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Quick ReviewLet us take a review on numbers.

z Natural numbers : The numbers which are used for counting are known as natural numbers. For example, 1, 2, 3, 4, ... etc., are all natural numbers. 1 is the smallest natural number. The collection of all natural numbers is denoted by N and defined as :

N = {1, 2, 3, 4, 5, ...}Moreover natural numbers are always represented on the right side of zero on the number line.

z Whole numbers : If we include zero in the collection of natural numbers, then the new collection we get is whole numbers. 0 is the smallest whole number. The collection of whole numbers is denoted by W and defined as : W = {0, 1, 2, 3, 4, ...}

z Integers : In the collection of whole numbers, if we include negative of natural numbers, i.e., ..., –3, –2, –1, then the collection obtained is called as collection of integers. The collection of integers is denoted by Z or I and defined as :I or Z = {..., –3, –2, –1, 0, 1, 2, 3, ...}, where Z is taken from “Zahlen” which is a German word, which means “to count”. On the basis of definition, integers are classified in two types : Positive integers and Negative integers. Positive integers are natural numbers i.e., 1, 2, 3 … and negative integers are negatives of natural number i.e., –1, –2, –3 ….

Note :(i) 0 is neither positive nor negative.(ii) Every natural number is a whole number but every whole number is not a natural number.(iii) Every natural number and whole number is an integer.

Number Systems

1CHAPTER

In earlier classes, we have learnt about various types of numbers (natural numbers, whole numbers, integers and

rational numbers) and their representation on number line. In this chapter, we will review them and study more about rational numbers and the existence of irrational numbers, their representation on number line and some mathematical operations on it.

TOPICS Review of Rational Numbers Irrational Numbers Real Numbers and their

Decimal Expansions Representing Real Numbers

on the Number Line Operations on Real Numbers Laws of Exponents for Real

Numbers

3.765

3.773.763.761 3.762 3.763 3.764 3.765 3.766 3.767 3.768 3.769

3.73.71 3.72 3.73 3.743.75 3.763.77 3.78 3.79

3.8

433.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

TOPIC 1 : REVIEW OF RATIONAL NUMBERSAs we know that, the numbers which can be written in the form of p/q, where p, q are integers and q ≠ 0 are called as rational numbers. And a rational number p/q is said to be in simplest form, if p and q are integers having no common factor other than 1, i.e., p and q are co-prime. The set of rational numbers is denoted by Q and defined as :

Q pq

p q q= ≠

, , where are integers and 0 or Q = − −{ }..., ,56

34

12

...,0..., , 1, ...

Note :(i) Every rational number can also be represented on number line.(ii) 0 is a rational number because 0 can be expressed as 0/1 i.e., in the form of p/q.(iii) Every natural number and whole number is a rational number.

POINTS TO

REMEMBER " The rational numbers are closed under the operation of addition, subtraction and multiplication but not

division because division by 0 is not defined.

" The operations addition and multiplication are commutative as well as associative for rational numbers.

" For rational numbers, 0 is the additive identity and 1 is the multiplicative identity i.e., 0 + a = a and a ⋅ 1 = a, where a is in the form of p/q.

" The additive and multiplicative inverse of a rational number ab are −ab and b

a respectively, where a, b ≠ 0.

" Two rational numbers are said to be equivalent, if numerators and denominators of both rational numbers are in same proportion.

Method to Determine Rational Numbers between Two Rational NumbersAs we know that there are infinitely many (countless) rational numbers between two rational numbers, when we try to find out these rational numbers, then two cases arise :Case I : When we have to find only one rational number : In this case, add the given two rational numbers and then divide the sum by 2, the resulting number is the required rational number which lies between given numbers.

Let a and b be two rational numbers such that a < b, then q a b1 2= + such that a < q1 < b, where q1 is the required

rational number which lies between a and b.Case II : When we have to find two or more rational numbers : We have three methods to find two or more rational numbers.

Method-1 : Let a and b be two rational numbers such that a < b, then q a b1 2= + , a < q1 < b, where q1 is the rational

number between a and b.

Now, q a q a q q b21

2 12= + ⇒ < < < , where q2 is the rational number between a and q1.

Now, q q b a q q q b31

2 1 32=

+⇒ < < < < , where q3 is the rational number between q1 and b.

In this manner, we can find infinite rational numbers between two distinct rational numbers.Method-2 : Let x and y be two rational numbers, such that y > x and we need to find n rational numbers between x and y. Then, n rational numbers lying between x and y are

(x + d), (x + 2d), (x + 3d), ..., (x + nd), where d y xn

=−+ 1

.

Method-3 : Let x and y be two rational numbers such that y > x, then we can find n rational numbers between x and y by following the steps given below :(a) If x and y are rational number with different denominator then make their denominator same by taking L.C.M.(b) Now, multiply the numerator and denominator of x and y by (n + 1), i.e.,

x x n

nx nn

y y nn

y nn

= × ++

= ++

=× +

+=

++

( )( )

( )( )

( )( )

( )( )

11

11

11

11

and

(c) Let x1 , x2 , x3 ... xn are n rational numbers lying between x and y.

Then, xx nn1

11

1= ++

+

( )( )

, x2 = (x1 + 1), x3 = (x2 + 1), ..., xn = (xn – 1 + 1), i.e., x < x1 < x2 < x3 ... < xn – 1 < xn< y.

2 100PERCENTMathematics Class-9

ILLUSTRATIONS

Find a rational number between −23

14

and .

Sol. A rational number lying between −23

14

and is − +( )÷ = − +( )×23

14

2 23

14

12

= − × + ×

× = − +( ) = −( )2 4 3 1

1212

8 324

524

Thus, the required rational number is −524

i.e., − < − <23

524

14

Find five rational numbers between 3/5 and 2/3.Sol. Let x = 3/5 or 9/15, y = 2/3 or 10/15 and n = 5.\ x < y

Now, we have, d y xn

= −+

=−

+=( )

( ) ( )1

23

35

5 11

90Thus, five rational numbers between 3/5 and 2/3 are (x + d), (x + 2d), (x + 3d), (x + 4d) and (x + 5d)

i.e., 35

190

35

290

35

390

35

490

+

+

+

+

, , ,

and 35

590

+

i.e., 55

905690

5790

5890

5990

, , , ,

i.e., 35

5590

5690

5790

5890

5990

23

< < < < < < .

Find four rational numbers between –7 and –8.Sol. Here, –7 > –8 and n = 4Since, we need to find four rational numbers between –7 and –8.So, multiply the numerator and denominator by (4 + 1) = 5, of –7 and –8, i.e.,

− = − × = − − = − × = −7 7 55

355

8 8 55

405

and

Thus, four rational numbers between –7 and –8 are − − − −36

5375

385

395

, , and such that

− > − > − > − − > −355

365

375

385

395

405

> .

or − > − > − > − − > −7 365

375

385

395

8> .

YOURSELF1. Find a rational number lying between 1

315

and . (Ans. 4/15)

2. Find five rational numbers between 4 and 5. Ans. 256

266

276

286

296

, , , ,

3. Find three rational numbers between − −16

17

and . Ans. − − −25168

26168

27168

, ,

For YOURSELF Solutions visit http://bit.ly/mtg-100-mt-tryyourself-9

1. Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0?2. Find six rational numbers between 3 and 4.3. Find five rational numbers between 3/5 and 4/5.

4. State whether the following statements are true or false. Give reasons for your answers.(i) Every natural number is a whole number.(ii) Every integer is a whole number.(iii) Every rational number is a whole number.

For Solutions visit http://bit.ly/mtg-100-mt-ncertfocus-9

3Number Systems

EXERCISE - 1.1

Objective Type QuestionsMCQs (1 Mark)1. Every rational number is(a) a natural number (b) an integer(c) a real number (d) a whole number

(NCERT Exemplar)2. Between two rational numbers(a) there is no rational number(b) there is exactly one rational number(c) there are infinitely many rational numbers(d) there are only rational numbers and no irrational numbers (NCERT Exemplar)

Fill in the Blanks (1 Mark)3. ______ is the smallest whole number.4. For every number except ______, rational numbers are closed under division.

VSA Type Questions (1 Mark)5. Find a rational number between the numbers 19

29

and .

Short Answer Type QuestionsSA Type I Questions (2 Marks)

6. Find a rational number lying between 0.75 and 1.2.

SA Type II Questions (3 Marks)

7. Find two rational numbers between 0.26 and 0.27.

8. Find three rational numbers between –2/5 and 2.

9. Find three rational numbers between –2 and –3.

Long Answer Type QuestionsLA Type Questions (4 Marks)

10. Insert 6 rational numbers between 2.2 and 2.3.

11. Find nine rational numbers between 0 and 0.1.

1. (c) : Since, real numbers is a collection of rational and irrational numbers. Hence, every rational number is a real number.2. (c) : We know that, between two rational numbers, there are infinitely many rational numbers.3. 0 is the smallest whole number.4. For every number except zero, rational numbers are closed under division.

5. A rational number lying between 19

29

and is 19

29

2+

÷ .

= +

× = × = × =1 2

912

39

12

13

12

16

Thus, the required rational number is 16

19

16

29

i e. ., .< <

6. Let a = 0.75 = 75100

and b = 1.2 = 1210

.

A rational number between 75100

and 1210

is 75100

1210

2+

÷

= +

× = +

× = +

×

75100

1210

12

34

65

12

15 2420

12

= × =3920

12

3940

i.e., 0.75 < 3940

< 1.2

7. Let a = 0.26 = 26100 and b = 0.27 = 27

100


27100

and is

26100

27100

2+

÷

= +

× =26 27

10012

53200

i.e., 26100

53200

27100

< <


27100

and is

53200

27100

2+

÷ =

+ ×

× =

+

× =

53 27 2200

12

53 54200

12

107400

i.e., 26100

53200

107400

27100

< < < .


8. Here, 2 25

> − , so let x = −25

, y = 2 and n = 3.

Now, dy xn

=−+

=− −

+=

+= +

×= =

12 2 5

3 1

21

25

410 25 4

1220

35

( / )

Thus, three rational number between 2 and –2/5 are (x + d), (x + 2d) and (x + 3d)

i.e., − +

− + ×

− + ×

25

35

25

2 35

25

3 35

, and

i.e., − +

− +

− +

2 35

25

65

25

95

, and

i.e., 15

45

75

, and

i.e., − < < < <25

15

45

75

2.

9. Since, we need to find 3 rational numbers between –2 and –3. Thus, multiply the numerator and denominator of –2 and –3 by 4

i.e., − × = − − × = −2 44

84

3 44

124

and

Thus, three rational numbers between –2 and –3 are − − −94

104

114

, , i.e., − > − > − > − > −84

94

104

114

124

.

10. Let a = 2.2 = 2210

and b = 2.3 = 2310

A rational number lying between 2210

2310

and is

2210

2310

2+

÷ =

× =45

1012

4520 i.e., 22

104520

2310

< < .


4520

and is 2210

4520

2+

÷

= +

× =44 45

2012

8940

i.e., 2210

8940

4520

2310

< < < .


2310

and is 4520

2310

2+

÷

= +

× =45 46

2012

9140

i.e., 2210

8940

4520

9140

2310

< < < < .


4520

and is 8940

4520

2+

÷

= +

× =89 90

4012

17980

i.e., 2210

8940

17980

4520

9140

2310

< < < < < .


