number calculated as percentage and the denominator is 100. In case of ratio if the denominator is arranged to be 100, then the first quantity can simply be expressed as the percentage of the second. For

example, 15 percent means 15100 i. e., 15 t 100.

Example 13. Express 17 t 25 as a percentage.

Solution : 17 t 25 = 1725 =

17 × 4 10025 1 × 100 =

68100 = 68%

∴ First quantity is 68% of the second quantity. Ans : 68%

43

Example 14. Express 32 t 60 as a percentage.

Solution : 32 t 60 = 3260 =

32 × 10060 × 100 =

1603 × 100 =

1603 % = 53

13 %

∴ First quantity is 5313 % of the second quantity.

Ans : 5313 %.

Example 15. If one quantity is 64% of another quantity, then what is the ratio of two quantities?

Solution : 64% = 64100 =

1625 = 16 t 25

∴ Ratio of the two quantities is 16 t 25 Ans : 16 t 25

Example 16. If one quantity is 12313 % of another quantity, then what is the ratio

of two quantities?

Solution : 12313 % =

3703 % =

3703 × 100 =

3730 = 37 t 30.

∴ Ratio of the two quantities is 37 t 30. Ans : 37 t 30. Some Problems Relating to Percentage

Example 17. There are 450 pupils in a school and 18% of them are girls. How many girls are there in that school?

Solution : 18% of 450 pupils = 450 × 18

100 pupils = 81 pupils.

∴ The required number of girls = 81. Ans : 81 girls. Example 18. Rangan got 576 marks out of 800 marks in the examination. Find his marks in percentage.

Solution : 576 is 576800 of 800.

44

Expressing the fraction, 576800 as percentage, we get

576800 =

⎝⎜⎛

⎠⎟⎞576

800 × 100 % = 72%

∴ The required percentage marks = 72 Ans : 72% Second method :

Out of 800 marks he got 576 marks

∴ " 1 " " " 576800 "

∴ " 100 " " " 576 × 100

800 " or, 72 marks.

∴ The required percentage marks = 72 Ans : 72%. Example 19. 10% of monthly salary of Mr. Zafar was deducted for provident fund and then he received Tk. 2700.00. How much was his monthly salary? Solution : If Mr. Zafar' s monthly salary is Taka 100, then after deducting Taka 10 for provident fund he receives Tk. (100 − 10) or, Tk. 90 When Mr. Zafar receives Tk. 90 then his monthly salary is Tk. 100

∴ " " " " " 1 " " " " " " 10090

∴ " " " " " 2700 " " " " " " 10

100 × 2700300

9091

or, Tk. 3000 ∴ The required monthly salary is Tk. 3000. 2nd method : If Tk. 10 percent is deducted for provident fund, Mr. Zafar receives from his monthly salary Tk. (100 – 10) or, Tk. 90 percent. Now 90% of Mr. Zafar's salary = Tk. 2700

" 1% " " " " = Tk. 270090

" 100% " " " " = Tk. 270090 × 100 or, Tk. 3000

∴ Required monthly salary is Tk. 3000. Ans : Tk. 3000.

45

Example 20. A book of mathematics is sold for Tk. 68 at 85% of the actual price. What is the actual price?

Solution : 85% of the actual price of the book = Tk. 68

∴ 1% " " " " " " " = Tk. 6885

∴ 100% " " " " " " " = Tk. 4

68 × 10020

17185 or, Tk. 80.

∴ Required actual price Tk. 80

2nd method : If the actual price of the book is Tk. 100, then the selling price is Tk. (100 – 15) = Tk. 85. Now if the book's selling price is Tk. 85, actual price is Tk. 100

∴ " " " " " " Tk. 1 " " " Tk. 10085

∴ " " " " " " Tk. 68 " " " Tk. 4

68 × 10020

17185

or, Tk. 80. ∴ Required actual price Tk. 80. Ans : Tk. 80.

Example 21. The mortality rate per annum of a village is 4% of its population. If the total population of that village is 1250, how many people die per annum?

Solution : 4% of 1250 persons = 1250 × 4

100 persons = 50 persons.

∴ Required number of persons = 50. Ans : 50 persons.

Example 22. When the price of tea leaves decreased by 25%, what is the present price of tea leaves per kg. which was Tk. 72 per kg. before decrease?

46

Solution : 25% of Tk. 72 = Tk. 18 72 ×

125

10041

= Tk. 18

∴ Required price = Tk. (72 – 18) = Tk. 54. Ans : Tk. 54.

Example 23. Rajat's monthly income increases from Tk.6600 to Tk. 7260. Find percentage of the increase in his income. Solution : Increase in Rajat's income = Tk. (7260 – 6600) = Tk. 660.

∴ Required precentage increase in income =Tk. 660 × 100

6600 = Tk. 10

Ans : 10%

Example 24. For 15% increase in Bakar's salary it becomes Tk. 5750. What was his salary before?

Solution : Salary of Bakar increases to (100 + 15)% or, 115%. Previous salary of 115% = Tk. 5750

∴ " " 1% = Tk. 5750115

∴ " " 100% = Tk. 250

5750 × 10020

231115 or, Tk. 5000

∴ The required previous salary is Tk. 5000. Ans : Tk. 5000

Example 25. When the price of pulse is increased by 25% then at what rate in the use of pulse be reduced so that there would be no change in the expenditure for use of pulse?

Solution : For 25% increase in the price of pulse it becomes (100 + 25)% or, 125%. In order to keep the expenditure for pulse unchanged that quantity of pulse in the increased price obtained by the amount equivalent to the amount as before is to be used.

After the increase of price be 25% 125 expenditure of previous can provide Tk. 100 use of previous

47

∴ 1 " " " " " Tk. 100125 " " "

∴ 100 " " " " " Tk. 100 × 100

125 " " "

= Tk. 80 " " "

∴ Reduction in use of pulse = (100 – 80)% = 20% That is, 20% use of previous should be reduced for the no change in expenditure. Ans : 20%

Example 26. Mr. Farid paid 15% per annum income tax amounting to Tk. 8250 . What was the annual income of Mr. Farid?

Solution : 15% of Mr. Farid's income = Tk. 8250.

∴ 1% " " " " = Tk. 825015

∴ 100% " " " " = Tk. 2750

8250 × 10020

3115 or, Tk. 55000.

∴ Required annual income = Tk. 55000. Ans : Tk . 55000. Example 27. The population of a town increases by 4% per annum and the population of that town was 2000000 persons. What will be the population of that town after 3 years?

Solution : If the number of persons is 100 at the beginning of the year, number of persons become 104 at the end of the year. Therefore, 100 persons increase to 104 at the end of one year

∴ 1 persons increase " 104100 " " " " " "

∴ 2000000 persons increase at 104100 × 2000000 persons

Similarly, 104100 × 2000000 persons increase to

104100 ×

104100 × 2000000

or, ⎝⎜⎛

⎠⎟⎞104

1002 × 2000000 person at the end of second year.

48

∴ At the end of third year population becomes ⎝⎜⎛

⎠⎟⎞104

1003 × 2000000 persons.

or, 104 × 104 × 104 × 2000000

100 × 100 × 100 persons

or, 2249728 persons. ∴ The required population = 2249728 persons. Ans : 2249728 persons.

EXERCISE 2.1 1. Express as percentages:

(a) 38 (b)

710 (c) 2

625 (d) 4

1116 (e)

56 (f) 2

712 (g) 2

932 (h) 0⋅06

(i) 0⋅567 (j) 4⋅67 (k) 8⋅459 (l) 0⋅0003 (m) 7⋅045 (n) 9⋅5 (o) 3⋅46 (p) 9t10 (q) 7 t 12 (r) 13 t 25 (s) 25 t 32 (t) 65 t 125. 2. Express in simple fractions :

(a) 25% (b) 125% (c) 12356 % (d) 25

23 % (e) 342

37 % (f) 16

23 %

(g) 3114 % (h) 37

12 % (i) 66

23 %.

3. Express in decimal fractions :

(a) 12% (b) 37% (c) 1212 % (d) 8

14 % (e) 123

18 % (f) 115

58 %

(g) 2258 % (h) 14

332 % (i) 42

316 %

4. (a) What is 25% of 124? (c) What is 4% of 250? (e) What is 12% of 550? (g) How many is 20% of 60 pencils? (i) How much is 75% of 3 kg? (k) How much is 80% of 10 kilometers?

(b) What is 20% of 300? (d) What is 10% of 670? (f) What is 32% of 750%? (h) How much 25% of Tk. 150 ? (j) How many is 80% of 600 students?

5. (a) What percentage is 20 of 100? (b) What percentage is 40 of 80? (c) What percentage is 60 of 90? (d) What percentage is 125 of 75? (e) What percentage is 20 pencils of 150 pencils?

N.M.G. -8

49

(f) What percentage is Tk. 25 of Tk. 60? (g) What percentage is 5 kilometers of 8 kilometers? (h) What percentage is 20 kg. of 120 kg? (i) What percentage is Tk. 10 of Tk. 250? 6. (a) Find the number of which 50% is 25. (b) Find the number of which 20% is 64. (c) Find the number of which 60% is 72. (d) Find the number of which 55% is 132. (e) Find the amount of which 16% is Tk. 20 (f) Find the distance in km. of which 30% is 24 km.

(g) Find the weight in kg. of which 6623 % is 160 kg.

(h) Find the amount of which 30% is Tk.70 (i) Find the weight in kg. of which 15% is 45 kg. 7. Percentage of 1st quantity as the 2nd quantity is given below. Find the ratio of

the two quantities: (a) 25% (b) 32% (c) 48% (d) 60% (e) 125% (f) 350%. 8. The monthly salary of Mr. Hanif is Tk. 5000. He saves 24% of his salary per

month. How much does he save in a year? 9. In a certain school number of male students is 55% of the number of pupils. If

the number of male students is 1100, then what is the number of pupils in that school?

10. Mr. Helal's monthly income and expenditure are Tk. 4200 and Tk. 2940 respectively. What is the percentage of expenditure of his income?

11. 15% commission is allowed for buying a book on Mathematics. The actual price of the book is Tk. 120. How much will be needed to buy the book?

12. In the annual examination Rasel obtained 600 out of 800 marks. What is the percentage of his marks?

13. The monthly salary of Mr. Belal is Tk. 4820. He has to deposit 10% of his salary to the provident fund. How much is deposited by him to the provident fund?

14. Mr. Bakar's salary increases from Tk. 4000 to Tk. 4500. Find the rate of increase in his salary.

15. Rahim Miah bought a cow for Tk. 5000 and sold it for Tk. 4000. How much his loss in percent?

50

16. Popy has to pay house-rent at 712 % of his basic pay. After deduction of his

house-rent she gets Tk. 3700 per month. What is her monthly basic pay? 17. The population of a village increased by 8% and become 21600. What was

the population of that village before? 18. Mr. Moin Uddin transferred 12% of his property to his wife, 20% to his son

and the rest of Tk. 8,16,000 to his daughter. What was the value of his whole property?

19. 68% of S.S.C Examinees in a school came out successful. The percentage pass would have been 75% if 14 more students passed. What is the number of the Examinees?

20. The number of male and female students in a school is 900. 4% male students left the school and 5% female students got themselves admitted afresh. Thus there was no change in the original total number of male and female students'. How many female students were in that school before and at present how many male students are there?

21. If 6% commission is allowed the amount to be paid for the price of a book is less by Tk. 15 than the amount to be paid for the price if 5% commission is allowed. What is the actual price of the book?

22. In a pooling booth, a candidate has been elected by getting 55% vote of the voters present. He got 10000 votes more than number of votes of his only competitor. How many voters were present in the pooling booth?

23. In an electoral college 75% voters were present. One of the two candidates got 55% votes of voters present and it was seen that his number of votes was more by 75000 than the number of votes of the other candidate. What was the total number of voters?

24. When the price of tea-leaves is reduced by 20%. Karim can buy tea-leaves of 15 kg. more than previous quantity by Tk. 4500. Find the present and the previous prices of per kg. tea-leaves.

25. When the price of banana is reduced by 1623 % then by Tk.75 five more

bananas can buy. What is the present price of each dozen of banana? 26. When the price of suger is increased by 10 percent, then what

percentage in the use of suger should be reduced so that there would be no change in the expenditure for use of sugar?

51

27. When the price of fish is reduced by 25%, then at what percentage in the use of fish should increased so that there will be no change of expenditure for use of fish in the family?

28. When the tax on tea-leaves is reduced by 10 percent, then at what percentage in the use of tea-leaves is to be increased so that the Government will get tax on tea-leaves at Tk. 8 percent more than the previous rate?

29. In an examination 52% and 42% examinees failed in English and Mathematics respectively. If 17% examinees failed in both the subjects, then at what percentage examinees were successful in both the subjects?

30. In an examination 60% and 50% examinees passed in Bangla and Mathematics respectively. If 40% examinees passed in both the subjects and 60 examinees failed in both the subjects, what is the total number of examinees?

31. In an examination 80% and 60% examinees passed in Bangla and Mathematics respectively and 160 examinees passed in both the subjects. If no body failed in both the subjects, then what is the total number of examinees?

32. The population of a city increases by 8% in each year. At present the population of the city is 50000. What will be the population of the city after two years?

33. In a hostel 25 percent seats are increased in each year. At present there are 512 seats. After 4 years, how many seat will be there?

2.2 Calculation of Interest

When money is deposited in a bank or money is lent to an institution or to a person or invested in a business, an excess sum of money depending on the total amount and period of time is obtained. This excess sum of money is called interest. The sum of money deposited or lent is called the Principal or Capital. The total sum of money received as interest and principal together is called Interest-Principal or Increased Principal. ...Interest + Principal = Amount... The sum of interest on a definite sum of money for a limited period of time is called the rate of interest. Generally sum of interest on Tk. 100 for a period of one year is called the rate of interest. This rate is mentioned as the rate of interest percent per annum. So if in the rate of interest percent yearly or monthly not mentioned in that case interest is to be calculated on the basis of percent per annum. If Tk. 4 is the rate of interest percent per annum, then it is written as 4%

52

interest. When interest is calculated on principal only, then it is called simple interest. The method by which interest of a definite sum of money for definite period of time determined is called calculation of interest. The problems relating to interest is generally solved by unitary method. In the discussion of calculation of simple interest, if three out of four data as principal, interest, rate of interest and time is known, then the 4th one can be found with the help of unitary method. To Find Interest and Amount

Example 1. How much is the interest on Tk. 850 in 5 years at 4% per annum? How much is the amount?

Solution : Interest on Tk. 100 in 1 year is Tk. 4

∴ " " " 1 " 1 " " Tk. 4

100

∴ " " " 1 " 5 " " Tk. 4 × 5100

∴ " " " 850 " 5 " " Tk. 41 × 51 × 170850

1002551

or, Tk. 170

∴ The required interest = Tk. 170 and amount = Tk. (850 + 170) or, Tk. 1020

Ans : Tk. 170 and Tk. 1020.

Example 2. How much is the interest on Tk. 600 in 3 years at 512 % per annum?

How much is the amount?

Solution : Interest on Tk. 100 in 1 year is Tk. 512 or Tk.

112

∴ " " " 1 " 1 " " Tk. 11

2 × 100

∴ " " " 1 " 3 " " Tk. 11 × 32 × 100

∴ " " " 600 " 3 " " Tk. 11 × 3 × 60063

21×1001 or, Tk. 99

∴ The required interest = Tk. 99 and amount = Tk. (600 + 99) or, Tk. 699. Ans : Tk. 99 and Tk. 699.

53

Remark : Amount of interest = rate of interest per annum × time × principal

100

[Time is always to be expressed in years.] Example 3. How much is the interest on Tk. 525 in 1 year 6 months at 12% per annum? How much is the amount? Solution : Here, time = 1 year 6 months = 1 year + 6 months

= 1 year + 12 year = 1

12 year =

32 year

Interest on Tk. 100 in 1 year is Tk. 12

∴ " " " 1 " 1 " " Tk. 12100

∴ " " " 1 " 32 " " Tk.

12 × 3100 × 2

∴ " " " 525 " 32 " " Tk.

312 × 3 × 52521

100251 × 2 or, Tk.

1892 or, Tk. 94⋅50

∴ The required interest = Tk. 94⋅50 and amount = Tk. (525 + 94.50) or, Tk. 619⋅50

Ans : Tk. 94⋅50 and Tk. 619⋅50 To Find the Principal Example 4. How much will be the principal for which interest is Tk. 200 in 10

yerars with rate of interest 212 % per annum?

Solution : Interest on Tk. 100 in 1 year is Tk. 212 or, Tk.

52

∴ " " " 100 " 10 " " Tk. 5 × 10

2 or, Tk. 25

So, if interest is Tk. 25 then principal is Tk. 100

" " " 1 " " " Tk. 10025

" " " 200 " " " Tk. 100 × 200

25 or, Tk. 800

∴ The required amount of principal = Tk. 800. Ans : Tk. 800

54

Remark : Principal = 100 × interest

time × rate of interest per annum

Example 5. How much sum of money at the interest of 612 % per annum in 3 years

4 months will be Tk. 788⋅40 as amount?

Solution : Here, time = 3 years 4 months = 313 years or,

103 years

and amount = Tk. 788⋅40 or, Tk. 788410

Interest on Tk. 100 in 1 year is Tk. 612 or, Tk.

132

∴ " " " 100 " 103 " " Tk.

1321

× 105

3 or, Tk. 653

So, the principal of Tk. 100 in 3 years 4 months amounts to Tk. (100 + 653 )

or, Tk. 3653

If amount is Tk. 3653 then principal is Tk. 100

∴ " " " " 1 " " " Tk. 100 × 3

365

∴ " " " " 788410 " " " Tk.

100102×3×7884108

365731× 101

or, Tk. 648

∴ The required principal = Tk. 648 Ans : Tk. 648

Remark : Principal = 100 × amount

(time × rate) + 100

To Find the Rate of Interest Example 6. How much will be the percentage rate of interest per annum if the

interest on Tk. 700 in 5 years is Tk. 105? Solution : Interest on Tk. 700 in 5 years is Tk. 105

∴ " " " 700 " 1 " " Tk. 1005

55

∴ " " " 1 " 1 " " Tk. 1055 × 700

∴ " " " 100 " 1 " " Tk.

321105 × 100

1

51 × 700

71

or, Tk. 3

∴ The required rate of interest is Tk. 3 Ans : 3%

Example 7. How much will be the percentage rate of interest per annum if Tk. 431 in 8 years becomes Tk. 862 as amount?

Solution : Here, interest on Tk. 431 in 8 years is Tk. (862 – 431) or, Tk. 431. Interest on Tk. 431 in 8 years is 431

∴ " " " 431 " 1 " " Tk. 4318

∴ " " " 1 " 1 " " Tk. 4318 × 431

∴ " " " 100 " 1 " " Tk. 1431 ×

25100

82 × 431

1

or, Tk. 252 or, Tk. 121

2

∴ The required rate of interest is Tk. 12 12

Ans: 12 12 %

Remark : ...Rate of Interest (in percentage) = 100 × interest

time × principal

To Find Time Example 8. In how many years interest on Tk. 175 at 5% per annum will be Tk. 87⋅50?

Solution : Interest = Tk. 87.50 or, Tk. 1752

Interest on Tk 100 in 1 year is Tk. 5

∴ " " " 1 " 1 " " Tk. 5

100

56

∴ " " " 175 " 1 " " Tk. 5 × 7175

1004 or, Tk.

354

∴ The required time = ⎝⎜⎛

⎠⎟⎞175

2 ÷ 354 years =

⎝⎜⎜⎛

⎠⎟⎟⎞5175

21 ×

42

351 years = 10 years.

Ans : 10 years.

Example 9. In how many years Tk. 450 at the interest of 6% per annum will be Tk.558 as amount ?

Solution : Here, interest = Tk. (558 – 450) = Tk. 108 Interest on Tk. 100 in 1 year is Tk. 6

∴ " " " 1 " 1 " " Tk. 6

100

∴ " " " 450 " 1 " " Tk. 6 × 450

100 or, Tk. 27.

∴ The required time = 4108271

years = 4 years.

Ans : 4 years. Second method :

Here, interest = Tk. (558 – 450) = Tk. 108 Interest on Tk. 100 in 1 year is Tk. 6

∴ " " " 1 " 1 " " Tk. 6

100

∴ " " " 450 " 1 " " Tk. 6 × 450

100 or, Tk. 27

On Tk. 450 interest becomes Tk. 27 in 1 year.

∴ " " 450 " " Tk. 1 " 127 year

∴ " " 450 " " Tk. 108 " 1 × 1084

271 years or, 4 years

∴ The required time = 4 years. Ans : 4 years.

N.M.G. -9

57

Remark : (a) Time = 100 × interest

rate × principal

(b) Generally it is considered that 30 days make 1 month, 12 months make 1 year and 365 days make I year . The number of days in a month (Bengali or English) is to be considered according to calendar if time is calculated from a given date to another given date. If a certain month consists of 31 days then 31 days are to be counted for that month. Similary, for months consisting of 28 days or 29 days, 28 days or 29 days are to be counted respectively. Miscellaneous Problems Example 10. How long would it take a certain sum of money to be double itself as amount at 5% per annum?

Solution : Let, the principal be Tk. 100. So, amount will be double of Tk. 100, i. e. Tk. 200 and the interest will be Tk. (200 – 100) or, Tk. 100. Interest on Tk. 100 in 1 year is Tk. 5. Tk. 100 yields an interest of Tk. 5 in 1 year

∴ " " " " " " Tk. 1 " 15 "

∴ " " " " " " Tk. 100 " 1 × 100

5 years or, 20 years

∴ The required time is 20 years. Ans : 20 years. Example 11. Amount of a certain sum of principal becomes Tk. 484 in 3 years and Tk. 540 in 5 years respectively. Find the rate of interest per annum and the principal. Solution : Here, in 5 years amount is Tk. 540 and 3 " " " Tk. 484 By subtracting, interest in 2 years = Tk. 56 (in both the cases the principal being

the same)

∴ " " 1 " Tk. 562

∴ " " 3 " Tk. 56 × 32 or, Tk. 84.

58

In 3 years amount = Tk. 484 " 3 " interest = Tk. 84 ∴ principal = Tk. 400 So, Interest on Tk. 400 in 3 years is Tk. 84

∴ " " " 400 " 1 " " Tk. 843

∴ " " " 1 " 1 " " Tk. 84

3 × 400

∴ " " " 100 " 1 " " Tk. 84 × 1003 × 400 or, Tk. 7

∴ The required rate of interest is Tk. 7 and the principal is Tk. 400. Ans: 7 % and Tk. 400. Example 12. As a consequence for reduction of rate of interest from 7% to 5% the income of Ranjit reduced by Tk.70 in 5 years. What was his capital?

Solution: For Tk. 100 income reduces in 1 year by Tk. (7 – 5) or, Tk. 2 " " 100 " " " 5 " " Tk. (2 × 5) or. Tk. 10

Therefore, if in 5 years time, income is reduced by Tk. 10 for the capital of Tk. 100

∴ " " " " Tk. 1 " " " " Tk. 10010

∴ " " " " Tk. 70 " " " " Tk. 100 × 70

10 or, Tk. 700

∴ The required principal is Tk. 700. Ans : Tk. 700. Example 13. If at the same rate of interest the total sum of interest on Tk. 300 in 4 years and interest on Tk. 500 in 5 years be Tk. 148. What is the rate of interest percent per annum.?

Solution : At the same rate of interest, interest on Tk. 300 in 4 years = At the same rate of interest, interest on Tk. 100 in (3 × 4) years or, 12 years. Again, at the same rate of interest, interest on Tk. 500 in 5 years = interest on Tk. 100 in (5 × 5) years or, 25 years.

59

The interest for both the principals is Tk. 148 (given), sum of interest = Tk. 148 which is the interest on Tk. 100 in (12 + 25) = 37 years. Interest on Tk. 100 in 37 years is Tk. 148

∴ " " Tk. 100 " 1 " " Tk. 14837 or, Tk. 4

∴ The required rate of interest is Tk. 4 Ans : 4%

EXERCISE 2.2

1. How much is the interest on Tk. 650 in 6 years at 7 % per annum?

2. How much is the interest on Tk. 800 in 4 years 3 months at 312 % per annum?

3. How much less will be interest on Tk. 500 in 3 years if the rate of interest is reduced from 6% to 4%?

4. How much is the amount if the principal of Tk.1800 yield interest for 6 years

at 612 % per annum?

5. How much will the amount if the principal of Tk. 1050 yields interest 4 years

at 514 % per annum?

6. How much will be the sum of money if that sum at 4% per annum for 5 years becomes Tk. 900 as amount?

7. How much will be the sum of money if that sum at 334 % per annum for 100

days becomes Tk. 737.50 as amount? 8. How much will be the sum of money if the daily interest be Tk. 1 on that sum

at the rate of interest of 4118 % per annum?

9. At what rate of interest percent per annum will the interest on Tk. 750 in 2 years be Tk.210?

10. At what rate of interest percent per annum will the interest on Tk. 1200 in 3 years be Tk. 216?

11. At what rate of interest percent per annum will the amount in 5 years be three times of any the sum of principal?

12. How much is the sum of principal which becomes Tk. 703 as amount in 9

years at 513 % per annum?

60

13. What is the rate of interest percent if the amount in 10 years be four times of any sum of the principal?

14. In how many years will Tk. 290 at 5% be Tk. 377 as amount? 15. In how many years will Tk. 500 at 6% be Tk. 800 as amount? 16. In how many years will Tk. 425 at 5% per annum be Tk. 510 as amount? 17. What is the time required for yielding interest of Tk. 18 on the principal of

Tk. 96 at 614 % per annum?

18. As a consequence for reduction of rate of interest from 8% to 6% Rahim's income reduced by Tk. 77.50 in 5 years. What was his principal?

19. How much will be the sum of money which becomes Tk. 575.00 as amount in 5 years at the rate of interest by which Tk. 450 be Tk. 504 as amount in 4 years?

20. As a consequence for increase of rate of interest from 8% to 10% Jalil's income increased by Tk. 128 in 4 years. What was his capital?

21. Walid borrowed Tk. 400 for 3 years and Tk. 600 for 5 years and gave a total of Tk. 168 as interest. If the rate of interest of both the cases be the same, then how much was the rate of interest?

22. In how many years will a certain sum of interest on Tk. 700 at 10% per annum be found if that certain sum of interest is found from Tk. 1050 in 5 years at 8% per annum?

23. Interest on Tk. 250 in 6 years at 3% per annum is equal to the interest on a sum of money for 4 years at 5% per annum. What is that sum of money?

24. Amount is Tk. 442 on any sum of principal in 6 years at 5% per annum. In how many years will its amount be Tk. 510?

25. How much sum of money is to be deposited in the bank at the rate of interest of 6% per annum so that in each day a sum of Tk. 60 as interest is found?

26. A sum of principal becomes Tk. 550 as amount after 5 years and interest is 38

of the principal. How much is the principal and rate of interest per annum?

27. The amount for a certain period of time is Tk. 450 and interest is 27 of the

principal. If the rate of interest be 347 % per annum, find the time.

28. The interest on Tk. 500 in 4 years at 6% is equal to the interest on a sum of

money in 412 years at 5%. How much is that sum of money?

29. If in 20 years amount becomes double of any sum of principal, then in how many years will it become thrice of that principal?

61

30. Any sum of principal after 3 years is Tk. 1452 as amount and after 5 years is Tk. 1620 as amount. Find the principal and the rate of interest.

31. In how many years will interest of any sum of money at 1212 % be equal to 34

of that sum of money? 32. Any sum of capital becomes its double as amount after 6 years at a particular

rate per annum. How much will be the sum of money which becomes Tk. 2050 as amount at the previous rate in 4 years? 33. A sum of principal becomes Tk. 560 as amount after 3 years and Tk. 600 as

amount after 5 years respectively. What is the rate of interest? 34. A sum of capital was invested for 3 years. If the rate of interest is Tk. 5 then

what is the part of interest in respect to the principal? 35. A sum of Tk. 750 at the interest of 8% and another sum of Tk. 1250 at 6% were

invested. What is the average percent rate of interest of the total sum of capital? 2.3 Profit and Loss

Transactions of money are made for buying and selling of commodities. Again money is invested for manufacturing commodities. The expenditure which is done for buying or manufacturing is the cost price and the price that is obtained by selling is the selling price. If the selling price is greater than the cost price then it is said that profit has been earned. If the selling price is smaller than the cost price then it is said that loss has been incurred. The amount by which the selling price is greater than the cost price is the amount of profit. The amount by which the cost price is greater than the selling price is the amount of loss. Selling price − cost price = profit or, selling price = cost price + profit. Cost price − selling price = loss or, selling price = cost price − loss. Similarly, cost price = selling price − profit and cost price = selling price + loss. For comparison profit or loss is expressed as percentage. Profit or loss is always calculated in terms of the cost price. Remark : The businessmen or the proprietors of factories determine the actual expenditure by adding rent of buildings, salary of employees, trasportation-cost of goods and wages of labourers etc. With the cost price of goods this actual expenditure is known as investment. This investment is taken as the cost price for determination of profit or loss.

62

Example 1. Hanif bought a watch for Tk. 575 and sold it Tk. 600. What is his profit or loss?

Solution : The cost price of the watch is Tk. 575 and its selling price is Tk. 600. Here, as the selling price is greater than the cost price, so he has earned profit. ∴ Profit = Tk. (600 − 575) = Tk. 25

∴ The required amount of profit = Tk. 25. Ans : Profit Tk. 25

Example 2. Raihan bought a parker-pen for Tk. 300 and sold it for Tk. 260. What is his profit or loss?

Solution : The cost price of the parker-pen is Tk. 300 and its selling price is Tk. 260. Here as the selling price is greater than the cost price, so he has incurred loss. ∴ loss = Tk. (300 − 260) = Tk. 40. ∴ The required amount of loss is Tk. 40. Ans : Loss is Tk. 40. Example 3. 4000 kg. of rice at Tk. 7 per kg. was bought and then 2500 kg. at Tk. 7.15 per kg. and 1500 kg. at Tk. 6.50 per kg. were sold. Thus how much will be profit or loss? Find the percentage of profit or loss. Solution : Cost price = Tk. (7 × 4000) = Tk. 28000

Selling price = Tk. (7⋅15 × 2500) + Tk. (6⋅50 × 1500) = Tk. 17875 + Tk. 9750 = Tk. 27625

Here as the cost price is greater than the selling price, so loss has been incurred. ∴ Loss = Tk. (28000 − 27625) = Tk. 375. ∴ The required amount of loss = Tk. 375

Again in Tk. 28000 loss is Tk. 375

∴ " " " 1 " " Tk. 375

28000

∴ " " " 100 " " Tk. 375 × 100

28000 or, Tk. 7556 or, Tk. 1

1956

∴ The required loss = 1 1956 %

Ans : Amount of loss Tk. 375 and 1 1956 %

63

Example 4 Rashid buys salt at Tk. 7 per kg. and sells at Tk. 7⋅50 per kg. and thus he makes a profit of Tk. 30. How much salt does he buy?

Solution : By selling one kg. of salt profit is Tk. (7⋅50 – 700) = Tk. 0⋅50 or, Tk. 12

Here, total amount of profit is Tk. 30

Profit of Tk. 12 is from 1 kg. of salt

∴ " " " 1 " " 1 × 2 kg " ∴ " " " 30 " " 1 × 2 × 30 kg " or, 60 kg of salt, ∴ The required quantity of salt = 60 kg. Ans : 60 kg. Example 5. Buying one dozen of bananas for Tk. 37⋅50 and then selling them for Tk. 39.75, how much will be the percentage of profit?

Solution : Here, cost price = Tk. 37⋅50 and selling price = Tk. 39⋅75 Profit = Tk. (39⋅75 – 37⋅50) = Tk. 2⋅25

For Tk. 37⋅50 profit is Tk. 2⋅25

∴ " " 1 " " Tk. 2⋅2537⋅50

∴ " " 100 " " Tk. 2.25 × 100

37⋅50 or, Tk. 6

∴ The required profit = 6% Ans : 6% Example 6. Buying 25 oranges by Tk. 100 and then selling 20 oranges for Tk. 100, what will be the percentage of profit? Solution : The cost price of 25 oranges is Tk. 100

∴ " " " " 1 " " Tk. 10025 or, Tk. 4

Again, 20 oranges are sold for Tk. 100

∴ " 1 " " " " Tk. 10025 or, Tk. 5

∴ Profit for one orange = Tk. (5 – 4) or, Tk. 1. So profit in Tk. 4 is Tk. 1

∴ " " " " 1 " Tk. 14

64

∴ " " " " 100 " Tk. 1 × 100

4 or, Tk. 25.

∴ The required profit = 25% Second Method : Suppose, any body buys oranges of Tk. 100. 25 oranges are bought by Tk. 100. Here, 20 oranges are sold by Tk. 100

∴ " 1 " " " " Tk. 10020

∴ " 25 " " " " Tk. 100 × 25

20 or, Tk. 125

So profit in Tk. 100 is Tk. (125 – 100) or, Tk. 25 ∴ The required profit = 25%

Ans: 25%

Example 7. Buying some litchies at the rate of litchies 10 for Tk. 10 and the same number at the rate of litchies 8 for Tk. 10 and then selling all of them at 9 for Tk. 10, how much will be the percentage of profit or loss?

Solution : The L. C. M of 10, 8 and 9 is 360. Suppose, 360 litchies of each kind is bought. 10 litchies are bought by Tk. 10

∴ 1 " is " " Tk. 1010

∴ 360 " are " " Tk. 10 × 360

10 or, Tk. 360.

Again, 8 litchies are bought by Tk. 10

∴ 1 " is " " Tk. 108

∴ 360 " are " " Tk. 10 × 360

8 or, Tk. 450

∴ (360 + 360) = 720 litchies are bought by Tk. (360 + 450) or, Tk. 810 But 9 litchies are sold by Tk. 10

∴ 1 " is " " Tk. 109

∴ 720 " are " " Tk. 10 × 720

9 or, Tk. 800

N.M.G. -10

65

So, the cost price of 720 litchies is Tk. 810 and the selling price is Tk. 800 ∴ Loss = Tk. (810 – 800) or, Tk. 10 ∴ In Tk. 810 loss is Tk. 10

∴ Tk. 1 " " Tk. 10810

∴ Tk. 100 " " Tk. 10 × 100

810 or, Tk. 10081 or, Tk. 1

1981

∴ The required loss 11981 %

Ans : 11981 %

Example 8. Mr, Rafiq buys a car for Tk. 3,50,000. What should be the selling price for a profit of 10%?

Solution : If the cost price be Tk. 100, then selling price is Tk. (100 + 10) or, Tk.110. If cost price be Tk. 100 then selling price is Tk. 110

∴ " " " Tk. 1 " " " " Tk. 110100

∴ " " " Tk. 3⋅50⋅000 " " " Tk. 110 × 350000100 or, Tk. 3,85,000

∴ The required selling price = Tk. 3,85,000. Ans : Tk. 3,85,000

Example 9. A profit of 20% is made by selling a bicycle for Tk. 7200. What is the cost price of the bicycle?

Solution : If the cost price be Tk. 100, then at the profit of 20% the selling price be Tk. (100 + 20) or, Tk. 120. If the selling price is Tk. 120 then the cost price is Tk. 100

∴ " " " " 1 " " " " " Tk. 100120

∴ " " " " 7200 " " " " " Tk. 100 × 7200120 or, Tk. 6000

∴ The required cost price =Tk. 6000 Ans : Tk. 6000.

66

Example 10. A loss of 10% is incurred by selling a bike for Tk. 18000. How much is it sold to make a profit of 12%? Solution : If the cost price be Tk. 100, then the selling price at a loss of 10% is Tk. (100 – 10) or, Tk. 90. Selling price at a profit of 12% = Tk. (100 + 12) or, Tk. 112. The selling price instead of Tk. 90 will be Tk. 112

∴ " " " " " " 1 " " Tk. 11290

∴ " " " " " " 18000 " " Tk. 112 ×18000

90 or, Tk. 22400.

∴ The required selling price = Tk. 22400. Ans : Tk. 22400. Example 11. A cow is sold at a loss of 12%. If the cow can be sold by Tk. 1200 more, then there has been a profit of 8%. What is the cost price of the cow? Solution : If the cost price of the cow be Tk. 100, then the selling price at a loss of 12% is Tk. (100 – 12) or Tk. 88. The selling price at a profit of 8% = Tk (100 + 8) or, Tk. 108.

Excess of the selling price for making profit = Tk. (108 – 88) or, Tk. 20. If excess of selling price be Tk. 20, then the cost price Tk. 100

∴ " " " " " Tk. 1 " " " " Tk. 10020

∴ " " " " " Tk. 1200 " " " " Tk. 100 × 1200

20

or, Tk. 6000. ∴ The required cost price = Tk. 6000. Ans : Tk. 6000. Example 12. Naser buys 8 dozen of eggs at Tk. 30 a dozen and 12 dozen of eggs at Tk. 25 a dozen. At what rate is eggs per dozen sold to make a profit of Tk. 3 a dozen in average? Solution : Cost price of 8 dozen of eggs at Tk. 30 a dozen = Tk. (30 × 8) or, Tk. 240 " " " 12 " " " " " 25 " " = Tk. (25 × 12) or, Tk. 300. By adding, cost price of 20 dozens of eggs = Tk. (240 + 300) or, Tk. 540

67

If a profit of Tk. 3 a dozen is made, total amount of profit for 20 dozen = Tk. (20 × 3) or, Tk. 60 Therefore, 20 dozen eggs are sold for Tk. (540 + 60) or Tk. 600

" 1 " " " " " Tk. 60020 or, Tk. 30

∴ The required selling price = Tk. 30 a dozen. Ans : Tk. 30.

Example 13. 180 tons of rod are manufactured in a steel mill, the raw materials per ton bilet is bought for Tk. 12000. The monthly incidental expenditure of that mill is Tk. 90000. How much is the selling price of per ton rod to make a profit of 10%?

Solution :

To manufacture 180 tons of rod, the cost price of 180 tons of bilet = Tk. (180 × 12000) = Tk. 2160000. The investment for manufacturing 180 tons of rod = Tk. (2160000 + 90000) = Tk. 2250000. " " " " 1 " " " = Tk. (2250000 ÷ 180) = Tk. 12500. To make a profit of 10% rod of Tk. 100 is to be sold by Tk. (100 + 10) or, Tk. 110. Rod of Tk. 100 is to be sold by Tk. 110

∴ " " " 1 " " " " " " 110100

∴ " " " 12500 " " " " " " 110 × 12500

100 or, Tk. 13750.

(... the cost price of per ton rod is Tk. 12500)

∴ The required selling price of per ton rod is Tk. 13750.

Ans : Tk. 13750.

68

EXCERCISE 2.3

1. How much is the sum of profit or loss if Rashid bought a parkcr pen for Tk. 325 and sold it for Tk. 300? 2. Habib buys 50 kg. of rice at Tk. 15 per kg. and sells the whole quantity at Tk. 15.25 per kg. As a result how much is the profit or loss? 3. Taher bought 40 kg. of rice for Tk. 300 and sold them at Tk. 7⋅75 per kg. How

much will be his profit or loss? 4. A box of apples was sold for Tk. 750 at a loss of Tk. 90. If that box is sold for

Tk. 850, how much will be his profit or loss? 5. Lozens are bought at 50 paisa each and sold them at 60 paisa each and for this

deal a total profit of Tk. 2.50 was made. How much lozens are bought? 6. One person buys some quantities of tea-leaves at Tk. 50 per kg. By selling the

whole quantity of tea-leaves at Tk. 45 per kg. he incurrs a loss of Tk. 300. How much were the quantity in kg. of tea-leaves he bought?

7. 80 metres of cloth were bought for Tk. 2000. 60 metres at Tk. 24 per metre and the rest at Tk. 27 per metre were sold. How much was profit or loss?

8. 100 books on Arithmetic were bought for Tk. 4500 and sold them at Tk. 45⋅25 each. How much will be profit or loss? 9. A person bought 140 kg. of pulse at Tk.32 per kg. for his family's

consumption. After 2 months, he finds that 28 kg. of pulse have been consumed. Now at what rate is the remaining quantity of pulse to be sold, so that he gets back the whole amount of cost price?

10. One person exchanged 22 kg. of pulse at Tk. 25 per kg. with 15 kg. of pulse at Tk. 35 per kg. How much was his profit of loss on the deal?

11. One person bought rice and found that the cost price of 20 kg. is equal to the selling price of 25 kg. What was his percentage profit or loss?

12. If an item is bought for Tk. 175 and sold for Tk. 189, then what is the percentage profit or loss?

13. If 20 metres of cloth were sold for the price by which 30 metres of cloth were bought, what was the percentage of profit or loss?

14. One person sells rice at Tk. 20 per kg. to make a profit of 25%. What is the cost price of per kg. rice?

15. One person bought 1000 mangoes at Tk. 250 a hundred and sold half of them at Tk. 300 a hundred. Then half of the rest at Tk. 250 a hundred and the remaining at Tk. 200 a hundred were sold. What was his percentage profit or loss?

69

16. Myrobalans are bought at the rate of 10 per Taka and sold at the rate of 8 per Taka. What will be the percentage profit?

17. If myrobalans are bought at the rate of 15 per Taka, how many is the number of myrobalans are to be sold per Taka to make a profit of 25% ?

18. If a thing is sold for Tk. 252, there is a loss of 16%. If it is sold for Tk. 312, then what will be the percentage profit or loss ?

19. A loss of 15% is incurred by selling a watch for Tk. 612. How much is sum of money by which it is sold to make a profit of 10%?

