maths book (1).docxd

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Edition 1

New Syllabus

Consultant

Sidra Javaid

Authors

Hania Asif, Maheen Touqeer

Zainab Fatima, Hafsa Tahir

Sadia Anwar, Sidra Javaid

Department of Elementary Education

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University of the Punjab, Lahore

Elementary Education is a department of Institute of Education and Research abbreviated as (IER) of Punjab University. The objectives of this University are worldwide known. This book has been selected as the Best Book by the Ministry of Education.

All rights are reserved with the Department of Elementary Education. This book has been approved by the Curriculum Authority. No parts of this publication can be copied or imitated in any form, without the permission of Department of Elementary Education.

© Institute of Education and Research

First Impression 2013

Printed in Pakistan

Published by


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PrefaceNew Syllabus Mathematics of grade 5 is an authentic and standardized book for 5th graders. This book follows the syllabus for Elementary Schools, implemented from 2007, by the Ministry of Education, Pakistan. This book covers the whole syllabus of for the Pakistan’s Primary Level Mathematics.

This new edition of 5th grade mathematics retains the goals and objectives of the previous edition and all the basic concepts have been revised to keep materials up – to – date

Examples and exercises have been carefully organized to aid students in progressing within and beyond each level.

This book includes very much interesting features for students

An interesting introduction at the beginning of each chapter includes variety of photographs and graphics to raise the level of interest among students.

Activities are also included to arouse the students’ interest in mathematics.

Extra information is also provided to enable students in more critical thinking.

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ContentsNumbers and Arithmetic Operations

HCF and LCM

Unitary Method

Fractions

Decimals and Percentages

Geometry

Perimeter and Area

Information Handling

5

31

59

120

161

204

214

Unit

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Unit 8

81

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After studying this unit, the students will be able to: Read numbers up to 1,000,000,000(one billion) in numerals

and in words. Write numbers up to 1,000,000,000(one billion) in

numerals and words. Add numbers of complexity and of arbitrary size. Subtract numbers of complexity and of arbitrary size. Multiply numbers up to 6 digits by 10,100 and1000. Multiply numbers up to 6 digits by a 2-digit and 3-digit

number. Divide numbers up to 6-digits by a 2-digit and 3-digit

number. Recognize BODMAS rule, using only parenthesis ( ). Carryout combined operations using BODMAS rule. Verify distributive laws.

Unit:1Numbers And

Arithmetic Operations

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1.1 Numbers up to one billion.1.1.1 Read the numbers up to one billion.We know that 100000000 in international place value system is written as:

Millions Thousands Ones

Hundred Millions

Ten Millions

Millions

Hundred Thousands

Ten Thousands

Ten Thousands

Hundreds

Tens

Ones

1 0 0 0 0 0 0 0 0

Commas are placed after three places from right most digits.For example:

1. The number 200,000,000 is written in figures. It can be written as

1.Two hundred million.

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2.The number 100,000,000 is written in figures . It can be written as: One hundred million.Similarly, 345,267,982 is read as: Three hundred forty five million, two hundred sixty seven thousand, nine hundred eighty two.100,000,000 and 999,999,999 are respectively smallest and largest 9-digits numbers.

999,999,999 is read as: Nine hundred ninety nine million, nine hundred ninety nine thousand, nine hundred ninety nine.One more than 999,999,999 is one thousand million.One thousand million is known as 1 billion.1 billion is written as 1,000,000,000.

1 Billion=1000 million

Example 1 : Read the following numbers and write in words.I. 14,020,748.

II. 843,658,952III. 435,525,243.

Solution:

I. 14,020,748: Fourteen million, twenty thousand, seven hundred and forty eight.

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II. 843,658,952: Eight hundred forty three million, six hundred fifty eight thousand, nine hundred and fifty two.

III. 435,525,243: Four hundred thirty five million, five hundred twenty five thousand, two hundred and forty three.

1.1.2 Write numbers up to one billion

Example 2:

Write the following numbers in figures.

I. Seven hundred eight million, two hundred seventy two thousand, nine hundred twenty six.

II. Eight hundred eighty six million , five hundred thousand, one hundred twenty six.

Solution

I. 708,272,926. I. 886.500,126.

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Exercise 1.1

1. Read the following numbers and write them in words:

I. 65,086,236 II. 345,666,780

III. 980,765,455

IV. 999,999,999

ActivityPlace the commas in following numbers

according to international place value system: 1.705366935 2.

211435126___________ ____________

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V. 455.980,650VI. 1,000,000,000

VII. 455,678,999.

2. Write the following numbers in figures:

I. One hundred twenty million three hundred six thousand four.

II. Ninety five million, sixteen thousand, fifty five.III. Sixty million.

IV. Five hundred ninety million three hundred five thousand five.

1.2 .Addition and Subtraction:

We have learnt addition of numbers up to 6-digit numbers. Addition of larger number (7,8 or 9-digit) is done similarly. That is, addition of numbers is done by addition of digits having same place values(starting from ones). We should be very careful while placement of digits of numbers and commas.

As an example, addition of 2,556,876,135 and 605,957

1.2.1 Add numbers of complexity and of

arbitrary size :

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is performed as under: 1 1 1

2, 5 5 6 , 8 7 6 , 1 3 5

+ 6 0 5 , 9 5 2

2 , 5 5 7 , 4 8 2 , 0 8 7

Example 3 : Add 24,571,344 and 5,690,345

2 4,5 7 1 ,3 4 4

+ 5, 6 9 0, 3 4 5

3 0 ,2 6 1 , 6 8 9

Activity

Answer the following in 10 seconds each!

208,070

+ 399,770

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422,134

+ 533,450

Exercise:1.2

Solve:

1. 3, 9 5 6, 2 3 4 + 8 9 0, 1 2 4

2. 5 7 8, 9 8 0

+ 1 2 3, 6 7 0

3. 2, 8 9 0, 6 7 8

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+ 7 8 3, 1 2 7

4. 6 ,7 3 4,5 6 7 + 1 2 3, 6 8 0.

5 . 4, 4 4 4, 4 4 4

+ 4 4 4,4 4 4

6 . 7 , 6 8 0 , 7 8 1

+ 5 6 1 , 6 7 0

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7 . 6 , 8 7 0 , 4 5 6

+ 6 7 5 , 3 2 1

8. 5 6 7 , 7 8 9 ,5 5 5

+ 2 3 4 , 5 6 7 , 6 5 7

Find the Sum:

9. 490,456 and 467,335.

10. 65,444,290 and 23,543,670.

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Same method is applied for subtraction of larger numbers (7, 8, 9) as we have done for subtraction of numbers upto 6 digit numbers . It is done by subtraction of digits having same place while placement of digits of the numbers and commas .

Let us take an example :

Subtraction of 6, 567, 239 from 4,456, 112 is as under:

6 , 5 6 7, 2 3 9

- 4 , 4 5 6 , 11 2

2 , 1 1 1 ,1 2 7

Example 4:

Subtract 4,567,887 from 5,678,997.

Solution:

1.2.2 Subtract numbers Of complexity and Of arbitrary size.

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5 , 6 7 8 , 9 9 7

- 4, 5 6 7 , 8 8 7

1, 1 1 1 , 1 1 0

Exercise: 1. 3

Solve:

1. 7 8 9, 9 9 9 , 8 6 7

_3 2 1 , 8 9 0 , 7 6 6

2. 5 8 9, 7 9 0, 9 8 8 _ 4 7 8 , 8 9 0 ,3 2 2

3. 8 8 , 9 8 0 ,8 7 9_ 6 7 , 7 8 0 , 3 4 5

4. 8 9 7, 8 9 9_ 3 4 5, 7 7 8

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5. 3 3 9, 9 9 8, 9 8 4 _ 2 2 8, 7 8 0 , 2 1 3

Solve:

6. 5 8 8 , 9 8 9 – 4 6 7, 8 1 2

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Multiplication by 10,100 and 1000.Let us see the product of a number 22222 with 10,000 and 1000.22222 × 10 = 222220.22222×100=2222200.22222×1000= 22222000.

Example:Multiply 3654 with 10,100 and 1000.

1.3.1. Multiply numbers up to 6 digits by 10 ,100 and 1000 :

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Solution: 3654×10=36540. 3654×100= 365400. 3654×1000=3654000.

Example 6 : Multiply 224222 by 14.

Solution: 1 2 2 4 2 2 2 × 1 4

8 9 6 8 8 8

1.3.2 Multiply numbers upto 6 digits by a 2 digits and 3 digits number

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2 2 4 2 2 2 0 3 1 3 9 1 0 8

Exercise: 1.4Multiply:

1. 512222 by 10. 2. 422116 by 13.3. 112333 by 114. 310333 by 12.

Example:

Divide 537809 by 35 .Find the quotient and remainder as well.

1.3.3 Division Of numbers upto 6

digits by a 2-digit and 3-digit number:

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Solution:

15365 Quotient

35 537809

-35

187 - 175

128 - 105

230

- 210

209

- 175

34 Remainder

Note: The remainder at every step would be less than the divisor.

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Exercise: 1.5Divide: 1. 425580 by 602. 304480 by 403. 209800 by 204. 500800 by 405. 566700 by 30

BODMAS stands for Brackets, Of, Division, multiplications, addition and subtraction. It helps remembering the order of preferences of mathematical operators. We solve brackets first if given in the question. In case of more brackets, we solve from inner most bracket to the outer most bracket.BODMAS given order of preferences as:

Preference No. Rule Operation Symbol 1 BODMAS Brackets (…) 2 BODMAS Of Of 3 BODMAS Division + 4 BODMAS Multiplication × 5 BODMAS Addition + 6 BODMAS Subtraction -

1.4 Order Of operations: BODMAS rule :

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Order of Preferences is as under:(…), Of, ÷ , × , + , - .Now we simply 16+ (12 + 9 ÷ 3) × 5, using BODMAS rule, 16+ (12+ 9÷3) ×5= 16 + (12+3) × 5 (First divide within bracket)=16+ 15×5 (After Addition remove Bracket)= 16+75. (Multiplication)= 91. (Addition)

Example 12: Solve (18+6) ÷2×4.Solution: = (18+6) ÷2×4 =24÷2×4. = 24÷8. = 3.

Exercise: 1.6 Solve: 1. 24× (12÷6).

1.4.1 Recognize BODMAS rule,using only parenthesis ( ).

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2. (5×10) ÷5. 3. (18÷2+6) ×3-2. 4. (8×5) ÷7+1.

We verify the distributive laws with the help of following examples.

Example : Verily Distributive laws.a. 10× (13+2) = (10×13) + (10×2).

Solution: L.H.S= 10× (13+2) = 10×15

= 150. (a)

Verification Of Distributive Laws:

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R.H.S=(10×13)+(10×2) =130+20 =150. (b)From (a) and (b)L.H.S=R.H.S.b. (6+4) ×4= (6×4) + (4 ×4).

Solution: L.H.S= (6+4) ×4. = (6+4) ×4 = 10×4. = 40.R.H.S= (6×4) + (4 ×4). = 24+16. =40.

Exercise:1.7

Verify distributive laws:1. 14 × (9+5) = (14×9)+ (14×5).

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2. (12+5) ×6= (12×6) + (5×6).3.16× (10+4) = (16×10) +(16×4).4. 11× (12-7) = (11×12)-(11×7).5.16× (12-6) = (16×12)-(16×6).

Review Exercise1. Read the following numbers and write them in words.

I. 101,001,017.II. 106,011,111.

2. Multiply:

I. 244664 by 24.II. 993399 by 33.

3. Divide:

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246444 by 22

4. Evaluate:

60-(40÷10-2) + 5.

5 .Multiple choice questions.

Encircle the correct answer:

1. 1 billion= ?

a. 100 million b.10 million. c. 1000 million d.500 million.

2. Eight hundred twenty three million five hundred thirty eight

thousand two hundred seventy eight.

a. 800,23,538,208 b.823,538,278 c.804,567,278 d.893,567,278.

3. 3566×100= ?

a. 355600 b. 356600 c. 356000 d.344600.

4. 7× (4+4) = (7×4) + (7×4) is

a. Distributive law b. Associative law c. Commutative law

d. BODMAS rule.