2310

and is 9140

2310

2+

÷

= +

× =91 92

4012

18380

i.e., 2210

8940

17980

4520

9140

18380

2310

< < < < < < .


8940

and is 2210

8940

2+

÷

= +

× =88 89

4012

17780

i.e., 2210

17780

8940

17980

4520

9140

18380

2310

< < < < < < < .

11. Here, 0.1 > 0, so let x = 0, y = 0.1 and n = 9

Now, d y xn

= −+

= −+

= =1

0 1 09 1

0 110

0 01. . .

Thus, nine rational numbers between 0 and 0.1 are(x + d), (x + 2d), (x + 3d), (x + 4d), (x + 5d), (x + 6d), (x + 7d), (x + 8d) and (x + 9d)i.e., (0 + 0.01), (0 + 2 × 0.01), (0 + 3 × 0.01), (0 + 4 × 0.01), (0 + 5 × 0.01), (0 + 6 × 0.01), (0 + 7 × 0.01), (0 + 8 × 0.01) and (0 + 9 × 0.01)i.e., 0.01 , 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08 and 0.09i.e., 0 < 0.01 < 0.02 < 0.03 < 0.04 < 0.05 < 0.06 < 0.07 < 0.08 < 0.09 < 0.1.

TOPIC 2 : IRRATIONAL NUMBERSThe numbers which are not rational are known as irrational numbers i.e, the numbers which can not be expressed in the form of p/q, where p, q ∈ Z and q ≠ 0 are called irrational numbers.In other words, we can say that every non-terminating and non-repeating decimal number is known as irrational number. For example : 2 7 53, , , p, 0.101001000 ... etc. are irrational numbers. There are infinitely many irrational numbers between any two irrational numbers.

Note : 227

is a rational number but p is an irrational number.

Representation of Irrational Numbers on Number LineTo represent n on the number line, where n ∈ N, we have to follow the following steps :Step 1 : Write n as the sum of squares of two numbers. i.e., n = a2 + b2.For example :

z If n = 5, then 5 = 12 + 22

z If n = 7, then 7 = 12 + ( 6 )2

Step 2 : Draw OA = a units on number line. OX′ XAa

b

B

C

5Number Systems

Step 3 : Draw AB = b units such that AB ⊥ OA. Join OB.Step 4 : Taking O as centre and OB as radius draw an arc, which cuts the number line at C, where point C represents the required irrational number on number line.Note : In the same way, we can locate n for any positive integer n, after locating n − 1 on the number line.

ILLUSTRATIONS

Locate 17 on the number line. Sol. Draw a number line X′OX and let O be the origin. Take OA = 4 units and draw AB = 1 unit such that AB ⊥ OA. Join OB. We get, OB

= + = + =OA AB2 2 2 24 1 17( ) ( ) units. WithO as centre and radius, OB = 17 units, draw an arc which intersects OX at P. Then, OP = OB = 17 units. Thus, the point P represents 17 on number line.

–1 O XX′ 1 2 3 4

1A

P

B

5 6

17

Represent each of the numbers 2 3 and on the number line. (NCERT)Sol. Draw a number line X′OX and let O be the origin.Take OA = 1 unit and draw AB = 1 unit such that AB ⊥ OA.Join OB. We get,

OB OA AB= + = + =2 2 2 21 1 2 units.

With O as centre and radius, OB = 2 units draw an arc which intersects OX at P.Then, OP = OB = 2 units.Thus, the point P represents 2 on the number line.Now, draw BC = 1 unit such that BC ⊥ OB.Join OC. We get,

OC OB BC= + = + =2 2 2 22 1 3( ) units

With O as centre and radius, OC = 3 units, draw an arc which intersects OX at Q. Then,OQ = OC = 3 units.Thus, the point Q represents 3 on the number line.

O 1

2

–1 A P Q

B1

1

C

XX’

23

Hence the points P and Q represent the numbers 2 and 3 respectively.

YOURSELF4. Locate 11 on number line.

5. Represent 6 7and on the number line.



1. State whether the following statements aretrue or false. Justify your answers.(i) Every irrational number is a real number.(ii) Every point on the number line is of the form

m , where m is a natural number.(iii) Every real number is an irrational number.

2. Are the square roots of all positive integersirrational? If not, give an example of the squareroot of a number that is a rational number.

3. Show how 5 can be represented on thenumber line.

EXERCISE - 1.2


TOPIC 3 : REAL NUMBERS AND THEIR DECIMAL EXPANSIONSReal NumbersThe collection of all rational and irrational numbers will form the real numbers. It is denoted by R.So, we can say that a real number is either rational or irrational.Since every number is represented by a unique point on the number line and every point on the number line represents a unique real number, therefore we call the number line as the 'real number line'.As we know that, any real number is either rational or irrational, so they can be expressed as a decimal.Here, we study about the decimal expansions of rational numbers and irrational numbers.

Decimal Expansion of Rational NumbersIt has two types of decimals :z Terminating Decimals Non-terminating recurring decimals

Terminating DecimalsIf the decimal expression of p/q terminates, i.e., comes to an end, then the decimal so obtained is called a terminating decimal. In other words, we can say that a rational number p/q is said to be terminating decimal if on dividing p by q, the remainder becomes zero after some steps.

POINTS TO

REMEMBER " A fraction p/q is a terminating decimals, only if q has prime factors of 2 and 5 only i.e., q is of the form

2m × 5n.

Non-terminating Recurring DecimalsA decimal in which a digit or a set of digits repeats periodically, is called a repeating or a recurring decimal. In other words, we can say that a rational number p/q is said to be non-terminating repeating (recurring) decimal if on dividing p by q, the remainder never becomes zero and a set of digits repeat in the same interval. In a recurring decimal, we place a bar over the first block of the repeating digits and omit the other repeating digits.Note : The decimal expansion of a rational number is either terminating or non-terminating recurring.

Method of Conversion of Decimal Expansions as Rational Numbers or a Fraction (in Simplest Form)Case I : Conversion of Terminating DecimalStep I : Remove decimal point from the numerator and put as many zeroes on the right hand side of denominator as the number of digits after the decimal in numerator.Step II : Convert the rational number obtained into its lowest term by dividing the numerator and denominator by their common factors.Case II : Conversion of non-terminating repeating decimalsStep I : Firstly, transform the non-repeated digit (s) (if exists) between decimal and recurring digits to the left side of decimal by multiplying both sides by 10r i.e., where ‘r’ is the number of digits between decimal and recurring digits.Step II : Count the recurring digits (say R) and then multiply the equation obtained in step I by 10R.Step III : Lastly, subtract these two equations obtained above.Step IV : Divide both sides of the equation in step III by the coefficient of x to get the required fraction.

7Number Systems

4. Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a linesegment OP1 of unit length. Draw a line segment P1P2, perpendicular to OP1 of unit length

2

1

P3 P2

1

1

P13

O

(see given figure). Now draw a line segment P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to OP3. Continuing in this manner, you can get the line segment Pn–1Pn by drawing a line segment of unit length perpendicular to OPn–1. In this manner, you will have created the points P2, P3, ... .., Pn, ...., and joined them to create a beautiful spiral depicting

2 3 4, , , .....

ILLUSTRATIONS Without actual division determine which of

the following fractions are terminating or recurring decimals. Verify your answer by actual division.(a) 13/50 (b) 54/56Sol. (a) 13

5013

2 52=×

Since, denominator of 13/50 is of the form 2m × 5n. Thus, the given number is terminating decimal.Verification : On dividing 13 by 50, we get\ 13/50 = 0.26 is terminating decimal

(b) 5456

2728

32 7

3

2= =×

Since, denominator of 54/56 is not in the form of 2m × 5n. Thus, the given number is non-terminating recurring decimal.Verification : 54.00000000056

–504360

240

160

480

320

400

80

240

–224

–112

–448

–280

–392

–56

–22416

–336

0.964285714...

54/56 = 0.9642 … is non-terminating recurring.

Express 2157625

in the decimal form.

Sol. We have,

–1875 2820 –2500 3200 –3125 750 –625 1250 –1250 0

625 2157.0000 3.4512

50 13.00–100

300

0–300

0.26

\ 2157625

3 4512= .

Express each of the following decimals in

the form p/q :

(i) 0 2. (ii) 23 43.Sol. (i) Let x = 0 2. . Then,

x = 0.222222 ... ...(i)

Mutliplying (i) by 10 both sides, we get

10x = 2.222222 ... ...(ii)

Subtracting (i) from (ii), we get

10x – x = (2.222 …) – (0.222...)

⇒ 9x = 2 ⇒ x = 2/9

Thus, 0.2 = 2/9

(ii) Let x = 23 43. . Then,

x = 23.434343 ... ...(i)

Multiplying (i) by 100, we get

100x = 2343.4343 ... ...(ii)

Subtracting (i) from (ii), we get

100x – x = (2343.4343 ...) – (23.4343 ...)

⇒ 99x = 2320

⇒ x = 232099

Thus, 23 43 232099

. =

Express 15 712. in the form of p/q.

Sol. Let x = 15 712. . Then,

x = 15.7121212 … ...(i)

Multiplying (i) by 10 and 1000, we get

10x = 157 12. = 157.121212 ... ...(ii)

⇒ 1000x = 15712.1212... ...(iii)

Subtracting (ii) from (iii), we get

1000x – 10x = (15712.1212...) – (157.1212...)

⇒ 990x = 15555

⇒ x = =15555990

103766

Thus, 1 5712 103766

. =


YOURSELF6. Convert 35/16 into decimal form. (Ans. 2.1875)

7. Express 0 163. in the form of p/q. (Ans. 9/55)

8. Without actual division determine which of the following fractions are terminating or recurring decimals. Verify your answer by actual division.

(i) 96/300 (ii) 169/91(Ans. (i) Terminating decimal, (ii) Recurring decimal)

9. Express 8.0025 in the form of p/q. (Ans. 3201/400)

10. Express each of the following in the form of p/q. (i) 0 585. (ii) 0 35. (Ans. (i) 65/111, (ii) 35/99)


Decimal Expansion of Irrational NumbersThe decimal expansion of irrational numbers is non-terminating non-repeating.

Method of Finding Irrational Numbers between two Rational NumbersLet a

bcd

and be any two rational numbers. To find an irrational number between ab

cd

and , firstly write ab

cd

and in their decimal form. Now, as we know, irrational number is non-terminating non-repeating decimal. We can find the required number of irrational numbers.

ILLUSTRATIONS Find an irrational number between 1/7 and

2/7. (NCERT)Sol. To find an irrational number, firstly we will divide 1 by 7 and 2 by 7.Now, 1.0000007

–730

20

60

40

50

1

–14

–56

–35

–49

–28

0.142857...

\ 1/7 = 0.142857

Also, 2.0000007–14

60

40

50

10

30

2

–35

–49

–7

–28

–56

0.285714...

\ 2/7 = 0.285714Thus, the required irrational number will lie

between 0.142857 and 0.285714. Also, the irrational numbers have non-terminating non-repeating decimals.Hence, the required number between 1/7 and 2/7 is 0.15015001500015...