20. There has been a loss of 25% by selling 8 bananas for Tk. 5. What was the cost price of a dozen of bananas?

21. The selling price of a house is 34 of its cost price. Find the percentage in profit or loss.

22. Potatoes are bought for Tk. 2⋅50 per kg. and sold at a loss of 10%. How much is the selling price of one kg. of potatoes?

23. If 12 things are bought for Tk.10 and sold 8 for Tk. 10. What is the percentage in profit or loss ?

24. If a thing is sold for Tk. 378, there is a loss of certain amount and if it is sold for Tk. 480, then it makes a profit of three times of that certain amount of loss. What is the cost price of the thing?

25. A pen is sold at a loss of 20%. If the selling price would have been Tk. 67⋅50 more, then there will be a profit of 10%. What is the cost price of the pen?

26. 5% of total number of fruits of a fruit-seller has been rotten and also 5% of them has been damaged during transportation. What is the percentage profit by which he sells the rest so that he profit 20% as a whole ?

27. A goat is sold at a loss of 10%; If the selling price is Tk. 45 more, there would have been a profit of 5%. What is the cost price of the goat ?

28. The price of a shirt and a trouser is Tk. 525⋅00. If the prices of a shirt and a trouser are increased by 5% and 10% respectively, then to buy those it costs Tk. 568⋅75. What is the cost price of each of a shirt and a trouser?

29. A person buys some bananas at Tk. 15 per dozen and the same number of bananas at Tk. 10 per dozen and then sells all of them at Tk. 14 per dozen. What is the percentage profit or loss?

30. The cost price of a horse and a cow is Tk. 10000. The horse is sold at a profit of 20% and the cow is sold at a loss of 15% and thus a total profit of 6% is made. What is the price of the horse and the cow separately?

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31. Some bananas at 10 for Tk. 20 and the same number at 15 for Tk. 20 were bought and then sold all of them at 12 for Tk. 20. What will be the percentage in profit or loss ?

32. Some myrobalans at the rate of 15 per Taka and the same number at the rate of 20 per Taka were bought and then sold all of them at the rate of 18 per Taka. What will be the percentage in profit or loss ?

33. Rahim buys 15 dozen of bananas at Tk. 21 per dozen and 20 dozen at Tk. 14 per dozen. What is the selling price of each dozen of banana if on the average he makes a profit of Tk. 5 per dozen?

34. Babul buys 20 dozen of eggs at Tk. 24 per dozen and 12 dozen at Tk. 32 per dozen. What is the selling price of each dozen of eggs if on the average he makes a profit of Tk. 6 per dozen?

35. Kamal buys 100 eggs at Tk. 3 per egg and 400 eggs at Tk. 3⋅25 per egg. If he wants to make 5% profit, then what is the selling price of an egg?

36. 'I'he manufacturer sells goods to a wholesaler at a profit of 20% the wholesaler sells to the retailer at a profit of 20% and the retailer sells to the customer at a profit of 20%. If the customer's price of that goods is Tk. 21.60 then what is the manufacturing cost of goods?

37. While selling a thing a commission of 10% on the written price in the list of things was allowed to make a profit of 20%. What was the percentage in terms of the cost price of the excess sum of money marked in the price list?

38. A businessman has to spend monthly Tk. 2000 as house rent. Tk. 5000 as salary of employees and Tk. 3000 as other expenses. He buys rice of Tk. 5 lacs and then sells in every month. If he wants to make 10% profit per month, what is the selling price of rice of Tk. 5 lacs?

39. In a factory 50,000 bags of cement are manufactured. The incidental charges in a month of that factory is Tk. 80,000 and Tk. 75,00,000 in a month as expenditure for raw materials. If 20% profit is to be made, what is the price of each bag of cement?

Multiple Choice Questions: [Mark (√) on the correct answer] 1. What percent is 30 of 50? (a) 30% (b) 50% (c) 55% (d) 60%

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2. Which one of the following is the conversion of 28t35 in percentage?

(a) 63% (b) 65% (c) 72% (d) 80%

3. Observe the following informations: (i) The 100% means the whole part.

(ii) The conversion of 53 into percent is 60%

(iii) It is used to write 15% for 15 percent.

Which one of the following is correct in view of the above informations?

(a) i and ii (b) ii and iii (c) i and iii (d) i, ii and iii 4. Which one of the following numbers is 65% of 39? (a) 100 (b) 65 (c) 60 (d) 55 Answer the questions (5−6) in view of the following informations: Mr. Rahim deposited Tk. 15,000 at the profit rate 8% per annum. 5. How much Taka would be yearly profit? (a) 1500 (b) 1200 (c) 1000 (d) 800 6. How much Taka would become with profit after 6 years? (a) 7200 (b) 16200 (c) 21200 (d) 22200 7. Buying15 oranges at Tk. 100 and then selling 12 oranges for Tk.100, what will be the percentage of profit or loss? (b) 20% loss (b) 20% profit (c) 25% loss (d) 25% profit Answer questions (8-9) in view of the following informations: Mr. Aziz paid Tk. 1080 for the electricity bill for the month of

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May. A VAT at the rate 15% on the amount of the electricity bill is fixed.

8. How much taka is the VAT? (a) 15 (b) 85 (c) 126 (d) 162 9. How much taka is the price of the unit consumed by Mr. Aziz? (a) 1065 (b) 995 (c) 918 (d) 854 CREATIVE QUESTIONS 1. The population of Gopaldy Union increased to 64800 at the rate of

8%. 75% of the present population of that area are voters.55% of the total voters are female. 75% of the total voters were present at union parisad election. Amongst the two participating candidate one was elected obtaining 60% votes out of the voters presented.

(a) What was the population of union before increased? (b) What percent is the male voters of the total voters at present

of that union? Calculate the numbers of male and female voters.

(c) In what difference of casted votes the elected candidate defeated the other candidate?

2. The retail price of a good is Tk. 30030. The producer gained 4%, whole seller gained 5% and retailer gained 10% by selling the good.

(a) If the retailer sale giving 15% commission on the listed price, then find the list price of the good.

(b) Find the buying price of whole seller. (c) How much more is the retail price than the production cost? 3. A person invested Tk. 750 for profit rate 8% and Tk. 1250 for the

profit rate 6%. (a) Find the profit for the principal Tk. 1250. (b) Find the average rate of profit. (c) In how many years the principal will be doubled?

N.M.G. -11

Chapter III

Measurement and Unit

3.1 Conception regarding measurement and the whole quantity in units and their uses. In our every day life it is required to measure the transactions in buying and selling and in different fields and also to find total population, total number of plants, total number of animals and birds. These are determined by counting or by measurement. Measurement is also used for determining area and volume. Total population, total number of plants and total number of animals and birds are determined by counting. To measure length, to find weight and to find volume of liquid materials, measurement used. There are different kinds of measurements which are used in different fields. Quantity of weight is used for things. But quantities of oranges, bananas, litchies, eggs, lemons etc. are expressed numbers by counting. Either in counting or for measurement unit is essential. For counting the first natural number 1 is unit. Adding with 1 the second number is found. Adding I with 2nd unmber 3rd number is found, proceeding as in the above way all the natural numbers are found. In order to measure the length any line is divided into some equal parts of which each part will be 1 unit. It will be one unit of length, similarly to measure the quantity of weight a body is divided into some pieces of equal quantites of weights of which the weight of one piece will be one unit . This will be one unit of weight. Again the unit of measurement for the volume of liquid materials can similarly be found for measurement of area a piece of paper is divided into some small equal pieces of square sizes. Each small piece of square size is taken as one unit. It will be one square unit. A Hali (a group of 4) or a dozen is used as unit for oranges, eggs, lemons etc. 4 oranges are called one hali of oranges. 12 oranges are called one dozen of oranges. Here a dozen is taken as one unit. So at the time of selling or buying, it is said to be one dozen, two dozens etc. Again for litchies hundred is known as unit, one hundred litchies called 'hundred', 20 fishes are called 'one kuri'. There are different systems for measurement. The unit in each system is different. The different units are also different in respect of their values, but for counting population and number of plants unit is the same. In all the fields an idea regarding the whole quantity with the help of units either in counting or in measurement can

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be obtained. The number of people that live in a village is known by counting. The quantity of rice in a store or in a godown is known by measuring weight. How much water are there in a reservoir is known by measuring the volume of liquid materials. Even for various problems relating to time, there are different units relating to time. 1 is taken as unit for counting in all the countries. But for measurement, there are different units in different countries.

3.2. Measurement in Metric System

As there are different units for measurement for use in different countries, so there is much difficulty in carrying out international trade and transactions. Therefore, for measurement in international trade and transactions metric system is used. The speciality of this measurment is that calculation is made in base ten. Measurement in this system can easily be expressed by decimal fractions. This system was first introduced in France in the Eighteenth century. The unit for measuring length is a metre. Millionth of longitudinal length from the north pole of the earth to the equator is considered to be one metre. Subsequently, this measurement of length was not considered to be convenient, so the length of a piece of rod made of platinum kept in the museum of Paris had been accepted as one metre. This length is taken as a unit for linear measurement. If the length is small, then it is taken as centimetre and if it is large, then it is expressed in kilometre. The length of the small line-segments are written in centimetres and the length of the roads are written in kilometres. The metric system has been name from the unit of length, metre. The unit of measurement of weight is a gramme. This is also unit of metric system. Less quantity of weight of a body is expressed in gramme. More quantity of weight of a body is expressed in kilogramme (kg). The unit for measuring the volume of liquid materials is a litre. This is also unit of metric system. Litre and kilolitre are used respectively for less quantity and more quantity of volumes of liquid materials. The metric system has been introduced in Bangladesh from 1st July, 1982. Now in each measurement of length, weight and volume this system is fully in use. Units for Measurement of Length in Metric System

10 millimetre (m.m.) = 1 centimetre (c.m.) 10 centimetre = 1 decimetre (decim.) 10 decimetre = 1 metre (m.) 10 metre = 1 decametre (deca m.) 10 decametre = 1 hectometre h.m.)

10 hectometre = 1 kilometre (k.m.)

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Unit of measuring length: Metre.... The relationship between the metric unit of length measurement and the principal unit of metre is given below :

Kilometre Hectometre Decametre Metre Decimetre Centimetre Millimetre

1000 m. 100 m.

10 m. 1 metre 110 m.

or, 0⋅1m.

1100 m.

or, 0⋅01m.

11000 m.

or, 0⋅001m.

Remark : (a) In metric system if any length is converted from lower to upper or from upper to lower units by writing the digits side by side the decimal point is to be shifted to the left or to the right according to the requirements. For example, 5km. 4h.m. 7 decam. 6m. 9decim. 2 cm. 3m.m. = 547692.3 cm = 5476⋅923 m. = 547⋅6923 decam. = 54⋅76923 h.m. = 5⋅476923 k.m. (b) The word representing multiple from Greek language and the word representing portion from Latin language have been used before the name of unit. In Greek language deca means 10 times, hecto means 100 times and kilo means 1000 times. In Latin language deci means tenth, centi means hundredth and milli means thousandth. Units for Measurement of Length in British (English) System

12 inches = 1 feet (ft) 3 feet = 1 yard (yd) 1760 yards = 1 mile (m)

Relation between Metric and British Systems in the Measurement of Length

1 metre = 39⋅37 inches (approx.) 1 km = 0⋅62 mile (approx.) 1 inch = 2⋅54 cm (approx.) l yard = 0⋅9144 metre (approx.) 1 mile = 1.6 kilometre (approx.)

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Remark : (a) It is not possible to find exactly the relationship between Metric system and British system. So the value obtained from a relationship is expressed approximately to some places of decimal or only to some places of decimal. (b) There is a scale for measurement of smaller lengths. It is 30 cm. or 1 feet in length. One side of it is marked upto 30 cm. by dividing each centimetre into 10 small divisions. The other side of the scale is marked upto 12 inches by dividing each inch into either 8 or 10 small divisions. Similarly a tape is used for the measurement of larger length. The tape is either 30 metre or 100 feet in length. One side of it is marked by dividing each centimetre into 10 divisions in centimetres and metres, the other side of it is marked by dividing each inch into either 10 or 8 divisions in inches and feet. Units for Measurement of Weight in Metric System

10 milligramme (m. gm.) = 1 centigramme (c. gm.) 10 centigramme = 1 decigramme (decigm.) 10 decigramme = 1 gramme (gm.) 10 gramme = 1 decagramme (decagm.) 10 decagramme = 1 hectogramme (h. gm.) 10 hectogramme = 1 kilogramme (kg.)

Unit of measuring weight : Gramme

1 kilogramme or, kg =1000 gramme or, gm

There are two other units in metric system for measurement of weight, they are given below. When the materials are larger in quantity then these two units are used, specially for measurement of rod and cement or for measurement of food grains.

100 kilogramme (kg) = 1 quintal 1000 kilogramme or, 10 quintal = 1 metric ton.

Remark : Balance and weighting materials of fixed quantity are used for measurement of weight in towns and villages. The weighting materials are of 1 gramme, 5 grammes, 10 grammes, 50 grammes, 100 grammes, 200 grammes, 500 grammes, 1 kg., 2 kg., 5 kg., 10 kg., 100 kg ........... etc. But in the town marked balance is used for measurement of weight, such balance is marked by dividing each kg. into 1000 divisions .

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Remark : In the local system the units of weights were chhatak, seer and maund. After the introduction of metric system this local system is not in use. Units for Measurement of Volume of Liquid Materials in Metric System

10 millilitre (m.l.) = 1 centilitre(c.l.) 10 centilitre = 1 decilitre (decil.)

10 decilitre = 1 litre (1.)

10 litre = 1 decalitre (deca1.)

10 decalitre = 1 hectolitre (hecto1.)

10 hectolitre = 1 kilolitre (k.l.)

1000 millilitre = 1 litre Remark : For measurement of the volume of liquid materials measuring container of litre is used. These measurirg container are angular shaped made of alumunium

or tin sheets which are of 14 ,

12 , 1,2, 3, 4, 5,------ etc. litres.

Again these may be vertical container made of transparent glass marked by 25, 50, 100, 200, 300, 400, 500, 600, 700, 800,900, 1000 millilitres. Remark : If the units in the metric system of one kind of measurement is known, then the rest of them can easily be remembered. If the units of measurement of length is known, then the other two are found by putting gramme or, litre respectively in place of metre. Measurement of land :

The area of a rectangular region = measurement of length × measurement of breadth

The area of a square region = (length of side)2

The area of a triangular region = 12 × length of base × length of height

Remark : In the measurement of length and breadth, they should be of the same unit. If length is 1 metre and breadth is 1 metre, then the area will be 1 square metre.

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Units for Measurement of Area in the Metric System

100 Square centimetre (sq. cm.) = 1 Square decimetre (sq.dccim.) 100 Square decimetre = 1 Square metre (sq.m.) 100 Square metre = 1 Ayor (sq. decametre) 100 Ayor = 1 Hector, or 1sq. hectometre

Principal unit for measurement of land : Square metre

Remark : The meaning of 2 square metre is not the same as that of 2 metre square. 2 metre square means a square region of which the length of each side is 2 metres and whose area is 4 square metres. 2 square metre means such a region whose area is 2 square metres. Units for Measurement of Area in the British System

144 square inches = 1 square feet 9 square feet = 1 square yard 4840 square yards = 1 Acre 100 decimals = 1 Acre

Units for Measurement of Area in Local System

1 square hand = 1 ganda 20 gandas = 1 chhatak 16 chhataks = 1 katha

20 kathas = 1 bigha

Relationship between Metric and British System in Measurement of Area

1 square centimetre = 0⋅16 square inch (approx.) 1 square metre = 10⋅76 square feet (approx.) 1 hector = 2⋅47 acre (approx.) 1 square inch = 6⋅45 square centimetre (approx.) 1 square feet = 929 square centimetre (approx.) 1 square yard = 0⋅84 square metre (approx.) 1 square mile = 640 acre.

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Relationship between Metric, British and Local systems in Measurement of Area

1 square hand = 324 square inches 4 square hand or, 4 ganda = 9 square feet = 0⋅836 square metres (approx.) 1 katha = 720 square feet = 80 square yards = 66⋅89 square metres (approx.)

1 bigha = 1600 square yards = 1337.8 square metres (approx.)

1 acre = 3 bigha 8 chhataks = 4046.24 square metres (approx.) 1 decimal = 435⋅6 square feet (approx) =1000 square kori (100 kori = 66 feet) 1 square mile = 1936 bigha

1 square metre = 4.78 ganda (approx.) = 0⋅239 bigha (approx.)

1 ayor = 23⋅9 bigha (approx.)

Volume ..Cubic units of a solid body is volume...

volume of a rectangular solid body = measurement of length × measurement of breadth × measurement of height. Remark : Expressing the measurements of length, breadth and the height in the same unit, measurement of volume is found in cubic unit. The volume of the body whose length 1 cm, breadth 1 cm and height 1 cm is 1 cubic centimetre. Units for Measurement of Volume in the Metric System

1000 cubic centimetre (c.c) = 1 cubic decimetre (c. decimetre) 1000 cubic decimetre = 1 cubic metre (c.m.) 1 cubic metre = 1 stayor 10 cubic stayor = 1 decastayor

Relationship between Metric and British Units in Measurement of Volume

1 stayor = 35⋅3 cubic feet ( approx.) 1 deca stayor = 13⋅08 cubic yards (approx.) 1 cubic feet = 2867 litre (approx.) Remark : Weight of 1 cubic centimetre of pure water at the temperature of 4° selcius is 1 gramme. 1000 cubic centimetres = 1 litre Weight. Weight of 1 litre of water = 1 kilogramme.

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Example 1. A seller of bananas has 13 dozen of bananas. How many are bananas in his possession? Solution : 1 dozen of bananas = 12 bananas 13 " " " = (13 × 12) or, 156 bananas Ans : 156 bananas

Example 2. A runner runs for 24 turns in a round track measuring 400 metres. How long does he run?

Solution : If he runs for 1 turn, the distance is 400 metres ∴ " " " " 24 " " " " (400 × 24) metres

or, 9600 metres or, 9 km. 6 h.m. ∴ The required distance = 9 km.6 h.m. Ans : 9 km. 6 h.m. Example 3. If Ranjit runs for 16000 metres, then how many kilometre did he run? Solution : 1000 metres = 1 kilometre

∴1 " = 1

1000 "

∴ 16000 " = 1 × 16000

1000 " or, 16 km.

∴ Ranjit runs = 16 kilometres. Ans : 16 kilometres. Example 4. Kalu Mia produces 500 kg. 700 gm. of potatoes in one piece of his land. How much will he produce potatoes in his 12 pieces of land of which the area of one piece is equal to that of the previous piece?

Solution : In 1 piece of land he produces 500 kg. 700 gm potatoes ∴ " 12 " " " " " (500 kg. 700 gm) × 12 = 6008 kg. 400 gm. = 6 metric tons 8 kg 400 gm

∴ The required production 6 metric tons 8 kg 400 gm. Ans : 6 metric tons 8 kg 400 gm. Example 5. 28 metric tons of paddy was produced in 16 acres of Liton's land. How much is the quantity of paddy produced in one acre of his land?

Solution : In 16 acres of land he produced 28 metric tons of paddy N.M.G. -12

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∴ " 1 " " " " " 2816 " " " "

or, 1 metric ton 750 kg. of paddy. ∴ The required production per acre = 1 metric ton 750 kg. Ans : l metric ton 750 kg. Example 6. 20,000 metric tons of rod per month are manufactured in a steel mill. How much is the daily production in that mill?

Solution : In 30 days production is 20000 metric tons of rod

∴ " 1 " " " 20000

30 " " " "

or, 666 metric tons 666 kg 66623 gm of rod.

∴ The required quantity = 666 metric tons 666 kg. 66623 gm.

Ans : 666 metric tons 666 kg. 66623 gm.

Example 7. A businessman sells 20 kg. 400 grammes of pulse in a certain day. How much is the quantity of pulse which he sells in a month on the basis of that certain day?

Solution : He sells in 1 day 20 kg. 400 gm. ∴ " " " 30 days (20 kg. 400 gm.) × 30 or, 612 kg. ∴ The required quantity 612 kg.

Ans : 612 kg.

Example 8. A motor car runs 80 km. by 10 litres of diesel. How much is the quantity of diesel required to run 1 km?

Solution: It runs 80 km. by 10 litres of diesel

∴ " " 1 " " 1080 litres or, 125 millilitres of diesel.

∴ The required quantity of diesel =125 millilitres. Ans: 125 millilitres. Example 9. A rectangular region is 80 metres in length and 60 metres in breadth. What is the area of the region?

Solution : We know area of a recgular region = length × breadth

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Here, length = 80 metres and breadth = 60 metres ∴ The required area = (80 × 60) square metres = 4800 square metres. Ans: 4800 square metres. Example 10. A rectangular region is 40 metres in length and 30 metres 30 cm. in breadth. What is the area of the region?

Solution :

Length = 40 metres = (40 ×100) cm. = 4000 cm. breadth = 30 metre 30 cm. = (30 × 100) cm. + 30 cm. = 3030 cm. ∴ The required area = (4000 × 3030) square cm. = 12120000 square cm. = 1212 square metre. = 12 ayor 12 square metre. Ans : 12 ayor 12 square metre. Example 11. A rectangular garden is 40 metre in length and 30 metre in breadth. l metre wide path passes arround the garden inside. What is the area of the path?

Solution : The length of the garden is 40 metres and breadth of " is 30 " ∴ The area of the garden = (40 × 30) square metres

or 1200 square metres. As the path is 1 metre wide, so excluding the path, length of inner portion of the rectangular garden = {40 − (1 × 2) } metres = 38 metres

and its breadth = {30 − (1 × 2)} metres = 28 metres

∴ Area of inner portion of the rectangular garden = (38 × 28) square metres = 1064 square metres

∴ Area of the path = (1200 − 1064) square metres =136 square metres. Ans : 136 square metres. Example 12. The measurement of the base of a triangle is 4 metres and its height 3 metres. What is the area of the triangular region?

1

1

1 1 1

1

1 1

38 metre

28 metre

40 metre 30 metre

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Solution : The area of the triangular region = 12 (base × height)

= 12 (4 × 3) square metres

= 6 square metres. Ans : 6 square metres. Example 13. If the area of a rectangular garden is 840 square metres and its length is 40 metre, what is its breadth? Solution : Area = 840 square metres and length = 40 metres

∴ Breadth = area ÷ length = (840 ÷ 40) metres = 21 metres

Ans: 21 metres. Example 14. The length of a box is 2 metres, breadth 1 metre 50 cm. and height 1 metre. What is the volume of the box?

Solution : Length = 2 metres = 200 centimetres Breadth = 1 metre 50 centimetres = 1 metre + 50 centimetres = 150 centimetres

and height = 1 metre = 100 centimetres ∴ Volume of the box = (length × breadth × height) = (200 × 150 × 100) cubic cm. = 3000000 cubic cm. = 3 cubic metres. Second method : Length = 2 metres,

breadth = 1 metre 50 centimetre = ⎝⎜⎛

⎠⎟⎞1 +

50100 metres

= ⎝⎜⎛

⎠⎟⎞1 +

12 metres =

32 metres

and height = 1 metre.

∴ Volume of the box = ⎝⎜⎛

⎠⎟⎞2 ×

32 × 1 cubic metres

= 3 cubic metres. Ans : 3 cubic metres. Example 15: The length of a reservoir is 3 metres, breath is 2 metres and height is 4 metres. How much in litre and in kilogramme of pure water will it contain ?

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Solution : Length = 3 metres = 300 centimetres Breadth = 2 metres = 200 centimetres and Height = 4 metres = 400 centimetres ∴ Volume of the reservoir = (300 × 200 × 400) cubic centimetres. = 24000000 cubic centimetres = 24000 litres [Q 1000 cubic centimetres = 1 litre] The weight of 1 litre pure water is 1 kilogramme. ∴ The weight of 24000 litre pure water is 24000 kilogramme. Ans : 24000 litre and 24000 kilogramme.

EXERCISE 3

1. The population of a town is 50000. In each day 20 persons die and there is birth of 30 babies. What will be the population in that town after one year?

2. The population of a village is 7000. In each day 3 persons go to the town in search of employment and 2 persons come back to the village. What will be the change in population after 1 month?

3. Rahim exchanges 5 dozen of hen's eggs with 7 dozen duck's eggs belonging to Karim. Rahim had 9 dozen of eggs before the exchange was made. How many is the number of eggs in possession of Rahim now?

4. The price of each lemon is Tk. 1. How much sum of money did Ratan get by selling 20 halis of lemons?

5. If the price of walking fish (Koi) is Tk. 50 a kuri, then what is the price of one walking fish (koi fish) ?

6. A seller of bananas has 35 dozen of bananas. How many is the number of bananas in his possession?

7. The distance between two places is 10 km. Express it in metres. 8. The circumference of a wheel of a vehicle is 6⋅25 metres. How many will be

the number of turns of the wheel if the vehicle moves for a distance of 40 kilometers?

9. In what way is the track to be made so that 16 turns is required for a distance of 10000 metres in a competition of race?

10. What will be the measurement in metres if 24 turns are completed in a circular track of 125 metres?

11 Starting from the same place Moula covers a distance of 1⋅6 kilometres in 15

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minutes and Mostafa covers a distance of 1 km 500 metres in 15 minutes. What will be the distance between them after 15 minutes?

12. A bus runs 325 km. in 6 hours. What is the speed of the bus in kilometres per hour?

13. Raihan produces 400 kg. of paddy from one acre of land. If he gets 700 grammes of rice out of 1 kg of paddy, then how much quantity of rice did he get?

14. If 400 metric tons of potatoes are produced in 15 acres of land; then how much quantity of potatoes will be produced in 1 acre of land?

15. If 20 kg 850 gm 350 milligramme of mustard are produced in a piece of land, then how much quantity of mustard will be produced in such 7 pieces of land?

16. In each day 5000 bags of cement are manufactured in a cement factory. If the weight of each bag of cement is 25 gm 30 centigramme, then what is the daily production of cement?

17. If the yearly production of rod in a steel mill is 300000 metric tons, what is the daily production of rod?

18. A businessman sells in a certain day 18 kg 300 gm of rice and 534 kg. of salt.

How much quantities of rice and salt does he sell in a month on the basis of the previous quantities?

19. A businessman has 500 metric tons of rice in his godown. He brings daily 2 metric tons 500 kg. of rice to his shop from the godown. How many days will be required to bring all of rice from the godown?

20. If a bus runs 125 km. by 80 litres of diesel, then how much quantity of diesel will be required for running 1 km.?

21. If a motor car runs 128 km. by 9 litres of petrol, then how much quantity of petrol will be necessary to run 1 kilometre?

22. A bus runs 16 km. by 4 litres of diesel. How much quantity of diesel will be necessary to run 64 kilometres?

23. The length of a rectangular garden is 8 metres and its breadth is 4 metres. What is its area in square centimetres?

24. The area of a rectangular region is 900 square metres and its length is 36 metres. What is the breadth of the region?

25. The length of a rectangular garden is 32 metres and its breadth is 24 metres. A path of 2 metres wide passes around the garden inside. What is the area of the path?

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26. The length of a pond is 60 metres and its breadth is 40 metres. If the wide of the bank of the pond is 3 metres, them find the area of the bank of the pond.

27. The area of a rectangular region is 10 acres and its length is 4 times of its breadth. What is the length in metres of the region?

28. The length of a rectangular room is 112 times of its breadth. If its area is 216

square metres, then what is its perimeter? 29. If the base of a triangular region is 20 metres and its height is 15 metres 50

centimeters, then find its area. 30. The length of a rectangular region is 48 metres and its breadth is 32 metres 80

centimetres. A path of 3 metres wide passes around the region outside. What is the area of the path?

31. The length of a square region is 200 metres and a path of 4 metres wide passes around it outside. What is the area of the path?

32. The area of a triangular region is 216 square metres. If its base is 18 metres, then find its height.

33. A reservoir contains 8000 litres of water. If its depth is 2⋅56 metres and breadth is 1⋅25 metres, then what is its length?

34. Gold is 19⋅3 times heavier than water. The length of a rectangular golden bar is 8⋅8 centimetres, breadth 6⋅4 centimetres and height 2⋅5 centimetres. What is the weight of the golden bar?

35. The length of a small box is 15 centimetres 2⋅4 millimetres, its breadth is 7 centimetre 6⋅2 millimetres and height is 5 centimetres 8 millimetres. What is the volume in cubic centimetres of the box?

36. The length of a room is 20 metres, breadth 15 metres 50 centimetres and height 4 metres. What is the volume of the room ?

37. Iron is 7⋅5 times heavier than water. The length of a piece of iron sheet is 3 metres, wide 2 metres and the thickness 1 centimetre. What is the weight of the iron sheet?

38. The length of a reservoir is 5 metres, its breadth is 4 metres and its height is 3 metres. If it is fully filled in by water, how much quantity of water in litres will contain the reservoir? How much is the weight of that water?

Multiple Choice Questions [Marks (√) on the correct answer] 1. Which one of the following is the basic unit of length measurement? (a) metre (b) centimetre (c) kilometre (d) millimetre

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2. The volume of a solid is- (a) area (b) square unit (c) cubic unit (d) cubic metre 3. There is a road around the boundary outside of a rectangular field.

The field is 50 metres in length, 40 metres in breadth and the road is 2 metres in width. What is the length of the field including the road?

(a) 54 (b) 50 (c) 46 (d) 44 4. Observe the following informations: (i) 1 kilometre = 100 metre (ii) 100 kilogram = 1 quintal (iii) mass of 1 liter water = 1 kg. Which one of the following is correct in view of the above

informations? (a) i (b) i and iii (c) ii and iii (d) i, ii and iii The length and breadth of a copper sheet are 4 metre and 3 metre

respectively. Answer the questions (5-7) in view of the above informations: 5. What is the area in square metre of the sheet? (a) 6 (b) 7 (c) 12 (d) 14 6. What is the perimeter in metres of the sheet? (a) 7 (b) 12 (c) 14 (d) 24 7. The length of the diagonal of the sheet is- (a) 7 metre (b) 5 metre (c) 4 metre (d) 3 metre. CREATIVE QUESTIONS

1. A house is 15 metres in length, 12 metres in breadth. There is a verandah of width 2 metres around the house.

a) Find the length and breadth of the house including the verandah.

(b) Find the perimeter of the house and the area of the verandah. (c) How many tiles would be require to set on a square shape

floor of perimeter equal to the perimeter of that house?

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2. A piece of iron-sheet is 3 metres in length, 2 metres in width and 0.01 meter in thickness. Wings of fan would be made by melting the iron sheet. The length of each wing of each fan is 4 metre, width 1 metre and thickness is 0.001 metre. There are three wings in each fan. The price of one fan is Tk. 10.25. (The mass of 5 cubic cm water is 5 gm and that of iron is 0.037 kg).

(a) How many times is iron heavier than water? (b) Find the mass of the iron sheet. (c) How many fan and wing of fan can be made using the iron-

sheet? Find the price of the fans. 3. The length is twice of the width of a rectangular plot and

perimeter is 120 metres. There is a pond with bank of equal breadth in side the plot. The breadth of bank is 2 metres. The cost of digging the pond is Tk. 25 per cubic metre and the depth of the pond is 5 metre.

(a) Find the width of the plot. (b) Find the perimeter and area of the pond. (c) How much would be expend to dig the pond and how much

kilogramme of water does the pond contain in the pond?

N.M.G. -13

Chapter IV

Statistics

4.1. Preliminary Concept Regarding Statistics Statistics is a informative science, some statisticians define it as a numerical informations, some others have defined it as the science for research with numbers. According to the first meaning statistics are numerical states i. e. informations of any fact or matter. For example, statistics of characteristics of birth and death regarding population, price of goods, wages etc. are denoted by numerical information's. Numerical index of such a definite characteristic or subject is called statistics of that fact. The informations expressed by numbers is called data. Statistics means the collection, classification, presentation, analysis and interpretation of information's systematically. With the help of it some light on any sphere of enquiry can be thrown by analysing a data. Statistics is a branch of applied mathematics which is applied in collecting and analysing numerical information's. 4.2. Purpose of Statistics: The main purpose of statistics is the research on numerical information's for implementation of a plan or making decision, policy and formation of procedure, method for any uncertain matter. Statistics presents a large number of information's in brief and in easily understandable way and also help in comparison the information among many characteristics. It helps to take decisions in taking managerial plan of different social, economical and commercial institutions. In statistics death and birth, import and export, production, wages, census, agricultural census and management of population etc. are discussed. It is used for states relationship income and expenditure, man power and bank insurance and industry in the commercial and economical sectors etc. The task of statistics is the explanation to make decisions for the future by induction and application of necessary formulae obtained on the basis of past experiences and information's regarding any matter. 4.3. Characteristics of Statistics

Information expressed by one number only can not be called statistics. But this can be named for some numerical information's. Data of statistics are expressed in numbers. The data are collected by statistical investigations. In collecting data measurement, observation and enumeration are necessary. Measurement and enumeration will their proper units are to be clearly determined so that there is no confusion in data.

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The unit of enumeration is as the base of collection of informations. The units should be necessary of the same kind and measurement. If there is no similarity, comparisons of information can not be judged and interpreted. 4.4. Data of Statistics

The data are collected in two ways: by direct observation or, from data collected by other institutions for their use or used data of other institutions. The ages or incomes of the inhabitants of certain cities, the maximum and the minimum temparatures of some places or the quantity of rain fall of any place in day or the daily income of a commercial institution etc. are those data when expressed in numbers. Data collected at random generally called non-arranged data. When the maximum and the minimum temparatures of each day or the quantity of rain fall of each day in a place were collected for some days, such collections are called non-arranged data. For proper presentation of data collected in such a way are classified conveniently. Data obtained by classification are called arranged data. To make classes of the same value a convenient number is taken as the lower limit of first class. Similarly, a convenient number is selected for determination of the upper limit. The number thus determined may not belong to informations. The difference between the number of upper limit and the number of lower limit in each class is called the class interval of that class. In classification the class intervals of all the classes are equal to one another. The class interval is obtained by dividing that number which is equal to the number of classes as found from the difference between the upper limit of the highest class and the lower limit of first class. 4.5. Non-arranged Data and Arranged Data

Concept regarding non-arranged and arranged data is given by the example below: Obtained the marks in Mathematics of 50 students in the annual examination (Marks are given according to roll numbers)

7 18 37 53 24 39 41 23 64 67 68 50 93

43 11 27 68 72 19 12 21 19 32 75 52 84 15 11 23 19 52 29 92 79 45 81 63 36 21 33 53 8 41 14 26 26 33 49 40 19. In the above way obtained marks are non arranged data. If the numbers from 1 to 100 are divided into 10 classes, then the classes will be as follows-

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l - 10, 11 - 20, 21 - 30, 31- 40, 41 - 50, 51 - 60, 61- 70, 71- 80, 81- 90, 91- 100. On the basis of the marks obtained by these 50 students, the number in each class of students who have got the marks included in that class can be known. The number of students belonging to a particular class is called frequency.

Table - 1 Classification of marks Number of students

1 - 10 2 11 - 20 10 21 - 30 9 31 - 40 7 41 - 50 6 51 - 60 4 61 - 70 5 71 - 80 3 81 - 90 2

91 - 100 2 Total = 50

In such a way the table formed by showing the obtained number of students in each class along with the classification of marks is the arranged data. 4.6 . Rules to convert non-arranged data into arranged data The obtained marks of 50 students in Mathematics in the annual examination are given according to their roll-numbers above. These are to be converted into arranged data. In this example, the lowest mark is 7 and the highest mark is 93. So all the marks will be divided into 18 classes if in each class there consists of 5 numbers beginning from 4. At first classes of numbers are written at the left side The given marks are considered one by one. The first mark is 7. It will remain in the class of 4 - 8. So a tally sign '|' is placed against that class. The 2nd mark is 18, so it will remain in the class of 14 - 18. Now a tally sign '|' is placed against that class. Proceeding this way, if four tally signs are placed in any class, the fifth tally sign is not to be placed separately after the fourth one rather is placed by a crossing of 4 tally signs. Then keeping a small space the next tally sign will be placed after every four tally signs. The fifth one will be placed in the way such as

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⎪⎪⎪⎪. It is easy to count if the fifth tally-sign is placed in such a way. The numbers of tally signs placed against a class is the numbers of data included in a class is the frequency of that class.

Table-2

Class of marks Tally-sign Frequency 4 - 8 ⎪⎪ 2

9 - 13 ⎪⎪⎪ 3

14 - 18 ⎪⎪⎪ 3

19 - 23 ⎪⎪⎪⎪ ⎪⎪⎪ 8

24 - 28 ⎪⎪⎪⎪ 4

29 - 33 ⎪⎪⎪⎪ 4

34 - 38 ⎪⎪ 2

39 - 43 ⎪⎪⎪⎪ 5

44 - 48 ⎪ 1

49 - 53 ⎪⎪⎪⎪ ⎪ 6

54 - 58 0 59 - 63 ⎪ 1

64 - 68 ⎪⎪⎪⎪ 4

69 - 73 ⎪ 1

74 - 78 ⎪ 1

79 - 83 ⎪⎪ 2

84 - 88 ⎪ 1

89 - 93 ⎪⎪ 2

Total = 50 4.7. Mean of nonarranged marks

Mean of obtained marks = Sum of marks

number of students = 202750 = 40⋅54

This is the mean of non-arranged marks

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4.8. Mean of classified marks In order to find the mean of classified marks first the mid value of each class is to be found out. The frequency of each class indicates the repeated number of the mid value of that class. Finding out the mid value of each class, it is to be multiplid by the frequency of that corresponding class. Dividing the sum of these products by the total number of students, the mean of these classified marks will be found.

Table - 3

Class of marks Mid value of class

Frequency Mid value of class × frequency

4 - 8 6 2 12 9 - 13 11 3 33 14 - 18 16 3 48 19 - 23 21 8 168 24 - 28 26 4 104 29 - 33 31 4 124 34 - 38 36 2 72 39 - 43 41 5 205 44 - 48 46 1 46 49 - 53 51 6 306 54 - 58 56 0 0 59 - 63 61 1 61 64 - 68 66 4 264 69 - 73 71 1 71 74 - 78 76 1 76 79 - 83 81 2 162 84 - 88 86 1 86 89 - 93 91 2 182

Total = 50 2020

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So, mean of classified marks = 202050 = 40⋅4

Mean of classified data = sfrequencie class of Sum

sfrequencie and class of valuemid of products theof Sum

Remark : Some informations have been lost for classifying the given 50 marks in the manner as done above. It is known only from the classified data in which class do a definite number of students get marks. It is not known how much are the marks exactly. In classification the frequency for a particular number among the included numbers of a class is not also known. On the other hand from the classified data the class at which the highest number of students has obtained marks is known. Let us find the mean of data given in Table -1

Table-4

Class-division Mid-value of class

Frequency Mid-value of class × frequency

1 - 10 5.5 2 11 11 - 20 15.5 10 155 21 - 30 25.5 9 229.5 31 - 40 35.5 7 248.5 41 - 50 45.5 6 273 51 - 60 55.5 4 222 61 - 70 65.5 5 327.5 71 - 80 75.5 3 226.5 81 - 90 85.5 2 171

91 - 100 95.5 2 191 Total = 50 2055

Therefore, Mean of classified marks as stated above = 205550 = 41⋅1.

Remark : The data of Table - 3 and Table - 4 has been formed from non-arranged marks obtained by 50 students. The actual mean of the marks obtained by 50 students is 40. 54. But from Table - 3 the obtained mean is 40⋅4 and from Table - 4

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the obtained mean is 41.1. As much larges the divisions of classes are the mean of classified marks will be such closer to the actual mean; that is, as much smaller the class interval the obtained mean in classification will be such nearer to the actual mean. 4.9. Determination of mean of classified data by easy method

The class in which mean may belong is to be assumed. It is assumed that the mean belongs to the class of 51- 60 in Table - 4. This class is situated almost at the middle of the table . A separate column is to be constructed in the table stated above and at the right side of this class 0 is to be written in the new column. Then from the middle of that class in the right side -1,-2, -3, ............ are to be written successively for the lower classes and 1, 2, 3, ........... are to be written successively for the upper classes in the new column. These numbers are called deviation numbers. Then the frequency of each class is to be multiplied by the related deviation-number. Then the sum of such products are to be determined. This sum is to be divided by the sum of frequencies and its result is multiplied by class interval. This product is to be added with the mid value of the assumed class. This will be the mean.

Table - 5

Class-division Mid-value of class

Frequency Deviation number

Frequency × deviation number

1 - 10 5. 5 2 - 5 - 10 11 - 20 15. 5 10 - 4 - 40 21 - 30 25. 5 9 - 3 - 27 31 - 40 35. 5 7 - 2 - 14 41 - 50 45. 5 6 - 1 - 6 51 - 60 55. 5 4 0 0 61 - 70 65. 5 5 1 5 71 - 80 75. 5 3 2 6 81 - 90 85. 5 2 3 6

91 - 100 95. 5 2 4 8 Total = 50 Sum = − 72

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Here, the class interval =10

Therefore, mean = the mid-value of assumed class + Sum of (frequency × deviation number)

Total frequencies × Class interval

= 55. 5 + – 7250 × 10

= 55 . 5 – 14. 4 = 41.1 The mean of classified data

= mid-value of assumed class +Σ(frequency × deviation number)

Total frequencies ×Class interval.

Remarks : (a) This is equal to the obtained mean from Table - 4 (b) Whatever may be the class at which mean is assumed to be situated, the mean will always be the same. (c) The sum is expressed by the symbol, 'Σ' Example 1 . The marks in Mathematics in the annual examination obtained by 50 students of any class are given below : 68, 87, 81, 72, 65, 69, 73, 84, 73, 48, 56, 72, 62, 93, 73, 83, 75, 56, 58, 87, 97, 81, 51, 61, 53, 72, 62, 79, 88, 47, 78, 76, 42, 79, 70, 58, 71, 67, 80, 78 , 85, 66, 76, 45, 66, 91, 77, 64, 71, 74. (a) What is the mean of marks? (b) Find the mean by classifying data. Solution : (a) The sum of given marks = 3540 and the number of marks = 50

∴ The required mean = 354050 = 70⋅8

Ans : 70⋅8. (b) The highest mark of the given marks is 97 and the lowest mark is 42. So beginning from 40 classes are formed taking class interval of 5 marks. The marks are classified by tally-signs.