5. 5668-1333=?

a. 4335 b.4512 c. 4412 d. 4125.

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Answers:

Exercise 1.11.

i. Sixty-five million, eighty-six thousand, two hundred and thirty-six.

ii. Three hundred forty-five million, six hundred sixty-six thousand ,seven hundred and eighty.

Answers

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iii. Nine hundred eighty million, seven hundred and sixty-five , four hundred and fifty-five.

iv. Nine hundred ninety-nine million , nine hundred ninety-nine thousand, nine hundred ninety-nine.

v. Four hundred and fifty-five, nine hundred and eighty, six hundred and fifty.

vi. One thousand million.vii. Four hundred and fifty-five, six hundred and seventy-

eight, nine hundred and ninety nine.

2.

i. 120,306,004.ii. 95,016,055.

iii. 60,000,000.iv. 590,305,005.

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Exercise: 1.2i. 4,846,358.

ii. 702,650.iii. 3,673,805.iv. 6,858,247.v. 4,888,888.

vi. 8,242,451.vii. 7,545,777.

viii. 802,357,212.ix. 957,791.x. 88,987,960.

Exercise: 1.3i. 468,109,101.

ii. 111,100,666.iii. 21,200,534.iv. 552,121.v. 111,218,771.

vi. 111,177.

Exercise: 1.4i. 5122220.

ii. 5487508.iii. 112344.iv. 3723996.

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Exercise : 1.5

1.

i. 7093.ii. 7612.

iii. 10490iv. 12520v. 18890.

Exercise: 1.6i. 48.

ii. 10.iii. 15.iv. 5.

Exercise: 1.7

i. 196.ii. 102.

iii. 224.iv. 55.v. 96.

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Review exercise

1.

i. One hundred and one million, one thousand and seventeen.

ii. Six million ,eleven thousand , one hundred and eleven.

2.

i. 5871936.ii. 32782167.

3.

11202.

4.

53.

5.

i. cii. b

iii. biv. av. a

HCFANDLCM

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CHAPTER-2

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Find the HCF of given two or three numbers up to 2 digits with the help prime factorization.Find the HCF of given two or three numbers up to 2 digits with the help of division method.Find the LCM of three or four numbers up to 2 digits with the help of prime factorization.Find the LCM of three or four numbers up to 2 digits with the help of division method.Solve real life problems.

After the instruction of this chapter students will be able to

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2.1 HCFThe term HCF stands for “Highest Common Factor”. It is defined as “the largest number with which if the given two or more numbers are divided the remainder will always be zero”.

HCF of two or more given numbers can be calculated by two methods:

Prime Factorization

Division Method

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PRIME FACTORIZATION:

To express a composite number as

the product of two or more prime

numbers or prime factors of that

number is called prime factorization

For Example:

10=2x5

8=2x2x2

CALCULATING THE HCF OF THREE NUMBERS UP TO 2 DIGITS WITH THE HELP OF PRIME FACTORIZATION METHOD:

EXAMPLE 1:

Using prime factorization find the HCF of 30, 40 and 60.

Note

Except the number 1 every other number is a prime number or a composite number.

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Solution:

Factors of 30=2x3x5

Factors of 40=2x2x2x5


Common factors of 30, 40 and 60=2x5=10

So the HCF of 30, 40 and 60=10

EXAMPLE 2:

Using prime factorization find the HCF of 12, 16 and 18.

Solution:

Factors of 12=2x2x3


Factors of 18=2x3x3

Common factors of 12, 16 and 18=2

So the HCF of 12, 16 and 18=2

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Using Prime Factorization find the HCF of the following:

1) 38, 42 and 63 2) 50, 14 and 22

3) 24, 48 and 84 4) 44, 96 and 80

5) 21, 52 and 63 6) 66, 33 and 55

7) 12, 14 and 16 8) 18, 63 and 15

9) 50, 60 and 70

Division Method:

Division method is a method in which we

divide the larger given number by the

smaller number to get the remainder, then

we divide the divisor by the remainder and

the process continues until the remainder

Note

Divisor is that number with which another given number is divided.

Exercise 2.1

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Becomes 0 and the last dividend is the HCF of the given number.

EXAMPLE 1:

With the help of division method find the HCF of 33 and 55.

Steps for finding the HCF of 33 33) 55 (1

and 55 by division method are: - 33

STEP 1: 22) 33 (1

Divide larger given number 55 with - 22

Smaller one 33. Remainder is 22. 11) 22 (2

STEP 2: -22

Divide 33 by first remainder we get 0

second remainder 11.

STEP 3:

Divide 22 with second remainder.

STEP 4:

The last divisor is 11 which is the HCF of 33 and 55.

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EXAMPLE 2:

Using the division method find the HCF of 16 and 20.

Steps for finding the HCF of 16

And 20 are:

Step 1:

Divide the larger number with 16) 20 (1

The smaller number. 4 is the last -16

remainder. 4) 16 (4

Step 2: -16

Divide 16 with first remainder 4. We 0

Get the answer 0.

As 4 is the last divisor so HCF of 16 and 20 is 4.

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EXAMPLE 3:

Using division method, find the HCF of 10, 15 and 25.

First find HCF of two numbers

Say 10 and 15. 10) 15 (1

-10

5) 10 (2

-10

0

The HCF of 10 and 15 is 5.

Now we find the HCF of 5 and 25. 5) 25(5

-25

0

So the HCF of 10, 15 and 25 is 5.

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Using division method, find the HCF of the following:

1) 18, 44 and 32 2) 33, 66 and 99

3) 48, 60 and 75 4) 24, 48 and 96

5) 18, 24 and 32 6) 27, 81 and 90

7) 39, 65 and 91 8) 34, 68 and 85

9) 18, 36 and 90

Exercise 2.2

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2.2LCM The term LCM stands for “Least Common Multiple”. LCM is the number which is the smallest common multiple of two or more numbers.

EXAMPLE:

Multiples of 4 shown on the line are 4, 8, 12, 16 and 20.

4 8 12 16 20

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

10 20

Multiples of 10 shown on the line are 10 and 20.

20 is common among the multiples of 4 and 10 represented by the line. So the LCM of 4 and 10 is “20”. 4 and 10 also have other common multiples such as 40, 60 etc. but they are not the common smallest multiples.

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STEPS:

Steps involved in finding the LCM of two or more numbers are:

EXAMPLE:

Using Prime Factorization find the LCM of the following numbers:

1) 72 and 32 2) 56 and 88

3) 16, 36 and 40

1) 72 and 32


Factorization of numbers

Product of common factors

Product of uncommon factors

Product of common factors x product of uncommon factors =

LCM

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Factors of 32=2x2x2x2x2

Product of common factors=2x2x2=8

Product of uncommon factors=2x2x3=12

=8x18=96

The LCM of 72 and 32 is 96.

2) 56 and 88




Product of uncommon factors=7x11=77

=8x77=616

The LCM of 56 and 88 is 616.

3) 16, 36 and 40





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Product of uncommon factors=2x3x3x5=40

=8x40=98

CALCULATING THE LCM OF FOUR NUMBERS UPTO 2 DIGITS WITH THE HELP OF PRIME FACTORIZATION METHOD:

EXAMPLE 1:

Using prime factorization find the LCM of 80, 70, 60 and 50.

Factors of 80=2x2x2x2x5

Factors of 70=2x5x7


Factors of 50=2x5x5


Product of uncommon factors=2x2x3x5x7=420

=20x420=8400

EXAMPLE 2:

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Using prime factorization find the LCM of 14, 16, 18 and 20.

Factors of 14=2x7


Factors of 18=2x3x3

Factors of 20=2x2x5

Product of common factors=2x2=4

Product of uncommon factors=2x2x3x3x5x7=1260

=4x12

Using prime factorization find the LCM of the following:

1) 25, 55 and 75 2) 32, 46 and 78 3) 10, 12 and 19

4) 18, 21, 27 and 30 5) 32, 36, 40 and 44

6) 18, 30, 42 and 54 7) 16, 18, 20 and 24

8) 66, 55, 44 and 33 9) 72, 62, 52 and 42

Exercise 2.3

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CALCULATING THE LCM OF NUMBERS UP TO 2 DIGITS WITH THE HELP OF DIVISION METHOD:

EXAMPLE 1:

With the help of division method find the LCM of 40, 68 and 70.

2 40, 68, 70

2 20, 34, 36

2 10, 17, 18

3 5, 17, 9

3 5, 17, 3

5 5, 17, 1

17 1, 17, 1

1, 1, 1

LCM=2x2x2x3x3x5x17=6120

EXAMPLE 2:

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With the help of division method find the LCM of 32, 46 and 54.

2 32, 46, 54

2 16, 23, 27

2 8, 23, 27

2 4, 23, 27

2 2, 23, 27

3 1, 23, 27

3 1, 23, 9

3 1, 23, 3

23 1, 23, 1

1, 1, 1

LCM=2x2x2x2x2x3x3x3x23=19872

EXAMPLE 3:

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With the help of division method find the LCM of 18, 20, 50 and 96.

2 18, 20, 50, 96

2 9, 10, 25, 48

2 9, 5, 25, 24

2 9, 5, 25, 12

2 9, 5, 25, 6

2 9, 5, 25, 3

3 3, 5, 25, 1

5 1, 5, 25, 1

5 1, 1, 5, 1

1, 1, 1, 1

LCM=2x2x2x2x2x3x3x5x5=7200

Using division method, find the LCM of the following:

Exercise 2.4

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1) 14, 15 and 16 2) 25, 56 and 63

3) 27, 68 and 92 4) 30, 72, 88 and 93

5) 33, 56, 65 and 81 6) 36, 46, 56 and 66

7) 50, 55, 60 and 65 8) 12, 22, 32 and 42

9) 38, 45, 56 and 63 10) 27, 35, 40 and 55

11) 39, 50, 63 and 77 12) 14, 54, 74 and 94

WORD PROBLEMS

EXAMPLE 1:

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Find the least number of chocolates required to make stacks of 4, 16, and 10 chocolates.

SOLUTION:

2 4, 16, 10

2 2, 8, 5

2 1, 4, 5

2 1, 2, 5

5 1, 1, 5

1, 1, 1

LCM=2x2x2x2x5=80

Therefore, the least number of chocolates=80

EXAMPLE 2:

There are 24 and 38 candies which are to be packed. How much maximum number of candies will have to be packed in each packet of candies to get packets of equal number?

SOLUTION:


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Factors of 38=2x19

Common factor=2

HCF=2

Therefore, the maximum numbers of candies=2

1) What is the least number of books distributed among 25, 30 and 35 students so that no book is left?

2) There is 42 and 58 number of people to stand in groups. How maximum number of people will have to stand equally in each group?

3) There are 6, 12 and 18 bottles to be filled with water. How many liters of water is required to fill all the bottles?

4) There are 24 and 36 pencils to be distributed among students. How many more number of students are

Exercise 2.5

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required so that all pencils are distributed equally in each student?

5) Two bells rang at an interval of 30 and 40 minutes. If the bells rang together at 4 am then at what time will they next ring together?

THINKING ACTIVITY 1:

THINKING ACTIVIES

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Q1) Using prime factorization, find the HCF of the following:

Product of two numbers=LCM x HCF

Product of n numbers=LCM of n numbers x HCF of n numbers

Take two numbers and show that the product of the two numbers is equal to the product of their HCF and

What is the highest 3 digit number which is exactly divisible by 3, 5, 6 and 7?

What is the greatest number which exactly divides 110, 154 and 242?

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1) 30, 40 and 50 2) 16, 18 and 20

3) 34, 42, 56 and 66

Q2) Using division method, find the HCF of the following:

1) 48, 60 and 75 2) 18, 24 and 32

3) 33, 66 and 99

Q3) Using prime factorization, find the LCM of the following:

1) 22, 44, 66 and 88 2) 30, 35 and 40

3) 12, 14, 16 and 18

Q4) Using division method, find the LCM of the following:

1) 20, 22, 26 and 28 2) 70, 80 and 90

3) 32, 40, 28 and 56

Q5) Find the least number of chocolates required to make stacks of 4, 16 and 10 chocolates.