Insert two irrational numbers between(i) 5 and 6 (ii) 2/3 and 3/4.Sol. (i) Given, two rational numbers 5 and 6.Now, two irrational numbers between 5 and 6 are 5.1010010001 … and 5.2020020002 … .(ii) To find an irrational number, firstly we will divide 2 by 3 and 3 by 4.Now, 3 2.000 0.666....

–1820

20

2

–18

–18

Now, 4 3.00 0.75–28

20

0–20

\ 2/3 = 0.6666 … \ 3/4 = 0.75Thus, the required irrational numbers will lie between 0.6 and 0.75. Also, the irrational numbers have non-terminating non-repeating decimals. Hence, the required irrational numbers between 2/3 and 3/4 are 0.71010010001 … and 0.72020020002 ….

9Number Systems

YOURSELF11. Find two irrational numbers between 0.6 and 0.66. (Ans. 0.61010010001…, 0.62020020002…)

12. Find an irrational number between 1/7 and 1/3. (Ans. 0.2101001000 ...)

13. Find two irrational numbers between 2/3 and 8/9. (Ans. 0.6101001000... and 0.7101001000...)



Objective Type QuestionsMCQs (1 Mark)1. Decimal representation of a rational number cannot be(a) terminating (b) non-terminating(c) non-terminating repeating

(d) non-terminating non-repeating(NCERT Exemplar)

2. The decimal representation of an irrational number is(a) always terminating(b) either terminating or repeating


EXERCISE - 1.3

1. Write the following in decimal form and say what kind of decimal expansion each has :(i) 36

100 (ii) 1

11 (iii) 4

18

(iv) 313

(v) 2

11 (vi) 329

400

2. You know that 17

0 142857= . . Can you predict

what the decimal expansions of 27

37

47

57

67

, , , , are, without actually doing the long division? If so, how?[Hint : Study the remainders while finding the value of 1/7 carefully.]3. Express the following in the form p/q, where p and q are integers and q ≠ 0.(i) 0 6. (ii) 0 47. (iii) 0 001.4. Express 0.99999... in the form p/q. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.6. Look at several examples of rational numbers in the form p

qq( ),↑ 0 where p and q

are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?7. Write three numbers whose decimal expansions are non-terminating non-recurring.8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.9. Classify the following numbers as rational or irrational.(i) 23 (ii) 225 (iii) 0.3796(iv) 7.478478... (v) 1.101001000100001...

(c) either terminating or non-repeating(d) neither terminating nor repeating

Fill in the Blanks (1 Mark)

3. Decimal expansion of 7/8 is ______.4. The decimal representation of 22/7 is _____ .

VSA Type Questions (1 Mark)

5. Identify 63 as rational or irrational number.6. Find one irrational number between 0.2101 and 0 2. .


7. Express the value of 1.999... in the form p/q, where p and q are integers and q ≠ 0.

(NCERT Exemplar)

8. Find the value of 2 8. in the form of p/q, where p and q are integers and q ≠ 0.

9. Express the number 0 27. in the form of p/q.10. Show that 0.142857142857... = 1/7. (NCERT Exemplar)

11. Write two irrational numbers between 0.202002000200002... and 0.203003000300003...

12. Find two irrational numbers between 1/5 and 2/5.


13. Write the following in decimal form and also find the nature of the decimal expansions :

(i) 561000

(ii) 3 67

14. In a survey, it was found that 9 out of every 11 households are donating some amount of their income to an orphanage or old age home or institutions for physically handicapped. What fraction of households are not donating? Write it in decimal form and find what type of decimal expansion it has?

15. Express the decimal number 2 218. in the form of p/q, where p and q are integers and q ≠ 0.


16. Express 1 32 0 35. .+ in the form p/q, where p and q are integers and q ≠ 0.

17. Express each of the following as a fraction in simplest form :(i) 0 324. (ii) 2 2554.

1. (d) : We know that the decimal expansion of rational numbers is either terminating or non-terminating but repeating.

\ Decimal representation of rational numbers cannot be non-terminating non-repeating.

2. (d)

3. On divide 7 by 8, we get

7.0008 0.875–64

60

40–56

–400

4. On dividing 22 by 7, we have7 22.000000 3.142857

–2110

30

60

1

20

50

40

–7

–14

–49

–28

–35

–56

\ 227

3 142857= . .

11Number Systems

5. Given, 63 3 3 7 3 7= × × = is an irrational number because we know that 7 is an irrational number and 3 is rational number and product of an irrational and rational number is always irrational. 6. Since, the irrational number have non-terminating non-repeating decimals.Hence, the required irrational number between 0.2101 and 0.222 ... is 0.210120010002.... .7. (c) : Let x = 1.999 ... ...(i)Multiply (i) by 10, we get10x = 19.999... ...(ii)On subtracting (i) from (ii), we get10x – x = (19.999...) – (1.999...) ⇒ 9x = 18 \ x = 28. Let x = 2.8. Then x = 2.888 ... ...(i)Multiplying (i) by 10, we get 10x = 28.888 ... ...(ii)Subtracting (i) from (ii), we get (10x – x) = (28.888 ... – 2.888 ...)

⇒ 9x = 26 ⇒ x = 26/9 \ 2 8 269

. =

9. Let x = 0.27. Then, x = 0.272727 ... ...(i)Multiplying (i) by 100, we get 100x = 27.272727 ... ...(ii)Subtracting (i) from (ii), we get (100x – x) = (27.272727 ...) – (0.272727 ...)⇒ 99x = 27 ⇒ x = 27/99 = 3/11 \ 0.27 = 3/1110. Let x = 0.142857142857... ...(i)Multiplying (i) by 1000000, we get1000000 x = 142857.142857... ...(ii)Subtracting (i) from (ii), we get1000000x – x = (142857.142857...) – (0.142857...)

⇒ 999999x = 142857 ⇒ x = =142857999999

17

\ 0 142857 17

. =

11. Given, two irrational numbers are 0.202002000200002 ... and 0.203003000300003 ... .Then, 0.202200120001 ... and 0.202200220002 ... lies between the given numbers.

12. Firstly, we have to calculate the value of 15

25

and .

Now, 1.05 0.2–10

0

∴ =15

0 2. and

2.05 0.4–20

0

∴ =25

0 4.

Thus, required irrational number will lie between 0.2 and 0.4.Also, the irrational number have non-terminating non-repeating decimal. Hence, the required irrational numbers between 1/5 and 2/5 are 0.21010010001 ... and 0.31010010001... .13. (i) We have, 56/1000 = 0.056\ The decimal expansion of 56/1000 is terminating.

(ii) We have, 3 67

277

=

Now, on dividing 27 by 7, we have27.00000007 3.8571428...

–2160

40

50

10

30

20

60

–56

–35

–49

–7

–28

–14

–564

\ 277

3 857142= .

Thus, the decimal expansion of 3 67

is non-terminating repeating.14. Total number of households = 11Donating households = 9Not donating households = 11 – 9 = 2\ Fraction of households which are not donating = 2/11On dividing 2 by 11, we have

11 2.0000 0.1818–11 90 –88 20 –11 90 –88 2

\ 2/11 = 0.18Thus, 2/11 has non-terminating repeating decimal expansion.15. Let x = 2.218. Thenx = 2.218181818... …(i)Multiplying (i) by 10 and 1000, we get 10x = 22.18181818... …(ii) 1000x = 2218.181818... …(iii)Subtracting (ii) from (iii), we get (1000x – 10x) = (2218.181818.......) – (22.181818........)

⇒ 990x = 2196 ⇒ x = =2196990

12255

\ 2 218 12255

. =

16. Let, x = 1.32Then, x = 1.32222… …(i)Multiplying (i) by 10 and 100, we get 10x = 13.222… …(ii) 100x = 132.222… …(iii)Subtracting (ii) from (iii), we get 100x – 10x = (132.222…) – (13.222…)


⇒ 90x = 119 ⇒ x = 119/90Again, let y = 0.35Then, y = 0.353535… …(iv)Multiplying (iv) by 100, we get100y = 35.3535… ...(v)Subtracting (iv) from (v), we get100y – y = (35.3535…) – (0.3535…) ⇒ 99y = 35 ⇒ y = 35/99

\ 1 32 0 35 11990

3599

. .+ = + = +x y

= × + × = + = =119 11 35 10990

1309 350990

1659990

553330

17. (i) Let x = 0.324. Then, x = 0.324324324 ... ...(i)Multiplying (i) by 1000, we get

1000x = 324.324324 ... ...(ii)Subtracting (i) from (ii), we get (1000x – x) = (324.324324 ...) – (0.324324324 ...)

⇒ 999x = 324 ⇒ = = ∴ =x 324999

1237

0 324 1237

.

(ii) Let x = 2.2554Then, x = 2.25545454 ... ...(i)Multiplying (i) by 100 and 10000, we get 100x = 225.545454 ... ...(ii) 10000x = 22554.545454 ... ...(iii)Subtracting (ii) from (iii), we get (10000x – 100x) = (22554.545454 ...) – (225.545454 ...)

⇒ 9900x = 22329 ⇒ = = ∴ =x 223299900

24811100

2 2554 24811100

.

TOPIC 4 : REPRESENTING REAL NUMBERS ON THE NUMBER LINEIn earlier classes, we learnt how to represent integers on number line. Now, decimals numbers can also be represented on a number line. The method of representing decimal numbers on number line is known as process of successive magnification. In this method, we divide the distance between two numbers again and again into 10 equal parts.Let us take few examples for brief understanding.

ILLUSTRATIONS

Visualise 4.765 on the number line, using successive magnification.Sol. 4.765 lies between 4 and 5.(i) Divide the distance between 4 and 5 into 10 equal parts.(ii) Mark the point 4.7, which is third from the left of 5 and seventh from the right of 4.(iii) 4.76 lies between 4.7 and 4.8. Now, divide the distance between 4.7 and 4.8 into 10 equal parts.(iv) Mark the point 4.76 which is sixth from the right of 4.7 and fourth from the left of 4.8.(v) The point 4.765 lies between 4.76 and 4.77.(vi) Divide the distance between 4.76 and 4.77 into 10 equal parts, mark the point 4.765 which is fifth from the right of 4.76 and fifth from the left of 4.77.

4.14 54.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9

4.71

4.7 4.84.724.73

4.744.75

4.764.77

4.784.79

4.761

4.76 4.774.7624.763

4.764

4.765

4.7664.767

4.7684.769

Visualise 3 26. on the number line, upto 4 decimal places.

Sol. We have, 3 26 3 262626. . ...= \ 3.262626… lies between 3 and 4.(i) Divide the distance between 3 and 4 into 10 equal parts.(ii) Mark the point 3.2, which is second from the right of 3 and eighth from the left of 4.(iii) 3.26 lies between 3.2 and 3.3. Now, divide the distance between 3.2 and 3.3 into 10 equal parts.(iv) Mark the point 3.26, which is sixth from right of 3.2 and fourth from left of 3.3.(v) 3.262 lies between 3.26 and 3.27. Now divide the distance between 3.26 and 3.27 into 10 equal parts.(vi) Mark the point 3.262, which is second from the right of 3.26 and eighth from left of 3.27.(vii) 3.2626 lies between 3.262 and 3.263. Now divide the distance between 3.262 and 3.263 into 10 equal parts.(viii) Mark the 3.2626 which is sixth from the right of 3.262 and fourth from the left of 3.263.