N.M.G. -14

97

Class division Tally Frequency Deviation number

Frequency × deviation number

40 - 44 ⎪ 1 - 6 - 6 45 - 49 ⎪⎪⎪ 3 - 5 - 15 50 - 54 ⎪⎪ 2 - 4 - 8 55 - 59 ⎪⎪⎪⎪ 4 - 3 - 12 60 - 64 ⎪⎪⎪⎪ 4 - 2 - 8 65 - 69 ⎪⎪⎪⎪ ⎪ 6 - 1 - 6 70 - 74 ⎪⎪⎪⎪ ⎪⎪⎪⎪ 10 0 0 75 - 79 ⎪⎪⎪⎪ ⎪⎪⎪ 8 1 8 80 - 84 ⎪⎪⎪⎪ 5 2 10 85 - 89 ⎪⎪⎪⎪ 4 3 12 90 - 94 ⎪⎪ 2 4 8 95 - 99 ⎪ 1 5 5

Total = 50 Sum = - 12 Suppose, the mean is in the class of 70 - 74. The mid value of this class is 72.

∴ The required mean = 72 + - 1250 × 5 = 72 - 1.2 = 70.8.

Ans: 70.8. Remark : Here the mean of the principal data is equal to the mean of classified data. But in a rare occasion this equality found. If the class interval is changed, there will be no equality between two means.

4. 10. Algebraic Formulae

(a) Let, n numbers of expressions or numbers (data) be given and those are x1, x2, ............, xn (in short xi, i = 1, 2, 3, ..............., n). The mean of these terms is expressed by x̄ . For indication of sum, the sign 'Σ' is used. i, e. x1 + x2 + .................. + xn = Σ xi , (i = 1, 2, ..........., n)

∴ x̄ = Σ xi

n , (i = 1, 2 ........, n) (formula - 1)

This formula is directly used for finding the mean.

98

(b) Let, n terms such as x1, x2, ..............., xn be given. Suppose, each term of them occurs f1, f2, ......., fn times respectively. (c) If those are classified, the class from which it begins is called first class, the next one is second class, the class next to the 2nd is the 3rd class. In such a way classes can be called 1st, 2nd , 3rd, ..........., nth class (if there are n terms). In statistics frequency is generally expressed by f. The frequency of i - th class is denoted by fi. The mid value of i - th class is denoted by xi and x expresses the

mean of all the data.

∴ x̄ = Σ fixiΣfi , (i = 1, 2 ........., n) (formula - 2)

(d) Algebraic formula for finding the mean by easy method : (1) Let, the actual mean of n terms such as x1, x2, x3 ,.............., xn be x̄ and its assumed mean be a.

Suppose, x11 = xi - a, x

1

2 = x2 - a, ............, x1

n = xn - a

∴ x̄ = a + Σfixi

n , (i = 1, 2 .............. n) (formula- 3)

(2) The actual mean of classified data is x̄ . Mid values of 1st, 2nd , 3rd, ............., n-th classes are respectively x1, x2, x3, ............, xn . Let, the mid value of r -th class be the assumed mean. Therefore, xr is the assumed mean. Let, d be the class-interval.

∴ x̄ = xr + ∑x1

ifi ∑fi × d (i = 1, 2,------, n)

Example 2. Marks obtained by 60 students are given below. Find the mean of their marks.

Marks Number of Students

52 10 57 15 60 14 62 12 65 9

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Solution :

Marks (xi) Number of students (fi) fixi 52 10 520 57 15 855 60 14 840 62 12 744 65 9 585

Total = 60 3544

∴ The required mean = 354460 = 59⋅07 (approximately) ( Applying formula - 2)

Second method:

Let the assumed mean be 60. Marks

xi Number of Students fi

x ′i fi x′i

52 10 − 8 − 80 57 15 − 3 − 45 60 14 0 0 62 12 2 24 65 9 5 45

Total = 60 − 56

∴ The required mean = 60 + −5660 = 60 − 0⋅93 = 59⋅07 (approximately) (Applying

formula - 3) Ans : 59⋅07 (Approx.)

4.11. Uses of Arithmetic Mean

The arithmetic mean is used in our daily life in the fields of our income, expenditure, production, rate of pass in the examination etc. This mean is widely used in social, economical and commercial fields. The use of this is the highest of all.

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EXERCISE 4.1

1. The daily wages (in Taka) of 40 workers are given below: 42, 49, 38, 50, 51, 44, 56, 53, 45, 43, 46, 42, 53, 54, 50, 46, 43, 36, 48, 56, 62,

49, 39, 47, 46, 54, 64, 40, 41, 74, 49, 50, 49, 57, 52, 49, 53, 48, 52 and 56. (a) Find the mean of daily wages (in Taka) of the workers. (b) Construct a table by taking a convenient class- interval and from the table find

the mean of daily wages (in Taka). 2. Marks in Statistics obtained by 25 students in the annual examination are given

below : 78, 72, 85, 78, 84, 69, 75, 88, 67, 80, 74, 77, 79, 69, 74, 83, 73, 65, 75, 63, 69,

75, 86, 66, 71. (a) Find the mean of the above marks by dividing them into 5 convenient classes. (b) Find the difference between the above mean and the mean obtained directly. 3. The weekly income of 120 families are given below :

Weekly income (in Taka) Number of families 251 — 500 22 501 — 750 25 751 — 1000 40

1001 — 1250 23 1251 — 1500 10

Find the mean of weekly income of the families. 4. Marks obtained by 50 students in an examination are given below : 40, 46, 48, 65, 34, 20, 15, 32, 38, 45, 48, 60, 82, 10, 30, 36, 44, 48, 60, 80, 20,

32, 38, 46, 48, 62, 20, 25, 29, 25, 34, 35, 34, 36, 44, 42, 44, 46, 44, 46,48, 46, 50, 68, 50, 70, 60, 75, 60 and 75.

Construct a table of classes with a class interval of 5 and then find the mean of them. Also calculate their mean directly.

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5. A table is given below. Find its mean. Obtained marks Number of students

6 – 10 5 11 – 15 17 16 – 20 30 21 – 25 38 26 – 30 35 31 – 35 10 36 – 40 7 41 – 45 3

6. Find the mean from the informations given below : Weight (kg) 20 30 40 50 60 70 80 90 Number of persons 5 14 27 16 13 3 1 1

7. Find the mean from the following table : House rent (in Tk. - hundred) Number of families

10 – 19 10 20 – 29 17 30 – 39 30 40 – 49 50 50 – 59 29 60 – 69 9 70 – 79 5

8. Find the mean from the following table : Daily sales (in Taka) 2210 2215 2220 2225 2230 2235 2240 2245 2250

Number of Shops 2 3 5 7 6 5 5 4 3

4.12. Concept Regarding Measures of Central Tendency

If informations are presented by classification, then it is seen that the frequencies of central classes are larger; i. e. numbers in a row of class show tendency to be accumulated to a mid-value and most of the numbers of this row are deviated from the mid-value by a regular and balanced way. The tendency to be accumulated to the same value is called the central tendency. It can be said easily that tendency for values of most of the data to be accumulated to the middle position roughly is called the central tendency.

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If it is wanted to know from a student about his monthly expenditure, generally he will give reply by a whole number. In reality his expenditur in each month is not the same. But his monthly expenditure remains close to that number. Thus the expenditure is presented by median value or central value, this tendency is konwn as central tendency. The purpose of measures of central tendency is to ascertain the middle place. Mean, median and mode are the measures of central tendency. But these have the common use. In addition to these there are other measures of central tendency. But common uses are not very large. But in some specific cases these may be very useful. We have discussed earlier elaborately about mean. So those are not repeated. 4.13. Median If the given data are arranged according to their values successively (from lower to higher or from higher to lower) the value which divides the data into two equal parts is called the median of the data. The median is then value of the middle term when the data are arranged successively, that is the term which remains at the middle. If the number of data be odd, then value of the middle term is the median. Again if the number of data be even, then the median will be the mean of the quantites of two middle terms. The median of 7 numbers such as 4, 5, 7, 8, 10, 12 and 13 is 8. Here the fourth term is the middle term.

The median of 8 numbers such as 4, 5, 7, 8, 10, 13, 16, and 19 is 8 + 102 or, 9. Here

the mean of the 4th and 5th terms. If n numbers of data are given, then the median

will be ⎝⎜⎛

⎠⎟⎞n + 1

2 th term, if n is odd or, the mean of n2 th and ⎝⎜

⎛⎠⎟⎞n

2 + 1 th terms, if n is

even. Example 1. Find the median of numbers such as 5, 7, 8, 19, 15, 17, 9, 12, 2, 20, 13, 4 and 18. Solution : The numbers are arranged according to their values successively from lower to higher as below : 2, 4, 5, 7, 8, 9, 12, 13, 15, 17, 18, 19 and 20. Here, n = 13 (an odd number)

∴ median = the value of ⎝⎜⎛

⎠⎟⎞13 + 1

2 th term = the value of 7th term = 12.

Ans : 12.

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Example 2. Find the median of the numbers. 12, 9, 15, 5, 20, 8, 25, 17, 21, 23, 13 and 11. Solution : The numbers are arranged according to their values successively from lower to higher as below : 5, 8, 9, 11, 12 , 13 , 15 , 17 , 20 , 21 , 23 and 25. Here, n = 12 (an even number)

∴ median = Sum of values of the two middle terms

2

= Sum of values of

122 th and ⎝⎜

⎛⎠⎟⎞12

2 + 1 th terms

2

= Sum of values of 6th and 7th terms

2 = 13 + 152 = 28

2 = 14.

Ans: 14. Example 3. The marks distribution of 50 students are given below. Find the median of their marks. / " ■

Obtained marks Number of students 40 8 50 13 56 10 60 8 68 7 70 3 75 1

Solution : The table for finding out the median : Obtained marks

( numbers) Number of students

(frequency) Cumulative frequency

40 8 8 50 13 21 56 10 31 60 8 39 68 7 46 70 3 49 75 1 50

n = 50 Here , n = 50 (an even number)

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∴ median = Sum of the values of

502 th and

⎝⎜⎛

⎠⎟⎞50

2 + 1 th terms

2

= Sum of the values of 25 th and 26th terms

2 = 56 + 56

2 = 56.

Ans : 56

Remark : Here, value of each from 22nd to 31st term is 56. 4.14. Determination of the Median of Classified Data

The value of n2 th term of the classified data is the median. The lower limit of that

class which contains the median is L and the cumulative frequency of the previous class is Fc. The frequency of the class which contains the median is fm and its class interval is d.

So, the formula for determination of median is as follows :

median = L + ⎝⎜⎛

⎠⎟⎞n

2 − Fc × dfm

Example 4. Find the median from the informations given below :

Daily wages Number of workers

30 ⎯ 35 3

36 ⎯ 41 10

42 ⎯ 47 18

48 ⎯ 53 25

54 ⎯ 59 8

60 ⎯ 65 6

N.M.G. -15

105

Solution : Daily wages (Class) Number of workers

(frequency) Cumulative frequency

30 - 35 3 3 36 - 41 10 13 42 - 47 18 31 48 - 53 25 56 54 - 59 8 64 60 - 65 6 70

n = 70

Here, n = 70 ∴ n2 =

702 = 35. So, value of 35 th term is the median.

35 th term is situated in the class of (48 - 53). So, the median is situated in the class of (48 - 53). Here, L = 48 , Fc = 31, fm= 25 and d = 6.

∴ median = L + ⎝⎜⎛

⎠⎟⎞n

2 - Fc × dfm

= 48 + (35 - 31) × 625 = 48 +

4 × 625 = 48 +

2425 = 48 + 0⋅96 = 48⋅96.

Ans : 48⋅96

4.15. Uses of Median

Median is used widely in case of measures of attributive characteristics. Very useful measurements with the help of median are found from those information which can not be expressed in numbers. Uses of the median is mentionable in cases of abnormally high and low values of informations. 4. 16. Mode

The number of the data which appears for the highest number of times is called the mode. When 5, 7, 9, 10, 15, 9, 7, 9, 11 and 6 are arranged systemetically from lower to higher it becomes 5, 6, 7, 7, 9, 9, 9, 10, 11 and 15. Here 9 appears for the highest number of times that is 3 times. So the mode of the given data (numbers) is 9.

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Example 5. Marks in Mathematics obtained by 40 students in the annual examination are given below. Find the mode of the data. 42, 31, 45, 27, 60, 61, 39, 41, 35, 58, 29, 53, 48, 39, 52, 38, 40, 47, 28, 51, 49, 78, 90, 52, 48, 36, 52, 39, 71, 64, 32, 49, 56, 33, 48, 33, 25, 37, 48 and 29. Solution : The data are arranged systemetically from lower to higher as below : 25, 27, 28, 29, 29, 31, 32, 33, 33, 35, 36, 37, 38, 39, 39, 39, 40, 41, 42, 45, 47, 48, 48, 48, 48, 49, 49, 51, 52, 52, 52, 53, 56, 58, 60, 61, 64, 71, 78, 90. Here, 29 appears for 2 times 33 " " 2 " 39 " " 3 " 48 " " 4 " 49 " " 2 " 52 " " 3 " and each of the rest numbers appears once. Here, 48 appears for the highest number of times, so the required mode is 48. Ans : 48.

4.17. Determination of the mode of classified data

The lower limit of mode-class is L, the difference of frequency between the mode - class and its previous (lower) class is f1, the difference of frequency between the mode-class and its next (higher) class is f2 and the class interval of mode-class is d. The formula for determination of the mode :

mode = L + f1

f1 + f2 × d

Example 6. Find the mode from the data given below : Daily savings (in taka) Number of workers

25 — 30 7 31 — 36 21 37 — 42 47 43 — 48 62 49 — 54 37 55 — 60 16 61 — 66 5

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Solution : In the given table the highest number of workers is in the class of (43 - 48). So, the mode is in this class. Here, L = 43, f1 = 62 − 47 =15, f2 = 62 − 37 = 25 and d = 6.

Now applying the formula, we get

mode = L + f1

f1 + f2 × d

= 43 + 15

15 + 25 × 6

= 43 + 9040 = 43 + 2⋅ 25 = 45.25

∴ The required mode =Tk.45⋅25 Ans : Tk. 45⋅25. Remark : (a) If any one of the numbers of the data does not appear more than once, there is no mode in such a case. (b) There may be mode more than one in the given data. (c) If in the classified data the 1st class is the mode-class, then the mode is to be calculated by taking 0 as the frequency of the previous class. Again if the last class is the mode-class, then the mode is also to be calculated by taking 0, as the frequency of the next class. Example 7. Find the mode of the numbers 2, 7, 5, 4, 6 and 10. Solution : Here, there is no mode, because no date appears more than once. Example 8. Find the mode from the data given below.

Class frequency

11 - 20 25 21 - 30 20 31 - 40 8 41 - 50 3

Solution : In the given table the highest number of frequency is in the class of (11 - 20). So the mode is in this class.

108

Here, L = 11, f1 = 25 - 0 = 25, f2 = 25 - 20 = 5, d = 10. Now applying the formula, we get,

Mode = L + f1

f1 + f2 × d

= 11 + 25

25 + 5 × 10 = 11 + 2530 × 10

= 11 + 253 = 11 + 8⋅33... = 19⋅33

Ans : 19⋅33

Example 9. Find the mode from the data given below.

Class frequency

6 — 10 4 11 — 15 10 16 — 20 15 21 — 25 20

Solution : In the given table the highest number of frequency is in the class of (21 - 25). So the mode is in this class. Here, L = 21, f1 = 20 - 15 = 5, f2 = 20 - 0 = 20 and d = 5. Now applying the formula, we get

Mode = L + f1

f1 + f2 × d

= 21 + 5

5 + 20 × 5

= 21 + 2525 = 21 + 1 = 22.

Ans : 22 4.18. Uses of Mode

We use mode ordinarily in our every day life. Mode is used in Biology, weather forecast, etc. Wages, rent, income, demand, supply, etc. are generally determined by the value of mode. Its use is remarkable in case of attributive informations.

109

EXERCISE 4.2

1. Find the mean, median and mode of the following numbers: 30, 12, 22, 17, 27, 25, 20, 24, 19, 2, 23, 32, 26, 29, 35, 21, 11, 28 and 19. 2. The marks obtained by 40 students in Mathematics in the annual

examination are given below: 53, 48, 65, 35, 55, 38, 50, 67, 25, 58, 62, 21, 76, 46, 57, 67, 55, 70, 44, 72, 60,

42, 64, 73, 38, 41, 55, 41, 34, 51, 40, 38, 28, 44, 36, 40, 42, 35, 22 and 30. Find the median and mode of the marks. Find the median and mode by

classifying them with class interval of 5. 3. The daily sales (in taka) of 25 grocer's shops are given below: 75, 63, 69, 86, 71, 66, 75, 65, 73, 80, 83, 74, 69, 79, 77, 69, 74, 85, 72, 78, 84,

69, 75, 88 and 67. Find the mean, median and mode of the distribution of daily sales. 4. The daily income (in taka) of 40 masons are given below : 155, 173, 166, 143, 168, 160, 156, 146, 162, 158, 159, 148, 150, 147, 132, 136,

156, 140, 155, 145, 135, 151, 141, 169, 140, 125, 122, 140, 137, 175, 145, 150, 164, 142, 156, 152, 146, 148, 157 and 167.

Find the mean, median and mode of their daily incomes. 5. Find the mean, median and mode from the table given below :

Weekly savings (in Taka) Number of workers 71 – 80 12 81 – 90 18 91 – 100 35 101 – 110 42 111 – 120 50 121 – 130 45 131 – 140 20 141 – 150 8

6. Find the mean, median and mode from the table given below : Daily wages (in Taka) Number of workers

51 – 55 7 56 – 60 25 61 – 65 76 66 – 70 32 71 – 75 17 76 – 80 12 81 – 85 3

110

7. Find the mean, median and mode from the data given below : Class Frequency

31 – 40 4 41 – 50 6 51 – 60 8 61 – 70 12 71 – 80 9 81 – 90 7 91 – 100 4

8. Find the mean, median and mode from the data given below : Age (in years) Number of students

5 – 6 25 7 – 8 27

9 – 10 28 11 – 12 31 13 – 14 29 15 – 16 28 17 – 18 22

9. The marks in Mathematics obtained by students according to their roll numbers in a class of a school are given below:

81, 80, 79, 78, 74, 74, 73, 71, 70, 69, 68, 68, 67, 66, 64, 64, 79, 64, 64, 64, 63, 62, 62, 61, 60, 60, 59, 57, 50, 49, 44 and 40.

What is the mean and median of their marks? What is the mark which is obtained by the highest number of students? 10. The table of daily wages of 100 workers working in a workshop is given

below. Daily wages (in Taka) Number of workers

51 – 55 6 56 – 60 20 61 – 65 30 66 – 70 15 71 – 75 11 76 – 80 8 81 – 85 6 86 – 90 4

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What is the mean and median of daily wages of workers? How much is the amount of daily wages which is earned by the highest number of workers? 11. The percentages of population according to different age-groups of population

as found in the census of 1991 are given below. Find the mean, median and mode of their ages.

Age - groups Population (%) 0 – 4 16⋅5 5 – 9 16⋅5

10 – 14 12⋅1 15 – 19 8⋅4 20 – 24 8⋅3 25 – 29 8⋅5 30 – 34 6⋅2 35 – 39 5⋅6 40 – 44 4⋅3 45 – 49 3⋅4 50 – 54 2⋅9 55 – 59 1⋅9 60 – 64 2⋅1 65 – 69 1⋅1 70 – 74 1⋅1 75 – 79 0⋅4 80 – 84 0⋅7

Total = 100 12. In the census of 1981 the population of Bangladesh was 87119965 and of them

males were 44919191 and females were 42200774. In the census of 1991 the population of Bangladesh become 106314992 and of them males were 54728350 and females were 51586642.

Find the percentage increase in population, percentage increase in males and percentage increase in females in the census of 1991. Find the average of males and also females of population as found in two censuses.

112

Multiple Choice Questions [Marks (√) on the correct answer] 1. (i) Statistics is an informative science. (ii) Statistics is a branch of Applied Mathematics. (iii) Statistics is the information expressed by one number only. Which one of the following is correct in view of the above

information? (a) i and ii (b) i and iii (c) ii and iii (d) i, ii and iii 2. By which one of the following does mean class interval mean? (a) Sum of first and last data among the data. (b) Sum of greatest data and smallest data of each of the class. (c) Difference of first and last data. (d) Difference between the greatest number and smallest number

of each class. 3. Which one of the following is the indicator of number of data in a

class? (a) frequency of the class (b) Mid value of class (c) class interval (d) Cumulative frequency. 4. Which one of the following is the average of 8,10,11,13 and 12? (a) 1.8 (b) 1.08 (c) 10.5 (d) 10.8 5. Which one of the following is the median of the numbers 15, 17,14,

and 10? (a) 13.5 (b) 14 (c) 14.5 (d) 15.5 6. Which one of the following is the mode of 5,11,13,6,13 and 11? (a) 13 and 5 (b) 11 and 6 (c) 5 and 6 (d) 11 and 13 7. Which one of the following is the formula to find the median? [Where L=lower limit of median class, n = total number of data, fm = frequency of median class, Fc = the cumulative frequency of the preceding class of median

N.M.G. -16

113

class, d= class interval.]

(a) L + m

c fdFn

×⎟⎠⎞

⎜⎝⎛ +

2 (b) L+

Cm F

dfn×⎟⎠⎞

⎜⎝⎛ −

2

(c) L+ m

c fdFn

×⎟⎠⎞

⎜⎝⎛ −

2 (d)

cm F

dfnL ×⎟⎠⎞

⎜⎝⎛ ++

2

Answer the questions (8-10) in view of the following informations: Classified marks in mathematics obtained by 40 students of class eight of Chalita Baria R.D. High school are given below:

class 36-41 42-47 48-53 54-59 60-65 Marks 6 7 14 8 5

8. Which one of the following is the class interval of the above classified data?

(a) 4 (b) 5 (c) 6 (d) 7 9. Which one of the following is the median of the third class interval? (a) 47.5 (b) 50.5 (d) 51.5 (d) 53.5 10. Which one of the following is the lower limit of the given mode

class? (a) 60 (b) 54 (c) 48 (d) 42

CREATIVE QUESTIONS

1. Marks in science obtained by 30 students of class eight in annual examination of Goga Kaliany Secondary School are given below:

42, 45, 60, 61, 58, 53, 48, 52, 51, 49, 73, 52, 57, 71, 64, 49, 56, 48, 67, 63, 70, 59, 54, 46, 43, 56, 69, 43, 68, 52.

(a) Find the number of class by taking class interval as 5. (b) Classify the unclassify data taking class interval 5 and find the

cumulative frequency. (c) Find the average of the classified data obtained in (b) using

algebraic formula.

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2. The table of daily wages of 50 labours of the village Isapur are given below: Daily income 41-50 51-60 61-70 71-80 81-90 91-100 Number of labour 6 8 13 10 8 5

(a) Construct a table of cumulative frequency. (b) Find the median from the table. (c) Find the mode and also find the difference between obtained

from (b) median and mode. 3. The marks obtained in evaluation Examination of 30 students of

class eight of Sonaton Kati Junior High School are given below: 62, 74, 80, 79, 81, 78, 74, 70, 73, 71, 69, 68, 67, 68, 64, 66, 64, 79,

64, 63, 64, 62, 60, 61, 59, 60, 57, 49, 50, 40. (a) Find the mean directly. (b) Find the median of these marks and also the mark which is

obtained by most of the students. (c) Find the mean constructing a table taking class interval as 5. 4. The weight of 25 students of class eight of Buruj Bagan Secondary

School are given in kg. below. 56, 46, 62, 46, 58, 48, 41, 54, 51, 47, 51, 40, 55, 63, 57, 47, 53, 54,

51, 43, 52, 53, 46, 53, 56. (a) Construct the frequency distribution using 40-44, 45-49, 50-54,

55-59, type of class interval. (b) Find the median of the frequency distribution. (c) Find the mean and mode of the weights of the students.


ALGEBRA

Chapter I

Algebraic Formulae and their

Application

1.1 Any general law expressed by algebraic symbols is known as algebraic formula or formula in short. The first four formulae and the corollary related to them have been elaborately discussed in class VII. Those are discussed below in short and many examples are given for acquiring knowledge regarding applications. 1.2. First Four Formulae Formula 1 : (a + b)2 = a2 + 2ab + b2. In words, the square of the sum of two quantities is equal to the sum of the squares of those two quantities and twice their product. Formula 2 : (a − b)2 = a2 − 2ab + b2. In words, the square of the difference of two quantities is equal to the result obtained by twice their product from the sum of the squares of those two quantities. Formula 3 : a2 − b2 = (a + b) (a − b)

In words, difference of squares of two quantities is equal to the product of the sum and the difference of those two quantities. Formula 4 : (x + a) (x + b) = x2 + (a + b) x + ab. In words, if the first term of two binomial expressions is the same, then their product will be equal to the square of the first term, product of the first term with the sum of second terms with their usual signs and product of two second terms with their usual signs. Corollary 1: a2 + b2 = (a + b)2 − 2ab. Corollary 2: a2 + b2 = (a − b)2 + 2ab. Corollary 3 : (a + b)2 = (a − b)2 + 4ab. Corollary 4: (a − b)2 = (a + b)2 − 4ab. Corollary 5 : 2 (a2 + b2) = (a + b)2 + (a − b)2: Corollary 6: 4ab = (a + b)2 − (a − b)2.

or, a b = ⎝⎜⎛

⎠⎟⎞a + b

22 −

⎝⎜⎛

⎠⎟⎞a − b

22

117

Square of trinomials : (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca. Example 1. Find the square of 7a + 9b.

Solution : (7a + 9b)2 = (7a)2 + 2 (7a) (9b) + (9b)2 = 49a2 + 126ab + 81b2

Answer : 49a2 + 126ab + 81b2. Example 2. Find the square of 4x3 − 5y2. Solution : (4x3 − 5y2)2 = (4x3)2 −2 (4x3) (5y2) + (5y2)2 = 16x6− 40x3y2 + 25y4. Answer : 16x6 − 40x3 y2 + 25y4

Example 3. If a + b = 5 and ab = 6, then find the value of a2 + b2. Solution: a 2 + b 2 = (a + b ) 2 – 2ab = 5 2 – 2 .6 = 25 – 12 = 13 Answer: 13. Example 4. If a – b = 7 and ab = 44, then find the value of a2 + b2. Solution : a2 + b2 = (a – b)2 + 2ab = 72 + 2.44 = 49 + 88 = 137 Answer: 137. Example 5. If x – y = a and xy = b, then what is the value of (x + y)2 ?

Solution : (x + y)2 = (x – y)2 + 4xy = a2 + 4b. Answer: a2 + 4b.

Example 6. If x + 1x = 2, then find the value of x2 +

1x2 .

Solution : x2 + 1x2 = (x)2 +

⎝⎜⎛⎠⎟⎞1

x2 =

⎝⎜⎛

⎠⎟⎞x +

1x

2 – 2x⋅

1x = 22 −2 = 4 – 2 = 2

Answer: 2. Example 7. If a = 21 and b = 9, then find the value of 16a2 – 72ab + 81b2 . Solution : 16a2 – 72ab + 81 b2 = (4a)2 – 2(4a) (9b) + (9b)2

= (4a – 9b)2 = (4.21 – 9.9)2 = (84 – 81)2 = 32 = 9. Answer: 9 Example 8. Find the square of 2a – 3b – 4c. Solution : (2a – 3b – 4c)2 = {2a – (3b + 4c)}2

= (2a )2 – 2 (2a) (3b + 4c) + (3b + 4c)2

= 4a2 – 12ab – 16ac + 9b2 + 24bc + 16c2

= 4a2 + 9b2 + 16c2 – 12ab + 24bc – 16ac. Answer : 4a2 + 9b2 + 16c2 – 12ab + 24bc – 16ac.

118

Example 9. Find the square of 4x + 5y – 7z. Solution : We know that, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca Assuming, a = 4x, b = 5y and c = – 7z, we get (4x + 5y – 7z)2

= (4x)2 + (5y)2 + (–7z)2 + 2 (4x) (5y) + 2 (5y) (–7z) + 2 (–7z) (4x). = 16x2 + 25y2 + 49z2 + 40xy – 70yz – 56zx. Answer : 16x2 + 25y2 + 49z2 + 40xy – 70yz – 56zx. Example 10. Simplify : (3a + 5b + 7c)2 + (7c – 4a + 5b)2 –2(3a + 5b + 7c) (7c – 4a + 5b). Solution : Let, x = (3a + 5b + 7c) and y = (7c – 4a + 5b)

∴ Given expression = x2 + y2 – 2xy = (x – y)2

= {(3a + 5b + 7c) – (7c – 4a + 5b)}2 [substituting the value of x and y] = (3a + 5b + 7c – 7c + 4a – 5b)2 = (7a)2 = 49a2 Answer : 49a2

Example 11. Resolve into factors : (8x2 + 9y3 – 1lz4)2 – (2x2 – 9y3 + 11z4)2

Solution : We know, a2 – b2 = (a + b) (a – b). ∴ Given expression

= {(8x2 + 9y3 – 11z4) + (2x2 – 9y3 + llz4)} {(8x2 + 9y3 – 1 lz4) – (2x2 − 9y3 + 11z4)}

= (8x2 + 9y3 – 11z4 + 2x2 – 9y3 + llz4) (8x2 + 9y3 – llz4 – 2x2 + 9y3 – llz4)

= 10x2 (6x2 + 18y3 – 22z4) = 20x2 (3x2 + 9y3 – llz4). Answer : 20x2 (3x2 + 9y3 – llz4). Example 12. Find the square of 9999. Solution : 9999 = 10000 −1 ∴ (9999)2 = (10000 – 1)2 = (10000)2 – 2.10000.1 + (1)2 = 100000000 – 20000 + 1 = 99980001. Answer: 99980001. Example 13. Express (4x − 3y) (6x + 5y) as the difference of two squares.

Solution : We know, ab = ⎝⎜⎛

⎠⎟⎞a + b

22 –

⎝⎜⎛

⎠⎟⎞a – b

22

∴ (4x − 3y) (6x + 5y) = ⎝⎜⎛

⎠⎟⎞4x – 3y + 6x + 5y

22 –

⎝⎜⎛

⎠⎟⎞4x – 3y – 6x – 5y

22

119

= ⎝⎜⎛

⎠⎟⎞10x + 2y

22 –

⎝⎜⎛

⎠⎟⎞– 2x – 8y

22

= (5x + y)2 – (– x – 4y)2 = (5x + y)2 – {(– 1) (x + 4y)}2 = (5x + y)2 – (x + 4y)2 Answer: (5x + y)2 – (x + 4y)2

Example 14. Find the product of (x − 12) and (x + 13) with the help of formula. Solution : We know, (x + a) (x + b) = x2 + (a + b) x + ab.

∴ (x – 12) (x + 13) = x2 + (–12 + 13)x + (–12)(13) = x2 + x – 156. Answer : x2 + x – 156. Example 15. Find the product of (5x2 – 16) and (5x2 – 11). Solution : We know, (x + a) (x + b) = x2 + (a + b) x + ab.

∴ (5x2 – 16) (5x2 – 11) = (5x2)2 + (– 16 – 11) (5x2) + (– 16) (– 11) = 25x4 – 27.5x2 + 176 = 25x4 – 135x2 + 176.

Answer : 25x4 – 135x2 + 176.

Example 16. Find the value of 9467 × 9467 – 2533 × 2533.

Solution : 9467 × 9467 – 2533 × 2533 = (9467)2 – (2533)2

= (9467 + 2533) (9467 – 2533) = 12000 × 6934 = 83208000. Answer: 83208000.

EXERCISE 1.1

1. Find the square of following expressions with the help of the formula: (i) 5a + 7b ; (ii) 7x – 9y ; (iii) 4ab + 5bc ; (iv) 6x2y – 5xy2 ; (v) 4x3 + 3y4;

(vi) x2 – 3 ; (vii) x3 + xy ; (viii) 11a – 12b; (ix) ax – by; (x) – x – y; (xi) – a + b; (xii) – xyz – abc ; (xiii) a2x3 – b2y4 ; (xiv) a – b + c ; (xv) x2 – y2 – z2 ; (xvi) x2yz – y2zx ; (xvii) xy + yz – zx ; (xviii) 301 ; (xix) 606; (xx) 597; (xxi) 999; (xxii) 1002; (xxiii) 5a – 6b – 7c; (xxiv) 7a2 + 8b2 – 5c2 ; (xxv) a – b – c – d. 2. Simplify:

i) (2a + 3b)2 – 2 (2a + 3b) ( 3b – a) + (3b – a)2

ii) (3x2 – 7y2)2 + 2(3x2 – 7y2) (3x2 + 7y2) + (3x2 + 7y2)2.

N.M.G. -17

120

iii) (5x – 9)2 – 2(5x – 9) (9 + 5x) + (9 + 5x)2. iv) (2a – 4b + 7c)2 + (2a + 4b + 7c)2 – 2(2a – 4b + 7c) (2a + 4b + 7c). v) (3x2 – y2 – 4z2)2 + 2(3x2 – y2 – 4z2) (6z2 + y2 – 3x2) + (6z2 + y2 – 3x2)2. vi) (5x2 – 3x – 2)2 – 2(5x2 – 3x – 2) (2 + 5x2 – 3x) + (2 + 5x2 – 3x)2. vii) 4569 × 4569 + 2 × 4569 × 5431 + 5431 × 5431

3. If a = 4, b = 6 and c = 3, then find the value of 4a2b2 – 16ab2c + 16b2c2. 4. If x = 4, y = – 8 and z = 5, then find the value of 25 (x + y)2 – 20 (x + y) (y + z) + 4 (y + z)2. 5. If m = 5, n = 7, then find the value of 16 (m2 + n2)2 + 56 (m2 + n2) (3m2– 2n2) + 49(3m2 – 2n2)2.

6. If x – 1x = 3, then find the value of x2 + 1x2 .

7. If a + 1a = 4, then find the value of a4 + 1a4.

8. If a – 1a = m, then show that, a4 + 1a4 = m4 + 4m2 + 2.

9. If a – 1a = 4, then show that, a2 + ⎝⎜⎛⎠⎟⎞1

a2

= 18.

10. If a + b = 8 and ab = 15, then find the value of (a – b)2 and a2 + b2. 11. If x – y = 7 and xy = 60, then find the value of (x + y)2 and x2 + y2. 12. If x + y = 12 and xy = 27, then find the value of (x – y)2 and x2 + y2.

13. If x – 1x = 5, then find the value of (x + 1x )2.

14. If a + b = 13 and a – b = 3, then find the value of (2a2 + 2b2) and ab. 15. If x + y = 19 and x – y = 11, then find the value of 2x2 + 2y2 and 4xy. 16. Multiply with the help of the formula :

(i) (x + y)(x – y); (ii) (5x + 7y) (5x – 7y); (iii) (7a – 11) (7a + 11); (iv) (a + b – c) (a – b + c); (v) (5a + 2b – 3c) (5a + 2b + 3c); (vi) (ax – by + cz) (ax + by – cz); (vii) (a2 + b2) (a2 – b2) (a4 + b4); (viii) (x + 10) (x – 14); (ix) (3a – 10) (3a – 5); (x) (6x + 17) (6x – 13); (xi) (ax + by + 5) (ax + by + 3).

17. Express as the square of the difference of two expressions. (i) (5a + 2b) (7a + 6b); (ii) (5x + 13) (5x – 13);

(iii) (3x + 5y) (7x – 5y); (iv) (6a + 9b) (7b – 8a).

121

1.3 Formula of Cubes and Their Corollaries Formula 5: (a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + b3 + 3ab (a + b) Proof : (a + b)3 = (a + b) (a + b)2 = (a + b) (a2 + 2ab + b2)

= a(a2 + 2ab + b2) + b(a2 + 2ab + b2) = a3 + 2a2b + ab2 + a2b + 2ab2 + b3 = a3 + 3a2b + 3ab2 + b3 = a3 + 3ab(a + b) + b3 = a3 + b3 + 3ab (a + b).

Corollary 1. a3 + b3 = (a + b)3 – 3ab (a + b) Formula 6 : (a – b)3 = a3 – 3a2b + 3ab2 +b =a3 − b3 − 3ab (a − b) Proof: (a – b)3 = (a –b)(a − b)2.

= (a – b)(a2 – 2ab + b2) = a(a2 – 2ab + b2) – b(a2 – 2ab + b2) = a3 – 2a2b + ab2 – a2b + 2ab2 – b3

= a3 – 3a2b + 3ab2 – b3 = a3 – 3ab (a – b) – b3 = a3 – b3 – 3ab (a – b).

Corollary 2. a3 – b3 = (a – b)3 + 3ab (a – b). Example 1. Find the cube of (2a + 3b)

Solution : (2a + 3b)3 = (2a)3 + 3 (2a)2 (3b) + 3 (2a) (3b)2 + (3b)3 = 8a3 + 3.4a2.3b + 3.2a.9b2 + 27b3 = 8a3 + 36a2b + 54ab2 + 27b3.

Answer : 8a3 + 36a2b + 54ab2 + 27b3. Example 2. Find the cube of (5a – 7b)

Solution : (5a – 7b)3 = (5a)3 – 3 (5a)2 (7b) + 3.(5a) (7b)2 – (7b)3 = 125a3 – 3.25a2. 7b + 3.5a. 49b2 – 343b3 = 125a3 – 525a2 b + 735ab2 – 343b3.

Answer : 125a3 – 525a2b + 735ab2 – 343b3

122

Example 3. Find the cube of (5x – 4y)

Solution : (5x – 4y)3 = (5x)3 – 3(5x)2 (4y) + 3. (5x) (4y)2 – (4y)3 = 125x3 – 3.25x2.4y + 3. 5x. 16y2 – 64y3 = 125x3 – 300x2y + 240xy2 – 64y3

Answer: 125x3 – 300x2y + 240xy2 – 64y3

Example 4. Find the cube of (a + b – c.)

Solution : (a + b – c)3 = {(a + b) – c}3 = (a + b)3 – 3(a + b)2. c + 3(a + b).c2 – c3 = (a3 + 3a2b + 3ab2 + b3) – 3 (a2 + 2ab + b2).c + 3(a + b). c2 – c3 = a3 + 3a2b + 3ab2 + b3 – 3a2c – 6abc – 3b2c + 3ac2 + 3bc2 − c3 = a3 + b3 – c3 + 3a2b + 3ab2 – 3a2c + 3ac2 – 3b2c +3bc2 – 6abc.

Answer: a3 + b3 – c3 + 3a2b + 3ab2 – 3a2c + 3ac2 – 3b2c + 3bc2 – 6abc.

Example 5. Simplify : (3x + 5y)3 + 3(3x +5y)2 (2x – 5y) + 3(3x+5y) (2x – 5y)2 + (2x – 5y)3.

Solution : Let, 3x + 5y = a and 2x – 5y = b. ∴ Given expression = a3 + 3a2b + 3ab2 + b3 = (a + b)3

= {(3x + 5y) + (2x – 5y)}3 = (3x + 5y + 2x – 5y)3 = (5x)3 = 125x3.

Answer : 125x3

Example 6. Simplify : (4x – 8y)3 – (3x – 9y)3 – 3 (x + y) (4x – 8y) ( 3 x – 9 y ) Solution : Let, (4x – 8y) = a and (3x – 9y) = b.

∴ a – b = (4x – 8y) – (3x – 9y) = 4x – 8y – 3x + 9y = x + y. Now given expression = a3 – b3 – 3 (a – b) a.b = a3 – b3 – 3 ab (a – b)

= (a – b)3 = (x + y)3 [putting the values of (a – b)] = x3 + 3x2y + 3xy2 + y3.

Answer : x3 + 3x2y + 3xy2 + y3. Example 7. If a + b = 8 and ab = 15, then find the value of a3 + b3. Solution : a3 + b3 = (a + b)3 – 3ab (a + b) = 83 – 3. 15.8 = 512 – 360 = 152. Answer: 152.

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Example 8. If x – y = 1 and xy = 30, then find the value of x3 – y3. Solution : x3 – y3 = (x – y)3 + 3xy (x – y) = 13 + 3.30.1 = 1 + 90 = 91. Answer: 91. Example 9. If x + y = 3, then what is the value of (x3 + y3 + 9xy)

Solution : x3 + y3 + 9xy = x3 + y3 + 3.3.xy = x3 + y3 + 3(x + y) xy; [Q 3 = x + y] = x3 + y3 + 3xy (x + y) = (x + y)3 = 33 = 27. Answer: 27. Example 10. If x = 2, then find the value of (27x3 – 135x2 + 225x –125) Solution : Given expression

= (3x)3 – 3.9x2.5 + 3.3x.52 – (5)3 = (3x – 5)3

= ( 3 . 2 – 5 ) 3 = ( 6 – 5 ) 3 = ( l ) 3 = l Answer : 1

Example 11. If x = 1, then find the value of 27x3 + 54x2 + 36x + 3 Solution : Given expression = (3x)3 + 3.9x2.2 + 3.3x.22 + (2)3 – 5 = (3x + 2)3 – 5 = (3.1 + 2)3 – 5 = 53 – 5 = 125 – 5 = 120 Answer: 120

Example 12. If a – 1a = 4, then find the value of a3 –

1a3

Solution : a3 – 1a3 = a3 –

⎝⎜⎛⎠⎟⎞1

a3

= ⎝⎜⎛

⎠⎟⎞a –

1a

3 + 3a .