Q6) there are 6, 12 and 18 bottles which are to be filled with water. How many liters of water is required to fill all the bottles?

QUESTIONS

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Q) Fill in the blanks:

1) HCF stands for

2) LCM of 15 and 20 is

3) LCM is the number which is the

of two or more numbers.

4) To express a composite number as the product of two or more prime numbers or prime factors of that number is called

5) HCF of 30 and 35 is

6) Factors of 12 are

7) LCM stands for

8) LCM of 12 and 14 is

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HCF stands for highest Common Factor. It is the largest number with which if the given two or more numbers are divided then the remainder will always be zero. HCF and LCM are calculated by two methods:Prime FactorizationDivision method To express a composite number as the product of two or more prime numbers or prime factors of that number is called prime factorization. LCM stands for Least Common Multiple. LCM is the number which is the smallest common multiple of two or more numbers.

EXERCISE 2.1:

1) 7 2) 11 3) 6 4) 2 5) 21

6) 11 7) 2 8) 3 9) 10

EXERCISE 2.2:

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1) 2 2) 33 3) 3 4) 24 5) 2

6) 9 7) 13 8) 17 9) 18

EXERCISE 2.3:

1) 825 2) 1088 3) 240

4) 1995 5)15840 6) 1890

7) 1564 8) 660 9) 203112

EXERCISE 2.4:

1) 1680 2) 12600 3) 42228

4) 122766 5)3243240 6) 127512

7) 42900 8) 7392 9) 47880

10) 83160 11) 450450 12) 657342

EXERCISE 2.5:

1) 1050 2) 2 3) 36 liters

4) 12 5) 6am

REVIEW EXERCISE:

Q1) 1) 10 2) 2 3) 2

Q2) 1) 3 2) 2 3) 33

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Q3) 1) 528 2) 1680 3) 6048

Q4) 1) 20020 2) 5040 3) 1120

Q5) 80 Q6) 36

Blanks) 1) Highest Common Factor

2) 60

3) Smallest Common Multiple

4) Prime factorization

5) 5 6) 2, 2, 3

7) Least Common Multiple

8) 84

Unitary

Method

Chapter 3

~ 61 ~

AFTER THE INSTRUCTION STUDENTS WILL BE ABLE TO

~ 62 ~

Define unitary method. Memorize the concept of unitary method. Explain unitary method. Recall how to use the value of one object to calculate the value of similar objects. Recall how to use the value of given number of same type of objects to find the value of another number of same type of objects. Calculate the value of similar objects when the value of one object is given. Calculate the value of another number of same type of objects when the value of given number of same type of objects is given.

UNITARY METHOD

~ 63 ~

Unitary Method is the method in which the value of a single or unit object is used to find or calculate the number of similar objects.

6.1 USE THE VALUE OF ONE OBJECT TO CALCULATE THE VALUE OF SIMILAR OBJECTS:Such as:

If the price of one ball is 10 rupees then the price of 5 balls will be 50 i.e.

Price of one ball=Rupees 10

Price of 5 such balls=Rupees 10 x 5=Rupees 50

EXAMPLE 1:

The cost of one ring is Rupees 250. What is the cost of 2 such rings?

Cost of one ring=Rupees 250

~ 64 ~

Cost of 2 such rings=Rupees 250 x 2=Rupees 500

EXAMPLE 2:

The price of a watch is Rupees 3000. What would be the price of 3 such watches?

Price of one watch=Rupees 3000

Price of 3 such watches=Rupees 3000 x 3=Rupees 9000

EXAMPLE 3:

The cost of a bag is Rupees 800. What would be the cost of 2 such bags?

Cost of one bag=Rupees 800

Cost of 2 such bags=Rupees 800 x 2=Rupees 1600

~ 65 ~

6.2 USE THE VALUE OF GIVEN NUMBER OF SAME OBJECTS TO FIND THE VALUE OF ANOTHER NUMBER OF SAME TYPE OF GIVEN OBJECTS:Such as:

The total cost of 2 books is Rupees 300. If we buy 3 such books we will have to pay four hundred and fifty rupees.

Cost of 2 books=Rupees 300

Cost of 1 such book=Rupees 300 ÷ 2=Rupees 150

Therefore, the cost of 3 such books=Rupees 150 x 3

=Rupees 450

EXAMPLE 1:

The price of 10 shirts is Rupees 1000. What would be the price of 15 such shirts?

Price of 10 shirts=Rupees 2000

Price of 1 such shirt=Rupees 2000 ÷ 10=Rupees 200

~ 66 ~

Price of 15 such shirts=Rupees 200 x 15=Rupees 3000

EXAMPLE 2:

The amount of water in 4 bottles is 12 liters. What would be the amount of water in 6 bottles?

Amount of water in 4 bottles=12 liters

Amount of water in 1 bottle=12 liters ÷ 2=3 liters

Amount of water in 6 such bottles=3 liters x 6=18 liters

1) If 3 men ate 12 cupcakes. How many cupcakes will be eaten by 6 men?

2) One compartment of a train has 24 seats. How many seats are there in 10 compartments?

3) 4 students have 24 books. How many books does each student have?

Exercise 6.1

~ 67 ~

4) Ali paid an electricity bill of rupees 240000 in one year. How much bill he paid per month?

5) Aleena did her homework of 50 pages in 2 hours. How much hours will it take to complete the homework of 150 pages?

6) There are 90 students in 3 classes. How many students are there in 5 classes?

7) A man ran 36 kilometers in 2 hour. How much distance will he cover in 5 hours?

8) The cost of a dozen mangoes is Rupees 60. What would be the cost of 8 mangoes?

~ 68 ~

ACTIVITY 1:

ACTIVITIES

Try to understand unitary method by every day experience. For example go to the nearest shop and ask the price of a dozen eggs and then find the price of each single egg. Do similar activity with other things and make a list of those

things and their prices and the price of a single thing which you have calculated.

~ 69 ~

ACTIVITY 2:

Q1) the rent of a house is 5000 for a month. What would be the rent of the house for a year?

Q2) the cost of 10 shirts is Rupees 2000. What would be the cost of 15 such shirts?

Q3) the price of 12 packs of chips is 180. Find the price of 20 such packets.

Make groups of students having 6 students in each group. Then ask the students to write down 10 examples

related to unitary method.

REVIEW

EXERCISE

~ 70 ~

Q4) there are 90 students in 3 classes. How many students are there in 5 classes?

Q5) the price of one chocolate is Rupees 15. What would be the cost of 5 such chocolates?

QUESTIONS

Q6) MCQs:

1) If 3 men ate 12 cupcakes. How many cupcakes will be eaten by 6 men?

a) 18 b) 24 c) 36

d) 72

2) If the price of one ball is Rupees 10. What would be the price of 5 such balls?

a) 55 b) 15 c) 50

d) 105

3) Ali reads 14 pages of a book in 2 days. How many pages will he read in 20 days?

a) 140 b) 40 c) 28

d) 120

~ 71 ~

4) A man ran 36 km in 2 hours. How much distance will he cover in 5 hours?

a) 86 km b) 50 km c) 10 km

d) 90 km

5) The price of 1 glass is Rupees 50. What would be the price of 6 such glasses?

a) Rupees 300 b) Rupees 350 c) Rupees 320

d) Rupees 340

6) The total cost of 2 books is Rupees 300. If we buy 3 such books how much do we have to pay?

a) Rupees 600 b) Rupees 450 c) Rupees 900

d) Rupees 650

~ 72 ~

Value of a single object can be used to find the value of similar objects.Unitary method is a method in which the value of a single or unit object is used to find or calculate the number of similar objects.We can use the value of a single object to calculate the value of similar objects similarly we can also use the value of given number of same objects to find the value of another number of same type of given objects.

1) 24 2) 240 3) 6 4) 20,000

~ 73 ~

5) 3750 6) 150 7) 90 8) 40

REVIEW EXERCISE:

Q1) 60,000 Q2) 3000 Q3) Rupees 300

Q4) 150 students Q5) Rupees 75

MCQs) 1) b 2) b 3) a

4) d 5) a 6) b

Unitary method

After studying this unit , the the students will be able to :

Describe the concept of unitary method.

Define ratio of two numbers.

Define and identify direct and inverse method.

Solve real life problems with help of unitary method.

~ 74 ~

Describe the concept of unitary method.

Define ratio of two numbers.

Define and identify direct and inverse method.

Solve real life problems with help of unitary method.

Direct and Inverse proportion

Ratio

A ratio is relationship between two quantities f the same kind. The ratio of the two quantities a and b (b is not equal to 0)and read as “a is to b” .a:b is also written as a/b.

Proportion

If a:b and c:d are two ratios , then the proportion between these two ratios is written as:

a:b::c:d or a:b=c:d

Example

~ 75 ~

6.2

X:6::70:14

x/6=70/14

x=70/14.6 x=30ans

Exercise no. 3.2

1. The price of 100 bags is RS. 3000. What would be the price of 40 bags?

2. What is the value of x in the proportion 10:20::4:x?3. The price of 20 toy cars is 600.How many cars can be

purchased for RS.1200?4. A motor cycle covers 100kmiin 4 liters of petrol .In how

many liters of petrol will it cover250km?5. 6 men can paint a house in 4 days. How would it take to

paint the house by 12 men?

~ 76 ~

Ex no. 3.1

1. 12. 9/43. 3/74. 15/8 5. 4/5

Question no. 2 1. 5/22. 5/23. 2/3

1. The price of 100 bags is RS. 3000. What would be the price of 40 bags?

2. What is the value of x in the proportion 10:20::4:x?3. The price of 20 toy cars is 600.How many cars can be

purchased for RS.1200?4. A motor cycle covers 100kmiin 4 liters of petrol .In how

many liters of petrol will it cover250km?5. 6 men can paint a house in 4 days. How would it take to

paint the house by 12 men?

Solution

Fraction

~ 77 ~

4. 3/25. 4

Question no. 3

1. 1 2. -23. 14. 2 5. -5/76. -4/3

Ex no 3.2

1. 1 11/182. 33/453. 13/154. 1 ¾5. 2 13/156. 37. 13/188. 1 1/119. 1 5/42

Ex no. 3.3

1. 1/32. -1/423. 1 15/364. 13/255. ½6. 5/8

~ 78 ~

7. 18. -1 13/219. -4/21

Ex no. 3.4

1. 1 1/32. 2/3 3. 1 3/54. 1 2/3 5. 1 1/36. 167. 1 7/28. 1 3/89. 3

Ex no. 3.5

1. 3/82. ¾3. 1 1/64. 4/215. 146. 5 4/77. 8/218. 19. 14

Ex no 3.6

1. 1/9

~ 79 ~

2. 63. 9/44. 1/85. 89 2/76. 3 59/60

Ex no. 3.8

1. 57962. 230.43. 15504. 99/45. 406. 607. 72

Ex no. 3.91. 1/92. 1/93. 1/34. 1 1/35. 16. 4/97. 1/88. 1/359. 1/1210. 1/611. 1/8812. 31/513. 16/25

Ex no. 3.10

~ 80 ~

1. 6/252. 2 2/93. 6/354. 36/1755. 46. 8/97. 1 1/38. 9/209. 8/910. 25/259211. 1 9/212. 4 4/ 1713. 1/1214. 2 1/6

Ex no 3.11

1. 62. 24 1/33. 1 hour 15 min 4. 245. 1/456. 5 11/15

Ex no. 3.12

1. 11/502. 1 1/193. 3 213/244. 485. 1 4/56. 137 1/12

~ 81 ~

7. 63 ½8. 28/2259. 4810. 31 ¼

1. 12002. 83. 404. 10km5. 2days

7500

Unitary Method

Chapter 4~ 82 ~

~ 83 ~

Specific Objectives:

In this chapter we will study and will be able to solve:

1. What is the Fraction.2. A definition of fraction.3. Addition and subtraction of the Fraction with the

same denominator.4. Addition and subtraction of different denominator.5. How to solve the Fraction with the help of a diagram.6. Multiplication of the Fractions.7. How to solve the mix Fraction 8. Solution of Fraction involving brackets.9. Verification of Communicative Law and associative

Law of Fraction.10. Problem involving Fraction.11. Division of Fraction.12. Application of Fraction in real life.13. use BODMAS rule.