3.13 43.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

3.21

3.2 3.33.223.23

3.243.25

3.263.27

3.283.29

3.261

3.26 3.273.2623.263

3.264

3.265

3.2663.267

3.2683.269

3.2621

3.262 3.2633.26223.2623

3.2624

3.2625

3.2626

3.26273.2628

3.2629

13Number Systems

Short Answer Type QuestionsSA Type I Questions (2 Marks)1. Visualise 10.28 on the number line, using successive magnification.SA Type II Questions (3 Marks)2. Visualise 5.38 on the number line upto 3 decimal places, using successive magnification.

3. Visualise 4.575 on the number line, using successive magnification.4. Visualise 13.625 on the number line, using successive magnification.Long Answer Type QuestionsLA Type Questions (4 Marks)5. Visualise 3.98 upto 4 decimal places on the number line using successive magnification.

YOURSELF14. Visualise the representation of 3.671 by method of successive magnification.

15. Visualise the representation of 1.3 on the number line upto 4 decimal places.

16. Visualise the representation of 5.37 on the number line upto 5 decimal places i.e., 5.37777.



1. 10.28 lies between 10 and 11.

10

10.2

10.1

10.21

10.2

10.22

10.3

10.23

10.4

10.24

10.5

10.25

10.6

10.26

10.7

10.27

10.8

10.28

10.9

10.29

11

10.3

(i) 10.2 lies between 10 and 11.

(ii) 10.28 lies between 10.2 and 10.3.


EXERCISE - 1.4

1. Visualise 3.765 on the number line, using successive magnification.

2. Visualise 4.26 on the number line, upto 4 decimal places.

2. We have, 5.38 = 5.3838…, which lies between 5 and 6.



(iii) 5.383 lies between 5.38 and 5.39.

5.15 65.2

5.35.4

5.55.6

5.75.8

5.9

5.315.3 5.45.32

5.335.34

5.355.36

5.375.38

5.39

5.3815.38 5.395.382

5.383

5.3845.385

5.3865.387

5.3885.389


4.57 4.571 4.572

4 4.1

4.5 4.51

4.573 4.574 4.575 4.576 4.577 4.578 4.579 4.58

4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5

4.52 4.53 4.54 4.55 4.56 4.57 4.58 4.59 4.6





13

13.6

13.62

13.1

13.61

13.621

13.2

13.62

13.622

13.3

13.63

13.623

13.4

13.64

13.624

13.5

13.65

13.625

13.6

13.66

13.626

13.7

13.67

13.627

13.8

13.68

13.628

13.9

13.69

13.629

14

13.7

13.63




5. We have, 3.98 = 3.9898 ..., which lies between 3 and 4.(i) Divide the distance between the points 3 and 4 into 10 equal parts.(ii) Mark the point 3.9, which is ninth from the right of 3 and first from the left of 4.(iii) 3.98 lies between 3.9 and 4.0. Now, divide the distance between 3.9 and 4.0 into 10 equal parts.(iv) Mark the point 3.98, which is eight from right of 3.9 and second from left of 4.0.(v) 3.989 lies between 3.98 and 3.99. Now, divide the distance between 3.98 and 3.99 into 10 equal parts.(vi) Mark the point 3.989, which is ninth from right of 3.98 and first from left of 3.99.(vii) 3.9898 lies between 3.989 and 3.99. Now, divide the distance between 3.989 and 3.99 in 10 equal parts.(viii) Mark the point 3.9898, which is eighth from right of 3.989 and second from left of 3.99.

3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4

3.9 3.91 3.92 3.93 3.94 3.95 3.96 3.97 3.98 3.99 4

3.98 3.981 3.982 3.983 3.984 3.985 3.986 3.987 3.988 3.989 3.99

3.989

3.9891

3.9892

3.9893

3.9894

3.9895

3.9896

3.9897

3.9898

3.9899 3.99

15Number Systems

TOPIC 5 : OPERATIONS ON REAL NUMBERSReal numbers are closed under addition, subtraction, multiplication and division (with non-zero divisor). In other words, we can say that if we add, subtract, multiply or divide two or more real numbers, the resulting number is also a real number.

Properties of Real Numbersz Irrational numbers satisfy the commutative, associative

and distributive laws for addition and multiplication.z The sum and difference of a rational and an irrational

number is always irrational.z The product and quotient of a rational (non-zero) and

an irrational number is always irrational.z The sum, difference, product and quotient of two irrational numbers is not necessarily an irrational number i.e.,

the resulting number may also be a rational number.

ILLUSTRATIONS

Classify the following numbers as rational or irrational with justification:

(i) 3 18 (ii) 28

343(iii) − 0 4. (NCERT Exemplar)

Sol. (i) We have, 3 18 3 3 3 2= × × = × =3 3 2 9 2 , which is an irrational number. We know that product of rational and irrational number is always irrational number.

(ii) We have, 28343

2 2 77 7 7

27

=× ×× ×

= , which is a

rational number.

(iii) We have, − = − = − ××

0 4 410

2 25 2

.

= − = −25

25

which is an irrational number

because 2 and 5 both are irrational numbers.

Subtract 3 2 7 3 2 5 3+( ) −( )from .

Sol. We have, ( 2 – 5 3 ) – (3 2 + 7 3 )= 2 – 5 3 – 3 2 – 7 3= ( 2 – 3 2 ) – 5 3 – 7 3 = –2 2 – 12 3

Divide 3 3 3 3 1+( ) +( ) by 2 .

Sol. We have, ( )( )

( )( )

3 3 32 3 1

3 3 12 3 1

32

++

= ++

=

147

75 is not a rational number as 147

and 75 are not rational. State whether it is true or false. Justify your answer.Sol. Since, 147 3 7 7 7 3= × × = , which is an irrational number and 75 5 5 3 5 3= × × = , which is also an irrational number.

Now, 14775

7 35 3

75

= = , which is a rational

number.Thus, the given statement is false.Q Quotient of two irrational numbers need not to be an irrational number.

YOURSELF17. Classify the following numbers as rational or irrational. (i) 8 20 3 20/ (ii) 6 3/ (iii) ( ) ( )3 2 7 3 2 7+ − − +

(Ans. (i) rational (ii) irrational (iii) rational)

18. Write the value of ( ) ( ).2 3 2 3+ + − (Ans. 4)

19. Add ( ).2 2 5 3 2 3 3+ −) and ( (Ans. ( )3 2 2 3+ )

20. Divide ( ) ( )9 7 72 3 7 24+ +by . (Ans. 3)


Real Numbers

IrrationalNumbers

Natural Numbers

Whole Numbers

Integers

Rational Numbers


Existence of Square Root for a given Positive Real NumberLet a be a real number such that a > 0. Then, a b= means b2 = a and b > 0. The value of b is called the positive square root of the positive real number a.Representation of a on the number line (a > 0)Let a be a given positive real number. To represent square root of a on number line, we shall follow the given steps :Step 1 : Finding a geometrically. Draw a line segment, AB = a units and extend it to C such that BC = 1 unit.Find the mid-point O of AC.With O as centre and OA as radius, draw a semicircle.Now, draw BD ⊥ AC, intersecting the semicircle at D.Now, we have to prove that BD = aStep 2 : Proving that BD = a .We have, AB = a units and BC = 1 unit. So, AC = (a + 1) unitsNow, OD = OC (radii of the same semicircle)

= = +12

12

1AC a( ) units

And, OB = (OC – BC) = + −

= −a a1

21 1

21 units units( )

\ BD OD OB a a2 2 2 2 214

1 14

1= − = + − −( ) ( ) ( ) = + − − = ×

=

14

1 1 14

42 2. {( ) ( ) }a a a a ⇒ BD = a units

Thus, we have proved that BD = a .Step 3 : Representing a on the real line.Take BC produced as the number line with the origin at B.Now, BC = 1 unit.So, B and C denote 0 and 1 respectively.With B as centre and BD as radius, draw an arc, meeting BC produced at E. Then, BE = BD = a units.Hence, the point E represents a.Thus, for every positive real number a, a is also a positive real number and hence a exists.Let a (> 0) be a real number and n be a positive integer, such that a an n1/ = is a real number, then n is called exponent, a is called radicand and ‘ ’ is called radical sign.Note : an is called a surd.

Some Important IdentitiesLet x and y be any two positive real numbers. Then, we have the following identities :

z x y xy⋅ = xy

xy

= ( )( )x y x y x y+ − = −2

z ( )x y x xy y± = ± +2 2 ( )( )x y x y x y+ − = −

z ( )( ) ,w x y z wy wz xy xz+ + = + + + where w and z are positive real numbers.

ILLUSTRATIONS

Express 5 42. geometrically and represent it on the number line.Sol. Draw a line segment, AB = 5.42 units and extend it to C such that BC = 1 unit and AC = 6.42 units. Mark M as mid-point of AC. Draw a semicircle with centre at M and radius = MA = MC = 6.42/2 = 3.21 units. Now, draw BD ⊥ AC and it meets the semicircle at D. In right angled DMBD, MD = 3.21 units and MB = MC – BC = 2.21 units.

A M B C E

D

5.42

5.42

2.21

3.21

5.42 (0)

Now, BD2 = MD2 – MB2

= (3.21)2 – (2.21)2 = (3.21 + 2.21) × (3.21 – 2.21) = 5.42

⇒ BD = 5 42. units

D

A Oa unit 1 unit

B C E

17Number Systems

To represent 5 42. on the number line, let us treat the line BC as the number line, with B as zero, C as 1, and so on. Now, taking B as centre and BD as radius = 5 42. units, we draw an arc which meets the line AC (produced) at E. BD = BE =

5 42. units.\ E represents 5 42. .

Solve : ( )7 3 2 2−Sol. ( ) ( ) ( )7 3 2 7 3 2 2 21 22 2 2− = + − × [ (x – y)2 = x2 + y2 – 2xy]

= + − = −49 18 42 2 67 42 2.

Simplify the following expressions :

(i) 3 6 2 6

3

+ (ii) (3 + 3 )(2 + 2 )

Sol. (i) Given, 3 6 2 63

5 63

+ =

= = =

5 63

5 2 Q

xy

xy

(ii) (3 + 3 ) (2 + 2 )= + + + ×6 3 2 2 3 3 2 = + + +6 3 2 2 3 6 [ x y xy⋅ = ]

If x y= + = −7 5 7 5 and , evaluate : (i) xy (ii) x2 + y2 (iii) (x2 – y2)2

Sol. Given, x y= + = −7 5 7 5, (i) xy = + × −( ) ( )7 5 7 5= −( ) ( )7 52 2 {∵ (x + y) × (x – y) = x2 – y2}= 7 – 5 = 2(ii) x y2 2 2 27 5 7 5+ = + + −( ) ( ) = + + × + + − ×{ } { }7 5 2 7 5 7 5 2 7 5

Q( )x y x y xy± = + ± 2 2= + + −( ) ( )12 2 35 12 2 35 [ ]Q x y xy⋅ == 12 + 12 = 24(iii) x y2 2 2 27 5 7 5− = + − −( ) ( )= + + × − + − ×{ } { }7 5 2 7 5 7 5 2 7 5

Q( )x y x y xy± = + ± 2 2= + − + =12 2 35 12 2 35 4 35Now, ( ) ( )x y2 2 2 24 35 16 35 560− = = × =

YOURSELF21. Express 4 5. geometrically and represent it on the number line.