1a ⎝⎜⎛

⎠⎟⎞a –

1a = 43 + 3.4 = 64 + 12 = 76

Answer: 76. 1.3.1. Some more Formulae related to Cubes Formula 7 : a3 + b3 = (a + b) (a2 – ab + b2) Proof : a3 + b3 = (a + b)3 – 3ab(a + b) = (a + b) {(a + b)2 –3ab} = (a + b) (a2 + 2ab + b2 – 3ab) = (a + b) (a2 – ab + b2) Conversely, (a + b) (a2 – ab + b2) = a3 + b3.

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Now. (a + b) (a2 – ab + b2) = a(a2 – ab + b2) + b(a2 – ab + b2)

= a3 – a2b + ab2 + a2b – ab2 + b3 = a3 + b3 The formula can be proved in this way also. Formula 8 : a3 – b3 = (a – b) (a2 + ab + b2) Proof : a3 – b3 = (a – b)3 + 3ab (a – b) = (a – b) {(a – b)2 + 3ab} = (a – b) (a2 – 2ab + b2 + 3ab). = (a – b) (a2 + ab + b2). Conversely, (a – b) (a2 + ab + b2) = a (a2 + ab + b2) – b(a2 + ab + b2) = a3 + a2b + ab2 – a2b – ab2 – b3 = a3 – b3

Example 13. Resolve into factors : 8x3 + 27y3. Solution : 8x3 + 27y3 = (2x)3 + (3y)3 = (2x + 3y) {(2x)2 – (2x) (3y) + (3y)2} = (2x + 3y) (4x2 – 6xy + 9y2) Answer : (2x + 3y) (4x2 – 6xy + 9y2). Example 14. Resolve into factors : 64a3 − 125b3

Solution : 64a3 – 125b3 = (4a)3 – (5b)3

= (4a – 5b) {(4a)2 + (4a) (5b) + (5b)2} = (4a – 5b) (16a2 + 20ab + 25b2). Answer : (4a – 5b) (16a2 + 20ab + 25b2). Example 15. Multiply (49a2 – 28ab + 16b2) by (7a + 4b) Solution : (7a + 4b) (49a2 – 28ab + 16b2)

= (7a + 4b) {(7a)2 – (7a) (4b) + (4b)2} = (7a)3 + (4b)3 = 343a3 + 64b3

Answer: 343a3 + 64b3.

Example 16. Multiply (9x2 + 6xyz + 4y2z2) by (3x – 2yz)

Solution : (3x – 2yz) (9x2 + 6xyz + 4y2z2) = (3x – 2yz) {(3x)2 + 3x (2yz) + (2yz)2} = (3x)3 – (2yz)3 = 27x3 – 8y3z3. Answer : 27x3 – 8y3z3.

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EXERCISE 1.2

1. Find the cube of the following expressions : (i) 3x + 4y; (ii) x3 + y2 ; (iii) a2b + c2d: (iv) ab + bc; (v) 8x2 + 11y3; (vi) 7m + 5n; (vii) x + y + z; (viii) x3 + y3; (ix) 2x – 5y; (x) p2 – q2; (xi) 11a – 12b; (xii) x3 + 2; (xiii) x6 − 8; (xiv) 2x – 3y − z; (xv) x2 – y2 + z2;

(xvi) a2b – b3c; (xvii) xy – 2yz; (xviii) a2b2 – c2d2; (xix) x3 – 2y3; (xx) 7x2 – 9y2. 2. Simplify: i. (3a + 5b)3 + 3(3a + 5b)2 (3a – 5b) + 3(3a + 5b) (3a – 5b)2 + (3a – 5b)3; ii. (x + 2y)3 – 3(x + 2y)2 (x – 2y) + 3(x + 2y) (x – 2y)2 – (x – 2y)3. iii. (3a – 8b)3 – (2a – 7b)3 – 3(3a – 8b) (2a – 7b) (a – b); iv. (7x – 6)3 − (5x – 6)3 – 6x(7x – 6)(5x – 6); v. (x + y – z)3 + (x – y + z)3 + 6x(x2 – y2 – z2 + 2yz); vi. (x – y)3 + (x + y)3 + 6x(x2 – y2). 3. If a + b = 10 and ab = 21, then find the value of a3 + b3. 4. If 2x + 3y = 13 and xy = 6, then find the value of 8x3 + 2zy3. 5. If x – y = 10 and xy = 30, then find the value of x3 – y3. 6. If 5a – 7b = 14 and ab = 21, then find the value of 125a3 – 343b3. 7. If x + y = 2, show that, x3 + y3 + 6xy = 8. 8. If a = –3 and b = 2, then find the value of 8a3 + 36a2b + 54ab2 + 27b3. 9. If x = 5, then find the value of x3 – 12x2 + 48x – 64. 10. If a = 7, then find the value of a3 + 6a2 + 12a + 1. 11. If x – 2y = 5, then find the value of x3 – 8y3 – 30xy. 12. If 3x – 2y = 8, then find the value of 27x3 – 8y3 –144xy. 13. If a2 + b2 = c2, then show that, a6 + b6 + 3a2b2c2 = c6.

14. If a + 1a = 3, then show that, a3 +

1a3 = 18.

15. If x – 1x = 5, then find the value of x3 − (

1x )3.

16. Find the product with the help of formula : (i) (a2 + b2) (a4 – a2b2 + b4) ; (ii) (x2 + 2) (x4 – 2x2 + 4) ; (iii) (2a + 3b) (4a2 – 6ab + 9b2) ; (iv) (7a + 4b) (49a2 – 28ab + 16b2)

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(v) (4a – 3b) (16a2 + 12ab + 9b2); (vi) (8x – 3y) (64x2 + 24xy + 9y2); (vii) (ax – by) (a2x2 + abxy + b2y2); (viii) (x2 + a) (x4 – ax2 + a2); (ix) (x + a) (x2 – ax + a2) (x – a) (x2 + ax + a2); (x) (2a – 1) (4a2 + 2a + 1) (8a3 + 1); (xi) (5a + 3b) (25a2 – 15ab + 9b2) (125a3 – 27b3) . 17. Resolve into factors : (i) a3 + 8: (ii) 8x3 + 216y3; (iii) 8x3 + 343; (iv) 27a3b3 + 64b3c3; (v) 8a4 + 27ab3; (vi) 64b3 – 125; (vii) 27a3 – 8; (viii) 24x3 – 81y3; (ix) 56x3 – 189y3; (x). 729a3 – 64b3c6. 1.4. Resolution into Factors of quadratic expression

Discussion regarding factors has been made in class VII. It is discussed briefly here again. When any expression is expressed as the product of two or more simple expressions, then it is said to have been resolved into factors and each of such simple expressions is called the factor of the first expression. For example, when the expression x2 + 5x is resolved into factors we get, x2 + 5x = x(x + 5), where x and x +5 are the factors. Rules of resolving expressions into factors are stated below : (a) Arranging conveniently : ax – cy + cx – ay is arranged as, ax + cx – ay – cy. Now, ax + cx – ay – cy = x(a + c) – y (a + c) = (a + c) (x – y)

(b) Expressing an expression in the form of square :

x2 + 8xy + 16y2 = x2 + 2.x.4y + (4y)2 = (x + 4y)2 = (x + 4y) (x + 4y)

(c) Expressing an expression as the difference of two squares and then applying the formula a2 – b2.

Adding y2 once with and again subtracting y2 from x2 +2xy –2y– 1 it becomes x2 + 2xy + y2 – y2 – 2y –1. Now, x2 + 2xy + y2 – y2 – 2y – 1 = (x2 + 2xy + y2) – (y2 + 2y + 1) = (x + y)2 – (y+1)2

= (x + y + y + 1)(x + y – y – 1) = (x + 2y + 1)(x – 1). (d) Applying the formula, x2 + (a + b) x + ab = (x + a) (x + b):

x2 + 5x + 6 = x2 + (3 + 2) x + 3.2 = (x + 3)( x + 2).

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(e) Expressing the expression in the form of cubes : x3 + 6x2 +12x + 8 = x3 + 3.x2.2 + 3.x.22 + 23 = (x + 2)3 = (x + 2) (x + 2) (x + 2). (f) Applying the formulae : a3 + b3 = (a + b)( a2 – ab + b2) and a3 – b3 = (a – b)(a2 + ab + b2). 8x3 + 27 = (2x)3 + (3)3 = (2x + 3) {(2x)2 – (2x)(3) + (3)2}

= (2x + 3)(4x2– 6x + 9)

Again, x3 – 64 = (x)3 – (4)3 =(x – 4){(x)2 + (x) (4) + (4)2}

= (x – 4) (x2 + 4x + 16)

1.4.1 Factors of the Expression in the Form of x2 + px + q. If the expression of left hand side of the formula, x2 + (a + b) x + ab = (x + a) (x + b) is compared with the expression, x2 + px + q, then it is found that in both the expressions three terms occur, the first one being x2 whose coefficient is 1 (one), the second or middle term contains x whose coefficients are (a + b) and p respectively and the third term is free from x, where there are ab and q respectively. x2 + (a + b) x + ab consists of two factors. Therefore, x2 + px + q has also two factors. Let, two factors of x2 + px + q are (x + a) and (x + b)

Hence, x2 + px + q = (x + a) (x + b) = x2 + (a + b ) x + ab. Then, p = a + b and q = ab. Now in order to find the factors of x2 + px + q, q is to be expressed in two such factors so that their sum (algebraic) is equal to p. If it is required to resolve x2 + 9x + 20 into factors, then the number 20 is to be expressed into two such factors whose sum is 9 and the product is 20. The possible pairs of factors of 20 are (1, 20), (2, 10) and (4, 5). Of them, the sum of the pair (4, 5) is (4 + 5) = 9 and the product is 4. 5 = 20. ∴ x2 + 9x + 20 = (x + 4) (x + 5). This method is called middle term distribution or middle term break -up. Remark : If p and q are both positive, then in order to resolve into factors of the expression in the form of x2 + px + q, x2 – px + q, x2 + px – q and x2 – px – q, both the factors of q will be of the same sign i. e. both the factors will be positive or negative since q is positive in the first and second expressions. In this case, if p is positive, then both the factors of q are positive and if p is negative, then both the factors of q are negative. In the third and fourth expression, q, is negative i.e.(–q) and hence two factors of q will be of opposite sign and if p is positive, the positive

N.M.G. -18

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number of two factors will be greater than the absolute value of the negative number and if p is negative, the absolute value of negative number of two factors will be greater than the positive number. Example 1. Resolve into factors : x2 + 18x + 72. Solution : We have to find two such positive numbers whose product is 72 and their sum is 18. The possible pairs of factors of 72 are (1, 72), (2,36), (3,24) (4, 18), (6, 12) and (8, 9). Of them, the sum of numbers of the pair (6, 12) is 6 + 12 = 18. ∴ x2 + 18x + 72 = x2 + 6x + 12x + 72

= x (x + 6) + 12 (x + 6) = (x + 6) (x + 12) Answer : (x + 6) (x +12). Example 2. Resolve into factors : x2 – 23x + 132. Solution : We have to find two such numbers whose product is 132 and their sum is – 23. Since the sum of two numbers is negative but the product is positive, therefore both the numbers will be negative. The possible pairs of factors of 132 are (– 1, –132), (– 2, – 66), (– 3, – 44), (– 4, – 33), (– 6, – 22) and (– 11, – 12). Of them, the sum of numbers of the pair (–11, – 12) is (– 11 – 12) = – 23. ∴ x2 – 23x + 132 = x2 – 11x – 12x + 132 = x(x – 11) – 12(x – 11) = (x – 11)(x – 12). Answer : (x – 11)(x – 12). Example 3. Resolve into factors : x2 + 15x – 54. Solution : We have to find two such numbers whose product is – 54 and their sum is 15. Since the sum of two numbers is positive but their product is negative, hence of two numbers, the number whose absolute value is greater than that of the other is positive and that number is negative whose absolute value is smaller than the other. The possible pairs of factors of (– 54) are (– 1, 54), (– 2, 27), (–3, 18),( – 6, 9). Of them, the sum of the numbers of the pair (– 3, 18) is (– 3 + 18) = 15. ∴ x2 + 15x – 54 = x2 + 18x – 3x – 54 = x(x + 18) – 3(x +18) = (x +18) (x – 3) Answer : (x +18)(x – 3).

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Example 4. Resolve into factors : x2 – 9x – 36. Solution : We have to find two such numbers whose product is (– 36) and their sum is (– 9). Since the sum of two numbers is negative and their product is negative, hence of the two numbers, that number is negative whose absolute value is greater than the other and that number is positive whose absolute value is smaller than that of the other. The possible pair of factors of ( – 36) are (1, – 36); (2, – 18), (3, – 12), (4, – 9) and (– 6, – 6). Of them, the sum of numbers of the pair (3, – 12) is ( – 12 + 3) = – 9. ∴ x2 – 9x – 36 = x2 – 12x + 3x – 36 = x(x – 12) + 3(x – 12) = (x – 12) (x +3). Answer: (x –12) (x + 3). Example 5. Resolve into factors : (x2 + 6x)2 + 5(x2 + 6x) – 84. Solution : Let, x2 + 6x = a. ∴ Given expression = a2 + 5a – 84. = a2 + 12a – 7a – 84 [Since 12( – 7) = – 84 and – 7 + 12 = 5.]

= a (a + 12) –7(a + 12) = (a + 12) (a – 7)

= (x2 + 6x + 12) (x2 + 6x – 7), [Putting the value of a.]

= (x2 + 6x + 12)(x2 + 7x – x – 7), [Since (–1). 7 = – 7 and (7 – 1) = 6]

= (x2 + 6x + 12) {x(x + 7) –l(x + 7)} = (x2 + 6x + 12)(x + 7)(x – 1). Answer : (x2 + 6x + 12) (x + 7)(x – 1). Remark : x2 + 6x + 12 can not be resolve into factors for any real value of x. 1.4.2. Factors of the Expression in the Form of ax2 + bx + c. Let, (rx + p )(sx + q) be the factors of the expression ax2 + bx + c. ∴ ax2 + bx + c = (rx + p) (sx + q) = rsx2 + (rq + sp) x + pq. Then, a = rs, b = rq + ps and c = pq. Hence, ac = rspq = rq. sp and b = rq + sp

Now to find the factors of ax2 + bx + c, the product of the coefficient of x2 and the term free from x is to be expressed in two such factors such that their sum is equal to b, the coefficient of x. To factorise 2x2 + 9x + 10, (2.10) = 20 is to be expressed into two such factors whose sum is 9 and their product is 20. The pairs of factors of 20 are (1, 20), (2, 10) and (4,5). Of them, the sum of the pair (4, 5) is (4+5) = 9 and their product is 4.5 = 20. ∴ 2x2 + 9x + 10 = 2x2 + 4x + 5x + 10 = 2x (x + 2) + 5 (x + 2) = (x + 2) (2x + 5).

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Remark : To factorise ax2 + bx + c, the rules which are followed for different values of p, q having positive and negative signs of x2 + px + q are also followed for different values of a, b, c having positive and negative signs. Here b for p and the product of a and c for q are to be considered. Example 6. Factorise 8x2 + 18x + 9. Solution : Here, 8.9 = 72. Now, 6.12 = 72 and 6 + 12 = 18. Therefore, 8x2 + 18x + 9 = 8x2 + 6x + 12x + 9 = 2x(4x + 3) + 3 (4x + 3)

= (4x + 3) (2x + 3) Answer : (4x + 3) (2x +3). Example : 7. Factorise 4x2 – 23x2 + 33. Solution : Here, 4.33 = 132. For the sum of (– 23), we get the pair (–11, –12)

where (– 11) + (– 12) = – 23 and (– 11) × (– 12) = 132. Therefore, 4x2 – 23x + 33 = 4x2 – 1lx – 12x + 33 = x(4x – 11) – 3(4x –11) = (4x – 11) (x – 3). Answer : (4x – ll) (x – 3). Example 8. Factorise 9x2 – 9x – 4. Solution : Here, 9.( – 4) = – 36. Now, 3.(– 12) = – 36 and (– 12) + 3 = – 9. ∴ 9x2 – 9x – 4 = 9x2 + 3x – 12x – 4 = 3x (3x + 1) – 4 (3x + 1). = (3x + l)(3x – 4). Answer : (3x + 1) (3x – 4). Example 9. Factorise 27x2+15x – 2. Solution : Here, 27. (– 2) = – 54. Now, (– 3). 18 = – 54 and (– 3 ) + 18 = 15. ∴ 27x2 + 15x – 2 = 27x2 – 3x + 18x – 2 = 3x(9x – 1) + 2(9x – 1)

= (9x – l) (3x + 2)

Answer : (9x – l)(3x + 2). 1.4. 3 H. C .F. and L. C. M. of Algebraic Expressions

The clear concept for finding H. C. F. and L. C. M. of two or three algebraic expressions having not more than three terms (including arithmetical coefficients) has already been given in class VII. A brief discussion is made here again.

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Common Factor : If two terms have the same factor, this is called a common factor. a is the common factor of a2b, ab and a2c; again (x + y) is the common factor of (x + y)3, (x + y)2 and x2 – y2. Highest Common Factor (H. C. F.)

The product of common prime factors of two or more expressions is called the Highest Common Factor or simply H. C. F. of those two or more expressions. The Highest Common Factor of three expressions, a3b4c3, a5b3c4 and a4b3c5 is a3b3c3. Again the Highest Common Factbr of three expressions, (x2 + y2)2, (x2 + y2)3 and (x4 – y4) is x2 + y2. Common Multiple

If any expression is completely divisible by two or more expressions, then the dividend is called the common multiple of those two or more divisors. The expression, a2b2c is the common multiple of ab, bc, ab2, a2c and b2c. The expression, (x + y)2 (x – y) is the common multiple of three expressions, (x + y), (x + y)2 and x2 – y2. Lowest Common Multiple (L. C. M.)

Among different multiples of two or more expressions the common multiple which consists of lowest number of prime factors is called Lowest Common Multiple or L.C.M. in short. The expression a2b2c is the L.C. M. of three expressions, a2bc; ab2 and abc. The expression, (x + y)3 (x – y) is the L.C.M. of three expressions, (x + y)2, ( x + y)3 and (x2 – y2). Example 1. Find the H. C. F. of 6a3b2c, 12a2b2c4 and 15a4b4c2. Solution : H.C.F. of 6, 12 and 15 = 3 and the common factors with highest common power among a3b2c, a2b2c4 and a4b4c2 are a2, b2 and c respectively. ∴ Required H. C. F. = 3a2b2c. Ans : 3a2b2c. Example 2. Find the H. C. F. of x3 + x2y, x2y + xy2, x3 + y3 and (x + y)3. Solution : Here, 1st expression = x3 + x2y = x2(x + y) 2nd expression = x2y + xy2 = xy (x + y) 3rd expression = x3 + y3 = (x + y) (x2 – xy + y2) and 4th expression = (x + y)3 = (x + y) (x + y) (x + y).

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Here the common factor of the expressions is (x + y) and this factor with the highest common power among the expressions is (x + y). ∴ Required H. C. F = ( x + y) Answer : ( x +y ). Example 3. What is the L.C.M. of 4x2y2z and 6xy2?

Solution : L.C.M. of 4 and 6 = 12

The highest factor between the given expressions are x2, y2 and z respectively. ∴ The required L. C. M. = 12x2y2z. Answer :12x2y2z. Example 4. Find the L. C. M. of 3(a4 + a3b) and a2b + 2ab2 + b3 Solution : Here, 1st expression = 3(a4 + a3b) = 3a3(a + b) and 2nd expression = a2b + 2ab2 + b3 = b (a2 + 2ab + b2) = b (a + b)2

Here the highest powers of the factors a, b and (a + b) are a3, b and (a + b)2

respectively. Again , the L. C.M. of the coefficients which are natural numbers = 3. ∴ Required L. C. M. = 3a3b(a + b)2. Answer : 3a3b(a + b)2. Example 5. Find the L. C. M of 24a2bc, 18ab3c2 and 54a4b2c2. Solution : L. C. M. of 24, 18 and 54 = 216

The factors, a, b and c with their highest powers among three given expressions are a4, b3 and c2 respectively. ∴ Required L. C. M. = 216a4b3c2 Answer : 216a4b3c2. Example 6. Find the L. C. M. of 4(x2 + ax)2, 6(x3 – a2x) and 14x3 (x3 – a3). Solution : L. C. M. of 4, 6 and 14 = 84. Now, part of 1st expression = (x2 + ax)2 = x2(x + a)2

2nd expression = (x3 – a2x) = x(x2 – a2) = x(x + a)(x – a) and 3rd expression = x3(x3 – a3) = x3(x – a)(x2 + ax + a2). Here the factors, x, (x + a), (x – a) and (x2 + ax + a2) with their highest powers are x3, (x + a)2, (x – a) and (x2 + ax + a2) respectively. ∴ Required L. C. M. = 84x3(x + a)2 (x – a) (x2 + ax + a2)

= 84x3(x + a)2 (x3 – a3). Answer : 84x3(x + a)2 (x3 – a3).

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Example 7. Find the L.C. M. of x3 + y3, x3 – y3 and x4 + x2y2 + y4. Solution: 1st expression = x3 + y3 = (x + y) (x2 – xy + y2) 2nd expression = x3 – y3 = (x – y) (x2 + xy + y2) 3rd expression = x4 + x2y2 + y4

= (x2)2 + 2x2y2 + (y2)2 – x2y2 = (x2 + y2)2 – (xy)2 = (x2 + xy + y2) (x2 – xy + y2)

∴ Required L. C. M. = (x + y) (x – y) (x2 + xy + y2) (x2 – xy + y2) = (x3 + y3) (x3 – y3) = x6 – y6. Answer : x6 – y6.

EXERCISE 1.3

Resolve into factors : (1) 4x2 – y2; (2) x2 – 144y2; (3) 3x – 75x3; (4) ax4 + 64a; (5) 4a4 + 81 ; (6) x2 – y2 + 2y – 1; (7) x4 – 6x2 + 1; (8) x4 + x2 + 1; (9) a2 – 2ab + 2b – 1; (10) a4 + a2b2 + b4; (11) 64x3 – 8y3; (12) (x –y)3 + z3; (13) a6 + b6; (14) a6 – b6; (15) x2 + x – 72; (16) x2 + 18x + 56; (17) x2 – 8x – 105; (18) 15 – 8x + x2; (19) x2 – 51x + 650; (20) 35 – 2x – x2; (21) x2 + 14x + 40; (22) a2 + 7ab + 12b2; (23) x2 + 7x – 120; (24) x2 + 11x – 42; (25) x2 + 3xy – 40y2; (26) p2 + 2pq – 80q2; (27) (x2 – x)2 + 3 (x2 – x) – 40; (28) (a2 + 7a)2 – 8(a2 + 7a) – 180; (29) (x2 + 6x)2 – 32 (x2 + 6x) – 320 ; (30) 6x2 – x – 15; (31) x2 + ax – (3a – 2)(4a – 2); (32) x2 + (3a + 4b)x + (2a2 + 5ab + 3b2); (33) (x + 5)(x + 13) – 9; (34) 3x2 + 11x – 4; (35) 8x2 + 2x – 15; (36) 3x2 – 16x – 12; (37) 2x2–9x – 35; (38) 2a2 + 7ab – 15b2; (39) 9a4 + 2a2b2 – 32b4; (40) 2(x + y)2 – 3(x + y) – 2; (41) 15x2 – 11xy – 12y2; (42) 2x2 – 5xy + 2y2; (43) 10p2 + 11pq – 6q2; (44) 2x2 + 5x– 42; (45) ax2 + (a2 + 1) x + a. Find the H. C. F. of the following : (46) 15a3b2c4, 25a2b4c3 and 20 a4b3c2; (47) 36a2b2c4d5, 54a5c2d4 and 90a4b3c2; (48) 18a3b4c5, 42a4c3d4, 60b3c4d5 and 78a2b4d3;

134

(49) 32x4y5z6, 40x5y4z3, 56x2y7z4 and 72x6y2z3; (50) x3 + x2y, x2y + xy2 and x3 + y3; (51) x2 – 3x, x2 – 9 and x2 – 4x + 3; (52) x2 – 3x + 2, x3 – 5x2 + 6x and x3 – x2 – 2x; (53) x2 – y2, x3 – y3 and (x – y)3; (54) (x + 2)2, (x2 + 2x) and (x2 + 5x + 6); (55) 18 (x + y)3, 24 (x + y)2 and 32 (x2 – y2); (56) a2b (a3 – b3), a2b2(a4 + a2b2 + b4) and a3b2 + a2b3 + ab4; (57) x3 – 3x2 – 10x, x3 + 6x2 + 8x and x4 – 5x3 – 14x2. Find the L. C. M. of the following : (58) 4xy, 9x2z, 3xzy2 and 12x2z3; (59) 3a2d3, 9d2b2, 12c3d2, 24a3b2 and 36c3d2; (60) 3x3y2, 4xy3z, 5x4y2z2 and 12xy4z2; (61) 24b3c2, 15c2a. 30a3b and 40a2bc; (62) 5a2b3c, 10ab2c3 and 15ab3c; (63) a3 + b3, (a + b)3, (a2 – b2)2 and (a2 – ab + b2)2; (64) x2 + 3x + 2, x2 – 1 and x2 + x – 2; (65) a3 – 1, a3 + 1 and a4 + a2 + 1; (66) (x + 2)2, (x2 + 2x) and (x2 + 5x + 6); (67) a2b (a3 – b3), a2b2 (a4 + a2b2 + b4) and (a3 + b3); (68) 6x2 – x – 1, 3x2 + 7x + 2 and 2x2 + 3x – 2.

Multiple Choice Questions [Mark (√) on the correct answer] 1. If a – b = 3 and ab = 4, then which one of the following is the value

of a2+b2? (a) 25 (b) 17 (c) 11 (d) 1 2. (i) x3–y3 = (x–y) (x2+xy+y2)

(ii) ab = 22

44⎟⎠⎞

⎜⎝⎛ −

−⎟⎠⎞

⎜⎝⎛ + baba

(iii) x3 + y3 = x3+y3+3xy (x+y)

135

Which one of the following is correct according to the above informations?

(a) i and ii (b) ii and iii (c) i and iii (d) i, ii and iii 3. Which one of the following is the L.C.M. of x3+x2y, x2y+xy2? (a) xy (b) x +y (c) xy (x+y) (d) x2y (x + y) Answer the questions (4 – 6) on the basis of the following

informations: If x + y = 8 and x – y = 2 then, 4. How much is the value of 2x2 +2y2? (a) 32 (b) 34 (c) 64 (d) 68 5. Which one of the following is the value of 4xy? (a) 68 (b) 64 (c) 62 (d) 60 6. Which one of the following is the value of x2 – y2? (a) 16 (b) 10 (c) 8 (d) 4 7. Which one of the following is the product of (x+8) and (x–7)? (a) x2 + x–56 (b) x2 –15x +56 (c) x2 +15x–56 (d) x2 – x +56 8. Which one of the following is the H.C.F. of 15 a3b2c4, 25a2b4c3 and 20 a4b3c2? (i) 5a2b2c2 (ii) 150 a4b4c4 (iii) 150a2b2c2 (iv) 5a4b4c4 Answer the questions (9 –11) on the basis of the following informations

If 51=−

xx then,

9. Which one of the following is the value of 21⎟⎠⎞

⎜⎝⎛ +

xx ?

(a) 29 (b) 25 (c) 21 (d) 5

N.M.G. -19

136

10. Which one of the following is the sum of x2 and 2

1x

?

(a) 27 (b) 25 (c) 23 (d) 10

11. Which one of the following is the value of ⎟⎠⎞

⎜⎝⎛ − 3

3 1x

x ?

(a) 15 (b) 110 (c) 125 (d) 140 CREATIVE QUESTIONS 1. x2+x−2, 2x3+12x2+24x+16, 2x2+3x−2 and 2x3−8x are four algebraic

expressions. (a) Factorise the first expression. (b) Factorise the 3rd and 4th expressions. (c) Find the H.C.F. and L.C.M. of the above expressions. 2. a2 + b2 and a6 + b6 + 3a2b2c2 are two algebraic expressions. (a) If a = c and b = –c, then find the value of 2nd expression. (b) If a +b = 5 and a–b =1, then find the value of ab and a2 + b2. (c) If a2 + b2 = c2 and c = 2, then show that the value of 2nd

expression is 64.

3. x

x 1+ , 2

2 1x

x + and 44 1

xx + are three algebraic expressions.

(a) If x

x 1+ =3, then find the value of 2

2 1x

x +

(b) If x

x 1+ = p, then prove that, 4

4 1x

x + = p4– 4p2+2

(c) If 44 1

xx + = 2, then find the value of

xx 1+ and 4

4 1x

x − .

Chapter II

Algebraic Fraction

2.1 Algebraic Fraction

In class VII the concept of algebraic fraction, reduction of the algebraic fractions in its lowest terms, making the fractions of common denominator, addition and subtraction of algebraic fractions have already been given. But here some of them with examples relating to algebraic fractions have been discussed again. 2.1.1. Algebraic Fraction

If p and q are two algebraic expression, then pq is an algebraic fraction. Here in the

fraction pq , p and q are called numerator and denominator of the fraction

respectively, where q ≠ o. xy ,

a + ba ,

x2 + y2

x + y etc. are algebraic fractions.

2.1.2. To express the Fraction in its Lowest Terms

If there are common factors in the numerator and denominator of any fraction and the numerator and denominator are divided by the highest common factor of numerator and denominator, then the new fraction formed by the quotients of numerator and denominator is the original fraction in its lowest term. x4y – x2y3

x5 – x4y = x2y(x2 – y2)

x4(x – y) = x2y(x + y)(x – y)

x4(x – y) = y(x + y)

x2

Here the numerator and denominator have been divided by x2(x – y) which is the highest common factor of numerator and denominator. 2.1.3. To Express with the Common Denominators

To express two or more fractions with the common denominators, L. C. M. of denominators is to be found out and then this L. C. M. is to be made the denominators of each of the fractions. At the time of expressing fractions with the common denominators and to keep the values of the fractions unchanged, the L. C. M. of the denominators is to be divided by the denominator of each of the fractions and the corresponding numerator is to be multiplied by the quotient thus obtained. This product will be the numerator of the corresponding fraction. For

138

example, the L. C. M. of the denominators, a, b, c of xa , yb and zc is abc. If abc is

divided by a, b and c, the quotients are bc, ca and ab respectively. Hence, if xa , yb

and zc are expressed as the fractions with the common denominators, then these are xbcabc ,

yacabc and

zababc respectively.

2.1.4. Addition of Fractions

To add two or more fractions it is required to express the fractions with the common denominators and then numerators are to be added. The sum of the given fractions will be a new fraction of which the numerator is the sum of numerators of the transformed fractions and denominator is the common denominator.

For example, xa +

yb +

zc =

xbcabc +

yacabc +

zababc =

xbc + yac + zababc

2.1. 5. Subtraction of the Fractions

To determine the difference of two fractions it is required to express two fractions with the common denominator and then two numerators are to be subtracted. The difference of two fractions will be a new fraction of which numerator will be the difference of two numerators and denominator will be the common denominator.

For example, xab –

ybc =

xcabc –

yaabc =

xc – yaabc

Remark : In case of addition and subtraction of the algebraic fractions, the fractions are to be expressed in its lowest terms if necessary.

For example, xpqapq +

yrsbrs +

zuvcuv =

xa +

yb +

zc =

xbc + yac + zababc

Example 1. Express 24a3b2x4

56a5bx6 in its lowest terms.

Solution : The common factors with the highest common power of the expressions, 24a3b2x4 and 56a5bx6 are respectively 23, a3, b and x4. Hence, the H. C. F. of two expressions is 8a3bx4.

∴ The given fraction = 3b.8a3bx4

7a2x2.8a3bx4 = 3b

7a2x2 [dividing numerator and

denominator by 8a3bx4]

139

∴ Required fraction in its lowest terms = 3b

7a2x2

Answer : 3b

7a2x2

Example 2. Express x(x2 + 2xy + y2) (x2 – xy + y2)

(x3 + y3) (x2 – y2) in its lowest terms.

Solution : Resolving numerator and denominator of the fractions into factors, we get,

x(x2 + 2xy + y2) (x2 – xy +y2) (x3 + y3) (x2 – y2)

= x(x + y) (x + y) (x2 – xy + y2)

(x + y) (x2 – xy + y2) (x + y) (x – y)

= x(x + y)2 (x2 – xy + y2)

(x + y)2 (x – y) (x2 – xy + y2)

Now dividing numerator and denominator by (x + y)2 (x2 – xy + y2) which is H. C. F. of numerator and denominator, we get x(x2 + 2xy + y2) (x2 – xy + y2)

(x3 + y3) (x2 – y2) = x

x – y

∴ The required fraction in its lowest terms = xx – y

Answer : xx – y

Example 3. Express the fractions a

a2 – 9b2 , b

a2 + 6ab + 9b2 , c

a3 – 27b3 with the

common denominators. Solution : Here, the denominators are first resolved into factors. Then the L. C. M of denominations are to be found. a2 – 9b2 = (a)2 – (3b)2 = (a + 3b) (a – 3b); a2 + 6ab + 9b2 = (a)2 + 2.a.3b + (3b)2 = (a + 3b)2; a3 – 27b3 = (a)3 – (3b)3 = (a – 3b) (a2 + 3ab + 9b2). Hence, the L. C. M. of denominators = (a + 3b)2 (a – 3b) (a2 + 3ab + 9b2) = (a + 3b)2 (a3 – 27b3)

140

Therefore, a

a2 – 9b2 = a(a + 3b)2 (a3 – 27b3)

(a2 – 9b2) (a + 3b)2 (a3 – 27b3)

= a(a + 3b) (a + 3b) (a – 3b) (a2 + 3ab + 9b2)

(a – 3b) (a + 3b) (a + 3b)2 (a3 – 27b3) = a(a + 3b) (a2 + 3ab + 9b2)

(a + 3b)2 (a3 – 27b3)

Again, b

a2+ 6ab + 9b2 = b (a + 3b)2 (a3 – 27b3)

(a + 3b)2 (a + 3b)2 (a3 – 27b3) = b (a3 – 27b3)

(a + 3b)2 (a3– 27b3)

and c

a3– 27b3 = c(a+ 3b)2 (a3 – 27b3)

(a3 – 27b3) (a + 3b)2 (a3 – 27b3) = c(a + 3b)2

(a + 3b)2 (a3 – 27b3)

Answer : a(a + 3b) (a2 + 3ab + 9b2)

(a + 3b)2 (a3– 27b3) , b(a3 – 27b3)

(a + 3b)2 (a3 – 27b3) , c(a + 3b)2

(a + 3b)2 (a3 – 27b3)

Example 4. Add : 1

a – b + a

a2 + ab + b2 + b2

a3 – b3

Solution : Here, 1

a – b = 1

a – b , a

a2 + ab + b2 = a

a2 + ab + b2

and b2

a3 – b3 = b2

(a – b) (a2 + ab + b2)

Therefore, L . C. M. of denominators = (a – b) (a2 + ab + b2) = a3 – b3.

Hence, 1

a – b + a

a2 + ab + b2 + b2

a3 – b3

= a2 + ab + b2

a3 – b3 + a(a – b)a3 – b3 +

b2

a3 – b3

= a2 + ab + b2 + a2 – ab + b2

a3 – b3 = 2a2 + 2b2

a3 – b3 = 2(a2 + b2)

a3 – b3

Answer : 2(a2 + b2)

a3 – b3

Example 5. Add : 2x

x2–2x–3 + x

x2 + x–12 + 3x

x2 + 5x + 4

Solution : Here, 2x

x2–2x–3 = 2x

x2 – 3x + x –3 = 2x

(x – 3) (x + l)

xx2 + x–12 =

xx2 + 4x – 3x–12 =

x (x + 4)(x – 3)

141

and 3x

x2 + 5x + 4 = 3x

x2 + 4x + x + 4 = 3x

(x + 4)(x + l)

Now L. C. M. of denominators = (x + 1) (x + 4) (x – 3).

Hence, 2x

x2–2x–3 + x

x2 + x – 12 + 3x

x2 + 5x + 4

= 2x

(x – 3) (x + l) + x

(x + 4)(x – 3) + 3x

(x + 4) (x + l) .** ■ ' ■

= 2x(x + 4) + x(x + 1) + 3x(x – 3)

(x – 3) (x + 1) (x + 4) = 2x2 + 8x + x2 + x + 3x2 – 9x

(x – 3) (x + 1) (x + 4)

= 6x2

(x – 3) (x + l) (x + 4)

Answer : 6x2

(x – 3) (x + l) (x + 4)

Example 6. Add : x

x3 + y3 + y

x3 − y3 + xy

x4 + x2y2 + y4

Solution : Here, x

x3 + y3 = x

(x + y)(x2 – xy + y2)

yx3 – y3 =

y(x – y) (x2 + xy + y2)

and xy

x4 + x2y2 + y4 = xy

x4 + 2x2y2 + y4 −x2y2 = xy

(x2 + y2)2 – (xy)2

= xy

(x2 + xy + y2) (x2 – xy + y2)

Therefore, L. C. M. of denominators = (x + y) (x – y) (x2 + xy + y2) (x2 – xy + y2) = (x3 + y3) (x3 – y3) = x6 – y6

Hence, x

x3 + y3 + y

x3 – y3 + xy

x4 + x2y2 + y4

= x (x3 – y3)

x6 – y6 + y(x3 + y3)

x6–y6 + xy (x + y) (x – y)

x6 – y6

= x4 – xy3 + yx3 + y4 + x3y – xy3

x6 – y6 = x4 + 2x3y – 2xy3 + y4

x6 – y6

Answer: x4 + 2x3y – 2xy3 + y4

x6 – y6

142

Example 7. Add : a

(a – b)(c – a) + b

(a – b) (b – c) + c

(b – c)(c – a)

Solution : Here, L.C.M. of denominators = (a – b) (b – c) (c – a)

Hence, a

(a – b) (c – a) + b

(a – b) (b – c) + c

(b – c) (c – a)

= a(b – c) + b (c – a) + c(a – b)

(a – b)(b – c) (c – a) = ab – ac + bc – ab + ac – bc

(a – b) (b – c) (c – a)

= 0

(a – b) (b – c) (c – a) = 0

Ans : 0

Example 8. Add : 2

(x – l) (x–2) + 3

(x – l) (x – 3) + 2(5 – 2x)

(x – l)(x – 2)(x – 3)

Solution : Here, L.C.M. of denominators = (x – 1) (x – 2) (x – 3).

Hence, 2

(x – l)(x – 2) + 3

(x – l)(x – 3) + 2(5 – 2x)

(x – l)(x – 2)(x – 3)

= 2(x – 3) + 3(x –2) + 2(5 – 2x)

(x – l)(x – 2) (x – 3) = 2x – 6 + 3x – 6 + 10 – 4x

(x – l)(x – 2)(x – 3)

= x – 2

(x – l)(x – 2) (x – 3) = 1

(x – l)(x – 3)

[dividing the numerator and denominator by (x – 2)]

Ans : 1

(x – l)(x – 3)

Example 9. Subtract : a

6x2yz5 − b

8x3y3z2

Solution : Here, L.C.M. of denominators = 24x3y3z5

Now, a

6x2yz5 = a.24x3y3z5

6x2yz5.24x3y3z5 = 4axy2

24x3y3z5

and b

8x3y3z2 = b.24x3y3z5

8x3y3z2.24x3y3z5 = 3bz3

24x3y3z5

Hence, a

6x2yz5 – b

8x3y3z2 = 4axy2

24x3y3z5 – 3bz3

24x3y3z5 = 4axy2 – 3bz3

24x3y3z5

Ans: 4axy2 – 3bz3

24x3y3z5

143

Example 10. Subtract: a + b

(a – b)2 – a

a2 – b2

Solution : Here, L.C.M. of denominator = (a – b)2 (a + b).

Hence, a + b

(a – b)2 – a

a2 – b2 = (a + b)2 – (a – b)a

(a – b)2 (a + b)

= a2 + 2ab + b2 − a2 + ab

(a – b)2 (a + b) = 3ab + b2

(a – b)2 (a + b) = b(3a + b)

(a – b)2 (a + b)

Ans: b(3a + b)

(a – b)2 (a + b)

Example 11. Simplify : 1

a + b + b

a2 – b2 – a

a2 + b2

Solution : Here L.C.M. of denominators = (a2 + b2) (a2 – b2) = a4 – b4

Hence, 1

a + b + b

a2 – b2 – a

a2 + b2

= (a – b) (a2 + b2) + b (a2 + b2) – a(a2 – b2)

a4 – b4

= a3 + ab2 – ba2 – b3 + a2b + b3 – a3 + ab2

a4 – b4 = 2ab2

a4 – b4

Second method : The sum of first two expressions is,

1

(a + b) + b

a2 – b2 = a – b + ba2 – b2 =

aa2 – b2

The third expression is subtracted from this sum,

∴ 1

a + b + b

a2 – b2 – a

a2 + b2 = a

a2 – b2 – a

a2 + b2

= a⎝⎜⎛

⎠⎟⎞1

a2 – b2 – 1

a2 + b2 = a ⎩⎪⎨⎪⎧

⎭⎪⎬⎪⎫a2 + b2 – a2 + b2

(a2 – b2) (a2 + b2) = a × 2b2

a4 – b4 = 2ab2

a4 – b4

Ans : 2ab2

a4 – b4

Example 12. Simplify : 1

x – 1 – 1

x + 2 + 1

x – 2 – 1

x + 1

Solution : Given expression = ⎝⎜⎛

⎠⎟⎞1

x – 1 – 1

x + 1 + ⎝⎜⎛

⎠⎟⎞1

x – 2 – 1

x + 2

N.M.G. -20

144

= x + 1 – (x – 1)

x2 – 1 + x + 2 – (x – 2)

x2 – 4

= 2

x2 – 1 + 4

x2 – 4 = 2 (x2 – 4) + 4(x2 – 1)

(x2 – 1) (x2 – 4)

= 2x2 – 8 + 4x2 – 4(x2 – 1) (x2 – 4) = 6x2 – 12

(x2 – 1) (x2 – 4) = 6(x2 – 2)

x4 – 5x2 + 4

Ans : 6(x2 – 2)x4 – 5x2 + 4

Remark: Simplification can be done within the shortest time of some expressions are arranged conveniently and then added or subtracted, afterwards this sum or difference are added to or subtracted from any other expression or added or subtracted from the sum or difference of some other expressions.