)

Fraction

Definition of Fraction

~ 84 ~

A numerical quantity in the form of a/b, b=0 is called Fraction.

Example 2/3, 4/5, 1 3/2 mixed Fraction.

It is read as one whole num three over two

“The type of ability test that describes what a person has learned to do.”

Addition and Subtraction of the same denominator

Examples of Fraction having same denominator

~ 85 ~

Denominator is same(6)

2/5, 9/5, 6/5, 3/5

Denominator is same (5)

1. 1/4+3/4 2.2=4

= 1+3

2/6, 5/6, 3/6, 11/6

Examples

Solve

~ 86 ~

4

=4/4 =1ans.

2. 1/6+2/6+3/6

1/6+2/6+3/6 2.3=6

= 1+2+3

6

=6/6 =1ans

3. 3 1/3 ,2/3, 4/3

1/3,2/3,4/3 3.1=3 = 1+2+4 =7/3ans

3

1. 1/2, 1/2.2. 3/4 ,6/4.3. 1/7 ,2/74. 6/8 ,9/85. 3/5 ,1/5

Add the following:

Solve the following:

Exercise no. 4.1

~ 87 ~

1. 1/2 ,3/2 , 1/22. 3/4, 1/4, 6/43. 2/9 , 3/9, 1/94. 3/6 , 2/6 ,4/65. 6/3 ,9/3 , 2/3

1. 1/2 – 2/2 +3/22. 3/2 -6/2 – 1/23. 3/4- 1/4 + 2/44. 1/5 + 2/5 +3/55. 2/7 -1/7- 6/76. -1/6 -3/6 -4/6

Solve the following:

~ 88 ~

Addition and Subtraction of different denominator

Examples

1. 1/3 + 1/4 LCM

3.4=12

= 4+3

12

= 7/12 ans

2. 5/3+ 2/7 LCM

3.7=21

= 35+21

21

= 56/21 2 2/3ans

~ 89 ~

1. 1/3 + 1/4 LCM

3.4=12

= 4+3

12

= 7/12 ans

2. 5/3+ 2/7 LCM

3.7=21

= 35+21

21

= 56/21 2 2/3ans

3. 1/4 -1/2

LCM

2.2=4

= 1-2

4

=-1/4 ans

4. 3/6 -4/2

LCM

2.3=6

= 3-12

6

= 9/6 3/2 ansExercise no. 4.2

~ 90 ~

Solve

1. 1/9+ 3/6+3/22. 2/15+1/9+2/53. 1/5+1/6+1/24. 3/5+2/5+3/45. 3/6+1/5+7/26. 1/6+5/2+3/47. 2/9+1/6+1/38. 2/6+1/4+1/39. 1/6+2/3+3/7

Subtraction of two and more Fractions with different denominator

Examples

~ 91 ~

1. 6/9-1/2

LCM 3.3.2=18

= 12-9

18

=3/18

=1/6 ans

2. 6/3-1/2

LCM

3.2=6

= 12-6

6 =1ans

Exercise no. 4.3

~ 92 ~

Solve

1. 19/21-4/72. 25/30-6/73. 6/3-3/12-2/64. 23/25-3/15-1/55. 1/2-3/6-7/46. 5/6-1/8-1/127. 7/2-6/3-2/48. 1/6-2/7-3/29. 1/6-2/7-3/1

1/2 2/4 3/4 4/4

Multiply a Fraction by a number with the help of diagram

Example

~ 93 ~

Use the diagram to solve the following:

1/4.3

=

1.8.3/6

3.8/6 = 24/6

=4ans

2.2/9.3

2.3/9 =6/9

2/3ans

Examples

Solve the following

~ 94 ~

1. 2/6.42. 3/9.23. 8/15.34. 2/6.55. 6/3.26. 4/2.87. 5/2.38. 5.5/89. 3/7.7

Solve

~ 95 ~

Solve

1. 3/2.1/4

Multiplying a Fraction by another form

Solution

1. 5/2.6/3

5.6/2.3 =30/6 =5ans

2. 3/4.2/7

3.2/4.7 =6/28 =3/14ans

Exercise no 4.5

~ 96 ~

2. 6/3.1/43. 7/9.3/24. 4/7.2/65. 3/7.1/66. 6/7.11/27. 16/21.3/68. 3/2.2.39. 14/3.6/2

Multiplying two or more Fraction involving brackets

Examples

~ 97 ~

Solve

1. 1/4.3/2.2/72. 2/4.6/1.4/23. 2/8.6/1.3/24. 3/2.4/8.1/65. (1 3/2.2. 6/7).2 5/2

Examples

1.(1/2.6/4).5/4 2. (4/2.6/3).1/6

= (1/2.6/4).5/4 = (4.6/3).1/6

= (1.3/4).5/4 = 8.1/6

= 3/4.5/4 = 8/6

= 15/16ans = 1 1/3ans

Exercise no. 4.6

~ 98 ~

6. (1 1/12. 3 .5/7). 2 1/5

Distributive Laws

Show that

1. 1/3(2/5+3/4)=(1/3.2/5)+(1/3.3/4)

LHS RHS

1/3(2/5+3/4) =(1/3.2/5)+(1/3.3/4)

1/3(8+15/20) = (2/15+3/12)

1/3(23/20)

~ 99 ~

8/2

1. 1/3(2/5+3/4)=(1/3.2/5)+(1/3.3/4)

LHS RHS

1/3(2/5+3/4) =(1/3.2/5)+(1/3.3/4)

1/3(8+15/20) = (2/15+3/12)

1/3(23/20)

2. 2 4/2.6/2.5/6

= 8/2.6/2.5/6

=(8/2.6/4)+(8/2.5/6)

=12+20/6

=32/6

2 5/6ans

3. ½.(3/4+5/6)=(1/2.3/4)+(1/2+5/6)

½(9+10/12)=(3/8+5/12)

19/24 =9+10/24

19/24 =19/24ans

~ 100 ~

Exercise no. 4.7

Verify Distributive Law

~ 101 ~

1. 3/5(1/2+2/7)=(3/5.1/2)+(3/5.2/7)2. 16/3(9/2+8/9)=(16/5.9/2)+(16/5.8/9)3. 1/2(3/4+5/6)=(1/2.3/4)+(1/2.5/6)4. 4/2(2/3-3/6)=(4/2.2/3)-(4/2.3/6)5. 16/2(9/2.-8/9)=(16/2.9/2)-(16/2.8/9)6. 2/4(9/2+8/9)=(2/4.9/2)+(2/4+8/9)

Solve the real life problems involving Multiplication of Fraction

Examples

1. Cost of 1 kg sugar is RS.60.Find the cost of 15 ½ kg sugar.

Cost of 1 kg sugar=RS.60 (15 ½ 31/2)

Cost of 15 1/2 kg sugar=RS. 60.31/2

=60.31/2

=930ans

~ 102 ~

1. Cost of 1 kg sugar is RS.60.Find the cost of 15 ½ kg sugar.

Cost of 1 kg sugar=RS.60 (15 ½ 31/2)

Cost of 15 1/2 kg sugar=RS. 60.31/2

=60.31/2

=930ans

Num of the boys=3/5 of 25

=3/5.25=15

Num of girls=25-15 =10ans

Exercise no. 4.8

1. The cost of a Mathematic book is RS. 92.Find the cost of 63 such books?

2. The cost of 1 meter cloth is RS. 16 .Find the cost of 14 2/5 meter of cloth?

3. The cost of 1 liter petrol is RS.100.Find the cost of 15 ½ liter of petrol?

~ 103 ~

7.

1. The cost of a Mathematic book is RS. 92.Find the cost of 63 such books?

2. The cost of 1 meter cloth is RS. 16 .Find the cost of 14 2/5 meter of cloth?

3. The cost of 1 liter petrol is RS.100.Find the cost of 15 ½ liter of petrol?

Parked at a time?

6. A race car covers 1300 km in an hour How much distance will it cover in 4 5/6 hour?

7. a bag can hold 108 kg sugar. How much sugar will be in the bag when it is filled 6/9?

~ 104 ~

Divide a Fraction by a number

Examples

1. 12/3 /6 2. 3/8 /9/4

=12/3 . 1/6 = 3/8.4/9

=2/3ans = 1/6ans

3 ¼ / ½ =1/4.2/1 =1/2ans

Exercise no. 4.9

~ 105 ~

1. 1/3 /3 10. 6/12 / 32. 2/6 / 3 11. 9/11 / 723. 4/3 / 8 12. 62/5 / 24. 8/2 / 3 13. 6/77 / 115. 6/2 / 3 14. 6/25 /5 6. 8/9 / 27. ¾ / 68. 5/7 259. 11/6 / 22

Divided a Fraction by another Fraction

~ 106 ~

Same method is used in case of division of a Fraction by another Fraction.

Solve

1. 36/5 / 9/2 2. 2/8 / 2/3

=36/5 / 2/9 =2/8 / 3/2

=24/155 =3/8ans

1 3/5ans

Exercise no. 4.10

1. 6/15 / 5/2 10. 25/36 / 72/12. 10/6 / ¾ 11. 85/6 / 5/33. 6/7 / 5/1 12. 12/17 / 1/64. 36/70 / 5/2 13. 3/6 / 12/25. 8/4 / ½ 14. 8/4 / 3/26. 36/25 / 2/5

~ 107 ~

1. 6/15 / 5/2 10. 25/36 / 72/12. 10/6 / ¾ 11. 85/6 / 5/33. 6/7 / 5/1 12. 12/17 / 1/64. 36/70 / 5/2 13. 3/6 / 12/25. 8/4 / ½ 14. 8/4 / 3/26. 36/25 / 2/5

Solve real life problem involving division of Fraction

Examples

A rope of length 37 ½ m s to be cut into pieces each of length 2 ½. How many pieces can be made?

Num of pieces=37 ½= 75/ 2

~ 108 ~

requir

A rope of length 37 ½ m s to be cut into pieces each of length 2 ½. How many pieces can be made?

Num of pieces=37 ½= 75/ 2

Exercise no. 4.11

1. Asma can iron 10 shirts in 60 min. How long will she take to iron one shirt?

2. A jug contains 3 liters milk. How many jugs are required to hold 24 liter milk?

3. A student takes 5 minutes to solve a problem. How much time will he take to solve 15 question?

4. One cahpati takes 3 min to cook.How many time 8 chapati will take to cook?

5. Iqra can read 30 pages of a book in 2/3 of an hour. How many minutes does she take to read one page?

6. Zainab walks 3/10km to the bus stop. Then she takes a bus to ride 5 1/3km followed by a walk of 1/10km. What is the distance of her school from home?

~ 109 ~

1. Asma can iron 10 shirts in 60 min. How long will she take to iron one shirt?

2. A jug contains 3 liters milk. How many jugs are required to hold 24 liter milk?

3. A student takes 5 minutes to solve a problem. How much time will he take to solve 15 question?

4. One cahpati takes 3 min to cook.How many time 8 chapati will take to cook?

5. Iqra can read 30 pages of a book in 2/3 of an hour. How many minutes does she take to read one page?

6. Zainab walks 3/10km to the bus stop. Then she takes a bus to ride 5 1/3km followed by a walk of 1/10km. What is the distance of her school from home?

Use of BODMAS Rule

As we know that BODMAS stands for Brackets, Division, Multiplication, Addition and Subtraction.