22. Simplify : ( ) ( )5 7 2 5+ + (Ans. 10 5 5 2 7 35+ + + )

23. Simplify : ( )11 5 2− (Ans. 16 – 2 55)

24. Simplify : 9 32

9 32

+

−

(Ans. 159/2)

25. If x y x y= + = − +2 5 3 2 5 3 2 2 and evaluate , . (Ans. 46)


Rationalising Real NumbersSuppose we are given an expression whose denominator is irrational (or contains square root). To convert the denominators in such cases free from square roots, we multiply both the numerator and denominator of the given number by a suitable factor. This factor is called the rationalising factor and this process is known as rationalising the denominator.Note : For rationalising the denominator, we multiply the numerator and denominator by conjugate of denominator to remove the radical sign from the denominator.If the product of two irrational numbers is rational then each one is called the rationalising factor of the other. i.e., if a and b are integers and x, y are natural numbers, then(i) ( ) ( )a b a b+ − and are R.F. of each other.(ii) ( ) ( )a b x a b x+ − and are R.F. of each other.

(iii) ( ) ( )x y x y+ − and are R.F. of each other.

Some General Rationalising Factors

Number R.F.

a x/ x

a a b x/( )± ( )a b x

1/( )a x b y± ( )a x b y


ILLUSTRATIONS

Rationalise the denominator of 1

7 3 2+.

(NCERT)Sol. Given, 1

7 3 2+On rationalising, we get 1

7 3 27 3 27 3 2+

× −−

= −−

= −−

= −7 3 27 3 2

7 3 249 18

7 3 2312 2( ) ( )

Simplify the expression 7 2

7 2

7 2

7 2

−+

− +−

.

Sol. We have, 7 27 2

7 27 2

−+

− +−

On rationalising, we get

= − × −+ × −

− + × +− × +

( ) ( )( ) ( )

( ) ( )( ) ( )

7 2 7 27 2 7 2

7 2 7 27 2 7 2

= −−

− +−

= −−

− +−

( )( ) ( )

( )( ) ( )

( ) ( )7 27 2

7 27 2

7 27 4

7 27 4

2

2 2

2

2 2

2 2

= − − +( ) ( )7 23

7 23

2 2

= + − × × − +13

7 2 2 2 7 72 2 2[{( ) ( ) } {( )

( ) }]2 2 2 72 + × ×

= + − − + +

= − − − =− −

= −

13

7 4 4 7 7 4 4 7

13

11 4 7 11 4 74 7 4 7

38 73

[{ } { }]

[ ]

If ‘a’ and ‘b’ are two rational numbers and 2 3

2 33

+−

= +a b , then find the values of

a and b.Sol. We have, 2 3

2 3+−

On rationalising, we get( )( )

( )( )

( )( )

( )( )

2 32 3

2 32 3

2 32 3

2 34 3

2

2 2

2+−

× ++

= +−

= +−

= + = + + × ×( ) [ ( ) ]2 3 2 3 2 2 32 2 2

= + + = +( ) ( )4 3 4 3 7 4 3

∴ +−

= +2 32 3

7 4 3 …(i)

Given, 2 32 3

3 7 4 3 3+−

= + ⇒ + = +a b a b( ) ( )…[Using (i)]

On comparing, we get a = 7 and b = 4

If and a b= +−

= −+

3 2

3 2

3 2

3 2, then find

the value of a2 + b2.

Sol. Given, a = +−

3 23 2

On rationalising, we get ( ) ( )( ) ( )

3 2 3 23 2 3 2+ × +− × +

= +−

= +−

= +( )( ) ( )

( ) ( )3 23 2

3 23 2

3 22

2 2

22

= + + × × = +3 2 2 3 2 5 2 6 ...(i)and

b = −+

= − × −+ × −

= −−

3 23 2

3 2 3 23 2 3 2

3 23 2

2

2 2( ) ( )( ) ( )

( )( ) ( )

= −−

= −( ) ( )3 23 2

3 22

2

= + − × × = −3 2 2 3 2 5 2 6 …(ii)Now, a2 + b2 = ( ) ( )5 2 6 5 2 62 2+ + − [Using (i) and (ii)]⇒ a2 + b2 = (5)2 + (2 6 )2 + 2 × 5 × 2 6

+ (5)2 + (2 6 )2 – 2 × 5 × 2 6⇒ a2 + b2 = 25 + 24 + 25 + 24 = 98

If x = +−

+ −+

+ −+

2 3

2 3

2 3

2 3

3 1

3 1, find the

value of xx

22253+

.

Sol. Given, x = +−

+ −+

+ −+

2 32 3

2 32 3

3 13 1


x = + × +− × +

+ − × −+ × −

( ) ( )( ) ( )

( ) ( )( ) ( )

2 3 2 32 3 2 3

2 3 2 32 3 2 3

+ − × −

+ × −( ) ( )( ) ( )

3 1 3 13 1 3 1

⇒ x = +−

+ −−

+ −−

( )( ) ( )

( )( ) ( )

( )( ) ( )

2 32 3

2 32 3

3 13 1

2

2 2

2

2 2

2

2 2

⇒ x = +−

+ −−

+ −−

( ) ( ) ( )2 34 3

2 34 3

3 13 1

2 2 2

⇒ x = + + − + −( ) ( ) ( )2 3 2 3 3 12

2 22

⇒ x = + + + + − + + −( ) ( )4 3 4 3 4 3 4 3 3 1 2 32

⇒ x = + − = + − = −14 4 2 32

14 2 3 16 3

\ 1 116 3

1 16 316 32 2x x

=−

⇒ = +−( ) ( )

= +−

= +16 3256 3

16 3253

\ 253 16 3x

= +( )

Now, xx

22

2 2253 16 3 16 3+ = − + +( ) ( )

= + − × × +

+ + × ×

( ) ( ) ( )

( )

16 3 2 16 3 16

3 2 16 3

2 2 2

2

= + = + = × =2 16 3 2 256 3 2 259 5182 2{( ) ( ) } { }

19Number Systems

Objective Type QuestionsMCQs (1 Mark)

1. 10 15× is equal to(a) 6 5 (b) 5 6 (c) 25 (d) 10 5 (NCERT Exemplar)

2. After rationalising the denominator of 7

3 3 2 2−, we get the denominator as

(a) 13 (b) 19(c) 5 (d) 35

(NCERT Exemplar)

YOURSELF26. Rationalise the denominator of 5

3 5−. Ans. − +

52

3 5( )

27. If 5 2 37 4 3

3++

= +a b , where ‘a’ and ‘b’ are rational numbers. Then find the values of a

and b. (Ans. a = 11, b = –6)

28. If x = −( ),4 15 find the value of (x2 + 1/x2). (Ans. 62)

29. If x = 1 – 2, find the value of

(i) xx

+ 1 (ii) xx

− 1 (iii) xx

22

1+ (Ans. (i) –2 2 (ii) 2 (iii) 6)




EXERCISE - 1.51. Classify the following numbers as rational or irrational :(i) 2 5− (ii) ( )3 23 23+ −

(iii) 2 7

7 7 (iv) 1

2(v) 2p

2. Simplify each of the following expressions :(i) ( )( )3 3 2 2+ + (ii) ( )( )3 3 3 3+ −

(iii) ( )5 2 2+ (iv) ( )( )5 2 5 2− +

3. Recall, p is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, p = c/d. This seems to contradict the fact that p is irrational. How will you resolve this contradiction?4. Represent 9 3. on the number line.5. Rationalise the denominators of the following:

(i) 1

7 (ii) 1

7 6−

(iii) 1

5 2+ (iv)

1

7 2−

3. Rationalise the denominator of 14 3( )

.−

(a) ( )2 3+ (b) ( )2 3−(c) 1 (d) 4 3+

Fill in the Blanks (1 Mark)

4. 3 × 5 is a/an ______ number.

5. The rationalising factor of 1

3 2 7 5− is

_____.


6. Divide ( ) ( ).3 7 3 7+ − by

7. Evaluate 6

12 3−.

8. Add ( ).3 2 5 3 2 2 3+ −) and (


9. Rationalise the denominator of 4 3 5 2

48 18

++

.

10. Simplify the expression ( ) ( ).4 5 3 2 3 5 5 2− +

11. Rationalise the denominator of 1

2 5 3( ).

−


12. If 3

3 1

5

3 13

++

−= +a b , where ‘a’ and

‘b’ are rational numbers, find a and b.13. If a = +8 7, then find the value of

aa

222

1− + .


14. If x = −+

7 3

7 3 and y = +

−7 3

7 3, then find

x xy yx xy y

2 2

2 2+ +− +

.

15. If x = −5 212

, then show that

x

xx

xx

x3

32

21 5 1 1 0+

− +

+ +

=

16. Simplify : 7 3

10 3

2 5

6 5

3 2

15 3 2+−

+−

+

17. Show that 1

3 8

1

8 7

1

7 6( ) ( ) ( )−−

−+

−

−−

+−

=1

6 5

1

5 25

( ) ( ).

18. Represent geometrically the number 2 3. on the number line.

1. (b) : We have, 10 15×= × × × = × × × = ⋅2 5 5 3 2 5 3 5 { }Using xy x y

= × × × =( ) ( )2 3 5 5 5 6

2. (b) : We have, 73 3 2 2−

On rationalising, we get 73 3 2 2

3 3 2 23 3 2 2−

× ++

= +−

7 3 3 2 23 3 2 22 2

( )( ) ( )

= +−

7 3 3 2 227 8

( ) = +7 3 3 2 219

( )

Thus, after rationalising the denominator of 73 3 2 2−

we get, 19 as denominator.

3. (d) : We have, 14 3−

On rationalising, we get1

4 34 34 3

4 34 3

4 34 3

4 32 2−× +

+= +

−= +

−= +

( ) ( )

4. Since product of a rational and an irrational number is an irrational number.

\ 5 3× is an irrational number.