EXERCISE 2.1

1. Express the following fractions in its lowest terms:

(i) 18x3y2z5

24x5yz7 (ii) 49a4b7c9

42a5b9c5 (iii) 45a4b3x7

36ab6x9 (iv) 121p3q9r10

176q7p3r7

(v) x3y2 + x2y3

x2y4 + xy5 (vi) 6a(a + b)2

9a2(a2 – b2) (vii) (a – b)(a2 – ab + b2)

(a3 + b3) (a + b)

(viii) a2 – 6a + 9

(a2 – 9) (ix)x2 – 6x + 5

x2 + 4x – 45 (x) 2x2 – x – 6

3x2 – 2x – 8

(xi) 14x4 – 9x2 + 121x4 + 4x2 – 1 (xii)

a2 – b2 – 2bc – c2

a2 + 2ab + b2 – c2 (xiii) 2a2 + ab – b2

a3 + a2b – a – b

2. Express the following fractions with the common denominators:

(i) a2

12bc , b

20ca , c2

28ab (ii) ab

4cd2 , ac

6bd3 , bc

10ab5

(iii) a

a – b , b

a + b , c

a(a + b) (iv) a – b

ab , b – c

bc , c – a

ca

(v) a + b

(a – b)2 , a – b

a3 + b3 , a

a2 – b2 (vi) a

a – b , b

a2 + ab + b2 , c

a3 – b3

(vii) 2

x2 – x – 2 , 3

x2 + x – 6 , 4

x2 + 6x + 9

(viii) c – a

(a – b) , a – b

(b – c) , b – c

(c – a)

145

3. Add :

(i) x + y

x + x – y

y (ii) z

xy + xyz +

yzx (iii)

aa – b +

b a + b

(iv) 2x + 3y 2x – 3y +

2x – 3y2x + 3y (v)

1(a + b)2 +

1a2 – b2 +

a – b(a + b)3

(vi) 1

x2 – 3x + 2 + 1

x2 – 5x + 6 + 1

x2 – 4x + 3

(vii) 1

x2 – x + 1 + 1

x2 + x + 1 + 1

x6 – 1

(viii) 1

x2 – 5x + 6 + 1

x2 – 4 + 1

x2 – x – 6

4. Subtract:

(i) x

x – 4 – x2

x2 – 16 (ii) 1

a(a + b) – 1

a(a – b)

(iii) 1

a + b – (a – b)2

a3 + b3 (iv) x2 + 9y2

x2 – 9y2 – x – 3yx + 3y

(v) 1

x – 2 – x – 2

x2 + 2x +4 (vi) 1

1 – x + x2 – 1

1 + x +x2

5. Simplify:

(i) a – b

ab + b – c

bc – c – a

ac (ii) a − ba + b +

b – cb + c –

c – a c + a

(iii) b

(a – b)(b – c) + a

(c – a)(a – b) + c

(b – c)(a – c)

(iv) a + ba – b –

a – ba + b +

2abb2 – a2 (v)

1x + 2y +

1x – 2y –

2xx2 – 4y2

(vi) x – 2yx + 2y –

x + 2yx – 2y –

8xyx2 + 4y2 (vii)

1a – b –

22a + b +

1a + b –

22a – b

(viii) 2

x – 1 – x

x2 + 1 – 1

x + 1 – 3

x2 – 1

(ix) 1

a – 2 – a – 2

a2 + 2a + 4 + 6a

a3 + 8

146

(x) 1

1 – x + x2 – 1

1 + x + x2 – 2x

1 + x2 + x4

(xi) 1

x – 2 – 1

x + 2 – 4

x2 + 4 + 32

x4 + 16

(xii) c – a

(a – b)(b – c) + b – a

(a – c)(b – c) + b – c

(a – c)(b – c)

(xiii) 1

a – b – c + 1

a – b + c – a

a2 + b2 – c2 – 2ab

(xiv) 1

2x – 3y – 1

2x + 3y + 6y

4x2 – 9y2

2.2. Multiplication of Fractions

The product of fractions of more than one is such a fraction whose numerator is equal to the product of numerators of the given fractions and denominator is equal

to the product of denominators of the given fractions. ab and c

d are two fractions

and their product is ab ×

cd =

acbd .

The product of the fractions, ab ,

cd ,

ef ,

gh =

ab ×

cd ×

ef ×

gh =

acegbdfh

Remark : The fraction obtained from the product is to be expressed in its lower terms. In the process of finding the product, the common factors of numerators and denominators can also be cancelled out. If an expression is a, then the numerator of the expression is 'a' and its denominator is 1.

For example, a × bc =

a1 ×

bc =

abc

Example 1. Multiply ab

c2d3 by a3b2

c2d2

Solution : Required product = ab

c2d3 × a3b2

c2d2 = ab.a3b2

c2d3.c2d2 = (a.a3)(b.b2)

(c2.c2)(d3 .d2)

= a4b3

c4d5

Answer : a4b3

c4d5

147

Example 2. Multiply : 8a3bc3

25d3e2f and 15e3fg

16a2b2pq

Solution : 8a3bc3

25d3e2f × 15e3fg

16a2b2pq = 8 × 1525 × 16 ×

a3

a2 × bb2 ×

c3

d3 × e3

e2 × ff ×

gpq

= 310 ×

a.c3egb.d3pq =

3ac3eg10bd3pq

Answer : 3ac3eg

10bd3pq

Example 3. Multiply x (x – y)

x2 + 2xy +y2 by y (x + y) x3 – y3

Solution : Required product = x (x – y)

x2 + 2xy + y2 × y (x + y)x3 – y3

= xy (x – y) (x + y)

(x + y)2 (x – y) (x2 + xy + y2) = xy

(x + y) (x2 + xy + y2)

Answer : xy

(x + y) (x2 + xy + y2)

Example 4. a2 – x2

a + y × a2 – y2

ax + x2 × 1

a – x = how much?

Solution : Given expression = (a + x) (a – x)

(a + y) × (a + y) (a – y)

x (a + x) × 1

(a – x)

= a – y

x

Answer : a – y

x

EXERCISE 2.2

Multiply :

1. 6a2b2

35c3d3 , 49c5d4

36a5b7 and 4a7b8

21c4d5 2. 20x2y2

21z2 , 28z4

9x3y4 and 3y7z10x

3. x2

yz , y2

zx and z2

xy 4. x + 1x – 1 ,

x2 + x – 2x2 + x and

x2

x2 + 5x + 6

148

5. a4 – b4

a2 – 2ab +b2 , a – b

a3 + b3 and a + b

a3 – b3 6. 1 – x2

1 + b , 1 – b2

x + x2 and ⎝⎜⎛

⎠⎟⎞1 +

x1 – x

7. x2 – 6x – 16x2 – 4x – 21 ,

x2 – 11x + 28x2 – 12x + 32 and

x + 3x + 2

8. 2x2 – 7x + 32x2 + 7x – 4 ,

3x2 + 11x – 43x2 + 8x – 3 and

2x2 + x – 152x2 – 11x + 15

9. ab

a – b × a3 – b3

a3 – a2b + ab2 × a3 + b3

a2b + ab2 + b3

10. 2x(1–b2)

y × 1

(x + y)2 × x2 – y2

x + bx × y

2x – 2y

11. a3 + b3

a2 – b2 × a + ba – b ×

a2 + ab(a + b)2

Simplify :

12. ⎝⎜⎛

⎠⎟⎞a

b + bc ⎝⎜⎛

⎠⎟⎞b

a + cd 13.

⎝⎜⎛

⎠⎟⎞1

1 + x + 2x

1 – x2 ⎝⎜⎛

⎠⎟⎞1

x – 1x2

14. ⎝⎜⎛

⎠⎟⎞1 –

zx + y ⎝

⎜⎛

⎠⎟⎞x

x + y + z + x

x + y – z 15. ⎝⎜⎛

⎠⎟⎞1

1 + x + x

1 – x ⎝⎜⎛

⎠⎟⎞1

1 + x2 – 1

1 + x + x2

16. ⎝⎜⎛

⎠⎟⎞x

3x – y + x

3x + y ⎝⎜⎛

⎠⎟⎞9 +

8y2

x2 – y2

2.3. Division of Fractions

The meaning of dividing a fraction by another one is the multiplication of the first

fraction by the reciprocal of the second fraction. If ab is divided by

cd , the quotient

is ab ×

dc =

adbc . Here,

dc is the reciprocal of

cd . The reciprocal of 'a' will be

1a .

If cd is divided by a, then quotient is

cd ÷ a =

cd ×

1a =

cad .

Example 1. Divide a2b3

c2d by a3b2

cd3

Solution : The reciprocal of the expression a3b2

cd3 is cd3

a3b2

149

Therefore, a2b3

c2d ÷ a3b2

cd3 = a2b3

c2d × cd3

a3b2 = a2b3cd3

a3b2c2d = bd2

ac

Answer: bd2

ac

Example 2. 9a4b2c3

10x3y4z ÷ 6a3bc5

5x2y5z2 = how much?

Solution : 9a4b2c3

10x3y4z ÷ 6a3bc5

5x2y5z2 = 9a4b2c3

10x3y4z × 5x2y5z2

6a3bc5 = 3ab × yz2x × 2c2 =

3abyz4c2x

Answer : 3abyz4c2x .

Example 3. Divide a

a – b by a + b

a3 – b3

Solution : a

a – b ÷ a + b

a3 – b3 = a

a – b × a3 – b3

a + b = a

a – b × (a – b) (a2 + ab + b2)

a + b

= a (a2+ ab + b2)

a + b

Answer: a (a2+ ab + b2)

a + b

Example 4. Divide a3 – b3

a + b by a4 + a2b2 + b4

a3 + b3

Solution : a3 – b3

a + b ÷ a4 + a2b2 + b4

a3 + b3 = (a – b) (a2 + ab + b2)

(a + b) × (a + b) (a2 – ab + b2)

(a2 + b2)2 – (ab)2

= (a – b)(a + b) (a2 + ab + b2) (a2 – ab + b2)

(a + b) (a2 + b2 + ab) (a2 + b2 – ab) = (a – b)

Answer. (a – b).

Example 5. Simplify : ⎝⎜⎛

⎠⎟⎞a

a + b + b

a – b ÷ ⎝⎜⎛

⎠⎟⎞a

a – b – b

a + b

Solution : Given expression = a (a – b) + b(a + b)

a2 – b2 ÷ a (a + b) – b(a – b)

a2 – b2

= a2 – ab + ab + b2

a2 – b2 ÷ a2 + ab – ab + b2

a2 – b2 = a2 + b2

a2 – b2 ÷ a2 + b2

a2 – b2 = a2 + b2

a2 – b2 × a2 – b2

a2 + b2 = 1

Answer : 1.

150

Example 6. Simplify : x2 + x – 2x2+7x+12 ÷

x2 – 3x – 10x2 + x –12 ×

x2 – 4x – 5x2 – 4x + 3

Solution : Given expression = x2 + x – 2x2+7x+12 ÷

x2 – 3x – 10x2 + x –12 ×

x2 – 4x – 5x2 – 4x + 3

= x2 + x − 2

x2 + 7x + 12 × x2 + x − 12x2 − 3x − 10 ×

x2 − 4x − 5 x2 − 4x + 3

= (x + 2)(x – 1)(x + 3) (x + 4) ×

(x + 4)(x – 3)(x – 5) (x + 2) ×

(x – 5)(x + l)(x – 3) (x – 1) =

(x + 1)(x + 3)

Answer : (x + 1)(x + 3)

Example 7. Simplify : x3 + y3

(x – y)2 + 3xy ÷ (x + y)2 – 3xy(x3 – y3) × (x + y)

(x – y)

Solution : Given expression = x3 + y3

(x – y)2 + 3xy ÷ (x + y)2 – 3xy(x3 – y3) × (x + y)

(x – y)

= (x + y) (x2 – xy + y2)

(x2 + xy + y2) × (x – y) (x2 + xy + y2)

(x2 – xy + y2) × (x + y)(x – y) = (x +y)2 = x2 +2xy + y2

Answer : x2 + 2xy + y2.

Remark : If there are +, – , × , ÷ and brackets, the rule of BODMAS is followed in case of simplification. where B = Bracket, O = Of, D = Division, M = Multiplication, A = Addition, S = Subtraction.

EXERCISE 2.3 Divide the first expression by the second expression:

1. 2a2

3b , 9b2

16ac 2. 4a2b2

3c2 , 12ab3

5c5 3. 7a2b2c2

12xyz , 21a4b4c4

4x3y3z3 4. 5ab ,

a + b5b

5. a2 + ab

4a2 , ab + b2

6b2 6. x2 – 49x2 – 25 ,

x + 7x + 5 7.

a3 – b3

a + b , a2 + ab +b2

a2 – b2

8. x2 – 4

x2 + 3x – 18 , x2 – 5x – 14

x2 – 36 9. x2 – x – 30

x2 – 36 , x2 + 6x – 72x2 + x – 56

Simplify :

10. ⎝⎜⎛

⎠⎟⎞2a + b

a + b – 1 ÷ ⎝⎜⎛

⎠⎟⎞1 –

ba + b 11.

⎝⎜⎛

⎠⎟⎞1 +

1x ÷

⎝⎜⎛

⎠⎟⎞1 –

1x2

151

12. ⎝⎜⎛

⎠⎟⎞x

x + y + y

x − y ÷ ⎝⎜⎛

⎠⎟⎞x

x – y – y

x + y 13. ⎝⎜⎛

⎠⎟⎞1 –

2aba2 + b2 ÷

⎝⎜⎛

⎠⎟⎞a3 – b3

a – b – 3ab

14. ⎝⎜⎛

⎠⎟⎞a + b –

a + ba – b ÷

⎝⎜⎛

⎠⎟⎞1 –

a + ba2 – b2 15.

⎝⎜⎛

⎠⎟⎞2a

a + b – b

a – b + b2

a2 – b2 ÷ ⎝⎜⎛

⎠⎟⎞1

a + b + a

a2 – b2

16. x3 + y3 + 3xy (x + y)

(x + y)2 − 4xy ÷ (x − y)2 + 4xy

x3 − y3 − 3xy(x −y)

17. ⎝⎜⎛

⎠⎟⎞a

b + ba – 1 ÷

⎝⎜⎛

⎠⎟⎞a2

b2 + ab + 1 ×

a2 + ab + b2

b(a2 – ab + b2)

18. x2 + 2x – 15x2 + x – 12 ÷

x2 – 25x2 – x – 20 ÷

x – 2x2 – 5x + 6

19. 3x2 – 4x – 43x2 – x – 2 ×

2x2 + 5x – 32x2 + x – 1 ÷

x2 + x – 6x2 – 1

20. ⎝⎜⎛

⎠⎟⎞a

a – b – a

a + b ÷ ⎝⎜⎛

⎠⎟⎞b

a – b – b

a + b + ⎝⎜⎛

⎠⎟⎞a + b

a – b + a – ba + b ÷

⎝⎜⎛

⎠⎟⎞a + b

a – b – a – ba + b ÷

⎝⎜⎛

⎠⎟⎞1 +

b2

a2

21. a3 + a2b + ab2 + b3

2a (a – b) ÷ a + b

a2 – ab – 2b2 ÷ a2 + b2

4a2 × a2 – b2

a – 2b

Multiple Choice Questions [Marks (√) on the correct answer]

1. If rz

qy

px ,, are expressed in fractions with common denominator

then the result would be-

(a) pqrzpq

pqrypr

pqrxqr ,, (b)

pqrzrp

pqryqr

pqrxpq ,,

(c) pqrzqr

pqrypq

pqrxrp ,, (d)

pqrzpr

pqrypq

pqrxqr ,,

2. Which one of the following indicates the product of ab

c2d3 and a3b2

c2d2 ?

(a) 54

33

dcba (b) 54

34

dcba

(c) 45

43

dcba (d) 55

24

dcba

N.M.G. -21

152

3. By which one of the following the algebraic fractions can be expressed in general?

(a) x × y (b) yx

(c) 2 ÷ 3 (d) 2 × 3

4. How much should be subtracted from the sum of yx −

1 and yx +

1 ,

so that the result would be 2?

(a) 22

22yx −

− (b) 22

22yx

x−

−

(c) 2222 −

− yx (d) 22

22 −− yxx

5. (i) The sum of yx +

1 and yx −

1 is equal to 1

22 −x

(ii) The H.C.F. of the denominators of the fractions

)()(

,)()(

,)()( yxxz

zxzzy

yzyyx

x−−−−−−

is 1.

(iii) The product of more than one fractions means the ratio of product of numerators and product of denominators.

Which one of the following is correct on the basis of the above

informations. (a) i and ii (b) i and iii (c) ii and iii (d) i, ii and iii Observe the following four algebraic expressions:

yx

yyx

xyx

yyx

x+−−+

,,,

Answer the questions (6 – 8) on the basis of the above informations. 6. Which one of the following is the L.C.M. of the above expressions? (a) x+y (b) x – y (c) x2+y2 (d) x2–y2

153

7. Which one of the following is the sum of the 1st and 2nd expressions?

(a) 22

22

yxyx

−+ (b) ( )

22

222yxyx

−+

(c) 22

22yx

x−

(d) 22

22yx

y−

8. Which one of the following is the ratio of the sum of first two

expressions and subtraction of last two expressions? (a) -1 (b) 0 (c) 1 (d) 2

CREATIVE QUESTIONS 1. Observe the following algebraic expressions:

⎟⎠⎞

⎜⎝⎛ −+ 1

ab

ba , ⎟⎟

⎠

⎞⎜⎜⎝

⎛+− 12

2

ba

ba , ⎟

⎠⎞

⎜⎝⎛

−−

+ 22

1ba

aba

(i) Find the ratio of 1st and 2nd expressions. (ii) How much should be added with the product of 1st and 3rd

expressions so that the result would be ba

a+

?

(iii) Convert the expressions into fractions with common denominator.

2. Observe the following algebraic expressions:

33 yxx+

, 33 yxy−

, 4224 yyxxxy

++

(a) Find the quotient of 1st expression and 3rd expression. (b) Find the subtraction of the 3rd expression from the sum of 1st

and 2nd expressions. (c) Convert the expressions into fractions with common denominator.

154

3. 183

42

2

−+−xx

x , 36

1452

2

−−−

xxx ,

56726

2

2

−+−+

xxxx are three algebraic

expressions. (a) Express the 1st expression in simple form. (b) Divide the 1st expression by the 2nd expression.

(c) If the result of (b) is divided by the 3rd expression, then how

much the quotient would be more than )12()3(

22

+−−xx

x ?

Chapter III

Linear Equations And Their Applications

3.1 Linear Equation The concept of linear equation and the solution of given linear equations of easy form containing one unknown quantity have been solved in class VII. We have also learnt the linear equation on the basis of real problems and their solution. Here it is discussed briefly again. If two expressions are connected by the sign of equality, then such relation is called an equation. There are two expressions in the equation of which one at the left and the other at the right side of the sign of equality. The expression in the left side is called left hand side and the expression in the right side is called right hand side. 8x + 2 = 3x + 12 is a linear equation. Here x is an unknow quantity. 8x + 2 is left hand side and 3x + 12 is right hand side. 5x + 4 = 14 is a linear equation. 3 × 5 + 4 = 11 + 8 is not an equation. Here there is no unknown quantity though there is sign of equality. The value of the unknown quantity for which both the sides of the equation become equal is called the root of the equation or solution of the equation. There will be only one root for one unknown quantity whose power is one. The process of finding the roots of the equation is known as solving the equation. If we put x = 2 in the equation 8x + 2 = 3x + 12, then value of both left hand side and right hand side become 18. Hence 2, the value of x is the root of the equation. In 5x + 3, x is an unknown quantity and it is called a variable. Here any value can be assingned to x. To find the roots of the equation, the following rules are followed :

(a) Interchange of sides of terms If any term of right hand side is transfered to left hand side, then sign before that term is to be changed. Similarly, if a term of left hand side is transfered to right hand side, then sign before that term also be changed, a + b = c will be a = c – b and a = b + c will be a – b = c.

(b) Cancellation law of addition The root of a linear equation is not changed if equal terms are added to or subtracted from the expressions of both the sides. If a = b, then a + c = b + c, a – c = b – c.

156

(c) Cancellation law of multiplication The root of a linear equation is not changed if any term not zero is divide or multiply by the expression of both the sides. If ac = bc and c ≠ 0, then a = b;

If ac =

bc and c ≠ 0, then a = b.

(d) Cross– multiplication

If ab =

cd , then ad = bc, where b ≠ 0 and d ≠ 0

Example 1. Solve : 7(3 – 2x) + 5(x – 1) = 34. Solution : 7(3 – 2x) + 5(x – 1) = 34 or, 21 – 14x + 5x – 5 = 34. or, – 14x + 5x = 34 – 21 + 5. [Interchanging the sides of the terms]

or, – 9x = 18 or, x = 18– 9 , [dividing both sides by – 9]

or, x = – 2. ∴ Required solution, x = – 2. Answer : x = – 2. Example 2. Solve : 5(x + 7) + 9(2x + 23) = 3(x + 6) – 8x. Solution: 5(x + 7) + 9(2x + 23) = 3(x + 6) – 8x. or, 5x + 35 + 18x + 207 = 3x + 18 – 8x. or, 5x +18x – 3x + 8x = 18 – 35 – 207.

or, 28x = –224 or, x = – 224

28 [dividing both sides by 28]

or, x = – 8. ∴ Required solution, x = – 8. Answer : x = – 8. Remark : In the process of solution, unknown quantities are kept in the left hand side and the numbers or known quantities are kept in the right hand side. 3.2. Solution of the Equations Reducible to Linear Form To find the solution of the equation involving fractions the equation is to be reduced to a linear equation free from fractions. If the denominators of the fractions can be removed then the equation involving fractions becomes free from fractions. In order to remove the denominators, the L. C. M. of the denominators is

157

to be determined and then each term of both sides is to be multiplied by this L. C. M. and thus the fractions become free from denominators.

Example 3. (a) Solve : 5x – 2

8 + x + 4

6 = 6x + 3

5 – x + 2

4

Solution : Here, L. C. M. of denominators, 8, 6, 5 and 4 is 120.

Now, 5x – 2

8 + x + 4

6 = 6x + 3

5 – x + 2

4

or, ⎝⎜⎛

⎠⎟⎞5x – 2

8 + x + 4

6 × 120 = ⎝⎜⎛

⎠⎟⎞6x + 3

5 – x + 2

4 × 120.

or, 5x – 2

8 × 120 + x + 4

6 × 120 = 6x + 3

5 × 120 – x + 2

4 × 120.

or, 75x – 30 + 20x + 80 = 144x + 72 – 30x – 60. or, 75x + 20x – 144x + 30x = 72 – 60 + 30 – 80,[Interchanging the sides of terms]

or, – 19x = – 38 or, x = 2, [dividing both sides by – 19]

∴ Required solution, x = 2. Answer : x = 2.

Example 3. (b) Solve : 3x4 + 3 =

x6 + 4.

Solution: 3x4 + 3 =

x6 + 4.

or, 3x4 –

x6 = 4 – 3, [interchanging the sides of terms]

or, 9x – 2x

12 = 1, [L. C. M. of 4 and 6 = 12 and applying algebraic subtraction]

or, 7x12 = 1. or, 7x = 12, [multiplying both sides by 12]

or, x = 127 [dividing both sides by 7]

∴ Required solution, x = 127

Examining correctness it is found that left hand side = right hand side.

Answer : x = 127

158

Example 3. (c) solve : 3x4 +

5x – 26 =

4x + 58

Solution : 3x4 +

5x – 26 =

4x + 58

or, 24 × 3x4 + 24 ×

5x – 26 = 24 ×

4x + 58 [L.C.M. of 4, 6 and 8 is 24 and

multiplying both sides by 24]

or, 6(3x) + 4(5x – 2) = 3(4x + 5) or, 18 x + 20x – 8 = 12x + 15

or, 38x –12x = 15 + 8, [interchanging the sides of terms]

or, 26x = 23 or, x = 2326 [dividing both sides by 26]


Answer : x = 2326

Example 4. Solve : 5

4x – 15 = 2

3x + 1

Solution : 5

4x – 15 = 2

3x + 1

or, 5(3x + 1) = 2(4x – 15), [by cross– multiplication]

or, 15x + 5 = 8x – 30

or, 15x – 8x = – 30 – 5, [interchanging the sides of terms]

or, 7x = –35, or, x = – 35

7 or, x = –5,

∴ Required solution, x = –5. Answer : x = –5.

Example 5. Solve : 2x + 52x + 3 –

2x – 32x – 5 =

2x – 52x – 3 –

2x + 32x + 5

Solution : 2x + 52x + 3 –

2x – 32x – 5 =

2x – 52x – 3 –

2x + 32x + 5

159

or, (2x + 5)(2x – 5) – (2x – 3)(2x + 3)

(2x + 3)(2x – 5) = (2x – 5)(2x + 5) – (2x + 3)(2x – 3)

(2x – 3)(2x + 5)

or, 4x2 – 25 – 4x2 + 9(2x + 3)(2x – 5) =

4x2 – 25 – 4x2 + 9(2x – 3)(2x + 5) or,

– 16(2x + 3)(2x – 5) =

–16(2x –3)(2x +5)

or, 1

(2x + 3)(2x – 5) = 1

(2x –3)(2x +5) [dividing both sides by – 16]

or, (2x – 3)(2x + 5) = (2x + 3)(2x – 5) [by cross-multiplication] or, 4x2 – 6x + l0x – 15 = 4x2 + 6x – l0x – 15 or, 4x2 + 4x – 4x2 + 4x = – 15 + 15 [interchanging the sides of terms]

or, 8x = 0 or, x = 08 , [dividing both sides by 8]

or, x = 0. [The quotient is zero if 0 is divided by any number] ∴ Required solution, x = 0. Answer : x = 0.

Example 6. Solve : 5

x – 1 + 4

x – 2 = 9

x – 3

Solution : 5

x – 1 + 4

x – 2 = 9

x – 3

or, 5

x – 1 + 4

x – 2 = 5

x – 3 + 4

x – 3 [Q 9 = 5 + 4]

or, 5

x – 1 – 5

x – 3 = 4

x – 3 – 4

x – 2

or, 5 (x – 3) –5 (x – 1)

(x – 1) (x – 3) = 4 (x – 2) – 4 (x – 3)

(x – 3) (x – 2)

or, 5x – 15 – 5x + 5(x – 1) (x – 3) =

4x – 8 – 4x + 12(x – 3) (x – 2)

or, –10

(x – l) (x – 3) = 4

(x – 3) (x – 2)

1. or, –5

x – 1 = 2

x – 2 [dividing and multiplying both sides by 2 and (x –3)

respectively]

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or, – 5(x – 2) = 2(x – 1) [by cross– multiplication]

or, – 5x + 10 = 2x – 2

or, – 5x – 2x = – 2 – 10 [interchanging the sides of terms]

or, – 7x = – 12 or, x = 127

Answer : x = 127 .

Remark : In the above example the denominators of algebraic fraction are different but there is a relation among the three numerators. The sum of the numerators of left hand side is 5 + 4 = 9, which is the numerator of right hand side. Hence the above method is easier to find solution. Different techniques are applied to solve the equations involving fractions. A particular technique is applied above. There are many such techniques of which some are applied below :

Example 7. Solve : x – ab + c +

x – bc + a +

x – ca + b = 3

Solution : x – ab + c +

x – bc + a +

x – ca + b = 3 or,

x – ab + c +

x – bc + a +

x – ca + b – 3 = 0

or, x – ab + c – 1 +

x – bc + a – 1 +

x – ca + b – 1 = 0

or, x – a – b – c

b + c + x – b – c – a

c + a + x – c – a – b

a + b = 0

or, x – a – b – c

b + c + x – a – b – c

c + a + x – a – b – c

a + b = 0.

or, (x – a – b – c) ⎝⎜⎛

⎠⎟⎞1

b + c + 1

c + a + 1

a + b = 0

or, x – a – b – c = 0 [Since ⎝⎜⎛

⎠⎟⎞1

b + c + 1

c + a + 1

a + b ≠ 0; so dividing both sides by it]

or, x = a + b + c

∴ Required solution, x = a + b + c. Answer : x = a + b + c. Remark : If AB = 0, then must be either A = 0 or, B = 0. Hence, if any equation AX = 0 and A ≠ 0, then the factor involving the unknown quantity must be equal to zero.

N.M.G. -22

161

Example 8. Solve : 2x

x – 4 + 8x + 32x + 1 = 6

Solution : 2x

x – 4 + 8x + 32x + 1 = 6 or,

2xx – 4 +

8x + 32x + 1 = 2 + 4

or, 2xx – 4 – 2 = 4 –

8x + 32x + 1 [interchanging the sides of terms]

or, 2x – 2x + 8

x – 4 = 8x + 4 – 8x – 3

2x + 1 or, 8

x – 4 = 1

2x + 1

or, 8(2x + 1) = x – 4 or, 16x + 8 = x – 4 or, 16x – x = – 4 – 8 [ interchanging the sides of terms]

or, 15x = –12 or, x = – 1215 =

– 45

∴ Required solution, x = – 45

Answer : x = – 45

Remark : If x of (x – 4) is multiplied by 2, then the numerator of the first 2x is obtained and also if 2x of (2x +1) is multiplied by 4, then numerator of second fraction, 8x is obtained. Hence 6 of right hand side is written as 2 + 4.

Example 9. Solve : 22x – 1 +

33x – 1 =

84x + 1

Solution : 22x – 1 +

33x – 1 =

84x + 1

or, 2

2x – 1 + 3

3x – 1 = 4

4x + 1 + 4

4x + 1

or, 2

2x – 1 – 4

4x + 1 = 4

4x + 1 – 3

3x – 1

or, 8x + 2 – 8x + 4

(2x – 1) (4x + 1) = 12x – 4 –12x – 3 (4x + 1) (3x – 1)

or, 6

(2x − 1) (4x + 1) = −7

(4x + 1) (3x − 1)

162

or, 6

2x – 1 = – 7

3x – 1 [multiplying both sides by (4x + l)]

or, 18x – 6 = – 14x + 7, [by cross-multiplication] or, 18x + 14x = 7 + 6, [interchanging the sides of terms]

or, 32x = 13 or, x = 1332

∴Repuired solution, x = 1332

Answer : x = 1332

Remark : If 4x of (4x + 1) is multiplied by the numerator 2 of the first fraction, it becomes 8x and if 2x of (2x – 1) of first fraction is multiplied by 4, it also becomes 8x. Again if 4x of (4x + 1) is multiplied by the numerator 3 of the second fraction, if becomes 12x and if 3x of (3x – 1) is multiplied by 4 it also becomes 12x. Here 8, the numerator of the fraction of right hand side is written as (4 + 4).

Example 10. Solve : x + 1x – 1 +

x + 2x – 2 =

10x + 185x – 6

Solution : x + 1x – 1 +

x + 2x – 2 =

10x + 185x – 6

or, (x – 1) + 2

x – 1 + (x – 2) + 4

x – 2 = 2(5x – 6) + 30

5x – 6

or, 1 + 2

x – 1 + 1 + 4

x – 2 = 2 + 30

5x – 6

or, 2

x – 1 + 4

x – 2 = 2 – 2 + 10 + 205x – 6

or, 2

x – 1 – 10

5x – 6 = 20

5x – 6 – 4

x – 2 , [interchanging the sides of terns]

or, 10x – 12 – 10x + 10

(x – 1) (5x – 6) = 20x – 40 – 20x + 24

(5x – 6) (x – 2)

or, –2

(x – 1) (5x – 6) = – 16

(5x – 6) (x – 2)

163

or, 1

x – 1 = 8

x – 2 , [dividing and multiplying both sides by – 2 and (5x – 6) respectively]

or, x – 2 = 8x – 8 or, x – 8x = – 8 + 2, [interchanging the sides of terms]

or, – 7x = – 6 or, x = 67

:. Required solution, x = 67

Answer : x = 67 .

Example 11. Solve : 4x2 + 72x – 1 +

9x2 + 53x – 1 =

5x2 – 3x + 4x – 1

Solution : 4x2 + 72x – 1 +

9x2 + 53x – 1 =

5x2 – 3x + 4x – 1

or, (4x2 – 1) + 8

2x – 1 + (9x2 – 1) + 6

3x – 1 = 5x2 – 3x + 4

x – 1

or, 4x2 – 12x – 1 +

82x – 1 +

9x2 – 13x – 1 +

63x – 1 =

5x2 – 3x + 4x – 1

or, (2x + 1) (2x – 1)

(2x – 1) + 8

2x – 1 + (3x + 1) (3x – 1)

3x – 1 + 6

3x – 1 = 5x2 – 3x + 4

x – 1

or, (2x + 1) + 8

2x – 1 + (3x + 1) + 6

3x – 1 = 5x2 – 3x + 4

x – 1

or, (5x + 2) + 8

2x – 1 + 6

3x – 1 = (5x + 2) (x – 1) + 6

x – 1

or, (5x + 2) + 8

2x – 1 + 6

3x – 1 = (5x + 2) + 6

x – 1

or, 8

2x – 1 + 6

3x – 1 = 6

x – 1 [cancelling (5x + 2) from both sides]

or, 8

2x – 1 + 6

3x – 1 = 4

x – 1 + 2

x – 1

or, 8

2x – 1 – 4

x – 1 = 2

x – 1 – 6

3x – 1

164

or, 8x – 8 – 8x + 4(2x – 1) (x – 1) =

6x – 2 – 6x + 6(x – 1)(3x – 1)

or, – 4

(2x – 1)(x – 1) = 4

(x – 1)(3x – 1)

or, – 1

2x – 1 = 1

3x – 1 , [multiplying by ⎝⎜⎛

⎠⎟⎞x – 1

4 ]

or, – 3x + 1 = 2x – 1

or, – 5x = – 2 or, x = 25


Answer : x = 25

EXERCISE 3. 1

Solve: 1. 15x – 9 = 11x – 25; 2. 8x + 5(x + 7) = 3 (x + 6) – 9 (2x + 23); 3. 3(4x + 1) + 9 = 5(3x + 2) + 4(2x – 5); 4. 2(x – 1) – 4(x – 3) = 3(x – 2);

5. x5 + 4 =

3x10 + 6; 6.

5x6 + 3 =

3x5 – 10; 7.

3x – 42 +

5x – 33 = 4 ;

8. 2x + 5

6 – 3x + 1

9 = 2x – 3

8 ; 9. 7x2 –

10x – 38 =

7x + 56 ;

10. 6x + 7

5 – 2x – 1

10 = 3x – 2

15 ; 11. x – 3

7 – 2x – 3

3 = 6x + 1

2 – x – 6

4 ;

12. 2x – 3

4 – x – 3

5 = 3x – 4

8 – 2x + 5

3 ; 13. 5x + 6

4 + 4x – 5

5 = 2x + 3

8 + 3x – 7

3 ;

14. 5

x + 2 + 7

x – 3 = 12

x – 1 ; 15. a

x – a + b

x – b = a + b

x – a – b ;

16. 6

x + 1 +5

x + 5 = 11

x + 3 ; 17. 8

2x – 1 + 9

3x – 1 = 7

x + 1 ;

18. 4

2x + 1 + 15

5x + 4 = 35

7x + 6 ; 19. 10

2x – 5 + 1

x + 5 = 18

3x – 5 ;

20. x – bc b + c +

x – ab a + b +

x – cac + a = a + b + c ;

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21. ax – a2

b + c + bx – b2

c + a + cx – c2

b + a = a + b + c ;

22. x + a

b + c + 2a + x + b

c + a + 2b + x + c

a + b + 2c = 3;

23. 3x – bcb + c +

3x – ca c + a +

3x – aba + b = a + b + c ;

24. 2x – 1x – 1 +

3x – 4 x – 2 =

5x – 12x – 3 ;

25. 4x – 74x + 5 +

15x + 115x + 7 =

12x + 13x + 4 ;

26. x2 – x + 1

x – 1 + x2 – 2x + 1

x – 2 = 2x2 – 6x + 2

x – 3 ;

27. 2x2 – 3x + 7

2x – 1 + 6x2 + 2x + 21

3x + 1 = 3x2 + 8x + 7

x + 3 ;

28. x – 2x – 3 +

x – 3x – 4 =

x – 1x – 2 +

x – 4x – 5 ;

29. 2x + 11x + 5 –

9x – 93x – 4 =

4x + 13x + 3 –

15x – 473x – 10 .

3.3 Applications of Equations in Solving the Problems Relating to Our Real Necessity. Problems relating to our real necessity can be solved by using rules of arithmetic. Those problems can be solved easily with the help of algebraic equations. There is no general rule for solving all the problems. In solving the problems the required term is denoted by x. Then the problem is arranged in the form of an equation according to the given conditions. The process of solution can be understood from the following examples : Example 1. Divide Tk. 420 between Farid and Zafar in such a way so that Farid gets Tk. 30 more than double of Zafar's share. How much each of Farid and Zafar would get?

Solution: Let, Zafar gets Tk. x. Therefore, Farid gets Tk. (2x + 30)

∴ According to the conditions, x + (2x + 30) = 420 or, x + 2x + 30 = 420.

or, 3x = 420 – 30 or, 3x = 390 or, x = 3903 = 130.

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Therefore, Zafar gets Tk. 130 and Farid gets Tk. (2x+30) = (2×130+30) = Tk. 290. Answer : Farid gets Tk. 290 and Zafar gets Tk. 130. Example 2. The present age of Razat is 9 times that of Anik. After 9 years, the age of Razat will be 3 times that of Anik. What are the present ages of Razat and Anik?

Solution : Let, the present age of Anik is x years. Therefore, present age of Razat is 9x years. After 9 years, the age of Anik will be (x + 9) years and the age of Razat will be (9x + 9) years. Hence, according to the conditions, 9x + 9 = 3(x + 9) or, 9x + 9 = 3x + 27.

or, 9x – 3x = 27 – 9 or, 6x = 18 or, x = 186 = 3

Therefore, the present age of Razat is 9x years = 9 × 3 years = 27 years and present age of Anik is x years = 3 years. Answer : 27 years and 3 years. Example 3. Hanif bought some mangoes as 15 mangoes at Tk. 100.00 sold 12 mangoes at Tk. 100.00 and thus he made a profit of Tk. 400.00. How many mangoes did he buy? Solytion : Let, Hanif bought x mangoes.

Therefore, the cost price of x mangoes is Tk. ⎣⎢⎡

⎦⎥⎤x

15 × 100 = Tk. 20x3

Again, the selling price of x mangoes is Tk. ⎝⎜⎛

⎠⎟⎞x × 100

12 = Tk. 25x3

Hence, according to the conditions, 25x3 –

20x3 = 400 [sales – cost = profit]

or, 25x − 20x

3 = 400 or, 5x3 = 400

or, x = 4005 × 3 = 80 × 3 = 240

∴ Required number of mangoes = 240. Answer: 240.

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Example 4. If Ranjit is in possession of 200 coins in fifty paisa and ten paisa and that amounts to Tk. 44, then what is the number of coins of each kind? Solution : Let, the number of coins in fifty paisa = x pieces. Then, the number of coins in ten paisa = (200 – x) pieces.

∴ x pieces of coins of fifty paisa amounted to Tk. x2 and (200 – x) pieces of coins

of ten paisa amounted to Tk. 200 – x

10

Hence, according to the conditions, x2 +

200 – x10 = 44 or,

5x + 200 – x10 = 44

or, 4x + 200 = 440 or, 4x = 440 – 200 or, 4x = 240 or, x = 60. Therefore, the number of coins in fifty paisa pieces, x = 60 and the number of coins of ten paisa pieces (200 – x ) = 200 – 60 = 140 pieces. Answer : 60 and 140. Example 5. Income of Ranjan is equal to that of Ranjit. Ranjan saves one fifth of his total income. But in one year Ranjit spends Tk. 4000 more than the amount which Ranjan spends. After four years it is found that Ranjit is indebted Tk. 11000. How much is the income of each of them? Solution : Let, each of them earns Tk. x in one year.

Then, Ranjan spends Tk. ⎝⎜⎛

⎠⎟⎞x –

x5 = Tk.

4x5 in a year.

Therefore, Ranjit spends Tk. ⎝⎜⎛

⎠⎟⎞4x

5 + 4000 in a year and in one year he is indebted

to Tk. 11000

4 = Tk. 2750

Hence, Ranjit spends Tk. 2750 more than the amount of his income yearly.