1. 1 ½+{(4 2/3 / 2)-2/3} =10/3

=5/3+{(14/3 / 2/1)-2/3} = 3 1/3ans

=5/3+{(14/3 / ½ )-2/3}

=5/3+{7/3-2/3}

=5/3+{7-2/3}

=5/3{5/3}

~ 110 ~

(

Exercise no. 4.12

1. ½ (1/2+ 3/5)/ 5/2 9. 24+[3.{10 ½-(5/6+1/3)}]2. (1/3+1/2)/ (2/3.1/8) 10. 4 ½ +{(3 3/5+1 ¾).5}3. 1 ¾(4/9+2/3)/(1 1/5. ½)4. 24+[3.{10 ½ -(5/6/1/3)}]5. 1 2/4 (4/3+2/3)/ 1 1/5 .1/2)6. (1 ¼ .2 2/3).(3 1/3+2. ½)7. 20+[5.{9-(1 2/3. 1/5)}]8. 1/3.(1/3+3/5)/5/2

Review Exercise

~ 111 ~

1. 1/6+2/6+3/62. 1/7+2/7+4/73. 1/3+1/44. 1/9+3/6+2/35. 1/6+2/3+3/76. 19/21-4/77. 25/30-6/78. 1/6-2/7-3/19. 8.3/610. 2/6.4


11./3(2/5+3/4)=(1/3.2/5)+(1/3.3/4)

~ 112 ~

MCQ’S


11./3(2/5+3/4)=(1/3.2/5)+(1/3.3/4)

1. 1/6+2/6+3/6=

a) 2 b)1 c)4 d)5

2. 1/3+2/3+4/3=a) 4/5 b)2/7 c) 7/3 d)3

3. 1/13-6/13+5/13=a) -23 b) -10/13 c)23 d)5/7

4. 7/2-6/3-2/4a)12/3 b)1 c)23 d)3

5. 3/4+3/4+3/4=

a) 3 b) 2 c) 4 d)6

6. 8/15.3=

a) 34 b)25 c)1 2/3 d) 76

7. 3/7.7=

a)45 b)78 c)3 d)9

8. 3/2.2/3=

a)2 b)2 c) 4 d)1

9. 16/21.3/6=

a)8/21 b)4 c)98 d)34

10. 6/3.2= a)2 b)1 ½ c)23 d)67

~ 113 ~

6. 8/15.3=

a) 34 b)25 c)1 2/3 d) 76

7. 3/7.7=

a)45 b)78 c)3 d)9

8. 3/2.2/3=

a)2 b)2 c) 4 d)1

9. 16/21.3/6=

a)8/21 b)4 c)98 d)34

10. 6/3.2= a)2 b)1 ½ c)23 d)67

Fill in the blanks

1. By adding Fraction wit different denominators, we change these Fraction with

Denominator.

2. In subtraction of Fraction we find

Fraction with their LCM as

3. We find common denominator which is divided by all the original

4. BODMAS Rule stands for

5. find the LCM 2/6+1/4+1/8

~ 114 ~

1. To add Fractions with different denominators first we change these Fractions equivalent Fraction with the same denominators.

2. To add two and more than two Fraction we need to find a common denominators which is divisible by all the original denominators.

3. To subtract Fraction with different denominators we find equivalent Fraction with LCM as small denominator.

4. To subtract more than two Fraction we need to a common denominator which is divisible by all the original denominators.

Summary

~ 115 ~

5. Two multiply a Fraction by a whole number multiply the numerator by the whole number and keep the denominator as it is.

6. To multiply one Fraction by another Fraction , multiply the numerators with numerators and the denominators with denominators.

7. We can some times simplify a product by cancelling the common Fraction.

8. To multiply more than two Fractional numbers we multiply their numerators to get numerator and multiply the denominators to get the denominator of the require product.

9. To divide a Fraction by a non-zero whole number, we multiply the Fraction by the reciprocal of the whole number.

10.BODMAS stands for brackets, addition, subtraction, multiplication and division.

~ 116 ~

Decimals and

Chapter:

~ 117 ~

Author:

Hafsa Tahir

2011-1436

Instructional Objective

After studying this chapter, student will be able to

1. Perform addition & subtraction.2. Multiply diverse levels of decimals.3. Renovate decimals into fractions and

vice versa4. Round off decimal numbers.

~ 118 ~

Any number containing a useful component specified by a decimal position is known as a decimal number or decimal.

Let’s suppose a number that is 2345 and in this we look at the value of the place of every digit.

After studying this chapter, student will be able to

1. Perform addition & subtraction.2. Multiply diverse levels of decimals.3. Renovate decimals into fractions and

vice versa4. Round off decimal numbers.

4.1 Decim

~ 119 ~

Thousands Hundreds Tens Ones2 3 4 5

The value of every place is 1/10th of each place to the left.

Now we look at example in decimal figures. E.g., 23.234

Tens Ones Decimal Tenths Hundredths Thousandths

2 3 . 2 3 4

The decimal point disconnects the whole number part from the fractional part.

So that according to above definition of Decimal it is shown that

“Every number containing a useful component specified by a decimal position”

Examples:-

3.2

133.2

123.32

902.2

1.234

These are examples of decimal numbers.

~ 120 ~

Examples:

1) .34=0.342) .23=0.233) .982=0.982

To add decimals we put in writing the figures one underneath the -supplementary, so that the decimal points are in row.

If a number consists of just one decimal part or a whole number part, then we obtain zero as whole number part or decimal part.

4.1.1 Add and Subtract

Adding Decimals

~ 121 ~

Solve it.

a. 23.67+5.9 b. 32.9 + 0.223

Solution Solution

23.67 32.9

+5.9 +0.223

29.57

Exercise 5.1

33.123

Examples

~ 122 ~

Solve.

1. 32.2 + 9.212. 23.4 + 22.233. 562.28 + 7.2834. 5.22 + 28.235. 2.113 + 83.226. 0.998 + 9.2937. 78.393 + 2.398. 22.90 + 0.9989. 3.90 + 23.76510. 24.22 + 0.876

Although subtracting decimals, be cautious to line up the decimal points. Then we carry out subtraction as typical.

Subtracting Decimals

~ 123 ~

Solve it.

i. 76.23 – 6.98 ii. 93.29 – 2.983

76.23 93.2900

- 6.98 -2.9837

We add zeros at the end if the other number have more digits than previous one.

Solve it.

Examples

69.28 90.3027

Exercise 5.2

~ 124 ~

1. 9.7 – 3.82. 0.82 – 0.433. 28.12 – 14.234. 4.12 – 1.15. 10.25 – 1.1656. 1.1157 – 1.11137. 8.4 – 1.048. 5.7 – 1.234

We identify that 2/9, 3/9, 4/9, 7/9 are like fractions, as all these are having the similar denominators.

2/3, 4/9, 7/8, 1/7 are unlike fractions, as they are having diverse denominators.

Equivalent fractions are also unlike fractions, just like 1/3 and 3/9 are equivalent fractions but they are unlike fractions.

Unlike fractions may or may not be equivalent fractions. Mean it considers both conditions.

5.1.3 Multiplying decimals by 10, 100 and 1000

5.1.2 Like and Unlike

~ 125 ~

Now there we do that how to multiply decimals with 10, 100 and 1000.

Look at the certain examples. Become aware of the example shown.

1. 0.00098 × 10 = 0.00982. 0.00098 × 100 = 0. 0983. 0.00098 × 1000 = 0.98

Multiplication by 10 transfers the decimal position 1 place toward right.

Multiplication by 100 transfers the decimal position 2 places toward right.

Multiplication by 1000 transfers the decimal position 3 places toward right.

We can load the vacant gap by zeros

1. 3.8 × 10 = 38.0 = 382. 3.8 × 100 = 380.0 = 3803. 3.8 × 1000 = 3800.0 = 3800

Solve it.

Exercise

~ 126 ~

1. 1.05 × 102. 2.15 × 103. 2.1 × 1004. 52 × 105. 10.53 × 1006. 35.16 × 10007. 12 × 108. 0.125 × 10

In previous exercise, we have done the multiplication with 10, 100 and 1000. Now we have done a division with all these same terms.

For this purpose, look at following examples.

1. 4.2 ÷ 10 = 0.422. 4.2 ÷ 100 = 0.0423. 4.2 ÷ 1000 = 0.0042

4.1.4 Dividing decimals by 10, 100, 1000

~ 127 ~

If we divide the number 10 it shifts decimal point towards left by 1 place.

If we divide the number 100 it shifts decimal point towards left by 2 places.

If we divide the number 1000 it shifts decimal point towards left by 3 places.

Solve these.

1. 1.5 ÷ 102. 120 ÷ 1003. 13.4 ÷ 10004. 4.24 ÷ 1005. 7.9 ÷ 106. 67 ÷ 1007. 765 ÷ 1008. 29.1 ÷ 10009. 9.872 ÷ 10010. 65.99 ÷ 10

Exercise 5.4

5.1.5 Multiply and divide a decimal by a

~ 128 ~

We had seen the multiplication and division of decimals with 10, 100 and 1000. Now let us do all these multiplication and division with whole number.

Multiplication of a decimal with a whole number

When we multiply a decimal by a whole number, there are as many decimal places in the product as there are in the decimals.

Let see some examples to clear the concept of it.

1. 0.2 × 3 = 3.22. 9.33 × 5 = 46.653. 0.008 × 4 = 0.032

Division of a decimal with a whole number

We can divide a decimal with the whole number in simple way which is usually used. But it should be necessary that all the decimal points come in a line.

Examples:

~ 129 ~

Solve:

a. 4.8 ÷ 5

Solution:

3.2

3 9.6

- 9

0 6

6

0

4.8 ÷ 5 = 3.2

~ 130 ~

1.Solve these by multiplication.

1. 0.9 × 32. 2.33 × 83. 0.78 × 54. 999.2 × 75. 29.3 × 12

2. Solve these by division.

1. 0.94 ÷ 92. 9.78 ÷ 83. 83.73 ÷ 24. 93.9 ÷ 65. 4.12 ÷ 6

Exercise 5.5

~ 131 ~

Multiplication of decimal by a decimal:

We can multiply a decimal with a decimal by the same process as that is in whole numbers.

For further understanding let see some of the examples.

Example:

Multiply 3.24 × 9.8

Solution

324

×98

2592

2916

5508

5.1.6 Multiply and Divide a decimal

~ 132 ~

Number of decimal places in 3.24 is 2 and in other 9.8 is also 1 so that total decimal places are 3.

As 2 + 1 = 3

So that answer is 5.508.

Division of decimal by a decimal:

We can divide a decimal by a decimal by converting it in fraction.

To understand it clearly let us do an example.

Example:

Solve 1.69 ÷ 1.3

Solution

1.69 ÷ 1.3= 1.69 × 1/1.3

= 169/100 × 10/13

= 13/10

= 1.3

Therefore, 1.69 ÷ 1.3 = 1.3

1. Solve.

1. 3.29 × 3.22. 0.234 × 4.2

Exercise 5.6

~ 133 ~

3. 98.22 × 9.84. 798.3 × 9.235. 3.98 × 9.87

2. Solve.

1. 32.908 ÷ 8.32. 0.8983 ÷ 44.53. 9.87 ÷ 5.74. 22.98 ÷ 9.83 5. 982.34 ÷ 5.94

We can simplify the decimal terms involving in brackets. We solve it step by step. In this we solve first of all small bracket then it comes curly bracket and then large bracket will be solved. While simplifying it, we will have to take great care in placing decimal points.

Example:

Solve: 2.9 - [6.8 + {2.2 ÷ (78.9 × 3.89)}]

= 2.9 - [6.8 + {2.2 ÷ 306.921}]

= 2.9 - [6.8 + 0.00717]

= 2.9 - 6.80717

= -3.90717

5.1.7 Simplifying decimal terms

~ 134 ~

Solve.

1. 5.4 - [8.762 + (92.8 × 4.22)]2. 3.9 + [1.876 + {5.2 + (78.9 × 3.89) }]3. {2.8 ÷ (3.99× 8.9) }h4. [7.7 + {3.99 - (5.4 × 3.89) }]5. 4.9 - {8.9 + (4.3 × 7.98) }

Exercise

5.1.8 Convert decimals to fractions

~ 135 ~

We have to convert decimals to fractions and convert fractions into decimals.

Alter decimals to fraction:

We have to convert the decimal to fraction. For this purpose we remove the point and write the value which comes after the removal of decimal point as a denominator. If both numerator and denominator are go at the same table. Then the value comes in fraction is the answer.

Let us understand it more clearly with the help of example.

Example:

Solve: 12.4

Solution:

12.4 = 124/10

= 62/5

~ 136 ~

It is in the lowest form.

Alter fractions into decimals:

For this purpose we have divide it completely and it converts into decimals.

To understand it let us do an example.

Example:

Solve: 2/10

Solution:

2/10 = 0.2

Exercise 5.8

~ 137 ~

Convert the following decimals into fraction

1. 0.892. 9.73. 0.774. 3.75. 0.006

Convert the following fractions into decimals.