5. Rationalising factor of 13 2 7 5

3 2 7 5−

= +

6. We have, 3 73 7+−


( )( )

( )( )

( )( ) ( )

3 73 7

3 73 7

3 73 7

2

2 2+−

× ++

= +−

= + + × ×−

( ) ( )3 7 2 3 73 7

2 2

= + + × ×−

= +−

3 7 2 7 34

10 2 214

= +−

5 212

21Number Systems

7. We have, 612 3

62 3 3

63−

=−

=

On rationalising, we get 63

33

6 33

2 3× = =

8. We have, ( ) ( )3 2 5 3 2 2 3+ + −

= + + − = +3 2 2 5 3 2 3 4 2 3 3

9. We have, 4 3 5 248 18

4 3 5 24 3 3 2

++

= ++

On rationalising, we get4 3 5 24 3 3 2

4 3 3 24 3 3 2

4 3 5 2 4 3 3 24 3 3 22 2

++

× −−

= + −−

( )( )( ) ( )

= × − × + × − ×−

4 3 4 3 4 3 3 2 5 2 4 3 5 2 3 248 18

= − + − = + = + = +48 12 6 20 6 3030

18 8 630

2 9 4 630

9 4 615

( )

10. We have, ( )( )4 5 3 2 3 5 5 2− += 4 3 5 5 4 5 5 2 3 3 2 5 3 5 2 2× × + × × − × × − × ×= 12 5 20 10 9 10 15 2× + − − ×= 60 11 10 30+ − = 30 11 10+

11. We have, 12 5 3−

On rationalising, we get1

2 5 32 5 32 5 3

2 5 32 5 32 2−

× ++

= +−

( )( )

( )( ) ( )

= +−

= +( )2 5 320 3

2 5 317

12. We have, 33 1

53 1+

+−

On rationalising both the terms, we get3

3 13 13 1

53 1

3 13 1( )

( )( ) ( )

( )( )+

× −−

+−

× ++

= −−

+ +−

3 3 13 1

5 3 13 12 2 2 2

( )( ) ( )

( )( ) ( )

= −−

+ +−

= − + +3 3 33 1

5 3 53 1

3 3 32

5 3 52

= − + + = + = +3 3 3 5 3 52

8 3 22

4 3 1 ...(i)

Now, 33 1

53 1

3+

+−

= +a b (Given)

⇒ 1 4 3 3+ = +a b [From (i)]On comparing, we get a = 1 and b = 4

13. Given, a = +8 7 ...(i)

\ 1 18 7a

=+

On rationalising, we get1 1

8 78 78 7

8 78 72 2a

=+

× −−

= −−( ) ( )

⇒ 1 8 78 7

8 7a= −

−= − ...(ii)

Subtracting (ii) from (i), we get

aa

− = + − −1 8 7 8 7( ) ( ) ⇒ aa

− =1 2 7

On squaring both sides, we get

aa

−

=1 2 7

22( ) ⇒ a

aa

a2

21 2 1 28+ − × × =

Hence, aa

22

1 2 28+ − =

14. We have, x = −+

7 37 3

and y = +−

7 37 3

⇒ 1 7 3

7 31

y yx= −

+⇒ =

On rationalising 'x' we get

x = −

+× −

−( )( )

( )( )

7 37 3

7 37 3

⇒ x = −−

= + − × ×−

( )( ) ( )

( ) ( )7 37 3

7 3 2 7 37 3

2

2 2

2 2

= + −7 3 2 214

= − = −10 2 214

5 212

\ 1 25 21x

=−


5 215 215 21x

=−

× ++

( )( )

= +−

= +−

2 5 215 21

2 5 2125 212 2

( )( ) ( )

( )

⇒ 1 2 5 214

1 5 212x x

= + ⇒ = +( )

\ y = +5 212

We know that, x2 + xy + y2 = (x + y)2 – xy and x2 – xy + y2 = (x – y)2 + xy

Now, x y+ = − + + = − + + =5 212

5 212

5 21 5 212

5

x y− = − − + = − − − = −5 212

5 212

5 21 5 212

21

xy = − × + = − = − = =5 212

5 212

5 214

25 214

44

12 2( ) ( )

Now, x xy yx xy y

x y xyx y xy

2 2

2 2

2

2+ +− +

=+ −− +

( )( )

= −

− +=

+= =( )

( )5 121 1

2421 1

2422

1211

2

2

15. We have, x = −5 212

\ 1 2

5 21x=

−=

−× +

+2

5 215 215 21

⇒ 1 2 5 215 21

10 2 2125 21

5 2122 2x

=+−

= +−

= +( )( ) ( )


\ xx

+ = − + + =1 5 212

5 212

5

Now, xx

xx

xx

xx

+

= + + × × +

1 1 3 1 133

3

⇒ 5 1 3 5 1 125 153 33

33( ) = + + × ⇒ + = −x

xx

x⇒ x

x3

31 110+ =

Also, xx

xx

xx

+

= + + × ×1 1 2 12

22

⇒ ( )5 2 1 1 232 22

22− = + ⇒ + =x

xx

xOn solving L.H.S. of given equation, we have

xx

xx

xx

33

22

1 5 1 1 110 5 23 5+

− +

+ +

= − × +

= 0 = R.H.S.

16. We have, 7 310 3

2 56 5

3 215 3 2+

−+

−+

On rationalising each term separately, we get

7 310 3

10 310 3

7 3 10 310 32 2( )

( )( )

( )( ) ( )+

× −−

= −−

= × − ×−

= − × = −7 3 10 7 3 310 3

7 30 7 37

30 3( )

2 56 5

6 56 5

2 5 6 56 52 2( )

( )( )

( )( ) ( )+

× −−

= −−

= × − ×−

= − = −2 5 6 2 5 56 5

2 30 101

2 30 10( )

and 3 2

15 3 215 3 215 3 2

3 2 15 3 215 3 22 2( )

( )( )

( )( ) ( )+

× −−

= −−

= × − ×−

= −−

3 2 15 3 2 3 215 18

3 30 183

=− −

−= −

3 6 303

6 30( )

Now, 7 310 3

2 56 5

3 215 3 2+

−+

−+

= − − − − −( ) ( ) ( )30 3 2 30 10 6 30

= − − + − +30 3 2 30 10 6 30 = –3 + 10 – 6 = 1

17. We have, L.H.S., ss

=−

−−

+−

−−

+−

13 8

18 7

17 6

16 5

15 2( ) ( ) ( ) ( ) ( )

By rationalising, we get1

3 83 83 8

18 7

8 78 7

17 6( ) ( ) ( )−

× ++

−−

× ++

+−

× +

+−

−× +

++

−× +

+7 67 6

16 5

6 56 5

15 2

5 25 2( )

= +−

− +−

+ +−

3 83 8

8 78 7

7 67 62 2 2 2 2 2( ) ( )

( )( ) ( )

( )( ) ( )

− +

−+ +

−( )

( ) ( )( )

( ) ( )6 5

6 55 2

5 22 2 2 2

= +−

− +−

+ +−

3 89 8

8 78 7

7 67 6

( ) ( ) − +−

+ +−

( ) ( )6 56 5

5 25 4

= + − − + + − − + +3 8 8 7 7 6 6 5 5 2= 3 + 2 = 5 = R.H.S.18. Draw a line segment AB = 2.3 units and extend it to C such that BC = 1 unit and AC = 3.3 units. Let O be the mid-point of AC.Now, we will draw a semicircle with centre O and radius

OA, where OA AC= =2

1 65. units.

Let us draw BD perpendicular to AC passing through a point B, intersecting the semicircle at a point D.In DOBD, BD2 = OD2 – OB2

⇒ BD2 = (1.65)2 – (0.65)2 = (1.65 + 0.65) (1.65 – 0.65)⇒ BD = 2 3. unitsTo represent 2 3. on the number line, let us treat the line BC as the number line, with B as zero, C as 1, and so on.Now, draw an arc with centre B and radius BD, meeting AC produced at E, such thatBE = BD = 2 3. units. Thus, E represents 2 3. .

TOPIC 6 : LAWS OF EXPONENTS FOR REAL NUMBERSLet x (> 0) be a real number and m, n be two rational numbers, then the following laws hold good for rational exponents :(i) xmxn = xm+n (ii) xm ÷ xn = xm–n (iii) (xm)n = xmn

(iv) xx

nn

− = 1 (v) xy

yx

m m

=

−(vi) xm/n = (xm)1/n = (x1/n)m i.e., x x xm n mn n m/ ( )= =

(vii) (xy)m = xmym (viii) xy

xy

m m

m

= (ix) x xy y y y nn= × × ... times

where x, y are positive real numbers and m, n are rational numbers.Note : For every non-zero real number a, a0 = 1.

23Number Systems

ILLUSTRATIONS

Evaluate the following :

(i) 5 8 271 3 1 3 3 1 4( )/ / /

+{ }(ii) [811/2(641/3+ 1251/3)3]1/4

Sol. (i) We have, 5 8 2713

13 3

14

( )+

= +

5 2 3313 3

13

314

( ) ( ) = +

× ×5 2 3

3 13

3 13

314

[ ( ) ]Q x xm n m n= ×

= + = ⋅ = ⋅ =

= =

+ +

×

5 2 3 5 5 5

5 5

314 3

14 1 3

14

414

4

( ) ( ) ( )

( )

Q x x xm n m n

114 5=

(ii) We have, [811/2(641/3+ 1251/3)3]1/4

= [(92)1/2{(43)1/3 + (53)1/3}3]1/4

= [92×1/2{43×1/3 + 53×1/3}3]1/4 [ (xm)n = xm × n]= [9(4 + 5)3]1/4 = [9 ⋅ (9)3]1/4

= [91+3]1/4 = [94]1/4 = 94×1/4 = 9.

Simplify the following :

(i) (81)3/4 (ii) 6425

3 2

− /

(iii) (0.00032)–2/5

Sol. (i) We have, ( ) ( ) ( ) ( )81 3 3 3 2734 4

34

4 34 3= = = =

×

[ (xm)n = xm × n]

(ii) We have, 6425

2564

32

32

=

=

− −

Q

xy

yx

m m

=

=

=

×58

58

58

2

2

32 2

32 2 3

2 [ (xm)n = xm × n]

= =58

125512

3

3

(iii) We have, (0.00032)–2/5

=

=

−32

100000100000

32

25

25 Q

xy

yx

m m

=

−

=

=

=

×10

2102

102

5

5

25 5

25 5 2

5 [ (xm)n = xm × n]

=

= = =10

2102

1004

252 2

2( )( )

Simplify :

( ) . ( . )( )

625 0 00001 729 0 008243 125

34 5 3 23

5 25

− −

−

× × ×{ }×{ }

Sol. We have,

( ) . ( . )

( )

625 0 00001 729 0 008

243 125

34 5 3 23

5 25

− −

−

× × ×{ }×{ }

={ } ×

× ×

−−

( ) ( )

(

5 1100000

9 81000

3

4 314

15 3

13

213

5515 3 2

155) ( )× { }−

=×

× ×

× − × × ×−

5 110

9 210

4 3 14

5 15 3 1

33 2

( )

113

5 15

3 2 153 5

× × − ××

( )

[ (xm)n = xm × n]

=× ( ) × ×

×

−× − ×

× − ×

( ) ( )( )

( )

5 110

9 210

3 5

33 2 1

3

3 2 15

=× × ×

×

=× × ×

−−

− ×

−( ) ( ) ( )5 110

3 210

3 5

5 110

3 102

3 22

6 15

3 222

6 53 5× − /

= × ×× ×

−

−( ) ( )

/5 3 510 3 5

3 2 2

6 5 = × × ×− + −5 3 3 510

3 2 2 1 6 5/

= ×− + −5 3

10

1 65 2 1

= × =5 310

3 510

1 5 5/

If x, y, z are rational numbers, then show

that aa

aa

x y z

y x z

y

x

z( )

( ) .−

− ÷

= 1

Sol. On solving L.H.S., we have aa

aa

x y z

y x z

y

x

z( )

( )

−

− ÷

=

÷

= ÷[ ] ÷ ÷[ ]

−

−− −a

aaa

a a a axy xz

yx yz

yz

xzxy xz yx yz yz xz

= [ ] ÷ [ ]− − + −a axy xz xy yz yz xz [ xm ÷ xn = xm–n]

= a–xz + yz ÷ ayz – xz = a–xz + yz – yz + xz = a0 = 1 = R.H.S.

If l, m, n are rational numbers, then simplify

xx

xx

xx

l

m

l m lm m

n

m n mn n

l

n l n

−

+ −

−

+ −

−

+ −

×

×

2 2 2 2 2 2 ll

.