∴ According to the conditions, ⎝⎜⎛

⎠⎟⎞4x

5 + 4000 = x + 2750

or, ⎝⎜⎛

⎠⎟⎞4x

5 + 4000 – x = 2750

or, 4x + 20000 – 5x

5 = 2750

or, 20000 – x

5 = 2750 or, 20000 – x = 13750.

or, x = 20000 – 13750 or, x = 6250. Therefore, yearly income of each of them is Tk. 6250.00 Answer: Tk. 6250.00

N.M.G. -23

168

Example 6. A businessman of banana bought some bananas each at Tk. 1.00 and the same number of bananas each at 50 paisa and sold all bananas each at 80 paisa. Thus he made a profit of Tk. 100.00. How many bananas did he buy? Solution : Let, the businessman bought x bananas. Half of the total number of x

bananas was bought each at Tk. 1.00 and therefore, its cost price is Tk. x2 .

Again the rest was bought each at 50 paisa or Tk. 12 , so its cost price is Tk.

⎝⎜⎛

⎠⎟⎞x

2 × 12

Hence, the total cost price is Tk. ⎝⎜⎛

⎠⎟⎞x

2 + x2 ×

12 = Tk.

⎝⎜⎛

⎠⎟⎞x

2 + x4 = Tk.

3x4 .

As the total number of bananas are sold each at 80 paisa or, Tk. 45 , the selling

price is Tk. ⎝⎜⎛

⎠⎟⎞4

5 x = Tk. 4x5

∴ According to the condition, ⎝⎜⎛

⎠⎟⎞4x

5 – 3x4 = 100. [selling price – cost price = profit]

or, 16x – 15x

20 = 100 or, x

20 = 100 or, x = 2000.

Hence, the number of bananas is 2000. Answer : 2000.

EXERCISE 3.2

1. The sum of two numbers is 100. Twice of a number as much smaller than 100 as three times of another number is that much greater than 160. What are those two numbers ?

2. The sum of two numbers is 61. Twice of a number is greater than two third of another number by 10. What are those two numbers ?

3. 10 years ago, father was five times as old as his son. After 20 years, father will be twice as old as his son. What are the present ages of father and son?

4. The sum of two numbers is 75. One third of greater number is as much smaller than 30 as four times of smaller number is that much greater than 50. Find those two numbers.

169

5. The denominator of a fraction is greater than its numerator by 1. If 2 is subtracted from numerator of the fraction and 2 is added to its denominator, the fraction becomes equal to 16 . Find the fraction.

6. Divide Tk. 104.00 among Chaity, Pew and Rafat in such a way so that double of Chaity's share, three times of Pew's share and four times of Rafat's share equal to one another.

7. Rahim bought a certain number of dates at Tk. 1⋅00 for two and the same number of dates at Tk. 1⋅00 for three dates and sold all of them at Tk. 2⋅00 for five dates. As a result he made a loss of Tk. 4. How many dates did he buy?

8. Divide Tk. 170 among 50 men and women such that each man gets Tk.3⋅50 and each woman gets Tk. 3⋅25. Find the numbers of men and women.

9. The number of male students is of the 56 students in a school and the rest are

female students. If the number of male students is more than that of female students by 120, find the number of students in the school.

10. Half of a pillar is under the earth, one third of it is within water and the rest of 2 metres is above water. What is the length of the pillar?

11. Monju bought a certain quantity of sugar at Tk. 40 per kg. and certain quantity of sugar at Tk. 30. Then he sold the sugar at Tk. 36 per kg. and these

made a profit of 1212 % per kg. Find the ratio that he bought both kinds of

sugar.

12. 100 coins in fifty paisa pieces and twenty five paisa pieces were found in a box and amounted to Tk. 45. How many coins of each kind were there?

13. Tongi is 4 kms. from Uttara. From Uttara Taher was riding by ricksha at 6km/hour and kalam was walking towards Tongi at 3 Km/hour from the same place. After reaching at Tongi Taher took rest for 10 minutes and then started towards Uttara. How far from Tongi will he meet Kalam?

170

!4. Farid and Zafar had 80 kg. of goods with them. They had to pay Tk. 125 and Tk. 75 respectively for carring cost of goods in addition to quantities which they can carry with them free of cost. If all goods belonged to one of them, he had to pay Tk. 300.00 for carring cost. How much weight of goods did each of them carry and how much weight of goods could each of them carry free of cost ?

Multiple Choice Questions [Mark (√) on the correct answer] 1. How many unknown quantities are there in a simple linear

equation? (a) 1 (b) 2 (c) 3 (d) Infinite 2. Which one of the following is the cancellation law of addition? (a) If a + c = b + c, then b + c = a + c, (b) If p + q = r, then p = r − q, (c) If m + n = p + n, then m = p (d) If ac = bc, c ≠ 0, then a = b

3. If dc

ba= , then for which law of the following is applicable for ad = bc?

(a) Side interchange (b) Cancellation law of addition (c) Cancellation law of multiplication (d) Cross multiplication

4. If 2

26

xx=+ , then which one of the following is the value of x?

(a) 3 (b) 4 (c) 5 (d) 6

5. One number is 211 times of the other and the sum of the two

numbers is 25. Which one of the following is the large number? (a) 20 (b) 15 (c) 12 (d) 10

171

6. The 5 times of the age of Akib is equal to the 3 times of the age of Arif. The sum of their age is 24 years, then which one of the following are the ages of Arif and Akib respectively?

(a) 14, 10 (b) 9, 15 (c) 15, 9 (d) 18, 6

7. The 65 part of a pillar is black coloured and the remaining part is

white coloured. If the difference of the parts of the pillar is 6 metre, then which one of the following is the length of the pillar?

(a) 9 metre (b) 8 metre (c) 6 metre (d) 5 metre. 8. The sum of the ages of Mr. Jagadish and his two sons is 36 years.

What will be the sum of their ages after 5 years? (a) 40 (b) 41 (c) 46 (d) 51

CREATIVE QUESTIONS

1. Mr. Khaleque bought some mangoes as 15 mangoes at Tk. 100.00 and sold 12 mangoes at Tk. 100.00 and thus he made a profit of Tk. 400.00.

(a) Form an equation in the view of the above informations. (b) Find the buying price and selling price of his mangoes. (c) If Mr. Khaleque paid the shopkeeper total price by 170 notes of Tk. 5.00

and Tk. 10.00, then how many notes of which type did he give? 2. Ranu, Rabu and Panu are three sisters. Taka 208 is distributed among them

so that the double of Ranu’s share is triple of Rabu’s share and 4 times of Panu’s share are equal to one another.

(a) Form an equation in view of the above informations. (b) Find the amount of taka of Ranu, Rabu and Panu. (c) How much amount of taka would be added with each of their amount so

that the summation will equal to three times of the amount of Rabu and Panu and four times of the amount of Ranu?

172

3. Lutfor and Jagadish are two friends. The earnings of both of them

are equal, Lutfor used to save 51 of his earnings. He distributed his

savings between his wife and daughter such that the seven times of daughter’s amount is equal to the three times of wife’s amount. But in three years Jagadish spends Tk. 12000 more than the amount which Lutfor spends. After 5 years it is found that Jagadish is indebted Tk. 13750.

(a) How much taka was indebted by Jagadish in one year and how much more taka did he spend than that of Lutfor?

(b) Find out the amount of yearly earnings of each of them. (c) Find the amount of taka of the wife and daughter of Lutfor.

Chapter IV

Simultaneous Linear Equations And Their Applications

4.1. Simultaneous Linear Equations

x – y = 4 is an equation. Here x and y are two unknown quantities. Each of the unknown quantities, x and y are of first degree. It is a linear equation containing these two unknown quantities. The equation will be satisfied by those two numbers whose difference is 4. For example, x = 5, y = 1; or, x = 6, y = 2; or, x = 7, y = 3; or, x = 8, y =4; or, x – 2, y = – 2; or, x = – 1, y = – 5; ...... etc. will satisfy the equation. There are innumerable values of two unknown quantities of the equation. x + y = 6 will be satisfied by innumerable pairs of values of two quantities, x and y. For example, x = 5, y = 1; or, x = 4, y = 2; or, x = 3, y = 3; or, x = 7, y = – 1; or, x = 8, y = – 2, --------etc will satisfy the equation. Now, if two equations x – y = 4 and x + y = 6 are taken together, then the innumerable pairs of values of x and y obtained from both the equations, the pair of those values x = 5 and y = 1 will satisfy both the equations. Hence, two equations x – y = 4 and x + y = 6 are simultaneously satisfied for only values of x = 5 and y = 1. If more than one equation are satisfied simultaneously by the values of unknown quantities, then the equations taken together are called simultaneous equations and if the unknown quantities are of first degree then simultaneous equations are called simultaneous linear equations. The values of the unknown quantities which satisfy simultaneous equations simultaneously are called the roots or the solution of the simultaneous equations. Here two equations, x – y = 4 and x + y = 6 are simultaneous equations. Its only solution, x = 5 and y = 1 is written shortly as (x, y) = (5,1). 4.1.1. Solution of Two Simultaneous Linear Equations Involving Two Unknown Quantities

There are three methods of finding the solution of two simultaneous linear equations. The methods are: (a) Method of substitution (b) Method of elimination and (c) Method of cross– multiplication.

174

Substitution – method and elimination – method are discussed below. (a) Method of Substitution

In this method, the value of any one of two unknown quantities from any one of two equations is expressed by the other and the value thus obtained is substituted for another equation and then the value of one unknown quantity is found by applying the method as used in case of linear equations. The value of the other unknown quantity is found and thus the values of two unknown quantities are obtained. In this way to find out the values of unknown quantities. Example 1. Solve : x – y = 5 x + y = 9 Solution : First equation, x + y = 9 ........... (1)

or, x = 9 – y, [interchanging the sides of y] ............ (2)

Second equation, x – y = 5 ............. (3)

or, 9 – y – y = 5 [ putting the value of x from (2)]

or, 9 – 2y = 5

or – 2y = 5 – 9 [ interchanging the sides of 9]

or, – 2y = – 4

or, y = 2 [dividing both the sides by – 2]

putting the value of y in equation (1), x + 2 = 9 or, x = 9 – 2 [interchanging the sides of 2] or, x = 7

∴ Required solution, x = 7 and y = 2. Answer : (x, y) = (7, 2). [Verification : If we put x = 7 and y = 2 in the given equations then in the first equation 7 + 2 = 9 = right hand side and in the second equation 7 – 2 = 5 = right hand side]

Remark : The value of x can be obtained by putting the value of y either in equation (2) or (3). Thus the same value is found. Example 2. Solve : 2x + y = 5

4x – 5y = 3. Solution : First equation, 2x + y = 5 ............ (1 )

or, 2x = 5 – y or, x = 5 – y

2 .............. (2)

175

Again, putting the value of x in the second equation we get, or, 4x – 5y = 3 ......... (3)

or, 4⎝⎜⎛

⎠⎟⎞5 – y

2 – 5y = 3 or, 2(5 – y) – 5y = 3

or, 10 – 2y – 5y = 3 or, –7y = 3 – 10 or, –7 y = –7. ∴ y = l. Putting the value of y in equation (2) we get,

x = 5 – 1

2 ; or, x = 42 = 2. ∴ x = 2

∴ Required solution , x = 2 and y = 1. Answer : (x, y) = (2, 1). [Verification : Putting x = 2 and y = 1, the left hand side of first equation = 2.2 +1 = 4 + 1 = 5 = right hand side and the left hand side of second equation = 4. 2 –5.1 = 8 – 5 = 3 = right hand side]

Example 3. Solve : 3x + 2y = 13

5x – 3y = 9. Solution : First equation, 3x + 2y =13 ............ (1) or, 2y = 13 – 3x

or, y = 13 – 3x

2 ............... (2)

Second equation, 5x – 3y = 9

or, 5x – 3. ⎝⎜⎛

⎠⎟⎞13 – 3x

2 = 9 [putting the value of y from (2)]

or, 2.5x – 3(13 – 3x) = 9.2 [multiplying both the sides by 2]

or, 10x – 39 + 9x = 18 or, 19x = 18 + 39 or, 19x = 57 or, x = 5719 = 3. ∴ x = 3.

Putting x = 3 in equation (2), we get

y = 13 – 3.3

2 = 13 – 9

2 = 42 = 2 ∴ y = 2

∴ Required solution, x = 3 and y = 2. Answer : (x, y) = (3, 2)

N.M.G. -24

176

(b) Method of Elimination

In this method, two equations are to be multiplied by two such numbers respectively so that the absolute values of the coefficients of any one of two unknowns obtained from two equations after multiplication are both equal. Then the last two equations are to be added to or subtracted from, so that only one unknown quantity exists in the equation obtained, the value of the unknown quantity can be determined from the equation thus obtained. Putting the value of this unknown quantity in any one of the equations, the value of the other quantity can be determined. Example 4. Solve : 3x – 2y = 5 5x + 3y = 21. Solution : Multiplying both the sides of the first equation by 3 it becomes 9x – 6y = 15 and multiplying both sides of the second equation by 2 it becomes l0x + 6y = 42. Then adding we get, 19x = 57 ∴ x = 3. Putting the value of x in the first equation, we get 3.3 – 2y = 5

or, 9 – 2y = 5 or, – 2y = 5 – 9

or, – 2y = – 4 or, y = 2. ∴ Required solution, x = 3 and y = 2 Answer : (x, y) = (3, 2)

[Verification : Putting x = 3 and y = 2, the left hand side of the first equation = 3.3. –2.2 = 9 – 4 = 5 = right hand side and the left hand side of the second equation = 5.3 + 3.2 = 15 + 6 = 21 = right hand side]

Example 5. Solve : 5x – y = 2. 7x – 4y = –5

Solution : Multiplying the first equation by 4, we get, 20x – 4y = 8 ........... (i) and 7x – 4y = – 5 ............ (ii)

Subtracting, 13x = 13. or, x = 1. Putting x = 1 in the first equation, we get, 5.1 – y = 2 or 5 – y = 2 or, – y = 2 – 5 = –3 or, y = 3. ∴ Required solution, x = 1 and y = 3. Answer : (x, y) = (1, 3).

177

Example 6. Solve : 8x + 11 y = 3

6x – 13y = – 19. Solution : Multiplying the first equation by 3 and the second equation by 4, we get, 24x + 33y = 9 and 24x – 52y = – 76

Subtracting, 85y = 85. or, y = 1. Now putting y = 1 in the second equation, we get, 6x – 13.1 = –19 or, 6x = – 19 + 13

or, 6x = – 6; or, x = – 1. ∴ Required solution, x = – 1 and y = 1 Answer : (x, y) = (– 1, 1).

EXERCISE 4.1

Solve by the method of substitution and verify the solutions: 1. x – y = 1 2. x + y = 5 3. x – y = 0 x + y = 3. 2x – y = 4. 2x + 3y = 10. 4. 3y = 4x + 1 5. x + 4y = 14 6. 3x + 5y = –7 4x – 5y = –7. 7x – 3y = 5. 5x + 4y = 10.

7. 7x – 5y = 11 8. x3 +

y2 = 6 9. 3x – 5y = – 9

3x + 2y = 13. x2 +

y4 = 4 5x – 3y = 1

10. – 7x + 8y = 9 11. x + ay = b 12. 2x +

3y = 2

5x – 4y = – 3 ax – by = c 3x –

2y =

56

Solve by the method of elimination and verify the solutions: 13. x – y = 3 14. 2x + 3y = 7 15. 4x + 3y = – 12 x + y = 7. 8x – 7y = 9 5x + 4y = 15. 16. 2x + 5y = – 14 17. 7x – 5y = 11 18. 8x – 9y = 20 4x – 5y = 17. 3x + 2y = 13. 7x – 10y = 9.

178

19. 3x – 2y = 0 20. x – y = 2a 21. x3 +

y2 = 6

17x – 7y = 52. ax + by = a2 + b2 x2 +

y4 = 5

22. x2 +

y3 = 3 23.

x3 –

2y = 1 24. ax + by = ab

x + y6 = 3

x4 +

3y = 3 bx + ay = ab

4.2 Applications of Linear Simultaneous Equations in Solving Problems Relating to Daily Necessity:

Many of the problems relating to daily necessity can be solved by using the concept of linear simultaneous equations. Then one unknown may exists in many of such problems. Different symbols are used for each of the unknowns and then the equations are formed in such a case the number of equations of which one is independent of another is exactly equal to the number of the unknowns used. Then by solving the equations the values of the unknowns are determined. Example 1. Divide Tk. 50 between two persons in such a way so that three times of one person's Taka is equal to seven times of another person's Taka. Solution : Let, one has Tk. x and another has Tk. y. Therefore, according to the conditions, x + y = 50 .............. (1) and 3x = 7y ............. (2)

from (l), y = 50 – x ............. (3)

Putting the value of y from (3) in equation (2), we get, 3x = 7(50 – x) or, 3x = 350 – 7x or, 3x + 7x = 350 or, 10x = 350 or, x = 35. Putting x = 35 in equation (3), we get, y = 50 – 35 = 15. ∴ Required shares : One gets Tk. 35 and another gets Tk. 15. Answer : Tk. 35 and Tk. 15. Example 2. 10 years ago, father was four times as old as his son. After 4 years father will be twice as old as his son. What are their present ages?

179

Solution : Let, the present age of the father be x years and that of the son be y years. Therefore, according to the first condition, x – 10 = 4(y – 10) .............. (1)

and according to the second condition, x + 4 = 2(y + 4) ................ (2)

From equation (1), x – 10 = 4y – 40 or, x = 4y – 40 + 10 or, x = 4y – 30 ............. (3)

Putting the value of x from (3) in equation (2), we get, 4y – 30 + 4 = 2y + 8. or, 4y – 2y = 8 + 30 – 4 or, 2y = 34 or, y = 17. From equation (3) we get,

x = 4y – 30 or, x = 4.17 – 30 = 68 – 30 = 38. Therefore, the present age of the father is 38 years and that of the son is 17 years. Answer : 38 years and 17 years. Example 3. Pia and Papia had some mangoes. If 20 mangoes from Pia are given to Papia, then the number of mangoes of Papia will be twice that of Pia. Again 10 mangoes from Papia are given to Pia, then the number of mangoes of Pia will be three times that of Papia. How many mangoes did each of them have? Solution : Let, x be the number of Pia's mangoes and y be the number of Papia's mangoes. Therefore, according to the first condition, 2(x–20) = y + 20 ............. (1)

and according to the second condition, 3(y – 10) = x + 10 .............. (2)

From the first equation, y = 2x – 40 – 20 or, y = 2x – 60 .............. (3)

Putting the values of y from (3) in equation (2), we get, 3(2x – 60 – 10) = x + 10 or, 6x – 180 – 30 = x + 10

180

or, 6x – x = 10 + 180 + 30 or, 5x = 220 or, x = 44. Putting the value of x in (3) we get, y = 2(44) – 60 = 88 – 60 = 28. Therefore, the number of mangoes of Pia is 44 and that of Papia is 28. Answer : 44 and 28. Example 4. If 5 is added to the sum of two digits of a number consisting to two digits, then the sum will be three times the digit of ten's place. But if 9 is subtracted from the number, then the places of digits interchanged. Find the number. Solution : Let, x be the digit of ten's place and y be the digit of one's place. ∴ The number is 10x + y. Therefore, according to the first condition, x + y + 5 = 3x or, 2x – y = 5 ........... (1)

According to the 2nd condition, (l0x + y ) – 9 = 10y + x or, 9x – 9y = 9 or, x – y = 1 .............. (2)

Subtracting equation (2) from equation (1) we get, x = 4

Putting the value of x in equation (2) we get, y = 3. ∴ Required number is 10.4 + 3 = 40 + 3 = 43. Answer : 43. Example 5. The length of a rectangular region is greater than its breadth by 12 metres. If the perimeter of the rectangular region is 136 metre, find its length and breadth. Solution : Let, the length of the rectangular region be x metres and its breadth be y metres. Therefore, according to the first condition, x – y = 12 ............. (1), And according to the second condition, 2x + 2y = 136 or, x + y = 68 ............... (2)

Adding equation (2) to the equation (1), we get 2x = 80 or, x = 40 Putting x = 40 in equation (1), we get – y = 12 – 40 or, – y = –28; or, y = 28. ∴ Required length = 40 metres and breadth = 28 metres. Answer : 40 metres and 28 metres.

181

EXERCISE 4.2

1. The sum and difference of two numbers are respectively 80 and 10. What are these numbers ?

2. The difference of two numbers is 45 and one is four times the other. What are these numbers?

3. If the first of two numbers is added to five times the second, then the sum is 52. But if the second is added to eight times the first, then the sum is 65. What are these numbers?

4. Father's age is 30 years more than son's age. After 10 years father will be twice as old as his son. What are their ages?

5. 20 years ago, father was five times as old as his son. After 5 years three times of father's age will be equal to five times of son's age. What are the present ages of father and son?

6. If 7 is added to the numerator of a certain fraction, then it becomes 2 and if 2 is subtracted from the denominator, then it becomes 1. Find the fraction.

7. If 1 is added to the numerator of a certain fraction, then it becomes 12 and if 1

is added to the denominator, then it becomes 13 . Find the fraction.

8. If 7 is added to the sum of two digits of a number consisting of two digits, then the sum is three times the digit of ten's place. But if 18 is subtracted from the number then places of two digits interchanged. Find the number.

9. If 9 is added to the sum of two digits of a number consisting of two digits then the sum is equal to six times the digit of one's place. But if nine times the digit of one's place is subtracted from the number then the digits are interchanged. Find the number.

10. Anik and Rafat had some mangoes. If Anik gives Rafat 30 mangoes then the number of mangoes of Rafat becomes twice that of Anik. Again if Rafat gives Anik 10 mangoes then it becomes three times the number of mangoes of Anik. How many mangoes did Anik and Rafat have?

11. The length of a rectangular region is greater than its breadth by 30 metres. If the perimeter of the rectangular region is 140 metres, find the length and breadth of the region.

182

12. The positions of two persons are 27 kms. away. If they move at the same time in the same direction, then they can meet after 9 hours. But if the move towards each other then they can meet after 3 hours. Determine speed of each of them.

13. A man and a boy can do a work in 8 days. Two men and four boys can do the same work in 3 days. In how many days a man or a boy can do the same work?

14. The same work which two men and two boys can do in six days can be done by 3 men and 8 boys in three days. In how many days a man or a boy can do that work?

15. Rahim sold to a buyer his 9 cows and 7 goats at Tk. 51000. He sold to another customer 6 cows and 16 goats at the same price. What is the price of each of cows and goats?

16. A man rowing in favour of current covers 70 kms and against the current he required 70 hours to come back. Determine the speeds of rowing and current.

17. A tank is filled up in 6 minutes by two pipes. If the second pipe is closed after 3 minutes of opening the two pipes at the same time, then it takes 5 minutes more to fill up the tank. How much time will it take to fill up the tank by each of the pipe?

Multiple Choice Questions [Mark (√) on the correct answer] 1. If x – y = 1, x + y = 3, then (x, y) = how much? (a) (1, 2) (b) (2,1) (c) (1, 3) (d) (3, 1) 2. If x – y = 2a, ax +by = a2+b2 then (x, y) = how much? (a) (a, b) (b) (b, a) (c) (a + b, a – b) (d) (a + b, b – a) 3. (i) 3x+2y = 5 is a simple equation. (ii) All simultaneous equations can be solved by substitution

method. (iii) The linear equation of one unknown variable has only one root.

Which one of the following is correct according to the above informations?

(a) i and ii (b) ii and iii (c) i and iii (d) i, ii and iii

183

Answer the questions 4, 5 in the view of the following informations: The length of a rectangular house is 4 metre more than the breadth. The perimeter of the house is 32 metre. 4. How much metre is the length of the house? (a) 6 (b) 8 (c) 10 (d) 12 5. How much square metre is the area of the house? (a) 60 (b) 96 (c) 160 (d) 192 6. The present age of the father is 30 years more than the present age

of son. After 10 years, father will be twice as old as his son. What is the present age of the son?

(a) 40 (b) 30 (c) 20 (d) 10 7. Father’s age is four times of the age of son. The sum of the age of

father and son was 52 years before 4 years. What is the present age of father?

(a) 32 (b) 38 (c) 48 (d) 52

CREATIVE QUESTIONS 1. A man rowing in favour of current, covers 35 km in 5 hours and against the

current he required 35 hours to come back. (a) Express the above informations in two equations. (b) Determine the speeds of rowing and current. (c) How much time would he spend to go and come back for 84 km.

2. The sum of two digits of a number consisting two digits is 7. If 27 is subtracted from the number then the digits interchange their position.

(a) Express the above informations in two equations. (b) Determine the number. (c) Add such a number with that number so that the obtained number is 1

more than five times of the sum of two digits. Determine the obtained number.

N.M.G. -25

184

3. A man has total 903 notes of Dolar and Euro currency. The value of the notes in Bangladeshi taka is Tk. 77373.00 (1 dolar = Tk. 71, 1 Euro = Tk. 110).

(a) If the number of notes of Dolar is x, then express the value of that notes in terms of x.

(b) How many notes of what type? Find the amount in Bangladeshi taka separately.

(c) If the value of Dolar and Euro increase by Tk. 0.85 and Tk. 1.22 respectively, then what would be the percentage of increase or decrease of the value of notes in Bangladeshi taka?

4. Jasim and Iqbal are two friends. Tk. 152.00 is divided between them in such

a way so that three times of Iqbal’s share and five times of Jasim’s share are equal.

(a) Express the above information in two equations. (b) Find the amount of taka of each person. (c) Iqbal gave some taka to Jasim, then five times of Iqbal’s share and three

times of Jasim’s share are equal. Now find the amount of taka of each of them.

Chapter V

Drawing Of Graphs And Their Applications

5.1. Number line, Co-ordinates of a point in a Plane and Plotting of Points

By drawing geometrical figures the points of geometrical figures are expressed by algebraic terms. Again, algebraic terms are also denoted by geometrical points. The number corresponding to the point is the coordinate of that point. Here coordinates of a point in the straight line and the plane are discussed. 5.1.1. Number Line

The middle point of the segment of a straight line is fixed up to correspond zero. The length of a small segment of line considered as unit and the points situated at the distances of 1, 2, 3, ..........units to the left and right of the point corresponding to zero are marked. The point that corresponds to zero is called origin. The points in the right of origin which denote 1, 2, 3, ........ are considered as positive 1, 2, 3, ........... and the points in the left of origin which denote 1, 2, 3, .............. are considered negative – 1, –2, –3, ............... The process of establishing the relation between the points of straight line and the straight line is called coordination of the straight line. Thus straight line coordinated is called number–line. Each real number is denoted by a point of that line. The position of the point A in number line is completely denoted by the number related to it.

If the point, A is to the right of origin then it denotes a positive number and if it is to the left of origin then it denotes a negative number. In the same distance from number line, there are innumerable points out side it. The position of all these points can not be determined by that number line only. In this case, the process of determination of coordinates of a point in the plane is to be adopted. Remark : For different origins or different units of length the coordinates of points will be different. There are one to one correspondence between all real numbers and the points on the straight line. 5.1.2 Coordination of Points on Plane. The position of all the points on the plane can be determined by using mutually intersecting lines. Let us consider two mutually perpendicular straight lines XOX' and YOY' on the same plane. They intersect at the point O. The point O is the

186

origin of both the lines. Taking conveniently small segment as the unit of length, let us make coordination of two lines, XOX' and YOY'. Now two lines XOX' and YOY' become number lines.

The points on the right side of the origin O of the line, XOX' are positive and the points on the left side are negative. The points on the above of the origin O of the line, YOY' are positive and the points below are negative. XOX' is called x–axis and YOY' is called y– axis. A is on the number line XOX' and if A is the corresponding point of 3 in that number line, then A corresponds to the pair of numbers (3, 0). Again in the number line XOX', B the corresponding point of – 1 corresponds to (– 1, 0). If C is on the number line YOY' and C is the corresponding point of 2 in that number line, then C corresponds to the pair of numbers, (0, 2). Again in the number line YOY', D the corresponding point of –3 corresponds to (0, – 3).

O

′

′

187

From A, AM and AN are drawn perpendiculars on XOX' and YOY' respectively. The corresponding point of 3 in the line XOX' is M and the corresponding point of

2 in the line YOY' is N. A corresponds to the pair of number (3, 2). Thus when each point, A in the plane of axes corresponds to the pair of number (x, y), then (x, y) is called the coordinate of A. In this case, if from A, AM is drawn perpendicular on the x – axis, then M will be

the corresponding point (x, o). Again, if from A, AN is drawn perpendicular on the y – a xis, then N will be the corresponding point (0, y). x of (x, y) is called x–coordinate or abscissa and y is called y– coordinate or ordinate. 5.1.3. Plotting of Points

If the coordinates of any point are given then the placement of that point in the plane of coordiantion is known as plotting of the points. Generally graph paper is used for plotting the points. Graph Paper

In graph paper there are some parallel lines which have been drawn

x-axis

y-ax

is

188

at equal distances and there are also some lines perpendicular to those parallel lines at equal distances. Thus the paper is divided into some equal small squares. Generally, one side of the small square is considered as the unit of length. The graph paper is also called drawing paper. To plot points in the graph paper, two mutually perpendicular straight lines are taken. They are denoted by XOX' and YOY'. Generally, XOX' is considered as horizontal line and YOY' is considered as vertical line. In the direction of OX of the line XOX' and in the direction of OY of the line YOY' are considered positive directions. Quadrant and Signs of Coordinates As a result of drawing, the line XOX' or x–axis and the line YOY' or y–axis on graph paper, the whole plane of the graph paper has been divided into four parts and each part is called quadrant. The first quadrant by OX and OY, second quadrant by OX' and OY, third quadrant by OX' and OY' and the fourth quadrant by OX and OY'. The x–coordinate and y–coordinate of any point in the first quadrant are both positive.

In the second quadrant x–coordinate of any point is negative while y–coordinate is positive. In the third quadrant x–coordinate and y–coordinate of any point are both negative and in the fourth quadrant the x–coordinate or abscissa of any point is positive, while y– coordiante or ordinate is negative. Plotting of a Point in the Graph Paper

Let us plot a point in the graph paper with respect to the axis of XOX' and YOY'. Suppose, the length of the side of the smallest square has been taken as unit in case of both the axes. Let the coordinate of the point A be (5, 6). Since the abscissa and ordinate of the point A are both positive, hence the point is situated in the first quadrant.

Ist Quadrant

4th Quadrant 3rd Quadrant

2nd Quadrant

189

If we move to a distance equal to the length of 5 side of a small square along OX to the right of the origin O and then a distance equal to the length of 6 sides of small square above, then the position of the point A is found. A dot is placed there and A (5, 6) is written. Let us suppose that coordinates of the point B be (–6, 5). The point B is situated in the second quadrant. The point move to a distance equal to the length of 6 side of a small square along OX' to the left of the origin O and from there move upward to a distance equal to the length of 5 sides of a small square, then the position of the point B will be found. A dot is placed there and B(–6, 5) is written at the side of it. Let, the coordinate of the point C be (– 4, –2). The abscissa and the ordinate are both negative. Therefore it is situated in the third quadrant. Then position of the point C can be located if we move along OX' to the left of the origin at a distance equal to the length of 4 sides of a small square and from there move to a distance equal to the length of 2 sides of a small square below. It is written as C (– 4, – 2) at the side of the dot placed there. Let the coordinates of the point D be (3, – 6). The abscissa of the point D is positive and ordinate is negative. Therefore, it is situated in the fourth quadrant. If we move along OX to the right of the origin O to a distance equal to the length of 3 sides of a small square and from there move to a distance equal to the length of 6 sides of a small square below, then we get the position of the point D. A dot is to be placed there and D (3, –6) is to be written at the side of it. Example 1. Plot the points A(3, 7), B(3, 0), C(–5, 4), D (0, 4), E (–4, –6), F(–6, 0) and G(0, – 5) in the graph paper. Solution : Let, XOX' and YOY′ represent x–axis and y –axis respectively. O is the origin. Let us suppose the length of each side of the smallest square as unit in both the axes.

190

To find the position of the point A, we have to move to the right of the origin O at a distance equal to the length of 3 sides of a small square along OX and then at distance equal to the length of 7 sides of a small square above OX. Thus the point found is A. To find the point B it is required to move to the right of the origin O to a distance equal to the length of 3 sides of a small square along OX. (Here as the ordinate is zero it is not required to move below or above OX).

To plot the point C, we have to move to the left of the origin O at distance equal to the length of 5 sides of small square along OX′ and then to move above OX′ to a distance equal to the length of 4 sides of small square. To plot the point D, it is required to move above from O along OY to a distance equal to the length of 4 sides of small square (Here the abscissa is zero and the abscissa of the origin O is also zero. As the ordinate is positive, we are to move above directly from O along OY). To plot the point E, we have to move along OX' left from the origin O to a distance equal to the length of 4 sides of small square and then to move below OX' to distance equal to the length of 6 sides of small square. To plot the point F, the point which is obtained by moving to a distance equal to the length of 6 sides of small square to the left of the origin along OX (Here as the ordinate is zero, it is not required to move above or below from OX'). To plot the point G, the point which is obtained by moving to a distance equal to the length 5 sides of a small square below the origin O along OY′ is the point G. Example 2. Plot the points (1, 3), (–1, 1) and (–3, –1) in the graph paper. Also show that, they are collinear (i.e. the points lie on the same straight lines). Solution : Let XOX' and YOY' represent x–axis and y–axis respectively and O is the origin. Let us suppose that the length of two sides of a small square is taken as unit along the x–axis and the length of three sides of a small square is taken as unit along the y–axis. Now, for abscissa 1 of the point (1, 3) we are to consider the

3.0

191

length equal to (1 × 2), or 2 sides of small square and for the ordinate 3 we have to consider the length equal to (3 × 3), or 9 sides of small square. To plot the point (1, 3), the point which is obtained by moving to a distance equal to the length of 2 sides of a small square to the right from the origin O along OX and then to a distance equal to the length 9 sides of a small square is the point (1, 3). To plot the point (–1, 1), the point obtained by moving to a distance equal to the length of (1 × 2), or 2 small sides of squares to the left from the origin O along OX' and to distance equal to the length of (1 × 3) or 3 sides of a small square is the point (–1, 1). To plot the point (–3, –1), the point obtained by moving to a distance equal to the length of (3 × 2), or 6 sides of a small square to the left from the origin O along OX' and then to a distance equal to the length of (1 × 3), or 3 sides of small square below is the point (–3, –1). Now let us join the first and third points by a straight line. The other point lies on this line. Therefore, the given point are collinear. Remark : (a) Any coordinate of a point is to be multiplied by that number which is equal to the length of number of sides of a small square that is considered as unit. Then the points are to be placed in the graph paper according to the co–ordinates obtained by these multiplication. (b) The unit is to be taken so that the points can be plotted in graph paper with the help of whole numbers if any co-ordinate is fraction.

EXERCISE 5.1

1. Plot the following points in the graph paper and mention the quadrants in which they lie:

A (2, 5); B (–2, 7); C ( 4, 0); D (–4, –5); E ( 5, – 6); F (0,8); G (0,0); H (0, – 6); I (–8,0); J (3,9) and K (0,7) .

N.M.G. -26

192

2. Taking the length of two sides of smallest square as unit, plot the following points in the graph paper.

A (3, 4); B ( –3, 5); C ( 0, 6); D ( –3, –4); E ( –3, 0); F (0, –3) and G (52 , 92 )

3. Taking the length of three sides of the smallest square as unit, plot the following points in the graph paper:

A ( –2, 3); B (23 , 43 ); C ( 0, – 6); D ( –3, –5); E ( 5, 0) and F (3, 4).

4. Plot the following points in the graph paper. Show that they lie on the same straight line.

(a) A (2, –1), B (6, 3) and C (–2, –5) (b) A (1, –1), B (–2, –7) and C (5, 7) (c) A (3, 1), B (6, 3) and C (–3, –3) (d) A (1, 3), B (0, 0) and C (–1,–3)

(e) A ⎝⎜⎛

⎠⎟⎞1

2 , 14 ; B

⎝⎜⎛

⎠⎟⎞1

2 , –54 and C

⎝⎜⎛

⎠⎟⎞1

2 , 74 ;

5. Show that the straight line joining the points (3, 6) and (–2, –4) passes through the origin.

6. Find the coordinates of the intersecting point at which the straight line joining the points (2, 4) and (– 4, –5) intersects with the straight line joining the points (–1,0) and (–5, –4).

7. Find the coordinates of the intersecting point at which the straight line joining the points (

13 , 3) and (

–12 ,

12 ) intersects with the straight line joining

the points (3, 3) and (–2, –2).

5.2. Graphs of Linear Simultaneous Equations

The existing relation between two unknown quantities, x and y can be expressed by a figure. This figure is called graph of that relation. The linear equation x – y = 2 is satisfied by the pairs of numbers (3, 1), (4, 2), (5, 3), (–1, –3), (1, –1) ....... etc. of (x, y). These pairs of numbers are considered as coordinates and their corresponding points are plotted. Then the figure formed by joining these points is the graph. The relation between x and y expressed by the above mentioned equation can be expressed by this graph. There are innumerable points in the graph of such a equation. If some points of the graph are plotted then a clear idea regarding the graph is found. The graph can be drawn if the points are joined only.

193

Of the equation x – y = 2 for different values of x, corresponding definite values of y can be obtained. Again for different values of y, different definite values of x can be determined. The values thus obtained can be written in the tabulated form as given below :

x 3 4 5 1 0 – 1 y 1 2 3 – 1 – 2 – 3

Remark : Generally these values are to be taken as whole numbers. Now let us draw axes, XOX' and YOY' in the graph paper. Let, in both the axes, the length of each side of the smallest square be unit. Let us plot the points A, B, C, D, E and F which correspond to the pairs of numbers (3, 1), (4, 2) (5, 3), (1,–1), (0, –2) and (–1, –3) mentioned above. If the points are joined then the straight line CF is found. This line, CF is the graph of the equation, x – y = 2

Remark : The graph of any linear simulations equation containing two unknown quantities is a straight line. The coordinates of each point of this line satisfy the equation. It is sufficient to draw this graph from two points only, but it is better to take three or more points. Example 1. Draw the graph of the equation, 3x + 2y = 5.

Solution : 3x + 2y = 5. or, 2y = 5 – 3x or, y = 5 – 3x

2 ∴ y = 5 – 3x

2

The coordinates of some points of the graph are determined from this relation :

x 1 3 – 1 – 3 y 1 – 2 4 7

194

Let, XOX′ and YOY′ be x - axis and y - axis respectively and O be the origin. In both the axes, the length of each side of the smallest square is taken to be unit. The points (1, 1), (3, –2), (–1, 4) and (–3, 7) are plotted in the graph paper. By joining these points a straight line is obtained. This is the graph of the equation, 3x + 2y = 5

5.3. Solution of Linear Simultaneous Equations containing Two Terms with the help of Graph. There are two linear equations in the linear simultaneous equations containing two unknown terms.

If graphs are drawn for two linear equations then two straight lines are obtained. The point of intersection lies on both the lines. The x – 1 coordinate and y – coordinate of this point of Intersection will be the roots of the given linear simultaneous equations. These two equations will be simultaneously

satisfied by the values of x and y. Therefore, the only solution of the pair of linear simultaneous equations is this abscissa and this ordinate of two straight lines. Remark : There will be no solution of the given simultaneous equations if the two straight lines are parallel. Example 2: Solve with the help of graphs : 2x + 3y = 8 3x – 4y = – 5

Solution : From the given first equation we get, 3y = 8 – 2x or , y = 8 – 2x

3

Let us find the coordinates of some points of the graph of this equation. x – 5 4 7 y 6 0 – 2

Again the second equation, we get, 3x = 4y – 5

195

or, x = 4y – 5

3

or,4y = 3x + 5

or, y = 3x + 5

4

Let us find the coordinates of some points of the graph of this equation : x – 3 – 7 5 y – 1 – 4 5

Let, XOX' and YOY' be x-axis and y-axis respectively and O be the origin. Let, in both the axes, the length of each side of the smallest square be unit. Let us plot the points corresponding to (−5, 6), (4, 0) and (7, –2) in the graph paper. By joining these points, a straight line is obtained. Let us extend the straight line in both directions. Therefore, this is the graph of the equation, 2x + 3y = 8. Again, let us plot the points corresponding to (–3, –1), (–7, –4) and (5, 5)in the graph paper. ' '■■ ■ ■ / ■ ■ % ■■ . * By joining these points, a straight line is obtained. Let the straight line be extended in both directions. Therefore, this is the graph of the equation 3x – 4y = –5. This straight line intersects the previous straight line at a point A. The point A is the solution point of both the lines. The coordinates of this point satisfy both the equations. It is found from the graph that the abscissa and ordinate of the point, A are 1 and 2 respectively. Hence the required solution, x = 1, y = 2, Answer : (x, y) = (1, 2).

196

5.3.1. Solution of Linear Equations with the help of Graph In order to find the solution of the linear equation, 5x + 7 = 8x + 1 with the help of graph the following rules are to be followed. First of all, it is to be expressed in linear simultaneous equations. Each of both the sides is taken to be equal to y, then two equations y = 5x + 7 and y = 8x + 1 are obtained. If the graphs of these two equations are drawn. then two mutually intersecting straight lines are obtained. As the point of intersection lies on both the lines, their coordinates will satisfy both the equations. Hence the abscissa of the point of intersection will be the root of the given linear equation, i.e. the value of x will be the solution of the equation. Example 3. Solve the equation, 2x + 1 = 3x – 2 with the help of graph. Solution : Let each side of 2x + 1 = 3x – 2 be equal to y. Therefore, y = 2x + 1 and y = 3x – 2. Let us find the coordinates of some points of the graph of the equation, y = 2x + l :

x 1 – 1 4

y 3 – 1 9

Again, let us find the coordinates of some points of the graph of the equation, y = 3x – 2

x 2 – 1 – 3

y 4 – 5 – 11

Let XOX' and YOY' be x– axis and y-axis respectively and O be the origin. Let in both the axes the length of each side of the smallest square be unit. Let us plot the points corresponding to (1, 3), (–1, –1) and (4, 9) in the graph paper. By joining these points, a straight line is obatined. Let the straight line be extended in both the directions. This is the graph of the equation, y = 2x + 1. Again, let us plot the points corresponding to (2, 4), (–1, –5) and (–3, –11) in the graph paper. By joining these points, a straight line is obtained.