1. 6/72. 5/93. 5/1004. 45/105. 35/1000

Activity

~ 138 ~

~ 139 ~

In this, statements are given through which we have to justify it that what operation has applied on it.

Let us understand it more clearly by the following example.

Example:

If the multiplication of two numbers are 5.43. If one number is 2.34 then find the other number.

Solution:

Multiplication of two numbers = 5.43

One number = 2.34

Other number = 5.43 ÷ 2.34

= 543/100 × 100/234

= 2.32051

5.1.9 Solve real life

Exercise 5.9

~ 140 ~

1. Fatima has 5.6 inches height while Arooj has 4.9 inches height. How much there is difference in both of them height?

2. Ali and Musa bought 4.1 kg bananas and 2 kg bananas respectively. Who bought more bananas? How much more bananas he bought?

3. Hadiqa has 3.8 liter petrol in her car. She has to go at khewra miles and for these propose she needs 5 liter petrol. How much petrol she needs to reach her destination.

4. Qaisar has 5.8 million dollars in his bank balance. He had purchased the shares in 1.2 million dollars. How many money is remained in his bank balance.

5. The price of dozen oranges is 20.3 rupees. Find the price of 60 oranges?

Percent is the value of any score or numbers from hundredth part of it.

~ 141 ~

Percentage is used to find the score of anything from hundred. Mostly it is used to measure the scoring of the students.

The symbol of percentage is %.

When we have to find the percentage of any fraction we multiply it with 100. As the answer come is its percentage.

Example:

Change 3/9 into percentage.

= 3/9 ×100

= 33.34%

33% means 33 out of 100 i.e., 33/10067% means 67 out of 100 i.e., 67/10043% means 43 out of 100 i.e., 43/100

When we have a value in a percentage and we have to convert it into fractions then we divide the percentage letter with 100. Then the operation of division is done. The answer comes in the form of fraction.

Let us understand it with more details with the help of example.

5.2.1 Percentage

5.2.2 Convert a percentage to a fraction

~ 142 ~

Example

1. Convert 40% into fraction. 40/100 = 2/5

2. 56%56% = 56/100

= 28/50 = 14/25

When we will have to convert a fraction into a percentage. We have to multiply the fraction with 100 to convert into percentages.

Write some percentages of ur own and convert it into fractions.

Activity:

5.2.3 Convert a fraction into a percentage

~ 143 ~

Let us do some examples to understand it more appropriately.

Examples

Convert the following into percentage

a. 4/5= 4/5 × 100= 80%

b. 1/7= 1/7 × 100=14.28%

1. Convert the following into fractions.

i. 34%

ii. 25%

A fraction is converted into percentage by

multiplying it with 100

Exercise 5.10

~ 144 ~

iii. 87%

iv. 98%

v. 19%

2. Convert the following into percentages.

i. 5/10

ii. 7/20

iii. 32/70

iv. 12/80

v. 7/100

If we have a percentage. We have to convert it into a decimal. We will remove the percentage sign first. Then we divide it with 100. Then the answer comes in decimals.

Let us consider it with some examples.

5.2.4 Convert a percentage into

a decimal

~ 145 ~

Examples

1. 33%= 33/100= 0.33

2. 45% = 45/100=0.45

In order to convert a decimal into percentage we should have to multiply this decimal with hundred.

Examples

1. 0.5×100=5/10 ×100

A percentage sign is replaced by

1/100

Convert a decimal to

~ 146 ~

=50%2. 1.2×100

=12/10×100=120%

1. Convert a following percentage into decimals

a. 75%b. 65%c. 38%d. 45%e. 99%

Percentage is the 1/100 part of a

number

Exercise 5.11

~ 147 ~

2. Convert a following decimal into percentage

a. 0.8b. 1.7c. 8.2d. 1.24e. 4.25

Interchange the following into percentages, decimals and fractions.

Activity

~ 148 ~

Solve real life problems relating

~ 149 ~

In this statements are given, according to which we have to justify that what the statement reflect. How we can solve it by applying varieties of operations including percentage.

Examples:

1. If Ayan bought 50 oranges and 15 oranges become spoiled. How much percentage of oranges will remain?

Solution:

Ayan bought oranges = 50 oranges

Oranges which spoiled = 15 oranges

Percentage of remaining oranges = 15/50 × 100

= 30%

2. Hamza has 200 cars if 20% cars are made of Japan then how many cars will from Pakistan.Solution:

Total cars hamza has = 200

How many made of Japan = 20% = 20/100

Pakistani cars are = 200 × 20/100

= 40 cars

Exercise 5.12

~ 150 ~

1. If 4/5 of the students in a school are present. What percentage of the students is not present?

2. If 82% of the houses have refrigerator in their houses. What percentage do not have refrigerator?

3. If 30% of the students in the school are boys. Then how many girls are in the school. If total students are 75.

4. If a book consists of 30 pages. Arooj reads 20 pages. How many percentage of pages will remains as unread.

1. Fill in the blanks.

Review Exercise

~ 151 ~

i. Any number containing a useful component specified by a –––––––––– is called a decimal.

ii. A percentage sign is replaced by ––––––––––iii.We can divide a decimal by a decimal by converting it in

––––––––––iv. To convert decimal in a fraction it will be in its –––––––– form.v. The decimal point separates the whole number part from a

–––––––––– part.

2. Multiple Choices.i. If a number have only a whole part number then we take

decimal asa. 1 c. 3b. 0 d. 5

ii. Like fractions are a. 3/9,2/3 c. 2/3,6/9b. 4/9,8/9 d. 7/7,8/8

iii. 9.6 ÷ 3 =a. 4.9 c. 3.2b. 2.3 d. 5

iv. Percent in the word ofa. Greek c. Persianb. Latin d. English

v. 60% is equal toa. 60/100 c. 6/10b. 1/100 d. 3/5

3. Solve these.i. 3.4 × 3.2ii. 9.87 ÷ 5.78iii. 7.98 + 3.98

~ 152 ~

4. Solve the following.i. 0.899 – 98.8ii. 3.98 + 4.87

5. Convert the following into decimals and fractions.i. 20%ii. 65%

6. Convert the following into percentages and fractions.i. 4.2ii. 3.8

7. Convert the following into decimals and percentages.i. 3/7ii. 2/9iii. 14/80

Answers Chapter: 5

Exercise 5.1

1. 41.412. 45.633. 569.563

~ 153 ~

4. 33.455. 85.3336. 10.2917. 80.7838. 23.8989. 27.66510. 25.096

Exercise 5.2

1. 5.92. 0.393. 13.894. 3.025. 9.0856. 0.00447. 4.466

Exercise 5.3

1. 10.52. 21.53. 2104. 5205. 10536. 351607. 1208. 1.25

Exercise 5.4

1. 0.15 2. 1.2

~ 154 ~

3. 0.01344. 0.04245. 0.796. 0.677. 7.658. 0.02919. 0.0987210. 6.599

Exercise 5.5

1. 1. 2.72. 18.643. 3.94. 6994.45. 3501.62.1. 0.1042.2. 1.22252.3. 401.862.4. 15.652.5. 0.686

Exercise 5.6

1. i.10.528 ii.0.9828iii. 962.556iv. 7368.309v. 39.2826

2. i. 3.96482

ii. 0.02019

~ 155 ~

iii. 1.731

iv. 2.33v. 165.378

Exercise 5.7

1. -394.9782. 317.8973. 12.68254. 85.42335. 38.314

Exercise 5.8

1. i. 89/100

ii, 97/10

iii. 77/100iv. 37/10v. 6/1000

2. i. 0.8571ii, 0.5555

iii, 0.05

iv,4.5

v, 0.035

Exercise 5.9

1. 0.72. Ali,2.13. 1.2

~ 156 ~

4. 4.6 million dollar5. 4.06

Exercise 5.10

1. i.17/50iii. ¼iv. 87/100v. 49/50vi. 19/100

2. i. 50%ii. 35%iii. 45%iv. 15%v. 07%

Exercise 5.11

1. i. 0.75ii. 0.65

iii, 0.38

iv, 0.45

v, 0.99

2. i. 80ii, 170iii, 820iv, 124v, 425

Exercise 5.12

~ 157 ~

1. 80%2. 18%3. 22%4. 67%

Review Exercise

1. i. decimal point ii. 1/100

iii, fraction v. deciaml

vi. fractional2. i. (b) iii. (b)

ii. (c) v. (d)iv, (b)

3. i. 10.88vi. 1.70761vii. 11.964. i. -97.901

ii. 8.855. i. 1/5, 0.2

ii. 13/20, 0.656. i. 420%, 21/5

ii. 380%, 19/57. i. 0.428, 42%

ii.0.22, 22%iii, 0.175, 17.5%

~ 158 ~

H A P T E RC

6

GEOMETRY

In this chapter you will learn how to:

Identify different types of angles Calculate unknown angles Define triangle and its types Construct adjacent angles Recognize the kinds of quadrilateral (square, rectangle,

parallelogram, rhombus, and trapezium) Use protractor and compass to construct various types of

triangles

~ 159 ~

A line is the path described by a moving point. It is a straight path passing through two points and extends in both directions forever. Here it is labeled as QR

Identify different types of angles Calculate unknown angles Define triangle and its types Construct adjacent angles Recognize the kinds of quadrilateral (square, rectangle,

parallelogram, rhombus, and trapezium) Use protractor and compass to construct various types of

triangles

HERE MR. DONALD DUCK IS GOING TO TEACH YOU ABOUT LINES, RAYS, LINE SEGMENTS

LINES

~ 160 ~

The diagram drawn below show parts of lines with only one end-point and it always extend in one direction. We call these rays.

The following diagram drawn above is ray EF.

A straight line segment is formed when we use a ruler to join two points, say. C and D.

We call the line segment CD or DC.

RAYS

Line segments

~ 161 ~

A flat surface without boundaries,labeled by naming three nonlinear points on the plane. Planes are of two types,

Vertical plane Horizontal plane

PLANE

Our dear Mickey Miney is asking that can you give the examples of vertical plane and horizontal plane?

~ 162 ~

The given below is an example of plane GHI.

Lines that lie on the same plane and never intersect, labeled as

Our Math book has answered confidently that yes I can. The floor of classroom is an example of a horizontal plane and the wall of the classroom is an example of vertical plane

Parallel lines

~ 163 ~

Lines that intersect at a 90° angle, labeled as

Perpendicular lines

Intersecting lines

~ 164 ~

The given below figure shows two lines AB and CD, on the same plane having a common point X. We say that the two intersect at X. So the point X is called as point of intersection.

C

A

O

D B

An angle is the space or distance between two rays at the point at which they meet. The main point at which they meet is known as vertex.

ANGLES

~ 165 ~

B

angle

O A

The angle is called as angle AOB or angle BOA.

How many different angles are given in the following figure?

For your information

The system of naming angles was first used

By the Babylonians (3000 – 2000 BC)

Explorations

~ 166 ~

By definition, one complete rotation about a point has an angle of 360⁰

In the above figure the picture of protractor is given. It is showing with the blue line that the sum of all angles in a protractor measures to 180⁰. It is used to measure angles. The angle of 60⁰ has been

The use of protractor

~ 167 ~

measured in the above figure. To measure an angle, place the protractor so that its center B is at the vertex of the angle and its base BA along one side of the angle. Note under which graduation the other side is passing. Thus the following grade is of 60⁰.

ACute Angle

An acute angle is less than 90⁰

O

P Q

OPQ is an acute angle

Right angle

A right angle is equal to 90⁰. It is denoted by a square inside the angle.

~ 168 ~

Obtuse angle

An obtuse angle is larger than 90⁰ but less than 180⁰

Complementary Angles

Two angles are called as complementary angles if their sum is equal to 90⁰. In the given below figure these are called as complementary, because the sum of 40⁰+ 50⁰= 90⁰

http://0.tqn.com/d/create/1/0/z/w/B/-/rightangle.jpg

~ 169 ~

Supplementary Angles

Two angles are called as supplementary angles, if their sum is equal to 180⁰

1.