Sol. We have,

xx

xx

xx

l

m

l m lm m

n

m n mn n

l

n l n

−

+ −

−

+ −

−

+ −

×

×

2 2 2 2 2 2 ll

= × ×+ + − + + − + + −( ) ( ) ( )( ) ( ) ( )x x xl m l m lm m n m n mn n l n l nl2 2 2 2 2 2

[ xm ÷ xn = xm – n]


= × ×+ + +x x xl m m n n l3 3 3 3 3 3

[∵ (a + b) (a2 + b2 – ab) = a3 + b3]

= + +x l m n2 3 3 3( ) [∵ xm × xn = xm+n]

If 9 3 27

3 2

1729

1 2 2

3

n n n

m

+ − −× −×

=( )

( ),

/

prove that

m – n = 2.

Sol. We have, 9 3 273 2

1729

1 2 2

3

n n n

m

+ − −× −×

=( )( )

/

⇒ ( ) ( )( ) ( )

( / ) ( )3 3 33 2

13

2 1 2 2 3

3 3 6

n n n

m

+ − × −× −×

=

[ (xm)n = xm × n, (x × y)m = xmym]

⇒ 3 3 33 8

13

2 1 3

3 6

( )n n n

m

+ × −×

=

⇒ 3 33 8

32 1 3

36

( )n n n

m

+ +−−

×= [ xm × xn = xm + n]

⇒ 3 33 8

33 2 3

36

n n

m

+−−

×=

⇒ 3 3 1

3 83

3 2

36

n

m( )−

×= −

⇒ 3 83 8

33

36

n

m××

= − ⇒ 33n–3m = 3–6

On comparing the power of 3, we get3n – 3m = –6⇒ n – m = –2 ⇒ m – n = 2

YOURSELF

30. Simplify : ( ) ( )( ) ( )

/ /

/ /

25 24316 8

3 2 3 5

5 4 4 3

××

(Ans. 153 × 2–9)

31. Prove that 2 3 4

10 5

3 5

4 610

1 2 1 3 1 4

1 5 3 5

4 3 7 5

3 5

/ / /

/ /

/ /

/ .× ×

×÷ ×

×=−

−

−

32. If a bx

x

x

x= =−

−

−

+2

2

2

2

1

2 1, and a – b = 0, then find the value of x. (Ans. –1)

33. Simplify : 6 6

6

2 3 73

63

/ × (Ans. 6)

34. Prove that x x xa ba c

b cb a

c ac b1

11

11

1

1−−

−−

−−

=

( ) ( ) ( ). . .


EXERCISE - 1.6


25Number Systems

1. Find : (i) 641/2 (ii) 321/5

(iii) 1251/3

2. Find : (i) 93/2 (ii) 322/5

(iii) 163/4 (iv) 125–1/3

3. Simplify :

(i) 22/3 · 21/5 (ii) 1

33

7

(iii) 11

11

12

14

(iv) 71/2 · 81/2

Objective Type QuestionsMCQs (1 Mark)

1. 2234 equals

(a) 216

− (b) 2–6 (c) 2

16 (d) 26

(NCERT Exemplar)2. Value of ( )81 24 − is(a) 1/9 (b) 1/3 (c) 9 (d) 1/81

(NCERT Exemplar)3. (125)–2/3 =

(a) 5 (b) –5 (c) 152 (d)

−15

(NCERT Exemplar)Fill in the Blanks (1 Mark)

4. The value of ( )22 7 0 is ______.


5. Simplify : 5 2−


6. Simplify : 3125243

4 5

− /

7. Find the value of (256)0.16 × (256)0.09.


8. Simplify : ( ) ( )

( ) ( )

/ /

/ /

36 1024

81 125 2

3 2 3 5

5 4 4 3 5

×× ×

9. Simplify : ( ) ( )2 5 8 2 24 345 5 3 445− +

10. Simplify : 16 2 4 216 2 2 2

1

2 2

× − ×× − ×

+

+ +

n n

n n

11. Simplify : 2564

256625

164

125

3

1 4

2 3+

+

− /

/

12. Assuming that x, y are positive real numbers,

then simplify ( ) ( ) ./ /x y x y− −÷2 3 4 1 2

13. If 2x = 25y = 1000z, then show that z xyx y

=+

23 6

.


14. Simplify : 252 5 6 294 3 16

− + −

15. Simplify : 8136

259

52

3 4 3 2 3

×

÷

− − −/ /

1. (c) : We have, 2 2 2234 23 1 4 2 1 3 1 4= =[ ] [( ) ]/ / /

[ ]/Q x xn n= 1

= = =⋅

[ ]/ /2 2 22 3 1 423

14

16 [Q (xm)n = xm×n]

2. (a) : We have, ( )81 24 − = 181 24

( ) Q xx

nn

− =

1

= 181 2 4( ) / Q x xn n= 1/

= 192 1 2( ) / = 1

9[ ( ) ]Q x xm n m n= ×

3. (c) : We have, (125)–2/3 = (53)–2/3 = 53×(–2/3) [Q (xm)n = xm × n]

= 5–2 = 1/52 Q xx

nn

− =

1

4. We have, [(22)7]0 = [22 × 7]0 [ (xm)n = xm × n]= [214]0 = 214 × 0 = 20 = 1 [Q x0 = 1]

5. 5 5 52 2 1 2 1− − −= =( ) / [ (xm)n = xm × n]

= 15

Q x

xn− =

1

6. We have, 3125243

2433125

45

45

=

− Q x

yyx

m m

=

−

=

=

35

35

545 4

[ (xm)n = xm × n]

= 81625

7. We have, (256)0.16 × (256)0.09

= (256)0.16 + 0.09 [∵ (a)m × (a)n = (a)m+n]

= = =( ) ( ) ( ).256 256 2560 2525

10014

= [(4)4]1/4 = 4 [∵ (am)n = (a)mn]


8. We have, ( ) ( )

( ) ( )

36 1024

81 125 2

6 4

3 5 2

32

35

54

43 5

2 32

5 35

4 54

3 43 5

×

× ×

= ×

× ×

× ×

× ×

[ (xm)n = xm × n]

= ×× ×6 4

3 5 2

3 3

5 4 5

= × ×× ×

= × ×× ×

( ) ( )2 3 23 5 2

2 3 23 5 2

3 2 3

5 4 5

3 3 6

5 4 5[ (x × y)m = xm × ym]

= × = ×− + − −2 35

2 35

3 5 6 3 5

4

4 2

4 [ xm × xn = xm + n ; xm ÷ xn = xm – n]

=×

=29 625

165625

4

9. We have, ( ) ( )2 5 8 2 24 345 5 3 445− +

= − +× × × ×

2 5 2 2 24 3 1

45 35 3 4 1

45

Q a a a an n m n m n= = ×1/ ;( )

= − × + ×× × × × × ×

2 5 2 2 24 3 1

514

35

3 4 15

14 = − × + ×2 5 2 2 2

35

35

35

= − + = − = − = −2 1 5 2 2 2 2 2 2 835

35 35 5( ) ( )

10. We have, 16 2 4 216 2 2 2

1

2 2× − ×× − ×

+

+ +

n n

n n

= × − ×× − ×

= −−

+

+ +

+ +

+ +2 2 2 22 2 2 2

2 22 2

4 1 2

4 2 2

5 2

6 3

n n

n n

n n

n n [ xm × xn = xm + n]

= −−

= −−

=+ +

+ +

+ +

+ +2 2

2 2 2 22 2

2 2 212

5 2

5 2

5 2

5 2

n n

n n

n n

n n. . ( )

11. We have, 2564

256625

164125

3

1 4

2 3+

+

− /

/

= ×× ×

+

+

=−5 5

4 4 4625256

125643

1 4 2 3/ /Q

xy

yx

m

m

= +

+

( )( )

/

/

/ /54

54

54

2 1 2

3 1 3

4

4

1 4 3

3

2 3

= + +

= + +5

454

54

54

54

2516

2= + + =20 20 25

166516

12. We have, ( ) ( )/ /x y x y− −÷2 3 4 1 2

=−

−( ) ( )

( ( ))

/ / /

/ /x yx y

1 2 2 3 4 1 2

1 2 1 2 Q x xn n= 1/

=× − ×

−x yx y

( / ) ( / ) ( / )

/ / /( )

1 2 2 3 4 1 2

1 2 1 2 1 2 [ (xm)n = xm × n]

= =−

− ×

−

−x y

x yx yx y

1 3 2

1 2 1 2 1 2

1 3 2

1 2 1 4

/

/ ( / ) ( / )

/

/ / =+

+y

x

2 14

1 3 1 2/ /

[ xm ÷ xn = xm – n]

=yx

9 4

5 6

/

/

13. Let 2x = 25y = 1000z = k\ 2 = k1/x, 52y = k ⇒ 5 = k1/2y ...(i)and (2 × 5)3z = k ⇒ 23z × 53z = k [ (x × y)m = xm × ym]

⇒ (k1/x)3z × (k1/2y)3z = k [Using (i)]⇒ k3z/x × k3z/2y = k [ (xm)n = xm × n]

⇒ k kzx

zy

3 32

+= [ xm × xn = xm + n]

⇒ k kz y xxy

3 22( )+

=On comparing the power of k from both sides, we get3 2

21 3 2

2z y xxy

z xyy x

( )+= ⇒ =

+ \ z

xyx y

=+

23 6

14. Given, 252 5 6 294 3 16

− + −

= × × − + × × −

2 3 7 5 6 2 7 3 3 1

62 2 1

12 1 2 1

12( )

Q a an n= ( ) /1

= × × − × + × × − ×−

( ) ( ) ( )2 3 7 5 6 2 7 3 3 62 212

12 2 1

12

12

Q

1a

amm=

−

= × × − × + × × − ×−

2 3 7 5 6 2 7 3 3 612

12

12

12

12 [ (am)n = am × n]

= × − × + × × − ×−

6 7 5 6 2 3 7 3 612

12

12

12( )

Q a b a bm m m× = × ( )

= × − × + × − ×−

6 7 5 6 6 7 3 612

12

12

12

= × − − − ×−

6 7 6 5 7 3 612

12

12( )

= × + × − × = × + − ×−

−6 7 2 6 3 6 6 7 6 2 3 612

12

12

12

12 1( )

= × + −

= × + ×6 7 6 2 1

26 7 6 3

2

12

12

12

12 = +6 7 3

26

15. We have, 8136

259

52

34

32

3

×

÷

− − −

=

×

÷

− −−9

653

52

2

2

34 2

2

32 3

=

×

÷

× − × − −96

53

52

2 34

2 32

3 [ (xm)n = xm × n]

=

×

÷

− − −96

53

52

32

3 3

= × × ×

×

−−

− −( ) [( ) ]3 2 3 3

525

23

2 13

23 3

Q xy

yx

m m

=

−

= × × × ×− − ×− − × − −

−3 2 3 3 2

5

3 1 32

1 32 3 3

3 3

= × × × ×− −3 2 3 3 25

332

32 3 3

0 = × ×− −3 2 3

1

3 332

3 32

= × × =

−

3 2 3 32

03

232

32

Q

ab

ab

m

m

m=

= =32

278

3

3

27Number Systems

Decimal Expansion of Rational Numbers• The decimal expansion of a rational number is

either terminating or non-terminating.• Terminating Decimals : If the decimal expression

of p/q terminates, i.e., comes to an end, then the decimal so obtained is called a terminating decimal. A fraction p/q is a terminating decimal, only if q has prime factors of 2 and 5 only i.e., q is of the form 2m × 5n.