197

Let us extend this straight line in both the directions. This is the graph of the equation, y = 3x – 2. This straight line intersects the previous line at the point A. The point A is the common point of both lines. The coordinates of the point satisfy both the equations. It appears from the graph that the abscissa of the point, A is 3. Hence, the required solution, x = 3. Answer : x = 3.

EXERCISE 5.2

1. Draw the graphs of the following equations : i) x + y = 4: ii) 2x + y = 8; iii) 4x + 3y = 6; iv) 2x – y = 10; v) 3x – 2y = 9; vi) y = 3x – 7; vii) 3x + 4y = 0: viii) x – 2y – 5 = 0; ix) 2x = 6–3y; (x) 2x – 5y + 12 = 0 2. Solve with the help of graph : i) x + y = 5 ii) x + 4y = 11; iii) 3x – y = 5 x – y = 3, 4x – y =10 3x – 2y = 4 iv) 3x + 2y = 12, v) 3x + 2y = 6. vi) 3x + 4y = – 1 2x – 3y = – 5 4x + y = 8. x + 2y = 1 vii) 3x – 4y = 0, viii) 4x + 3y = 8. ix) 5x – 3 = 8. 2x – 3y = –1 3x + 2y = 6. 10x + 6y = 4. 3. Solve with the help of graph : i) 3x + 4 = 5x; ii) 3x + 2 = x – 2: iii) 7x – 5 = 4x + 1; iv) 5x – 3 = x + 1; v) 8x – 3 = 5x – 3; vi) 3x – 7 = 3 – 2x;

vii) 1 – 3x = 7 – 3x; viii) 12 x – 1 =

13 x + 1

Multiple Choice Questions [Make (√) on the correct answer]

1. In which quadrant the point P(-3, 21 ) is situated?

(a) 1st (b) 2nd (c) 3rd (d) 4th 2. In how many unit of distance the poins A (2,3) is situated from the

x-axis? (a) 1 (b) 2 (c) 3 (d) 4 3. Which line joining the following pair of points passes through the

origin? (a) (2,0), (0,2) (b) (-2, 0), (0, 2) (c) (1, 0), (-1, 0) (d) (1,0), (0,1)

198

4.

A(0,4)

C(0,6)

O

D(2,0) B(4,0) (0,0)

Which one of the following indicates the point of intersection of the lines AB and CD?

(a) (3, 1) (b) (2, 3) (c) (3, 2) (d) (1, 3)

5. (i) The point P(-2, 61

− ) is situated in the 3rd quadrant.

(ii) There is a one-one correspondence between each point on the real line and each real number.

(iii) Using two straight lines it is not possible to determine the position of all points.

Which one of the following is correct in the basis of the above

informations? (a) i and ii (b) i and iii (c) ii and iii (d) i, ii and iii

199

O (0,0) B (5,0)

D(5,5)

A(0,5)

The equation of the straight line AB is x+y = 5 Answer the questions 6-8 on the basis of the above informations. 6. Which one of the following points is situated on the line x+y=5? (a) (1,3) (b) (0,0) (c) (-1, -3) (d) (6, -1) 7. Which one of the following is the co-ordinate of intersecting point

of lines AB and OD?

(a) ⎟⎠⎞

⎜⎝⎛

25,0 (b) ⎟

⎠⎞

⎜⎝⎛

25,

25

(c) ⎟⎠⎞

⎜⎝⎛ 0,

25 (d) (5, 5)

8. What type of triangle is shown in the figure above? (a) Equilateral (b) isosceles (c) Right-angled (d) acute-angled

CREATIVE QUESTIONS 1. Observe the following points: A(2, -1), B(1, -1), C(-2, -5), D(5,7) (a) In which quadrant the two points A and D are situated? (b) Show that the points P(6,3) and Q (-4, -5) are situated on the

line AC and BD respectively using graph. (c) Find the co-ordinates of the intersecting points of the two lines

with the axes.

N.M.G. -27

200

2. Observe the following equation: x+3y = 3 (a) Find the co-ordinates of three points from the equation. (b) Show that the obtained points are situated on the same line. (c) Find the point of intersection of the obtained line and lines

joining the points (2,4) and (-4, -5); (-1,0) and (-5, -4). 3. Observe the two equations: 2x + y = 3 ------ (i) 3x + 2y = 5 ------ (ii) (a) Show that, the root of the equations (i) and (ii) is (1,1). (b) Find the co-ordinates of three points from each graph of each

equation. (c) Draw the graphs of the two equations and show that the area of

the triangle formed by the line of equation (i) with the axes is 9 square unit.

Junior Secondary Mathematics GEOMETRY

Chapter I

Parallelism of lines, Congruency

Of triangles

1.1 Parallelism of Lines

If any straight line does not intersect the other when each of two or more straight lines lying in the same plane are extended in both the directions, then they are said to be parallel lines to one another.

AB and CD are two lines parallel to each other. The parallelism is denoted by the symbol ║. Here: AB ║ CD, If any two line segments are taken from two parallel lines, then those line segments are parallel to each other. The perpendicular distances from any point of line of two parallel lines to the other line are always equal to another. The perpendicular distance or distance between AB and CD is PQ. Only one straight line can be drawn parallel to a straight line through if point–not lying on this line. Uclid's Postulates If a point is not situated on the same line, then one and only one parallel line draw through this point. Playfaire's Postulates. If two straight lines intersect then both the lines can not be parallel of a third straight line.

1.2 Congruency of Triangles

Each triangle has six parts – three sides and three angles. Of two triangles, if one is placed on the other in a proper way and if the sides and the angles of two triangles coincide completely, then those two triangles are congruent. Two congruent triangles will be of the same size.

203

Two triangles will be congruent if the following parts of those two triangles are respectively equal. (a) Two sides and their included angle (b) Three sides (c) Two angles and one side (d) One angle is a right angle, hypotenuse and one side. Remarks : (a) Two triangles may not be congruent if three angles of one are respectively equal to that of the other. Three angles of innumerable triangles of different sizes may be equal. These triangles are called similar triangles. (b) If two triangles are congruent, then their areas will also be equal. But two triangles may not be congruent though their areas are equal. A D

E F B C

1.3 Symbols The following symbols are used : Symbol Meaning Symbol Meaning ∴ Hence, therefore > greater than Q Since < less than = Equal to ≥ Greater than or equal to ≤ less than or equal to ∠ angle ≠ not equal to Δ triangle ⊥ perpendicular to ≅ Congruent ║ parallel to 1.4 For the same of continuity, the theorems studied in the previous class and discussed again and proofs of some theorems are also included.

204

Theorem 1

If another straight line meets a line segment at its end point then the sum of two adjacent angles thus formed is equal to two right angles.

The line AB meets the line segment CO at its end point O. Here ∠AOC and ∠BOC are two adjacent angles ∴ ∠AOC + ∠BOC = Two right angles.

Theorem 2

If the sum of two abjacent angles is equal to two right angles, then their two exterior sides lie in the same straight line.

∠AOC and ∠BOC are two adjacent angles and ∠AOC + ∠BOC = two right angles. OA and OB are their exterior sides. Hence OA and OB lie in the same straight line.

Theorem 3

If two straight lines intersect each other, then the vertically opposite angles formed at the point of intersection are equal to each other.

The lines AB and CD intersect at O. Hence, ∠AOD = vericaliy opposite ∠BOC and ∠AOC = vertically opposite ∠BOD.

205

Theorem 4 If a straight line intersects two other straight Lines, (a) two alternate angles will be equal to each other. (b) two corresponding angles will be equal to each other. (c) the sum of the two interior angles in the same side of the bisector will be equal to right angles.

Proposition : Let the straight line EF intersects the parallel straight lines AB and CD at G and H respectively, (a) It is to be proved that ∠AGH = alternate ∠GHD.

Proof : If ∠AGH is not equal to ∠GHD, then let us suppose that, ∠KGH = ∠GHD. They are alternate angles. Hence KG and CD are parallel. But AB and CD or, AG and CD have been taken to be parallel to one another. Each of AG and KG is parallel to CD though they intersect. But it is impossible according to Playfaire's Postulate, Therefore, ∠AGH and ∠GHD are not unequal, i.e. ∠AGH = ∠GHD. (b) It is required to prove that, ∠EGB = ∠GHD

Proof : ∠EGB = vertically oopposite ∠AGH and ∠AGH = alternate ∠GHD ∴ ∠EGB =∠GHD. (c) It is required to prove that, ∠BHG + ∠GHD = two right angles.

Proof: ∠EGB+ ∠BGH = two right angles, but ∠EGB = corresponding ∠GHD. ∴ ∠BGH + ∠GHD = two right angles (proved). Corollary. If a straight line is perpendicular to any one of two parallel lines, then it is perpendicular to the other.

206

Theorem 5

If a straight line intersects another two straight lines and if (a) the alternate angles are equal to each other or, (b) the corresponding angles are equal to each other. . ■ * . ■ ■ ' . . ■ •' or, (c) the sum of the two enterior angles in the same side of the bisector be equal to those two right angles then two lines are parallel. (a) Proposition : Let the straight line EF intersects AB and CD at G and H respectively so that ∠AGH = alternate ∠GHD and ∠BGH = alternate ∠GHC. It is required to prove that, AB and CD are parallel. Proof : If AB and CD are not parallel then draw PQ straight line through the point G and to the line CD (By uclid's postulate) Now PQ ║ CD ∴ ∠PGH = alternate ∠DHG But given, ∠AGH = ∠DHG ∴∠PGH =∠AGH But it is impossible, because between two angels one is the part of the other. So the straight line AB and CD are parallel. b) Proposition : Let, the straight line EF intersects AB and CD at the points. G and H respectively so that it becomes exterior ∠EGB = interior opposite ∠GHD. It is required to prove that, AB and CD are parallel.

Proof : ∠AGH = ∠EGB (being vertically opposite angles) and ∠EGB = ∠GHD (given) ∴ ∠AGH = ∠GHD. But they are alternate angles. ∴ AB and CD are parallel. (C) Proposition : Let the straight line EF intersects the straight lines AB and CD at the points G and H respectively, so that in the same side of EF, interior ∠BGH

A B

E Q

D

F

C

P

H

G

207

+ interior ∠GHD = two right angles. It is required to prove that, AB and CD are parallel.

Proof: Now, ∠AGH + ∠BGH = two right angles (being adjacent angles). ∠BGH + ∠GHD = two right angles (given) ∴ ∠AGH + ∠BGH = ∠BGH + ∠GHD. Now subtracting ∠BGH from both the sides we get, ∠AGH = ∠GHD. But they are alternate angles. ∴ AB and CD are parallel (proved). Corollary . If each of two or more straight lines is perpendicular to given straight line then they are parallel to one another.

Theorem 6

The straight lines which are parallel to the same straight line are parallel to. one another.

The straight lines AB and EF are parallel and CD and EF are parallel. Hence, the straight lines AB and CD are parallel.

N.M.G. -28

C D

208

Theorem 7

If in two triangles each of two sides of one is equal to the corresponding side of the other respectively and the included angle of those sides of one is equal to that of the other, then the triangles are congruent.

Proposition: Let, in ΔABC and ΔDEF, AB = DE , AC = DF and included ∠A = included ∠D. It is required to prove that, ΔABC ≅ ΔDEF. Proof: Let us place ΔABC upon ΔDEF so that A falls upon D. The side AB lies along the side DE and the point C falls on that side at which side of DE, F lies. Now, since AB = DE, the point B must fall upon the point E. Again, since the side AB lies completely on the side DE and ∠A = ∠D, the side AC will be along the side DF. Now, since AC = DF, the point C must fall upon the point F. Since the points B and C fall upon E and F respectively, the side BC must coincide completely with EF. Therefore, ΔABC is superimposed on ΔDEF. ∴ ΔABC ≅ ΔDEF (proved).

Theorem 8

If two sides of a triangle are equal to each other, then their opposite angles are also equal to each other. The sides AB and AC of the triangles ABC are equal to each other. Hence, ∠ACB = ∠ABC.

Theorem 9

If two angles of any triangle are equal to each other, then their opposite sides are equal to each other.

In the ΔABC, ∠ABC = ∠ACB. Hence, AC = AB

209

Theorem 10

If three sides of a triangle are respectively equal to the corresponding three sides of another triangle, then the triangles are congruent.

Proposition : Let, in ΔABC and ΔDEF, AB = DE, AC = DF and BC = EF. It is required to prove that, ΔABC ≅ ΔDEF. Proof: Let, BC and EF are the greatest sides of ΔABC and ΔDEF respectively. Let us place ΔABC upon ΔDEF in such a way that B falls upon E, the side BC lies along the side EF and the point A falls on the opposite of that side at which side of EF, D lies. Now, since BC = EF, the point C falls upon the point F. Let, the new position of the point A be denoted by G. As a result ΔGEF is the new position of ΔABC. Hence, EG = BA, FG = CA and ∠EGF = ∠BAC. Let us join D,G. Now, in ΔEGD, EG = ED [Q EG = BA = ED] ∴∠EDG = ∠EGD. Again, in ΔFGD, FG = FD, [Q FG = CA = FD] ∴ ∠FDG = ∠FGD. Hence, ∠EDG + ∠FDG = ∠EGD + ∠FGD or, ∠EDF = ∠EGF i.e. ∠BAC = ∠EDF. Therefore, in ΔABC and ΔDEF, AB = DE, AC = DF and included ∠BAC = included ∠EDF ∴ ΔABC ≅ ΔDEF. (proved)

Remark : If three angles of a triangle are equal to three corresponding angles of another triangle, then the triangles may not be congruent. AB F DE, AC F DF and BC F EF. Here, three angles of ΔABC are equal to three corresponding angles of ΔDEF.

D

A

E FB C

210

Theorem 11 If one side is greater than another side of a triangle the angle opposite to the greater side is greater than the angle opposite to the smaller side.

Proposition : Let, in ΔABC, AC >AB It is required to prove that, ∠ABC > ∠ACB. Construction : Let us cut off AD from AC such that, AD = AB and join B and D. Proof : In ΔABD, AD = AB ∴ ∠ABD = ∠ADB [Theorem 8] But in ΔBDC, exterior ∠ADB > ∠BCD ∴ ∠ABD > ∠BCD or, ∠ABD > ∠ACB. But ∠ABC > ∠ABD.[Q ∠ABD is a part of ∠ABC.] Hence, ∠ABC > ∠ACB (proved).

Theorem 12

If one angle is greater than another angle of a triangle, the side opposite to the greater angle is greater than side opposite to the smaller angle. In ΔABC, ∠ABC > ∠ACB. Hence, AC > AB.

Theorem 13

The sum of any two sides of a triangle is greater than its third side. In ΔABC, the side BC is greater than both AB and AC. Therefore, BC is the greatest side. Then, AB + AC > BC.

211

Theorem 14 Of all the line segments that are drawn to a straight line from any external point of the straight line, then perpendicular is the shortest. OC is the perpendicular to AB from any external point O of AB and OP is any other line segment from O to any point on AB. Then, OC < OP.

Theorem 15 The sum of three angles of a triangle is equal to two right angles.

∠ABC, ∠ACB and ∠BAC are the three angles of ΔABC. Hence, ∠ABC + ∠ACB + ∠BAC = two right angles.

Theorem 16

If two angles and a sides of a triangle are equal to the corresponding angles and side of another triangle respectively, then two triangles are congruent. A D B C E F

Proposition : Let, in ΔABC and ΔDEF, ∠A = ∠D, ∠B = ∠E and the side BC = corresponding side EF. It is required to prove that, Δ ABC ≅ ΔDEF. Proof : In ΔABC, ∠A + ∠B + ∠C = 180° [ Theorem 15]

212

and in ΔDEF, ∠D + ∠E + ∠F = 180° [Theorem 15] ∴ ∠A + ∠B + ∠C = ∠D + ∠E + ∠F. ... ∠A = ∠D and ∠B = ∠E, ∴ ∠C = ∠F. Now, place ΔABC on ΔDEF in such a way that the point B lies on E, the side BC lies along EF and the point A falls on that side at which side of EF, D lies. Since, BC = EF, therefore, the point C falls on the point F. Now since ∠B = ∠E, the side BA will lie along the side ED and since ∠C = ∠F. the side CA will lie along FD. Hence, the common point A of the sides BA and CA will fall on the common point D of the sides ED and FD. Hence, ΔABC is superimposed on ΔDEF. ∴ ΔABC ≅ ΔDEF (proved).

Theorem 17

If two hypotenuses of two right angled triangles are equal to each other and also one side of one of those triangles is equal to the corresponding side of another, then two triangles are congruent

Proposition : Let, in the right angle triangles ABC and DEF, the hypotenuse AC = hypotenuse DF and AB = DE. It is required to prove that, ΔABC ≅ Δ DEF. Proof: Let us place ΔABC upon ΔDEF in such a way that the point B lies on the point E, the side BA lies along the side ED and the point C falls on the opposite of that side at which side of DE, F lies. Let, the new position of the point C be denoted by G.

213

Since, AB = DE, therefore, the point A lies on the point D. Hence, ΔDEG is the new position of ΔABC. Therefore, DG = AC = DF, ∠DGE = ∠ACB and ∠DEG = ∠ABC = one right angle. Again, since ∠DEF + ∠DEG = one right angle + one right angle = two right angle, therefore, EF and EG lie on the same straight line [Theorem 2] Now, since in ΔDGF, DG = DF, therefore, ∠DFE = ∠DGE. [Theorem 8] Hence, ∠DFE = ∠ACB. Now, in ΔABC and ΔDEF, ∠ABC = ∠DEF, [each being one right angle] ∠ACB = ∠DFE and side AB = corresponding side DE. ∴ ΔABC ≅ ΔDEF (proved).

EXERCISE 1

1.

Given that, in ΔABC, ∠ABC = ∠ACB and DE F BC; prove that, ∠ADE = ∠AED.

2.

Given that, DC F AB and DC = AB; prove that, AD = BC.

3.

Given that, AB and CD intersect each other at the point O; AC ⊥ CD and BD ⊥ CD, prove that, ∠CAO = ∠OBD.

4.

Given that, AD is a median of ΔABC; prove that, AB + AC > 2AD.

214

5.

Given that, in ΔABC, AB = AC and D is any point on the extension of BC; prove that, ∠ABD > ∠ADB.

6.

In ΔABC, ∠BAC = ∠ABC : D is any point on BC; prove that, ∠CDA > ∠CAD.

7.

In ΔABC, AB = AC and ∠BAD = ∠CAE; Prove that, ΔADB ≅ ΔACE.

8.

In ΔABC, AB = AC and BE and CD are the bisectors of ∠ABC and ∠ACB respectively, prove that, ΔBDC ≅ ΔBCE.

9. If any straight line intersects two parallel straight lines then prove that, two bisectors of two alternate angles are parallel to each other.

10. Prove that, sum of any two sides of a triangle is greater than twice its median drawn on the third side.

11. D is a point inside ΔABC. Prove that, AD + BD + CD > 12 (AB + AC + BC).

12. The median AD of ΔABC is half of BC. Prove that, ∠A is a right angle. 13. If one of two acute angles of a right angled triangle is twice the other, prove

that, the smallest side is half of hypotenuse. 14. Prove that, the distances of any point on the bisector of any angle from the

sides are equidistant. 15. Prove that, two perpendiculars drawn from two terminal points of the base of

an isosceles triangle to their opposite sides are equal to each other.

215

16. Prove that, if three perpendiculars drawn from three angular points of a triangle to their opposite sides are equal to one another the triangle is an equilateral triangle.

17. Prove that, if one angle of a triangle is equal to the sum of its other two angles then the triangle is a right angled triangle.

18. Prove that, the sum of three medians of a triangle is smaller than its perimeter.

19. Standing on the same bank of the river, find its breadth.

Multiple Choice Questions [Mark (√) on the correct answer] 1. (i) If two straight lines intersect one another then they cannot be

parallel with a third straight line. (ii) More than one straight line can be drawn through two

particular points. (iii) Two similar triangles always will be congruent. Which one of the following is correct in view of the above

informations? (a) i (b) ii (c) ii and iii (d) i, ii and iii

2. Which one of the following indicates the sign of congruent? (a) = (b) ≥ (c) ≅ (d) || 3.

If DE || BC in ΔABC and ∠ADE =

60°, then which one of the following is the value of ∠B?

(a) 30° (b) 45° (c) 60° (d) 90°

A

E D

B C

N.M.G. -29

216

Answer the questions (4 − 6) on the basis of the following informations:

In the figure, AB || CD || EF, PQ is the intersector of them and ∠MNF = 60°, ∠MLR = 45°.

4. Which one of the following is the value of ∠LMR? (a) 30° (b) 45° (c) 60° (d) 120° 5. Which one of the following is the value of ∠ALM? (a) 30° (b) 45° (c) 60° (d) 90° 6. Which one of the following is the value of ∠LRD? (a) 45° (b) 90° (c) 105° (d) 120° 7. By which three line segments of the following it is impossible to

draw a triangle? (a) 5 cm, 8 cm, 3 cm, (b) 4 cm, 5 cm, 3 cm (c) 2.5 cm, 3.5 cm, 3 cm (d) 6.5 cm, 4.5 cm, 5.5 cm. 8. If ∠ACB > ∠ABC in ΔABC then, which one of the following is

correct? (a) AC > AB (b) AC ≥ AB (c) AB > AC (d) AB = AC 9. OP is the perpendicular to MN from the external point O and OC,

OD are two line segments. Which one of the following is correct? (a) OP < OC (b) OP > OC (c) OD < OP (d) OD > OP

A

C

E

B

D

F

P

R

L

M

N

Q

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10. Which one of the following is the condition for congruence of two triangles?

(a) Two sides and one angle equal, (b) Three angles are equal to three angles of another traingle. (c) Two angles and one similar side are equal. (d) The area of two triangles are equal. CREATIVE QUESTIONS 1. AB || CD, PQ is the intersector. The line PQ intersects AB and CD

at the points E and F respectively. (a) Draw the figure on the basis of the above informations and

write two alternate angles. (b) Using the above figure show that, ∠BEP = ∠PFD (c) If two bisectors of ∠BEF and ∠DFE intersect at the point G

then prove that, ∠EGF = 1 right angle. 2. ABC is a triangle. AD, BE and CF are medians. (a) Draw a figure with brief descriptions on the basis of the above

informations. (b) Show that, AB + AC > 2AD (c) If AB = AC, then prove that, ΔBCF ≅ ΔBCE 3. The median AD of ΔABC is the half of BC. (a) Draw the figure in view of the above informations. (b) Prove that, ∠A = 1 right angle. (c) If AD ⊥ BC, then prove that, AB = AC.

Chapter II

Theorems Related to Quadrilaterals

2.1 Quadrilateral The closed figure bounded by four line segments is a quadrilateral. The region is quadrilateral region. The quadrilateral has four sides. The four line segments by which the region is bounded are the sides of the quadrilateral.

Any three points of the points A, B, C, D are not collinear. The union of four line segments AB, BC, CD and DA is the quadrilateral ABCD. AB, BC, CD and DA are the sides of the quadrilateral. The points A, B, C and D are the angular points of vertices. ∠ABC, ∠BCD, ∠CDA and ∠DAB are four angles of the quadrilateral. The quadrilateral ABCD is sometimes denoted by the symbol " ABCD". The vertices A and B are the opposite vertices of C and D respectively. AB and CD are called opposite sides to each other and AD and BC are called opposite sides to each other. Two sides which meet at the same vertex are adjacent sides. AB and BC are two adjacent sides. The line segments AC and BD are the diagonals of ABCD.

2. 2. Different kinds of Quadrilaterals

Parallelogram : If the opposite sides of a quadrilateral are parallel to each other, then it is called a parallelogram. The region bounded by the parallelogram is called parallelogram region.

Parallelogram Rectangle Rectangle : The quadrilateral in which each of all the angles be a right angle

219

is called a rectangular–quadrilateral or rectangle. The region bounded by a rectangle is called a rectangular region. Remark : A rectangle is also a parallelogram. If one of the angles of a parallelogram is a right angle, the parallelogram will be a rectangle. Square : The quadrilateral whose sides are equal to each other and. each of its angles is a right angle is called a square. The region bounded by a square is called a square region.

Square Rhombus

Remark : If the sides of the rectangle are equal to each other then it becomes square. Rhombus : The quadrilateral whose sides are equal to each other but none of its angles is a right angle is called a rhombus. Remark : If the sides of a parallelogram are equal to each other then it becomes a rhombus. Trapezium : The quadrilateral whose two opposite sides are parallel to each other but its other two sides are not parallel to each other is called trapezium.

Trapezium Quadrilateral Perimeter : The total length of the line segments of a certain bounded region is called its perimeter. The perimeter of quadrilateral ABCD is equal to the length of (AB + BC + CD + DA).

220

2.3. Two Theorems Related To Quadrilaterals Are Given Below :

Theorem 18

If two sides of a quadrilateral are respectively equal and parallel to each other, the other two sides are also respectively equal and parallel to each other.

Proposition : Let, the sides AB and DC of the quadrilateral ABCD are respectively equal and parallel to each other. It is required to prove that, the sides BC and AD are respectively equal and parallel to each other. Construction : Let us join A and C. Proof : Since AB and DC are parallel to each other and AC is their interceptor, hence, ∠BAC = ∠DAC. [ being alternate angles]

Now, in ΔABC and ΔADC, AB = DC, AC is their common side and included ∠BAC = included ∠DCA. Therefore, ΔABC ≅ ΔADC [Theorem 7]

So, BC = AD and ∠ACB = ∠CAD. . Now since two alternate angles produced by the interceptor AC of the sides BC and AD are equal to each other, the sides BC and AD are parallel to each other. Therefore, two sides BC and AD are respectively equal and parallel to each other. [proved]

Corollary : If two sides of a quadrilateral are respectively equal and parallel to each other, then the quadrilateral is a parallelogram.

Theorem 19

The opposite sides and opposite angles of a parallelogram are respectively equal to each other and each diagonal divides the parallelogram into two congruent triangles.

221

Proposition : Let, ABCD be a parallelogram and AC and BD are its two diagonals. It is required to prove that, (a) The side AB = the side CD, the side AD = the side BC

(b) ∠BAD= ∠BCD, ∠ABC = ∠ADC; (c) ΔABC ≅ ΔADC, ΔABD ≅ ΔBCD. Proof : Since AB ║ CD and AC is their interceptor, therefore, ∠BAC = ∠ACD, [being alternate angles]

Again, since BC ║ AD and AC is their interceptor, therefore, ∠ACB = ∠DAC ; [being alternate angles]

Now, in ΔABC and ΔADC, ∠BAC = ∠ACD, ∠ACB = ∠DAC and AC is the common side. ∴ ΔABC ≅ ΔADC [Theorem. 16]

Therefore, AB = CD, BC = AD and ∠ABC = ∠ADC. Similarly, it can be proved that, ΔABD ≅ ΔBDC. Therefore, ∠BAD = ∠BCD [proved]

Corollary 1. If one angle of a parallelogram is a right angle, then each of its angels will be a right angle.

Let, in the parallelogram ABCD, ∠A is a right angle. Since AB and DC are parallel to each other and AD is their interceptor, therefore, ∠A + ∠D = 2 right angle [Theorem 4]

Hence, ∠D = one right angle.

222

Since ABCD is a parallelogram, therefore, ∠C = ∠A = one right angle and ∠B = ∠D = one right angle. Corollary 2. If one angle of a parallelogram is a right angle, then it is a rectangle. Corollary 3. If two adjacent sides of a rectangle are equal to each other, then it is a square. Corollary 4. Two diagonals of a parallelogram bisect each other.

Let, two diagonals AC and BD of the parallelogram ABCD intersect each other at O. It is required to prove that, AO.= CO, BO = DO. . Proof : Since AB and DC are parallel to each other, AC and BD are their interceptor, therefore, ∠BAC = alternate ∠ACD and ∠BDC = alternate ∠ABD. Now in ΔAOB and ΔCOD, ∠OAB = ∠OCD, ∠OBA = ∠ODC and AB = DC. Therefore, ΔAOB ≅ ΔCOD [ Thorem 16]

∴ AO = CO and BO = DO. . Corollary 5. The line segment found by joining points of two sides of a triangle is parallel to the third side and it is half length of the third side.

The points D and E are respectively the points of the sides AB and AC of ΔABC. Let us join D and E and produced so that DE = EF. C and F are joined. ΔADE ≅ ΔCEF, CF || BD, DF || BC and DF = BC.

∴ DE = 12 BC and DE || BC.

223

EXERCISE 2

1.

Given that, the median CO of ΔABC is produced upto D so that CO = DO. Prove that, ACBD is a parallelogram.

2.

Given that, in the quadrilateral ABCD, AB = CD and ∠ABD = ∠BDC. Prove that, ABCD is a parallelogram.

3.

Given that, in the parallelogram ABCD, BM is the bisector of ∠ABC and DN is the bisector of ∠ADC. Prove that, BMDN is a parallelogram.

4. Given that, in the parallelogram ABCD, AP = CR and DS = QB; prove that, PQRS is a parallelogram.

5.

Given that, in the parallelogram ABCD, AM and CN are both perpendicular to DB; prove that, ANCM is a parallelogram.

6. Prove that, sum of four angles of a quadrilateral is four right angles. 7. Prove that, if opposite angles of a quadrilateral are equal to each other, then it

is a parallelogram. 8. Prove that, if two diagonals of a parallelogram are equal to each other, then it

is a rectangle. 9. Prove that, if two diagonals of a quadrilateral are equal to each other and

bisect each other at right angle, then it is a square. 10. Prove that, two diagonals of rhombus bisect each other at right angle. 11. Prove that, if the quadrilateral formed by joining the mid-points of adjacent

sides of a rectangle, then it is a rhombus. 12. Prove that, the bisector of any two opposite angles of a parallelogram are

parallel to each other. 13. Prove that, the bisectors of any two adjacent angles of a parallelogram are

perpendicular to each other. N.M.G. -30

224

Multiple Choice Questions [Mark (√) on the correct answer] 1. In the figure, AD = BC and AD || BC. Again OA and OD are bisectors of ∠A and ∠D respectively, then ∠AOD = How much? (a) 90° (b) 60° (c) 45° (d) 30° 2. A and F are the mid points of DE

and CE respectively. If DC = 8 cm, then which one of

the following is the value of AF?

(a) 4 cm (b) 8 cm (c) 12 cm (d) 16 cm. 3. (i) The sum of the four angles of quadrilateral is 4 right angles. (ii) If the two adjacent sides of a rectangle are equal then the

rectangle is a square. (iii) Every rhombus is a parallelogram. Which one of the following is correct according to above

information? (a) i and ii (b) i and iii (c) ii and iii (d) i, ii and iii Answer questions (4 − 6) in view of the following figure: AB || DC, AD || BC and AB

≠ AD, ∠ABF = ∠FBC in quadrilateral ABCD. ∠ADF = ∠FDC.

D A

B C

O

D C

B

E

F A

C

B

F

O E

D

A

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4. How many parallelogram are there in the figure? (a) 2 (b) 4 (c) 6 (d) 8 5. If AC = BD, then what would be the parallelogram? (a) Square (b) Rectangle (c) Rhombus (d) Trapezium. 6. (i) ∠ADC and ∠BCD are two adjacent angles of the

Parallelogram ABCD. (ii) ∠EDF and ∠EBF are opposite angle in the parallelogram

BEDF. (iii) The two bisectors of ∠ADC and ∠ABC are perpendicular to

one-another. Which one of the following is correct in view of the above

information? (a) i and ii (b) i and iii (c) ii and iii (d) i, ii and iii 7. If PA = CQ and PA || CQ in

quadrilateral PAQC. If the bisectors of ∠A and ∠C are AB and CD, then what is the name of the region ABCD?

(a) parallelogram (b) rhombus (c) rectangle (d) square

CREATIVE QUESTIONS

1.

In above figure, ABCD is a quadrilateral.

C B

P D A

Q

A

B C

D

226

(a) For what conditions the quadrilateral ABCD become a parallelogram?

(b) If quadrilateral ABCD is a parallelogram, then show that, ΔABC ≅ ΔACD and hence show that, ∠ABC = ∠ADC.

(c) If ∠B= 90° in the parallelogram ABCD, then prove that, ABCD is a rectangle.

2. In the figure, ABC is a triangle. (a) Prove that, ∠BDF + ∠DFE + ∠FEB+∠EBD = 4 right angles. (b) If D and F are middle points of AB and AC, then prove that,

DF || BC and DF = 21 BC.

(c) ΔABC is equilateral and the middle points of sides AB, AC and BC are D, F and E, then prove that, BEFD is a rhombus.

3.

A

D C

B

O

In figure, AB = CD and AB || CD (a) Write the name of two triangles with base AB. (b) Prove that, AD and BC are equal and parallel. (c) Show that, OA = OC and OB = OD.

A

F

CB E

D

Chapter III

Area

3.1 Area

The measurment of a plane region is called area. Measurement of Area

If each side of a square region is one unit, then area of the square region is one square unit. If the length of one side of a square region is one centimetre, then the area of the square region is one square centimetre. 3.1.1 Area of Rectangular Region

Area of rectangular region = the length of the rectangular region × the breadth of the rectangular region. i.e. Area of the rectangular region = Product of the lengths of two adjacent sides of the rectangular region. ∴ The area of the rectangular region ABCD = AB × BC

3.1.2 Some Theorems Related to Areas are Included Below.

Theorem 20

If a triangular, region and a parallelogram-region stand on the same base and lie between two parallel lines, then area of the triangular region will be half of that of the parallelogram region. Proposition : Let, the triangular region, ABC and the parallelogram region EBCD stand on the same base BC and between two parallel lines BC and ED. It is required to prove that,

triangular region ABC = 12 (Parallelogram region EBCD)

228

fig (a) fig.(b)

Construction : Let us draw a line AF parallel either to EB or DC through A; the line AF intersects either the base BC (Fig. a) or BC produced (Fig. b) at F. Proof: AB is a diagonal of the parallelogram EBFA ;

∴ Δ region ABF = 12 (parallelogram region EBFA).

Again, AC is a diagonal of parallelogram region AFCD;

∴ Δ region AFC = 12 ( region AFCD).

In fig. (a), Δ region ABF + Δ region AFC

= 12 ( region EBFA) +

12 ( region AFCD)

i.e. Δ region ABC = 12 ( region EBCD)

Again, in fig. (b), (Δ region ABF) – (Δ region AFC)

= 12 ( region EBFA – region AFCD).

i.e. Δ region ABC =12 ( region EBCD)

Therefore, triangular region ABC = 12 (parallelogram region EBCD) (proved)

Corollary: If a triangular region and a rectangular region stand on the same base and lie between two parallel lines then the area of the triangular region is half of the rectangular region.

229

Area of the triangular region : A triangular region ABC stands on the base BC and AF is its height. Let us draw a rectangular region BCDE of height equal to AF on the base BC.

Then, Δ region ABC = 12 (rectangular region BCDE)

= 12 × BC × CD =

12 BC × AF

Remark : Hence, as a formula it can be stated :

Area of the triangular region = 12 × base × height.

Theorem 21

Areas of all the triangular regions standing on the same base and between two parallel lines are equal to one another.

Proposition : Let, triangular regions ABC and DBC stand on the same base BC and lie between two parallel lines BC and AD. It is required to prove that, Δ region ABC = Δ region DBC. Construction: Let us draw perpendiculars BE and CF at the points B and C of the line BC. They intersect the line AD or AD produced at the points E and F respectively. Proof : According to the construction, EBCF is a rectangular region. Since Δ region ABC and rectangular region EBCF stand on the same base BC and lie between two parallel lines BC and ED.

Therefore, Δ region ABC = 12 (rectangular region EBCF) [ Theorem 20]

Similarly, Δ region DBC = 12 (rectangular region EBCF)

∴ Δ region ABC = Δ region DBC (proved).

230

Corollary : Areas of all triangular regions standing on bases equal to one another and between the same two parallel lines are equal to one another.

Theorem 22

All triangular regions of areas equal to one another standing on the same base and on the same side of it will lie between the same two parallel lines.

Proposition : Let, the triangular regions ABC and DBC stand on the same base BC and lie on the same side of it and the areas of the triangular regions are equal to each other. AE and DF are the altitudes of ΔABC and ΔDBC respectively. It is required to prove that, AD F BC.

Proof : Δ region ABC = 12 (BC × AE) and Δ region DBC = 12 ( BC × DF)

∴ 12 (BC × AE) = 12 (BC × DF).

Therefore, AE = DF. Moreover, AE F DF, [Because both are perpendicular on BC] The opposite sides of the quadrilateral AEFD, AE and DF are equal and parallel to one another.

∴ AD F EF, (Theorem 18)

i.e. AD F BC (proved).

Corollary : All triangular regions of areas to one another standing on equal bases from the same straight line and lying on the same side will lie between the same two parallel lines.

231

3.1.3 Pythagoras' Theorem

Theorem 23

In a right-angled triangle the square on the hypotenuse is equal to the sum of the square regions on the other two sides.

Proposition : Let, ABC be a right–angled triangle. Its ∠C = one right angle. Hence, AB is the hypotenuse. It is required to prove that, the square region on AB = the square region on AC + the square region on BC i.e. AB2 = AC2 + BC2.

Construction : Let us draw square regions ABED, ACGF and BCHK on the sides AB, AC and BC respectively. CL is drawn through C parallel either to AD or BE. Let us suppose that it intersects AB and DE at M and L respectively. C, D and B, F are joined. Proof: ∠BAD = ∠CAF [each being one right angle]

Adding ∠BAC to both sides, we get ∠BAD + ∠BAC = ∠CAF + ∠BAC. ∴ ∠CAD = ∠BAF. Now between ΔCAD and ΔBAF, CA = AF, AD = AB and included ∠CAD = included ∠BAF, ∴ ΔCAD ≅ ΔBAF. [Theorem 5]

Since each of ∠ACB and ∠ACG is a right angle. ∴ BCG is in the same straight line.

N.M.G. -31

232

Now, since triangular region CAD and rectangular region ADLM stand on the same base AD and lie between the parallel lines AD and CL, hence rectangular region ADLM = 2 (Δ region CAD) ................ (1)

Again, since triangular region BAF and the square region ACGF stand on the same base AF and lie between the parallel lines AF and BG, hence square region ACGF = 2 (Δ region BAF) ..............(2)

∴ Rectangular region ADLM = Square region ACGF ............... (3)

Similarly, joining C, E and A, K, it can be proved that, Rectangular region BELM = Square region BCHK ....................(4)

Adding (3) and (4) it is obtained, Square region ABED = Square region ACGF + Square region BCHK. i.e. the square region on AB = the square region on AC + the square region on BC. ∴ AB2 = AC2 + BC2 [proved]

N.B. This theorem is known as the Theorem of Pythagoras. Alternative proof of Pythagoras Theorem:

Proposition : Let, in the triangle ABC, ∠B = 90°, hypotenuse AC = b, AB = c and BC = a. It is required to prove that, AC2 = AB2 + BC2 i.e. b2 = c2 + a2. Construction : Let us produce BC upto D, so that CD = AB = c. Let us draw perpendicular DE at D on BC produced, so that DE = BC = a. Let us join C, E and A, E. Proof : Between ΔABC and ΔCDE, AB = CD = c, BC = DE = a and included ∠ABC = included ∠CDE (each being a right angle). Hence, ΔABC ≅ ΔCDE. ∴ AC = CE = b and ∠BAC = ∠ECD.

233

Again, since AB ⊥ BD and ED ⊥ BD, therefore, AB F ED. Therefore ABDE is a trapezium. Moreover, ∠ACB + ∠BAC = ∠ACB + ∠ECD = one right angle. ∴ ∠ACE = one right angle. Now, area of the trapezium region ABDE = area of (Δ region ABC + Δ region CDE + Δ region AEC)

[Q Area of trapezium region = 12 × sum of the parallel sides × distance between

the parallel sides]

or, 12 × BD ( AB + DE) = 12 ac + 12 ac + 12 b2

or, 12 (a + c) (a + c) = ac +

12 b2 or,

12 (a2 + 2ac + c2) = ac +

12 b2

or, 12 a2 + ac +

12 c2 = ac +

12 b2 or,

12 a2 +

12 c2 =

12 b2

or, a2 + c2 = b2. (proved) Theorem 24

If the square-region on a side of any triangle is equal to the sum of the square–regions on other two sides of it, then the angle included between the last two sides is a right angle.

Proposition : Let in ΔABC, AB2 = AC2 + BC2. It is required to prove that, ∠C = one right angle. Construction : Let us draw such a triangle so that ∠F = one right angle, EF = BC and DF = AC. Proof: DE2 = EF2 + DF2 [because in ΔDEF, ∠F = one right angle]

= BC2 + AC2 = AB2 [supposition] ∴ DE = AB.

234

Now between ΔABC and ΔDEF, BC = EF, AC = DF and AB = DE. ∴ ΔABC ≅ ΔDEF [Theorem-10] ∴ ∠C = ∠F = one right angle (proved)

EXERCISE 3

1. O is any point inside the parallelogram ABCD. It is required to prove

that, Δ region AOB + Δ region COD = 12 (parallelogram region ABCD)

2. Prove that, any median of a triangle divides the triangular region into two triangular regions of areas equal to each other.

3. In ΔABC, D and E are the mid-points of the sides AB and AC respectively.

Prove that, Δ region CDE = 14 (Δ region ABC)

4. In ΔABC any straight line parallel to BC intersects AB and AC at points D and E respectively. Prove that, Δ region DBC = Δ region EBC

and Δ region BDE = Δ region CDE.