Exercise6.1

Identify the type of angles give below in

~ 170 ~

2.

o 20⁰o 157⁰o 197⁰o 242⁰

Identify the type of angles give below in

Use a protractor to draw the following angles :

~ 171 ~

o 320⁰o 287⁰

3.

o 46⁰o 53⁰o 64⁰o 7⁰

4.

o 36⁰o 102⁰o 171⁰o 88⁰

Find the measure of complementary angles of following angles given below

Find the measure of the following supplementary angle of each of the following angle

~ 172 ~

o Concept of Triangles

o Types of different Triangles

oConstruction of triangles

Triangles

~ 173 ~

Mr. Donald Duck is giving you a very important information he is explaining that

1. All the three angles in an equilateral triangle are equal in size

2. The two base angles of an isosceles triangle are equal

3. All the three angles in a scalene triangle are different in size.

~ 174 ~

Triangles can be classified according to:

(a)The number of equal sides

The triangle in which all the three sides are equal in length is called as Equilateral triangle

A triangle in which two sides are equal in length, while the base is unequal is called Isosceles triangle

C

Equilateral triangle

Isosceles triangle

~ 175 ~

A B

The ∆ ABC is isosceles triangle.

The triangle having all of its sides unequal in length is called Scalene triangle

(b) The types of angles

The angle in which all three angles are acute is called as Acute angled triangle

Scalene triangle

Acute angled triangle

~ 176 ~

The triangle in which one angle is obtuse, is called as Obtuse triangle

The triangle in which one angle is of 90⁰ is called as Right angled triangle

Obtuse angled triangle

Right angled triangle

~ 177 ~

1. The following are the base of isosceles triangles. In each case, find the third one of the isosceles triangle:o 42⁰o 82⁰o 18⁰o 54⁰

2. Construct the following triangles:i. ∆ ABC, when m AB = 5cm, m BC = 7cm, m CA = 3cm

ii. ∆ OPQ, when m OP= 4cm, m PQ = 4.5cm, m OQ = 5cmiii. ∆ ABC, when m AB = 5cm, m BC = 5cm, m AC = 6cm

EXERCISE 6.2

~ 178 ~

3. In the triangle ABC, A = 50⁰, C = 26⁰ and AB is produced to D. So you have to find the ∆ ABC and ∆ CBD.

Construction of different triangles with the help of

compass

~ 179 ~

Example 1Construct triangle ABC such that line AB = 6 cm, BAC = 80⁰ and BC = 7cm. Measure and write down the length of AC.

Solution

Construction steps:

1. Use ruler to draw a line AB = 6 cm

Our MR. Donald Duck is now going to teach us a very interesting thing i.e,

construction of triangles with the help of compass

~ 180 ~

2. Use a protractor to draw an angle BAL = 60⁰3. Join BL and produce. Now BAL = 60⁰.4. With A as the center and radius 6 cm, make an arc with the help

of compass to cut BL at C.5. Join BC and measure its length, AC = 4 cm

Example 2Use a protractor and ruler to construct ∆ ABC with AB = 7 cm, ABC =25⁰, BAC = 60⁰.

Solution

Constructions of a triangle when two angles and one side is given

~ 181 ~

Constructing steps:

1. Draw a line AB = 7 cm with the help of ruler2. Use a protractor to draw an angle of 25⁰ at ABL3. Use a protractor to draw an angle of 60⁰ at BAM and cut the arm

BL at C4. Join AC and measure its length, AC = 2.3 cm5. Hence the ∆ ABC is the required triangle

Example 3With the help of compass and ruler, construct ∆ ABC, when AB = 3 cm, BC = 4cm and AC = 5 cm

Solution

Construction of triangle when all three sides are

given

~ 182 ~

Constructing steps:

1. Draw a line AB = cm2. From B draw an arc of radius 4 cm with the help of compass3. With A as the center, draw an arc of radius 5 cm, to cut the arc

drawn in step 2 at C4. Then join AC and BC at the point of intersection of these two arcs

(point of intersection = center point)5. Thus the triangle ABC is the required triangle.

ExplorationWork in groups

Construct a line parallel to BC and passing through A.

Explain why the line you have drawn is really parallel to BC?

What tools you used for this activity? Can your group think of other ways to construct the parallel line?

At the end compare and contrast you results with other groups

This given below construction is the key to start your activity

~ 183 ~

C

A B

1. Construct a ∆ PQR where PQ = PR = 10 cm and QR = 9 cm. Measure and write down the size of PQR.

2. Construct a ∆ ABC where AB = 9.5 cm AC = 8.5 cm, and BC = 9.8 cm. Measure and write done the size of ABC. Construct the perpendicular bisector of AC.

3. Construct a ∆ LMN where LM = 5 cm, MLN = 45⁰ and LMN = 60⁰. Construct the perpendicular bisector of LM. If the perpendicular bisector of LM meets LM at O, measure and write down the length of MO.

4. Construct ∆ XYZ with XY = 9.4 cm, XZ = 8.8 cm and XYZ = 60⁰. Measure and write down the length of YZ.

Exercise

6.3

~ 184 ~

5. Construct ∆ PQR with PQ = 9.8 cm, QR = 6.5 cm and PQR = 90⁰. Measure and write down the size of QPR giving your answers correct to the nearest degree.

6. Construct ∆ ABC where AB = 4 cm, ABC = 60⁰, ACB = 60⁰, also measure and write down the length of BC

7. Construct ∆ ABC such that AB = 10.5 cm, ABC = 60⁰ and BC = 7.5 cm. Measure and write down the length of AC.

8. Construct a ∆ PQR in which PQ = 11 cm, QR = 8 cm, and PQR = 102⁰ also measure and write down the length of PR.

9. Construct the ∆ ABC in which AB = 9 cm, BC = 7 cm and ABC = 50⁰. Moreover, using compass and ruler only, construct the angle bisector of BAC to meet BC at O

10. Construct ∆ LMN in which LM = 5cm, MN = 6 cm, and LMN = 90⁰

~ 185 ~

~ 186 ~

A quadrilateral is defined as a closed plane bounded by four line- segments is called quadrilateral

There are four vertices and four angles in a quadrilateral.

Above pasted diagram is quadrilateral

Quadrilaterals

Types of Quadrilaterals

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There are four types of quadrilaterals

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Parallelogram

Our dear Mickey Mouse has taken this picture of building built in Thailand in sea. This is based on the structure of parallelogram

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Exploration

Now collect some more pictures of buildings or other thing of daily life based on structure of parallelogram and make a project of them

Rhombus

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Trapezium

Following picture is an interesting example of kite quadrilateral in which children are flowing it cheerfully

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Example With the help of compass and protractor construct a parallelogram ABCD such that BC = 4.2 cm, ABC = 115⁰ and BCD = 65⁰.

Solution

Constructing steps:

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1. Draw a line BC = 4.2 cm2. With the help of protractor measure XBC = 115⁰ and BCY = 65⁰3. Then with the help of compass draw an arc of 3.5 cm and cut the

arm BX4. Now again with the help of compass draw an arc of 3.5 cm with c

as a center and cut the are CY5. Now join the arc A and D6. The required shape of quadrilateral is a parallelogram

Example 2Construct a quadrilateral ABCD where AD = 5 cm, DC = 5 cm, BC = 6cm, AB = 4 cm and diagonal AC = 8 cm. Measure the angles of D, C and B

Solution

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Constructing steps:1. Draw a line segment AC = 8 cm2. With A and C as centers and radii 5 cm and 5.5 cm respectively,

draw two arcs to cut at D3. With A and C as centers and radii 4 cm and 6 cm, draw two arcs to

cut at B4. Join AD, DC, CB and AB, with ruler5. By measurement with the help of protractor D = 86⁰, C = 78⁰ and

B = 81⁰

Exercise 6,4

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1. Construct a quadrilateral WXYZ given that XY = 65 cm, YZ = 46 cm, WZ = 58 cm. Angle W = 105⁰ and X = 120⁰. Measure WY and XZ.

2. Construct a quadrilateral ABCL where AB = 5.3 cm, BC = 6.3 cm. CL = 6.7 cm with angle B = 75⁰ and C = 60⁰. Measure AL.

3. Construct a parallelogram PQRS with PQ = 10 cm, QR = 11.2 cm, Q = 80⁰. Measure QS

4. Construct a rhombus LMNO where LM = 7.5 cm and LN = 12 cm. Measure the length of other diagonal

5. Construct a quadrilateral ABCD where AD = 6 cm, CD = BC = 9cm and ADC = BCD = 110⁰. Measure AB.

6. Construct the quadrilateral PQRS in which the base PQ = 10 cm, QR = 6 cm, PS = 3.5 cm, PQR = 45⁰ and QPS = 60⁰

7. Construct a quadrilateral ABCD in which the base AB = 5.6 cm, B = 80⁰, C = 95⁰, BC = 6.2 cm and CD = 9.2 cm. Measure AD

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8. Construct a rhombus ABCD where AB = 6 cm, and B = 115⁰. Measure the length of the diagonals

9. Construct a parallelogram ABCD in which AB = 9 cm, BC = 6 cm and ABC = 115⁰ also measure and write down the length of AC

10. Construct the trapezium ABCD where AB = 5 cm, BAD = 120⁰, AD = 6 cm, DC = 9.8 cm and AB parallel to DC

11. Construct a quadrilateral ABCD where AB = 6.5 cm, BC = 4.8 cm, CD = 8.5 cm, B = 75⁰ and C = 98⁰. Measure the length of AD and the angles A and D

12. Construct a quadrilateral LMNO where LM = 4.5 cm, MN = 5.6 cm, NO = 6.1 cm, LO = 4.3 cm and diagonal LN = 6.9 cm. Measure the angles of M, N and O

1. Define the following:

REVIEW QUESTIONS

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I. Acute angleII. Right angle

III. Obtuse angleIV. Complementary anglesV. Supplementary angles

VI. Acute angled triangleVII. Obtuse angled triangle

VIII. Right angled triangleIX. QuadrilateralX. Isosceles triangle

XI. Scalene triangle

2. Construct ∆ ABC with BAC = 60⁰, AB = 7.5 cm and AC = 10 cm3. Construct a parallelogram PQRS where PQ = 8 cm, PQR = 120⁰ and

QR = 5.5 cm. You have to measure and write down the length of PR and QS.

4. Construct ∆ ABC where AB = 8 cm, AC = 5cm, and BC = 6 cm.5. Construct ∆ ABC with sides each of length 10.8 cm. Mark a point P

on AB so that PA = 2.4 cm. On BC, mark a point Q so that QC = 8.6 cm.

6. Draw a rhombus of side 6 cm and one of its diagonals 9 cm. Measure the length of the other diagonal.

7. In ∆ PQR, PQR = 64⁰, PQ = PR and QR is produced to O. Calculate the angles of QPR and PRO.

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8. Construct a parallelogram with diagonals 7.5 cm and 10.2 cm, and with the shorter sides 3, 6 cm long. Measure the length of longer sides.

9. Construct a parallelogram of sides 6.9 cm and 11.1 cm, it’s longer diagonal is 14.4 cm. Determine the length of other diagonal.

10. Construct ∆ ABC such that AB = 10 cm, BC = 9 cm and CA = 8 cm. A point X on AC is 2 cm from C. Draw a line through X which should be parallel to CB to AB at Y. Measure AY.

Mr. Donald Duck is getting ready to solve these questions

ANSWERS

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Exercise 6.11.

I. ObtuseII. Acute

III. AcuteIV. AcuteV. Right angle

VI. ObtuseVII. Acute

VIII. AcuteIX. AcuteX. Right angle

XI. ObtuseXII. Acute

2. (a) 44⁰ (b)37⁰ (c) 26⁰

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(d) 83⁰

4. (a) 144⁰ (b)78⁰ (c) 9⁰

(d) 92⁰

Exercise 6,2

1. (a) 96⁰ (b) 16⁰ (c) 144⁰

(d) 52⁰

3. 104⁰, 76⁰

Exercise 6.31. 63⁰2. 62⁰3. 3.5 cm4. 8.0 cm5. 34⁰6. 8 cm7. 9.5 cm

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8. 14.9 cm9. 7 cm, 50⁰10. 4.5 cm

Exercise 6.41. 97 mm, 82⁰2. 1.7 cm3. 16.9 cm4. 9 cm5. 14.4 cm6. 4.5 cm7. 7.0 cm8. 6.4 cm, 10.1 cm9. 12.8 cm10. 5.5 cm11. 56⁰12. 81⁰

Review Exercise2. 6.5 cm, 46⁰3. 11.8 cm, 7.1 cm4. 4.1 cm

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5. 8.6 cm6. 52⁰, 116⁰7. 8.2 cm8. 11.6 cm9. 7.5cm

PARIMETER AND AREA

Chapter 7

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After studying this unit, the students will be able to:

Find out region of a close figure. Compare perimeter and area of a region. To sort out the units for measurement of perimeter and area. Apply formulas to find out perimeter and area of a square and

rectangular region. Solve appropriate problems of perimeter and area.

7.1.1 Recognize region of a close figure

Region: The region of a closed shape is the space which is occupied by the boundary of that shape.

Difference between perimeter and area of region:

Perimeter:

“The perimeter of a closed shape is the distance along all the sides of that shape.”

Shapes with the same area can have different perimeter:

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(a)

(b)

(a) Area = 4cm² Perimeter = 10cm(b) Area = 4cm² perimeter = 8cm

Area:

“The amount of a surface a shape cover is called its area”

Area and perimeter:

Look at this rectangle. It is made up of 8 small squares, each with an area of 1cm².

Area = 8cm perimeter = 12cm

There are 2 rows of 4 squares each.

There is a very simple way of finding the area and perimeter of a rectangle.

To find the area we multiply the length (ȴ) of the rectangle by the width.

Here, ȴ × b = 4cm × 2cm = 8cm²

The area of the rectangle = 8cm²

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To find the perimeter we add ȴ and b and multiply the sum by 2:

ȴ×b = 4cm + 2cm = 6cm

(ȴ×b) ×2 = 6cm ×2 = 12cm

The perimeter of the rectangle = 12cm

If the area of a rectangle = 96cm²

and its width is 8cm, what is its length???

96cm² ÷ 8cm = 12cm

Length = 12cm

Write the formulas for perimeter and area of a square and rectangle

Example:

Find the perimeter and area of square with length of each side as 3cm.

Perimeter =3×ℓ

REMEMBER

Area is written as cm² and perimeter is written as cm.

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=3×4

=12cm

Area = 3cm × 3cm =9cm²

So the Perimeter of the square is 12cm and area is 9cm².

7.3 Application of formulas to find perimeter and area of a square and rectangular region.

Find the perimeter and area of rectangle with length 4cm and breadth 2cm

Solution:

Given that ℓ=4cm and b=2cm

= 2× (ℓ+b)

= 2× (4cm + 2cm)

= 2 × (6cm)

Perimeter =12cm

Area = ℓ × b

=4cm × 2cm

=12cm

So here the perimeter of the rectangle is 12cm and area is 12cm.

Example:

REMEMBER

In a rectangle,

The area=ℓ×b unit²

The perimeter= 2(ℓ×b) units.

If the perimeter of a rectangle is 72cm and its breadth is 20cm, what is its length?

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Find the perimeter of the square with following measurement of side:

(a) 5cm (b) 4cm

(a) length of the side=ℓ=5cm

Perimeter of the square =4×ℓ

=20cm

Area of a square= ℓ×ℓ =5×5cm² = 25cm²

(b) Length of the side =ℓ=7cm

If the perimeter of a rectangle is 72cm and its breadth is 20cm, what is its length?

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Perimeter of the square =7×ℓ =7×7cm =49cmArea of the square=ℓ × ℓ = 7cm × 7cm =49cm

Find the area of a road with length 21cm and breadth 19cm.

ℓ = 21m b = 19m ℓ × b 21 × 19 =399m²

Exercise # 7.1

1. Find the perimeter and area of squares having the following measurements of each side.

(a) 4cm (b) 7cm (c) 9cm (d) 11cm

2. Find the perimeter and area of the rectangles having the following measurements of adjacent side.

(a) 6cm, 8cm (b) 11cm, 12cm (c) 9cm, 10cm (d) 5cm, 7cm

3. The length of each side of a wall is 10 meters and breadth is 14m. Find

the area of the wall.

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4. Find the area of a floor of a room with length 7cm and breadth is 3cm.

Draw these rectangles.

Then find the area and perimeter of each:

(a) ℓ = 4cm, b = 3cm(b) ℓ = 7cm, b = 4cm(c) ℓ = 5cm, b = 1cm(d) ℓ = 8cm, b = 2cm

The length of a rectangular field is 75m and its breath is 26m.

Reem ran round the field 3 times.

(a) Find the perimeter (b) How far did Reem run

ACTIVITY

1

ACTIVITY

2

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Tick the shape with the smaller area

(a)

(b)

(c)

REVIEW EXERCISE

1. Write formulas for perimeter and area of a rectangle?

ACTIVITY

3

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2. Find the perimeter and area of the square having the following measurement of each style.

(a) 19cm,

(b) 21cm

3. Find the perimeter and area of the rectangles having measurements of adjacent side.

(4cm, 7cm)

(15cm, 18cm)

ANSWERS:

1. (a) 16cm, 16cm² (b) 28cm, 49cm²(c) 36cm, 34cm² (d)44cm, 121cm²

2. (a)perimeter: 36cm,area: 64cm² (b) Perimeter: 121cm, area: 144cm²

(c) perimeter: 81cm, area: 100cm²

3. 140m²4. 21m²

REVIEW EXERCISE ANSWERS

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1. 2(ℓ + b); ℓb 2. (a) 76cm; 361cm² (b) 84cm, 441cm² 3. (a) 5cm, 28cm²

(b) 66cm, 288cm²

Chapter 8

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“INFORMATION HANDLING” Find an average of numbers

Solve real life problem including average

Draw block graph or column graph

Study a simple bar graph given in vertical and horizontal

8.1 AVERAGE

8.2 AVERAGE (Arithmetic mean)

“Average is sum of quantities which is divided by number of quantity.”

Average= sum of the quantities

Number of quantities

Example 1: Find an average of given numbers

Find the average of 21, 22, 23, 24, 25

Solution: Sum of the given numbers = 21 + 22 + 23 + 24 + 25

Average = 21+22+23+24+25 =115

5 5

= 115 = 31

5

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Average=31

Thus the average of the given numbers is 31.

Example 2: Find an average of given numbers 25, 35, 45, 55

Solution: Sum of the given numbers =25+35+45+55

Average = 25+35+45+55 = 225

5 5

=225 = 45

5

Average=45

Thus the average of the given number is 45.

8.3 Solve real life problem involving average.

Example 1: The 7 days salary of a worker is given below

Monday Tuesday Wednesday Thursday Friday Saturday Sunday

500 400 300 600 700 800 900

Solution:

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= 500 + 400 + 300 + 600 + 700 + 800 + 900 = 4200

Average income = total income

Number of days

Average income = 4200 =60

7

Thus average daily salary of the worker is Rs.=600

Example 2: A motorbike covers a distance of 60km in the first hour, 80km in the second hour, and 90km in the third hour and 100km in the fourth hour, find the average speed of the car per hour.

Solution:

Distance cover in the 1st hour =60km

Distance cover in the 2nd hour=80km

Distance cover in the 3rd hour =90km

Distance cover in the 4th hour =100km

60km

80km

90km

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+ 100km

330km

Total numbers of hours = 4

330 = 82km

4

Average distance cover in one hour = 82km

EXERCISE 8.1

1. Find the average of the following numbers.

(i) 11, 22, 33, 44, 55 (ii) 44, 48, 55, 68, 77

(iii) 5o, 60, 70, 8o, 90 (iv) 39, 56, 80, 97, 98, 100

(v) 12, 17, 29, 30, 49, 55

2. Ahmed had 40 pens, Hassan had 20 pens, Sarmad had 25 pens, and Ashen had 30 pens. What is the average number of pen each one had.

3. The result of accounting students is given below,

What is the average number of each student?

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Ahmed 90

Hassan 70

Hadi 66

Irfan 45

Abdullah 33

4. A bus moves at a speed of 60km per hour in the 1st hour,70km per hour in the second hour, 80km per hour in the third hour, 90km per hour in the 4th

hour and 100 per hour in the 5th.what was the average speed of the bus?

5. Ashen has 10 goats, Ali has 50 goats, and Aslam has 40 goats, find the average of goats among the three?

8.3 Draw block graph and column graph

Block graph:

The graph in which we select an appropriate icon to represent each part of the information is called a block graph.

Favorite food of children in class 5th

55

55

45

35

25

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45

35

25

0 cake Chocolate cookies pastries

Now answer these questions.

How many children like chocolate the best?

Which food is least like?

For how many children are cookies the favorite food?

How many children are there in class 5th?

Bar graph:

“In a bar graph we represent each part of the

information in the shape of bars. It may

be vertical or horizontal.”

The presence of 21 different stationary is shown in the vertical bar graph. Look at the below graph given below and answer the question.

Pencil Eraser Sharpener Scale Colour box pens

4 3 2 6 1 5

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pencil eraser scale sharpener colour box

pen0

1

2

3

4

5

6

7

How many pencils are more than eraser? How many scales are more than sharpener? How many stationary are altogether? How many pens are fewer than colour box?

The following table shows the marks of different history students.

sara

kiran

aruj

fiza

0 10 20 30 40 50 60 70 80

Series 3Series 2Series 1

How many students gain 40 marks?

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How many marks aruj got?

How many total marks did they get?

How many marks Sara got?

Interpret a simple bar graph given in vertical form. Look at the graph below:

Hannan has 50 hens Junaid has 20 chickens Ahmed has 40 parrots Shahid has 30 pigeon

EXERCISE 8.2

Here is a column graph for 30 students, each unit of which stands for 5 children.

hannan junaid tammor shahid0

10

20

30

40

50

60

Series 1Series 2Series 3

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The numbers of student’s present in different days in a school are shown in the following table. Consider the graph and answer the given question.

From the above graph we can answer the following question:

How many children were absent on Friday? How many more children were absent on Monday than Thursday? How many children were not in school on Wednesday and Tuesday?

2. See the horizontal bar graph and answer these questions and

Children’s favorite fruits

Monday Tuesday Wednesday Thursday Friday Saturday0

5

10

15

20

25

30

35

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o How many children do like

mangos?o How many children are there who like fruits?

o How many children like banana?

o Which fruit is least like?

The number of chairs in 5 rooms is shown in the table.Interpret following vertical bar graph.

room 1 room 2 room 3 room 4 room 50

5

10

15

20

25

30

35

40

45

Series 1

Series 3

Series 4

apple

mango

peach

orange

banana

strawberry

0 10 20 30 40 50 60

Series 1

Series 1

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Review exercise

1. Find the average the following number

(a) 19, 23, 29, 33, 38, 47

(b) 26, 36, 46, 56, 66

(c) 79, 89, 99, 199,299

2. Tuba has 2 dolls, Alisha has bears and Mishaal has 16 cars. Find their average.

3. A shopkeeper has 4 cylinders two thermometers and 15 fans. How many things he has in his shop. Find their average.

4. The temperature in a weak per day in different cities are given in table. Consider the graph and answer the given question.

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

0 10 20 30 40 50 60

Series 4Series 3Series 2Series 1

What is the temperature of Monday?

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What is the temperature of Tuesday and Wednesday?

Which day has the highest temperature?

Which day has the lowest temperature?

ANSWERS:

1. (a) 35, (b) 58, (c) 70, (d) 78, (e) 322. 283. 604. 80km per hour5. 33.9

Review exercise

1. (a) 31.5 (b) 46 (c) 156

2. 7.2

3. 7

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CONTRIBUTION OF MEMBERS

HANIA ASIF NUMBERS AND ARITHMETIC OPTIONS

MAHEEN TOUQEER HCF AND LCM

UNITARY METHOD ( 59 – 71)

ZAINAB FATIMA FRACTIONS

UNITARY METHOD (72 – 80)

HAFSA TAHIR DECIMALS AND PERCENTAGES

SIDRA JAVAID GEOMETRY

SADIA ANWAR PERIMETER AND AREA

INFORMATION HANDLING

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