5. D and E are the mid points of AB and AC respectively of ΔABC.

Prove that, Δ region ADE = 14 (Δ region ABC) .

6. Prove that, the diagonals of the parallelogram divide the parallelogram region into four triangular regions equal to one another.

7. Prove that, any square region is half of the square regions drawn on its diagonal.

8. In a triangle ABC, ∠A = one right angle. D is a point on AC. Prove that, BC2 + AD2 = BD2 + AC2.

9. In a triangle ABC, ∠A = one right angle. If D and E are the mid–points of AB and AC respectively, then prove that, DE2 = CE2 + BD2.

10. In ΔABC, AD is perpendicular to BC and AB > AC. Prove that, AB2 – AC2 = BD2 – CD2.

11. In ΔABC, AD is perpendicular to BC and P is any point on AD and AB > AC; prove that, PB2 – PC2 = AB2 – AC2.

235

Multiple Choice Questions [Mark (√) on the correct answer] 1. If AB = CD, AB || CD and AD =

BC, AD || BC in the quadrilateral ABCD, which one of the following is area of the triangle ABD?

(a) 21 (CD × BE) (b)

21 (DE × BE)

(c) 21 (AB × AD) (d)

21 (CD × BC)

2.

AB || CD, AC ≠ BD Which one of the following indicates

the quadrilateral ABCD? (a) Square (b) Parallelogram (c) Trapezium (d) Rectangle

A B

C

D

QF

E

P

In ABCDE polygon AE || BC, CF ⊥ AE and DQ ⊥ CF. ED = 10 mm, EF = 2mm, BC = 8mm, AB = 12 mm. Answer the questions (3 - 6) on the basis of the above informations:

B

E C D

A

A B

CD

236

3. How much is square mm. area of quadrilateral ABCF? (a) 64 (b) 96 (c) 100 (d) 144

4. Which one of the following indicates the area of ΔFPC? (a) 32 (b) 48 (c) 72 (d) 80 5. Which one of the following expresses the length of CD? (a) 22 (b) 4 (c) 24 (d) 8 6. Which one of the following indicates the difference of the area of

ΔFPC and ΔDQC? (a) 46 square unit (b) 48 square unit (c) 50 square unit (d) 52 square unit. 7. i. The area of a triangle and rectangle with equal bases and equal

height.

ii. If an angle is 90° of a triangle, then the triangle would be a right-angled.

iii. Area of rectangle = the product of the length of two adjacent sides of the rectangle. Which one of the following is correct on the basis of the above

informations? (a) i and ii (b) i and iii (c) ii and iii (d) i, ii and iii

237

CREATIVE QUESTIONS

A P D

C B

M N

45°

60°

In figure, ABCD is a parallelogram. The point M is mid-point of BP

and MN || BC. (a) Find the value of ∠BPC

(b) Show that Δ region PBC = 21 (Parallelogram region ABCD)

(c) Prove that, Δ region PBN = 2 (Δ region PMN). 2.

T

a

SbRaQ

b c

P

c

(a) What type of quadrilateral PQST? Explain logically in favour.

(b) Show that, ΔPRT is a right-angled triangle. (c) Prove that, PR2 = PQ2 + QR2

238

3.

A

D

a

B

C

a

a

a

F

G

H

E

In the above figure, ∠BAD = 90° AE = BF = CG = DH = b BE = CF = DG = AH = c (a) Show that, ΔABE ≅ ΔBCF (b) Show that, EFGH is a square. (c) Prove that, a2 = b2+c2.

Chapter IV

Problems

4.1 Construction of Quadrilaterals : A quadrilateral consists of four sides, four angles and two diagonals. In case of a triangle if three sides are given, then the definite triangle can be constructed. But if four sides of a quadrilateral are given, then the definite quadrilateral can not be constructed.

Of two quadrilaterals ABCD and ABEF, AB = AB, BC = BE, CD = EF and DA = FA. But if four sides and an angle of any quadrilateral are given then the definite quadrilateral can be constructed. For drawing a quadrilateral five data independent of one another are essential. The definite quadrilateral can be constructed if the following five data are known : (a) four sides and an angle; (b) four sides and a diagonal; (c) three sides and two diagonals; (d) three sides and their two included angles; (e) two sides and three angles, or (f) parts of intercepts of two diagonals and one included angle of two diagonals. Some times quadrilateral can also be constructed from such a given data when five data can be obtained by logical argument. A square can be constructed if one side is given. Here the four sides are equal to one another and an angle is a right angle. A rectangle can be constructed if two adjacent sides are given. Here the opposite sides are equal to each other and one angle is a right angle. A rhombus can be constructed if a side and an angle are given. Parallelogram can be constructed if two adjacent sides and the included angle between them are given.

N.M.G. -32

240

4.1.1. Proof of constructions of some problems in relation to quadrilaterals including the drawings are given.

Problem 16

The lengths of four sides and an angle of any quadrilateral are given. It is required to construct the quadrilateral.

Let, the lengths of four sides of a quadrilateral be a, b, c, d and the included angle between a and b be x. The quadrilateral is to be constructed.

Construction : From any line BE, let us take BC = a, ∠EBF = ∠x is drawn at the point B. Let us take BA = b from BF. Taking radius equal to c and d respectively let us draw two axes within ∠ABC with the centres at A and C respectively. They intersect at the point D. Let us join A, D and C, D. Then, ABCD is the required quadrilateral. proof : According to the construction, AB = b, BC = a, AD = c, DC = d and ∠ABC = ∠x

∴ ABCD is the required quadrilateral.

241

Problem 17

Lengths of four sides and one diagonal of any quadrilateral are given. The quadrilateral is to be constructed.

Let, the lengths of four sides of quadrilateral be a, b, c, d and the length of its diagonal be e, where a + b > e and c + d > e. The quadrilateral is to be constructed.

Construction : Let BD = e be taken from any straight line BE. Taking radius equal to a and b respectively, let us draw two arcs with B and D as centres respectively. These arcs intersect at the point A. Again, taking radius equal to d and c respectively, let us draw two arcs with B and D as centres respectively to the side of BD opposite to A. These two arcs intersect each other at C. A and B, A and D, B and C, C and D are joined respectively. Then, ABCD is the required quadrilateral. Proof : According to construction, AB = a, AD = b, BC = d, CD = c and diagonal BD = e. Therefore, ABCD is the required quadrilateral.

242

Problem 18

The lengths of three sides and two included angles of any quadrilateral are given respectively. The quadrilateral is to be constructed.

Let, the lengths of three sides of a quadrilateral be a, b, c and two angles adjacent to the side of a and b, a and c given as x and y. The quadrilateral is to be constructed. Construction : Let us take BC = a from any line BE. ∠CBF at the point B and ∠BCG at the piont C equal to ∠x and ∠y respectively are drawn. Let us take BA = b from BF and CD = c from CG. A, D are joined. Then, ABCD is the required quadrilateral. Proof : According to construction, AB = b, BC = a, CD = c, ∠ABC = ∠x and ∠DCB = ∠y, Hence, ABCD is the required quadrilateral.

Problem 19

The length of a side of any square is given. The square is to be constructed.

Let, a be the length of a side of any square. The square is to be constructed.

243

Construction : Let us take BC = a from the line BE. BF ⊥ BC at B is drawn. Let us take BA = a from BF. Taking radius equal to a and centres at A and C respectively two arcs are drawn within ∠ABC. They intersect each other at the point D. A, D and C, D are joined. Then, ABCD is the required square. Proof : In the quadrilateral ABCD, AB = BC = CD = DA = a and ∠ABC = one right angle. Hence, ABCD is a square. Therefore, ABCD is the required square.

Problem 20

The length of a diagonal of square is given. The square is to be constructed.

Let, a be the length of a diagonal of a square. The square is to be constructed. Constuction : Let us take BD = a from the line BE. ∠EBF = 45° is drawn at B. Let us draw ∠BDG = ∠FBD at D. BF and DG intersect each other at A. Taking radius equal to BA and with centres at B and D respectively, let us draw two arcs to the opposite of A. They intersect each other at the pont C. C, B and C, D are joined. Then, ABCD is the required square. Proof: In ΔABD, ∠ABD + ∠ADB = 45° + 45° = 90° ∴ ∠BAD = 90° [ Q ∠BAD + ∠ABD + ∠ADB = 180° ]

Now in the quadrilateral ABCD, AB = BC = CD = AD and ∠BAD = one right angle. ∴ The quadrilateral ABCD is a square. Its diagonal BD = a. Hence, ABCD is the required square .

244

Problem 21

The lengths of two adjacent sides of a parallelogram and the included angle between these two sides are given. The parallelogram is to be constructed.

Let, two adjacent sides of the parallelogram be a and b and the included angle between them be ∠x. The parallelogram is to be constructed.

Construction : Let us take BC = a from the line BE. ∠EBF = ∠x at the point B is drawn. Let us take BA = b from BF. Taking radius equal to a, b respectively and with the centres at the points A and C respectively, two arcs are drawn within ∠ABC. They intersect each other at the point D. A, D and C, D are joined. Then, ABCD is the required parallelogram. Proof: Let us join A and C.

Between ΔABC and ΔADC, AB = CD = b, AD = BC = a and AC is the common side.

∴ ΔABC ≅ ΔADC.

Therefore, ∠B AC = ∠DCA . But they are alternate angles.

∴ AB F CD. Similarly, it can be proved that, BC F AD. Hence, ABCD is a parallelogram.

Again according to construction, ∠ABC = ∠x. Therefore, ABCD is the required parallelogram.

E

245

Problem 22

To draw a parallelogram of which one angle is equal to a given angle and the region bounded by it is equal to the given triangular region (area).

Let, ABC be a given triangular region and ∠D be a given angle. A parallelogram is to be constructed of which one angle is equal to ∠D and the region bounded by it is equal to Δ region ABC. Construction : Let us bisect BC at the point E. ∠CEF equal to ∠D is drawn at E of the line segment EC. Through the point A, AG is drawn parallel to the side BC. It intersects EF at the point F. Let us draw the line CG parallel to EF through the point C. It intersects the line AG at the point G. Then, ECGF is the required parallelogram. Proof: Let us join A and E. Δ region ABE = Δ region AEC [ because, BE = EC and the altitudes of both the triangles are equal to each other.]

∴ Δ region AEC = 12 (Δ region ABC)

or, Δ region ABC = 2 (Δ region AEC).

Again, Δ region AEC = 12 (parallelogram region ECGF) [Q EC is the common

base and E C F A G ]

∴ Parallelogram region ECGF = 2 ( Δ region AEC)

∴ Parallelogram region ECGF = Δ region ABC. Moreover, in the parallelogram ECGF, ∠CEF = ∠D, ∴ The parallelogram ECGF is the required parallelogram.

246

Problem 23

To draw a triangle so that the region bounded by it is equal to a definite quadrilateral region ( area).

Let, ABCD be a definite quadrilateral region. A triangle is to be constructed so that the region bounded by it is equal to the quadrilateral region ABCD. Construction : Let us join B and D. Let us draw a line CE parallel to the line DB through the point C. It intersects the side AB produced at E. Let us join D, E. Then, ΔDAE is the required triangle. Proof: ΔBDC and ΔBDE stand on the same base BD and DB ║ CE. ∴ Δ region BDC = Δ region BDE. ∴ Δ region BDC + Δ region ABD = Δ region BDE + Δ region ABD Or, quadrilateral region ABCD = Δ region ADE. ∴ ΔADE is the required triangle.

Problem 24

To draw a parallelogram of which one angle is equal to a definite angle and the region bounded by it is equal to a definite quadrilateral region.

247

Let, ABCD be a definite quadrilateral region and ∠E is a definite angle. Such a parallelogram is to be constructed so that its one angle is equal to the given angle ∠E and the region bounded by it is equal to the region of ABCD. Construction : Let us join D and B. Through the point C a line CF parallel to the line DB is drawn. It intersects the side AB produced at the point F. Let us find the mid-point G of AF. Let us draw ∠GAK equal to ∠E at the point A of line segment AG and the line GH is drawn parallel to AK through the point G. Let us draw a line parallel to AG through the point D. It intersects the lines AK and GH at the points K and H respectively. Then, AGHK is the required parallelogram. Proof : Let us join D and F. According to construction, AGHK is a parallelogram whose ∠GAK = ∠E. Moreover, triangular region DAF = quadrilateral region ABCD

Parallelogram region AGHK = Triangular region DAF Therefore, AGHK is the required parallelogram.

EXERCISE 4

1. Construct quadrilateral from the following given data : a) The lengths of four sides are 3 cm., 3⋅5 cm., 2⋅8 cm., 3 cm. and one

angle is 45°. b) The lengths of four sides are 4 cm., 3 cm., 3⋅5 cm., 3⋅8 cm. and one

angle is 60° c) The lengths of four sides are 3⋅2 cm., 3⋅5 cm., 2⋅5 cm. 2⋅8 cm. and

one diagonal is 5 cm. d) The lengths of four sides are 3⋅2 cm., 3 cm., 3⋅5 cm., 2⋅8cm., and one diagonal

is 5 cm. e) The lengths of three sides are 3 cm., 3⋅5 cm., 2⋅5 cm and its two angles are

60° and 45°. f) The lengths of three sides are 3 cm., 4 cm., 4⋅5 cm. and diagonals are 5 cm.

and 6 cm. 2. The length of a side of a square is 4cm. Construct the square. 3. The length of the diagonal of a square is 5cm. Construct the square.

N.M.G. -33

248

4. The length of a side is 3⋅5cm and one angle is 75°. Construct the rhombus.

5. The length of the adjacent sides are 3cm. and 4cm. Construct the rectangle.

6. The point of intersection of the diagonals of the quadrilateral, four intercepts of two diagonals and an angle between them are given. Construct the quadrilateral. OA = 4⋅2cm, OB = 5⋅8cm, OC = 3⋅7cm, OD = 4⋅5cm and ∠AOB = 100°

7. The lengths of two adjacent sides are given. Construct the rectangle. 8. The length of the diagonal and a side are given. Construct the rectangle. 9. The length of one side and two diagonals are given. Construct the

parallelogram. 10. The length of one side and a diagonal are given. Construct the rhombus. 11. The length of two diagonals are given. Construct the rhombus.

Multiple Choice Questions [Mark (√) on the correct answer] 1. How many independent and unique data are necessary to draw a

quadrilateral? (a) 3 (b) 4 (c) 5 (d) 6 2. i. If two adjacent sides are given then rectangle can be drawn. ii. If four angles are given then a quadrilateral can be drawn. iii. If a side of a square is given then the square can be drawn. Which one of the following is correct in view of the above

informations? (a) i and ii (b) i and iii (c) ii and iii (d) i, ii and iii 3. Which one of the following is correct for parallelogram? (a) The opposite sides are non-parallel. (b) If any one angle is right-angle then it is rectangle. (c) The opposite sides are not equal. (d) The two diagonals are equal to one-another.

249

4. How much is the sum of the opposite angles of a square? (a) 90° (b) 120° (c) 150° (d) 180°

5. If ∠BCD=130° in the parallelogram ABCD, then ∠ABC=How much? (a) 40° (b) 50° (c) 90° (d) 130° 6. Which one of the following is the characteristics of a rhombus? (a) Diagonals are equal to one-another. (b) Every angle is a right angle (c) Two opposite angles are not equal. (d) All sides are equal. 7. A triangle and a parallelogram are situated on the same base and

between two parallel lines. The area of the parallelogram is 12 square cm.

Which one of the following is the area of the triangle? (a) 6 square cm (b) 9 square cm. (c) 12 square cm. (d) 24 square cm. Answer the questions (8–10) on the basis of the following informations: The diagonals AC and BD of the

rhombus ABCD have intersected at the point O. Diagonal AC = 8 cm and side AB = 5cm.

8. How much is the length of the side AO in cm? (a) 2 (b) 3 (c) 4 (d) 5 9. ∠AOB = How much? (a) 60° (b) 90° (c) 120° (d) 180°

O

D C

B A

O O

250

10. How much is the length of the side OB in cm? (a) 2 (b) 3 (c) 4 (d) 5

CREATIVE QUESTIONS

1. Two adjacent sides of a parallelogram are 4cm and 3cm respectively and their included angle is 60°.

(a) Express the above informations in a figure. (b) Draw the parallelogram with the description of drawing. (c) Draw a square with a diagonal equal to the largest diagonal of

the parallelogram mentioned above. Give the description of drawing.

2. The length of four sides of a quadrilateral are 4 cm, 5 cm, 3.5 cm, 3.8 cm and one angle is 75°.

(a) Express the given informations in figure. (b) Draw the quadrilateral giving description of drawings.

(c) Draw a triangle of area equal to the area of the above quadrilateral and verify it logically.

Answer Arithmetic EXERCISE - 1

1 (i) 0.5 (ii) 0.75 (ii) 1.6 (iv) 1.1875 (v) 0.7 (vi) 2.35 (vii) 2.28125 (viii) 0.95 (ix) 3.87

2. (i) 310 (ii)

14 (iii) 1

14 (iv) 3

47100 (v) 2

18

(vi) 2 140 (vii) 3

2740 (viii) 10

1200 (ix) 12

4512000

3. (i) 5.94 (ii) 12.443 (iii) 25.8 (iv) 505.51 (v) 4.212101 (vi) 60.6347 (vii) 7.9667 4. (i) 3.212 (ii) 1.7799 (iii) 3.0845 (iv) 5.321517 (v) 0.000365 (vi) 7.8514 5. (i) 14.2129 (ii) 0.0147 (iii) 6.543 (iv) 1.9473 6. (i) 25.92 (ii) 81 (iii) 8010 (iv) 0.64 (v) 4.16 (vi) 148 (vii) 40.544 (viii) 0.416 (ix) 0.00015 (x) 435.52 (xi) 0.735012 (xii) 0.125 (xiii) 1.920919 (xiv) 0.00004 (xv) 0.00016 7. (i) 2.45 (ii) 17.3 (iii) 10.8 (iv) 1.22 (v) 0.044843 (vi) 0.002 (vii) 0.12 (viii) 6.4 (ix) 2250 8. (i) 30.132 (ii) 3.0901 (iii) 0.015 (iv) 40.7

9. (a) 0.1.6 (b) 0.

.6 (c) 0.

.1 (d) 0.

.5 (e) 0.

.6

.3

(f) 1.8.3 (g) 3.

.2 (h) 4.91

.6 (i) 0.0

.6 (j) 3.5

.3

(k) 0.9.4 (l) 0.32

.6

.7

10 (a) 29 (b)

3599 (c)

152333 (d)

215 (e) 1

1945 (f) 3

7190

(g) 2 241990 (h) 8

893999 (i) 4

16495 (j) 6

11514995 (k) 4

611900

252

11. (a) 1. .3

.3, 3.

.3

.2 (b) 2.3

.3

.3, 5.2

.3

.5 (c) 7.26

.6, 4.23

.7

(d) 5. .77777

.7, 8.

.34343

.4, 6.

.24524

.5

(e) 3.23.33333

.3 , 9.23

.48484

.8, 1.25

.76576

.5 (f) 12.32

.0

.0, 2.19

.9

.9, 4.32

.5

.6

12. (a) 0. .5 (b) 1.

.6 (c) 0.36

.5 (d) 0.58

.9 (e) 17.58

.3

(f) 17.117.9 (g) 8.48

.58867

.9 (h) 0.94

.93730

.0 (i) 8.30

.12048

.4

13. (a) 1.3.1 (b) 1.6

.6

.5 (c) 8.

.75966

.8 (d) 3.67

.8 8597

.6 (e) 2.6

.5

.4

(f) 3.13.3

.4 (g) 6.11

.06

.2 (h) 1.92

.63

.1 (i) 4.84

.16

.3 (j) 3.

.389

.2

(k) 3.88.83073

.9

14. (a) 0.2 (b) 3.0.2 (c) 2 (d) 28.9

.6 (e) 0.0

.6

(f) 1.25 (g) 0.2.07

.4 (h) 12.1

.8

.5 (i) 0.1

.1

.2

15. (a) 0. .6 (b) 0.5 (c) 55 (d) 0.2 (e) 5.

.2195

.1

(f) 26. .3

.6 (g) 0.2

.8 (h) 0.

.4 (i) 5.

.7 (j) 7.

.3

(k) 0. .1

.2 (l) 4.

.8

16. (a) 3.4641, 3.464 (b) 1.1180, 1.118 (c) 0.7453, 0.745 (d) 0.5025, 0.503 (e) 1.1595, 1.160 (f) 2.6526, 2.653 (g) 3.7875, 3.788 (h) 2.6457, 2.646 (i) 1.8539, 1.854 (j) 2.2650, 2.265 17. (a) Rational (b) Rational (c) Rational (d) Irrational (e) Rational (f) Irrational (g) Irrational (h) Rational (i) Rational (j) Rational (k) Rational (l) Rational 18. (a) 9 (b) 71.96 (c) 5 (d) 8

253

EXERCISE - 2.1

1. (a)37.5% (b) 70 % (c) 224% (d) 468.75% (e) 83. .3 %

(f) 258. .3 % (g) 228.125% (h) 6% (i) 56.7%

(j) 467% (k) 854.9% (l) 0.03% (m) 704.5%

(n) 950% (o) 346% (p) 90% (q) 58. .3 %

(r) 52% (s) 78.125% (t) 52%

2. (a) 14 (b)

54 (c)

743600 (d)

77300 (e) 3

297700

(f) 16 (g)

516 (h)

38 (i)

23

3. (a) 0.12 (b) 0.37 (c) 0.125 (d) 0.0825 (e) 1.23125 (f) 1.15625 (g) 0.22625 (h) 0.1409375 (i) 0.421875 4. (a) 31 (b) 60 (c) 10 (d) 67 (e) 66 (f) 240 (g) 12 pieces (h) Tk.37.50 (i) 2 kg 250 g (j) 480 person (k) 8 km

5. (a) 20% (b) 50% (c) 66 23 % (d) 166

23 % (e) 13

13 %

(f) 41 23 % (g) 62

12 % (h) 16

23 % (i) 4%

6. (a) 50 (b) 320 (c) 120 (d) 240 (e) Taka 125

(f) 80 km (g) 240 kg (h) Taka 233 13 (i) 300 kg

7. (a) 1 t 4 (b) 8 t 25 (c) 12 t 25 (d) 3 t 5 (e) 5 t 4 (f) 7 t 2 8. Taka 14400 9. 2000 person 10. 70% 11. Taka 102

12. 75% 13. Tk. 482 14. 12 12 % 15. 20%

16. Taka 4000 17. 20000 person 18. Taka 1200000 19. 200 person 20. 400 female students and at present 480 male students. 21. Taka 1500 22. 100000 person 23. 10,00,000 person

24. Present cost Tk. 60 and previous cost Tk. 75 25. Tk. 30 26. 9 1

11 %

254

27. 33 13 % 28. 20% 29. 23% 30. 200 person 31. 400 person

32. 58320 person 33. 1250 pieces

EXERCISE - 2.2

1. Tk. 273 2. Tk. 119 3. Tk. 30 4. Tk. 2502 5. Tk. 1270.50 6. Tk. 750 7. Tk. 730 8. Tk. 9000 9. 14% 10. 6% 11. 40% 12. Tk. 475 13. 30% 14. 6 years 15. 10 years 16. 4 years 17. 3 years 18. Tk. 775 19. Tk. 500 20. Tk. 1600 21. 4% 22. 6 years 23. Tk. 225 24. 10 years 25. Tk. 365000

26. Capital Tk. 400 and rate of interest 7 12 % 27. 8 years 28. Tk. 533

13

29. 40 years 30. Tk 1200 and 7% 31. 6 years 32. Tk. 1230

33. 4% 34. 320 portion 35. 6

34 %

EXERCISE - 2.3

1. Tk. 25 Loss 2. Tk. 12.50 Profit 3. Tk. 10 Profit 4. Tk. 10 Profit 5. 25 Pieces 6. 60 kg 7. Tk. 20 Loss 8. Tk. 25 Profit 9. Tk. 40 10. Tk. 25 Loss 11. 20% Loss 12. 8% Profit 13. 50% Profit 14. Tk. 16 15. 5% Profit 16. 25% Profit 17. 12 Pieces 18. 4% Profit 19. Tk. 792 20. Tk. 10 21. 25% Loss 22. Tk. 2.25 23. 50% Profit 24. Tk. 396

25. Tk. 225 26. 33 13 % 27. Tk. 300 28. Shirt Tk. 175 and

pant Tk. 350 29. 12% profit 30. Horse Tk. 6000 and cow Tk. 4000

31. There will be no loss or Profit 32. 4 1621 % Loss 33. Tk. 22

34.Tk. 33 35. Tk. 3.36 36. Tk. 12.50 37. 33 13 %

38. Tk. 561000 39. Tk. 181.92

255

EXERCISE - 3

1. 53650 Person 2. 30 Person will be reduced 3. 132 pieces 4. Tk. 80 5. Tk. 2.50 6. 420 pieces 7. 10000 metre 8. 6400 times 9. 625 metre 10. 3000 metre 11. 100 metre towards the same direction and 3100 metre towards the opposite direction 12. 54.17 km(app) 13. 280kg 14. 26.67 metric ton (app) 15. 145 kg 952 gm 450mg 16. 1 quintal 26 kg 500gm 17. 821.92 metric ton (app) 18. 549 kg rice and 172.5 kg salt 19. 200 days 20. 0.64 litre (app) 21. 0.07 litre (app) 22. 16 litre 23. 320000 sq.cm. 24. 25 metre 25. 208 sq.m 26. 636sq.m. 27. 402.31 metre(app) 28. 60 metre 29. 155 sq.m 30. 520.8 sq.m 31. 3264 sq.m 32. 24 metre 33. 2.5 metre 34. 2717.44 gm 35. 673.5470 cu.cm 36. 1240 cu. metre 37. 450 kg 38. 60000 litre and 60000 kg

EXERCISE - 4.1

1. (a) Tk. 49.40 (b) Class interval 5, Tk. 49.13 2. (a) 75.02 (b) 0.02 3. Tk. 821.331 4. 45.2, 44.66 5. 23.31

6. 44.5 kg 7. Tk. 42.3.6 hundred 8. Tk. 2230.63

EXERCISE − 4.2

1. Mean 22.21, Median 23, Mode 19 2. Median 45, Mode 38 and 55; Median 46 and Mode 41 from class interval 3. Mean Tk. 74.64, Median Tk. 74 and Mode Tk. 69. 4. Mean Tk. 150.43, Median Tk. 150, Mode Tk. 140 and Tk. 156 5. Mean Tk. 110.93, Median Tk. 112.60 and Mode Tk117.15 6. Mean Tk. 65.18, Median Tk. 64.55 and Mode Tk. 63.68 7. Mean 66.1, Median 66.83 and Mode 66.714

N.M.G. -34

256

8. Mean 11.44 years, Median 11.97 years and Mode 12.2 years. 9. Mean marks 65.125, Median 64 and the marks obtained by the highest number of students is 64. 10. Mean Tk. 66.65, Median Tk. 65 and the daily wages of the maximum number of labourers is Tk. 63. 11. Mean 22.25 years, Median 17.92 years and Mode 5 years. 12. Population growth rate 22%, Annual population growth rate 2.2%, Growth rate of male population 21.8%, Growth rate of female population 22.2%, Mean of population 96717478.5, Mean of male population 49823770.5 and Mean of female population 46893708.

ALGEBRA

EXRCISE - 1.1

1. (i) 25a2+70ab+49b2 (ii) 49x2-126xy+81y2 (iii) 16a2b2+40ab2c+25b2c2

(iv) 36x4y2-60x3y3+25x2y4 (v) 16x6+24x3y4+9y8 (vi) x4-6x2+9

(vii) x6+2x4y+x2y2 (viii) 121a2-264ab+144b2 (ix) a2x2-2abxy+b2y2

(x) x2+2xy+y2 (xi) a2-2ab+b2 (xii) 2y2z2+2abcxyz+a2b2c2

(xiii) a4x6-2a2b2x3y4+b4y8 (xiv) a2+b2+c2-2ab+2ac-2bc

(xv) x4+y4+z4-2x2y2-2x2z2+2y2z2 (xvi) x4y2z2-2x3y3z2+y4z2x2

(xvii) x2y2+y2z2+z2x2+2xy2z-2x2yz-2xyz2 (xviii) 90601 (xix) 367236 (xx) 356409 (xxi) 998001 (xxii) 1004004

(xxiii) 25a2+36b2+49c2-60ab-70ac+84bc

(xxiv) 49a4+64b4+25c4+112a2b2-70a2c2-80b2c2

(xxv) a2+b2+c2+d2-2ab-2ac-2ad+2bc+ 2bd+2cd 2. (i) 9a2 (ii) 36x4 (iii) 324 (iv) 64b2 (v) 4z4 (vi) 16 (vii) 10,00,00,000 3. 576 4. 196 5. 18225 6. 11 7. 194 10. 4,34 11. 289,169 12. 36, 90 13. 29 14. 178, 40

257

15. 482, 240 16. (i) x2-y2 (ii) 25x2-49y2 (iii) 49a2-121 (iv) a2-b2+2bc-c2 (v) 25a2+4b2+20ab-9c2 (vi) a2x2-b2y2+2bcyz-c2z2 (vii) a8-b8 (viii) x2-4x-140 (ix) 9a2-45a+50 (x) 36x2+24x-221 (xi)a2x2+2abxy+b2y2+8ax+8by+15 17.(i) (6a+4b)2-(a+2b)2 (ii) (5x)2-(13)2 (iii) (5x)2-(2x-5y)2 (iv) (8b-a)2-(b+7a)2

EXERCISE - 1.2

1. (i) 27x3+108x2y+144xy2+64y3 (ii) x9+3x6y2+3x3y4+y6 (iii) a6b3+3a4b2c2d+3a2bc4d2+c6d3 (iv) a3b3+3a2b3c+3ab3c2+b3c3

(v) 512x6+2112x4y3+2904x2y6+1331y9 (vi) 343m3 + 735m2n+525mn2+125n3

(vii) x3+y3+z3+3x2y+3xy2+3x2z+3y2z+3xz2+3yz2+6xyz (viii) x9+3x6y3+3x3y6+y9 (ix) 8x3-60x2y+150xy2-125y3 (x) p6-3p4q2+3p2q4-q6 (xi) 1331a3-4356a2b+4752ab2-1728b3 (xii) x9+6x6+12x3+8 (xiii) x18-24x12+192x6-512 (xiv) 8x3-27y3-z3-36x2y+54xy2-12x2z-27y2z+6xz2-9yz2+36xyz (xv) x6-y6+z6-3x4y2+x2y4+3x4z2+3y4z2+3x2z4-3y2z4-6x2y2z2 (xvi) a6b3-3a4b5c+3a2b7c2-b9c3 (xvii) x3y3-6x2y3z+12xy3z2-8y3z3 (xviii) a6b6-3a4b4c2d2+3a2b2c4d4-c6d6 (xix) x9-6x6y3+12x3y6-8y9 (xx) 343x6-1323x4y2+1701x2y4-729y6 2. (i) 216a3 (ii) 64y3 (iii) a3-3a2b+3ab2-b3 (iv) 8x3 (v) 8x3 (vi) 8x3 3. 370 4. 793 5. 1900 6. 33614 8. 0 9. 1 10. 722 11. 125 12. 512 15. 140 16. (i) a6+b6 (ii) x6+8 (iii) 8a3+27b3 (iv) 343a3+64b3 (v) 64a3-27b3

(vi) 512x3-27y3 (vii) a3x3-b3y3 (viii) x6+a6 (ix)x6-a6 (x) 64a6-1 (xi) 15625a6-729b6 17. (i) (a+2) (a2-2a+4) (ii) 8(x+3y) (x2-3xy+9y2) (iii) (2x+7) (4x2-14x+49) (iv) b3(3a+4c) (9a2-12ac+16c2) (v) a (2a+3b) (4a2-6ab+9b2)

258

(vi) (4b-5) (16b2+20b+25) (vii) (3a-2) (9a2+6a+4) (viii) 3(2x-3y) (4x2+6xy+9y2) (ix) 7(2x-3y) (4x2+6xy+9y2) (x) (9a-4bc2) (81a2+36abc2+16b2c4)

EXERCISE - 1.3

1. (2x+y) (2x-y) 2. (x+12y) (x-12y) 3. 3x(1+5x) (1-5x) 4. a(x2+4x+8) (x2-4x+8) 5. (2a2+6a+9) (2a2-6a+9) 6. (x+y-1) (x-y+1) 7. (x2+2x-1) (x2-2x-1) 8. (x2+x+1) (x2-x+1) 9. (a-1) (a-2b+1) 10. (a2+ab+b2) (a2-ab+b2) 11. 8(2x-y) (4x2+2xy+y2) 12. (x-y+z) (x2-2xy+y2-xz+yz+z2) 13. (a2+b2) (a4-a2b2+b4) 14. (a+b) (a-b) (a2-ab+b2) (a2+ab+b2) 15. (x+9) (x-8) 16. (x+14) (x+4) 17. (x-15) (x+7) 18. (x-5) (x-3) 19. (x-26) (x-25) 20. (x+7) (5-x) 21. (x+10) (x+4) 22. (a+4b) (a+3b) 23. (x+15) (x-8) 24. (x+14) (x-3) 25. (x+8y) (x-5y) 26. (p+10q) (p-8q) 27. (x2-x+8) (x2-x-5) 28. (a-2) (a+2) (a+5) (a+9) 29. (x-4) (x+2) (x+4) (x+10) 30. (3x-5) (2x+3) 31. (x+4a-2) (x-3a+2) 32. (x+a+b) (x+2a+3b) 33. (x+14) (x+4) 34. (x+4) (3x-1) 35. (2x+3) (4x-5) 36. (x-6) (3x+2) 37. (x-7) (2x+5) 38. (a+5b) (2a-3b) 39. (3a-4b) (3a+4b) (a2+2b2) 40. (x+y-2) (2x+2y+1) 41. (3x-4y) (5x+3y) 42. (x-2y) (2x-y) 43. (2p+3q) (5p-2q) 44. (x+6) (2x-7) 45. (x+a) (ax+1) 46. 5a2b2c2 47. 18a2c2 48. 6 49. 8x2y2z3 50. x+y 51. x-3 52. x-2 53. x-y 54. x+2 55. 2 (x+y) 56. ab(a2+ab+b2) 57. x(x+2) 58. 36x2y2z3 59. 72a3b2c3d3 60. 60x4y4z2 61. 120a3b3c262. 30a2b3c3

63. (a-b)2 (a+b)3 (a2-ab+b2)2 64. (x2-1) (x+2) 65. (a6 -1) 66. x(x+2)2 (x+3) 67. a2b2(a6-b6) 68. (2x-1) (3x+1) (x+2)

259

EXERCISE - 2.1

1. (i) 3y

4x2z2 (ii) 7c4

6ab2 (iii) 5a3

4b3x2 (iv) 11q2r3

16 (v) xy2

(vi) 2(a+b)3a(a-b) (vii)

(a-b)(a+b)2 (viii)

(a-3)(a+3) (ix)

(x-1)(x+9) (x)

(2x+3)(3x+4)

(xi) (2x2-1)(3x2+1) (xii)

(a-b-c)(a+b-c) (xiii)

(2a-b)(a2-1)

2. (i) 35a3

420abc , 21b3

420abc , 15c3

420abc (ii) 15a2b2d3

60abcd5 , 10a2c2d2

60abcd5 , 6b2c2

60abcd5

(iii) a2(a+b)a(a2-b2) ,

ab(a-b)a(a2-b2) ,

c(a-b)a(a2-b2) (iv)

c(a-b)abc ,

a(b-c)abc ,

b(c-a)abc

(v) (a+b) (a3+b3)(a-b)2 (a3+b3) ,

(a-b)3

(a-b)2 (a3+b3) , a(a − b) (a2 − ab + b2)

(a-b)2 (a3+b3)

(vi) a(a2+ab+b2)

(a3-b3) , b(a-b)(a3-b3),

c(a3-b3)

(vii) 2(x+3)2

(x-2) (x+1) (x+3)2, 3(x+1) (x+3)

(x-2) (x+1) (x+3)2, 4(x-2) (x+1)

(x-2) (x+1) (x+3)2

(viii) (b-c) (c-a)2

(a-b) (b-c) (c-a), (a-b)2(c-a)

(a-b) (b-c) (c-a), (b-c)2(a-b)

(a-b) (b-c) (c-a)

3. (i) x2+y2

xy (ii) x2+y2+z2

xyz (iii) a2+2ab-b2

a2-b2 (iv) 2(4x2+9y2)

4x2-9y2

(v) 3a2+b2

(a-b) (a+b)3 (vi) 3

(x-1) (x-3) (vii) 2x4-1x6-1

(viii) 3(x-1)

(x-2) (x-3) (x+2)

4. (i) 4x

x2-16 (ii) -2b

a(a2-b2) (iii) ab

a3+b3 (iv) 6xy

x2-9y2 (v) 6x

(x3-8)

(vi) 2x

(x4+x2+1)

5. (i) 2(a-c)

ac (ii) 3a2b-3bc2+a2c-ac2+ab2-b2c

(a+b) (b+c) (c+a) (iii) -2c

(b-c) (c-a)

(iv) 2aba2-b2 (v) 0 (vi)

−16x3y(x4-16y4) (vii)

6ab2

(a2-b2) (4a2-b2)

260

(viii) 2x

(x4-1) (ix) 12a4

(a6-64) (x) 0 (xi) 64x4

(x8-256) (xii) 2

(c-b)

(xiii) (a-2b)

(a2+b2-c2-2ab) (xiv) 12y

4x2-9y2

EXERCISE - 2.2

1. 2a4b3

45c2d4 2. 8y5z3

9x2 3. 1 4. x

(x+3) 5. (a+b) (a2+b2)

(a3-b3) (a2-ab+b2)

6. (1-b)

x 7. 1 8. 1 9. a+b 10. (1-b)(x+y)

11. a(a2-ab+b2)

(a-b)2 12. (b2+ac) (bd+ac)

abcd 13.- 1x2 14.

2x(x+y+z)

15. x

(1+x) (1-x3) 16. 6x2

x2-y2

EXERCISE - 2.3

1. 32a3c27b3 2.

5ac3

9b 3. x2y2z2

9a2b2c2 4. 25aa+b 5.

3b2a

6. x-7x-5 7. (a-b)2 8.

x2-8x+12x2-10x+21 9.

(x+5) (x2+x-56)(x+12) (x2-36)

10. 1 11. x

x-1 12. 1 13. 1

a2+b2 14. a+b

15. a(2a-3b)

2a-b 16. x2-y2 17. 1a 18. x-3 19. 1

20. 3a2b 21. 2a(a+b)2

EXERCISE - 3.1

1. - 4 2. - 8 3. 2 4. 3 15 5. - 20

6. -55 57 7. 2

419 8.

7918 9.

1126 10. - 2

124

261

11. -2455 12.

-24271 13.

-29596 14. 73 15.

a+b2

16. - 25 17. 25 18. -

2331 19. 15 20. ab + ac + bc

21. a + b + c 22. a + b + c 23. ab+bc+ca

3 24. 32

25. -117 26.

53 27.

15 28.

72 29.

-56

EXERCISE - 3.2

1. 40 and 60 2.19 and 42 3. Present age of father is 60 years and the present age of son is 20 years.

4. 15 and 60 5. 34

6. Chaity's portion Tk. 48, Pio's Tk. 32 and Rafat's portion Tk. 24. 7. 240 pieces 8. 30 male and 20 female. 9. 180 person 10. 12 metre 11. The ratio of prices of sugar at Tk 40 and Tk. 30 is 1 t 4. 12. The number of 25 paisa coin is 20 pieces and 50 paisa coin is 80 pieces 13. 1km. away from Rampura 14. 45 Kg. with one person and 35 kg with another person. Each of both can carry 20 kg without any cost.

EXERCISE - 4.1

1. (2, 1) 2. (3, 2) 3. (2, 2) 4. (2, 3) 5.(2, 3) 6. (6, -5) 7. (3, 2) 8. (3, 10) 9. (2, 3) 10. (1, 2)

11.{ (b2+ac)(a2+ab) ,

(ab-c)(a2+ab) } 12. (2, 3) 13. (5, 2) 14. (2, 1)

15. (-3, 0) 16. ( 12 , -3) 17. (3, 2) 18. (7, 4) 19.(8, 12)

20. (a+b, b-a) 21. (6, -8) 22. (2, 6) 23. (6,2) 24. ( ab

a+b , ab

a+b )

262

EXERCISE - 4.2

1. 45,35 2. 60,15 3. 7,9 4. 50 years and 20 years

5. 45 year and 25 year 6. 35 7.

38 8. 53 9.63

10. 62 and 34 mangoes respectively; Total number of mangoes 96. 11. 50 and 20 metre. 12. 6 km and 3km 13. Men by 12 days or boys by 24 days. 14. Male by 15 days or boys by 60 days. 15. The price of cow is Tk. 4500 and the price of goat is Tk. 1500. 16. Speed of rowing is 4 km. and the speed of current is 3 km. 17. By 1st tube in 10 minutes and by 2nd tube in 15 minutes.

EXERCISE - 5.1

1. 1st Quadrant, 2nd Quadrant, x-axis to the right, 3rd Quadrant, 4th Quadrant, y- axis above, Origin, y-axis below, x-axis to the left, 1st Quadrant and y-axis above. 6. (0,1) 7. (-1,-1)

EXERCISE - 5.2

2. (i). (4,1) (ii). (3,2) (iii). (2,1) (iv). (2,3) (v). (2,0) (vi). (-3,2) (vii). (4,3) (viii). (2,0) (ix). (1,-1) 3. (i) 2 (ii) -2 (iii) 2 (iv) 1 (v) 0 (vi) 2 (vii) -1 (viii) 12

THE END

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