• Non-Terminating Decimals : A rational number p/q is said to be non-terminating repeating (recurring) decimal if on dividing p by q, the remainder never becomes zero and a set of digits repeat in the same interval.

Decimal Expansion of Irrational Numbers• Decimal expansion of an irrational number is non-

terminating non-recurring.

Method of finding irrational numbers between two rational numbers

• Let ab

cd

and be any two rational numbers. To

find an irrational number between ab

cd

and ,

firstly write ab

cd

and in their decimal form. Now,

as we know, irrational number is non-terminating non-repeating decimal. We can find as many as irrational numbers we require.

Representation of Irrational Numbers on Number Line

To represent n on the number line, where n ∈ N, we have to follow the following steps :Step 1 : Write n = a2 + b2

Step 2 : Draw OA = a units on number lineStep 3 : Draw AB = b units such that AB ⊥ OA. Join OB.

O Aa

b

B

CX′ XStep 4 : Taking O as centre and OB as radius draw an arc, which cuts the number line at C, where C is the required irrational number.

Real Numbers and their Decimal Expansions• Numbers which can be written in the form of p/q, where p, q are integers and q ≠ 0 are called as rational numbers. The set of rational numbers is denoted by Q and defined as :

Q pq

p q q= ≠

, , where are integers and 0

Review of Rational Numbers13

1

19

13

7

25

31

2

8

14

20

3

32

26

9

15

2127

4

22

16

10

28

511

17

23

Your Teeth

Lower left

Upper left

Lower right

Upper right

6

29

12

18

24

303.765

3.773.763.761 3.762 3.763 3.764 3.765 3.766 3.767 3.768 3.769

3.73.71 3.72 3.73 3.743.75 3.763.77 3.78 3.79

3.8

433.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.91

Magic Number Wheel

19 13 7 25 312

814

203

3226

91521273342216102834

511

1723

635

2912

18 24 30 36

• Numbers which can not be expressed in the form of p/q, where p, q ∈ Z and q ≠ 0 are called irrational numbers or every non-terminating and non-repeating decimal number is known as irrational number.

Irrational Numbers2

A number is a mathematical object used to count, label and measure.

Method of Conversion of Decimal Expansions as Rational Number or a FractionCase I : Conversion of Terminating DecimalStep I : Remove decimal point from the numerator and write 1 in the denominator and put as many zeroes on the right hand side of denominator as the number of digits after the decimal.Step II : Convert the rational number obtained into its lowest term by dividing the numerator and denominator by their common factor.Case II : Conversion of non-terminating repeating decimalsStep I : Firstly, transform the non-repeated digit (s) (if exists) between decimal and recurring digits to the left side of decimal by multiplying both sides by 10r i.e., where ‘r’ is the number of digits between decimal and recurring digits.Step II : Count the recurring digits (say R) and then multiply the equation obtained in step I by that power of 10(i.e., 10R).Step III : Lastly, subtract these two equations obtained above.Step IV : Divide both side of the equation in step III by the coefficient of x to get the required fraction.

• Real numbers are closed under addition, subtraction, multiplication and division (with non-zero divisor).

Operations on Real Numbers5

(i) xmxn = xm+n (ii) xm ÷ xn = xm–n (iii) (xm)n = xmn

(iv) xx

nn

− = 1 (v) xy

yx

m m

=

−

(vi) xm/n = (xm)1/n = (x1/n)m i.e., x x xm n mn n m/ ( )= =

(vii) (xy)m = xmym (viii) xy

xy

m m

m

= (ix) x xy y y y nn= × × ........ times

where x, y are positive real numbers and m, n are rational numbers.

Laws of Exponents for Real Numbers6

• The method of representing decimal numbers on number line is known as process of successive magnification. In this method, we divide the distance between two numbers again and again into 10 equal parts.

Representing Real Numbers on the Number Line4

Properties of Real Numbers• Irrational numbers satisfy the commutative,

associative and distributive laws for addition and multiplication.

• The sum and difference of a rational and an irrational number is always irrational.

• The product and quotient of a rational (non-zero) and an irrational number is always irrational.

• The sum, difference, product and quotient of two irrational numbers is not necessarily an irrational number i.e., the resulting number may also be a rational number.

If an expression whose denominator is irrational (or contains square root), then convert the denominators in such cases free from square roots, we multiply both the numerator and denominator of the given number by a suitable factor. This factor is called the rationalising factor and this process is known as rationalising the denominator. For rationalising the denominator, we multiply the numerator and denominator by conjugate of denominator to remove the radical sign from the denominator.

Existence of Square Root for a given Positive Real NumberLet a be a real number such that a > 0. Then a b=means b2 = a and b > 0. The value of b is called the positive square root of the positive real number a.

Some Important IdentitiesLet x and y be any two positive real number, then

• x y xy⋅ = • xy

xy

=

• ( )( )x y x y x y+ − = −2

• ( )x y x xy y± = ± +2 2

• ( )( )x y x y x y+ − = −

• ( )( ) ,w x y z wy wz xy xz+ + = + + + where w and z are positive real numbers.

1. Insert 100 rational numbers between−314

914

and .

Sol. We have, − =− ×

×= − =

××

=314

3 1014 10

30140

914

9 1014 10

90140

and

We know that–30 < –29 <–28 < ... < –1 < 0 < 1 < 2 < ... < 70

⇒ − < − < − < < − < < < < <30140

29140

28140

1140

0140

1140

2140

70140

.... ...

Hence, 100 rational numbers between − = −314

30140

and

914

90140

= are − − −29140

28140

1140

0140

1140

2140

70140

, , ..., , , , , ....,

2. Evaluate : 7 2 10 8 2 15+ − +{ }Sol. We have, 7 2 10 8 2 15+ − +{ }= + + × × − + + × ×( ) ( ) ( ) ( )5 2 2 5 2 5 3 2 5 32 2 2 2

= + − +( ) ( )5 2 5 32 2

= + − + = + − − = −( ) ( )5 2 5 3 5 2 5 3 2 3

3. If x =+

1

3 2 2, then find the value of

x3 – 4x2 – 11x + 7.

Sol. We have, x =+

13 2 2


x =+

× −−

= −−

13 2 2

3 2 23 2 2

3 2 23 2 22 2( )

( )( )

( )( ) ( )

= −−

= −3 2 29 8

3 2 2

\ x x= − ⇒ − = −3 2 2 3 2 2On squaring both sides, we get( ) ( )x x x− = − ⇒ + − =3 2 2 9 6 82 2 2

⇒ x x2 6 1 0− + = ...(i)Now, x3 – 4x2 – 11x + 7 = x(x2 – 6x + 1) + 2(x2 – 6x + 1) + 5 = x × 0 + 2 × 0 + 5 [Using (i)]⇒ x3 – 4x2 – 11x + 7 = 5

4. Evaluate : 15

10 20 40 5 80+ + − −, if

it is given that 5 2 236 10 3 162= =. . . and

Sol. We have, 10 20 40 5 80+ + − −

= + × + × − − ×10 2 5 2 10 5 2 52 2 4

= + + − −10 2 5 2 10 5 2 52

= + + − −10 2 10 2 5 5 4 5= + + − − = − = −( ) ( ) ( )1 2 10 2 1 4 5 3 10 3 5 3 10 5 ...(i)

\ 1510 20 40 5 80+ + − −

=−

=−

153 10 5

510 5( )

[From (i)]

Now, on rationalising, we get

=+

− +=

+−

5 10 510 5 10 5

5 10 510 52 2

( )( )( )

( )( ) ( )

[ (a – b)(a + b) = a2 – b2]

= +−

= +5 10 510 5

10 5( ) = 3.162 + 2.236 = 5.398

5. If xp q p q

p q p q=

+ + −+ − −

2 2

2 2, then show that

qx2 – px + q = 0.

Sol. We have, xp q p qp q p q

=+ + −+ − −

2 22 2


xp q p qp q p q

p q p qp q p q

=+ + −+ − −

×

+ + −+ + −

2 22 2

2 22 2

=+ + −( )

+( ) − −( )p q p q

p q p q

2 2

2 2

2

2 2 [ (a + b)(a – b) = a2 – b2]

=+( ) + −( ) + + −

+ − −p q p q p q p q

p q p q2 2 2 2 2

2 2

2 2

( ) ( )

=+ + − + −( ) ( ) ( )p q p q p q

q2 2 2 2

4

2 2

⇒ xp p q

qp p q

q=

+ −=

+ −2 2 44

42

2 2 2 2

⇒ 2 42 2qx p p q= + −

⇒ 2 42 2qx p p q− = −

On squaring both sides, we get

( )2 42 2 22

qx p p q− = −( )⇒ (2qx)2 + p2 – 2(2qx) p = p2 – 4q2

⇒ 4q2x2 – 4pqx = –4q2

⇒ q2x2 + q2 – pqx = 0On dividing both sides by q, we getqx2 + q – px = 0.


Multiple Choice Questions

1. If n is a natural number, then n is(a) always a natural number(b) always a rational number(c) always an irrational number(d) sometimes a natural or rational number and sometimes an irrational number

2. 2 2 323 4 12× × =(a) 2 (b) 2

(c) 2 2 (d) 4 2

3. Write the following in the ascending order of their magnitude 3 2 44 3 3, , .(a) 3 2 44 3 3< < (b) 4 3 23 4 3< <

(c) 2 3 43 4 3< < (d) None of these4. If g = t2/3 + 4t–1/2, then what is the value of g when t = 64?

(a) 312

(b) 332

(c) 16 (d) 25716

5. If 2 1 4142= . , then find the value of

2 12 1−+

.

(a) 1.4142 (b) 2.4142(c) 0.4142 (d) 3.4142

6. If a = −17 12 2, then find aa

22

1+ .

(a) 34 (b) 26(c) 38 (d) 28

7. If a = +−

7 4 37 4 3

, then find a2 (a – 14)2.

(a) 2 (b) 1(c) 3 (d) 4

8. The value of 2 7 3 3 55 12 212010 4020−( ) +( ) is(a) –1 (b) 0(c) 1 (d) 2

9. If a = 6 11 6 11− = + and b , then the value of (a + b) is(a) 22 (b) 2 11(c) 6 (d) 12

10. The simplified value of ( ) ( )12 8 3 2

5 24

− ++

is(a) 6 2− (b) 2 6−(c) 0 (d) 1

Match the ColumnsMatch column I with column II and select the correct answer by choosing an appropriate option.11. Column I Column IIP. 0 57. is 1. 98Q. If x = +3 2 2, then 2. an irrational

xx

+ 1 is number

R. If x = +5 2 6, then 3. 6

xx

22

1+

is

S. p is 4. a rational number(a) P-4, Q-3,4, R-1,4, S-2(b) P-3, Q-1, R-2, S-4(c) P-1, Q-4, R-2, S-1(d) P-3, Q-1, R-4, S-212. Column I Column II

P. 8116

3 4

=− /

1. 12

Q. ( ) ( )/256 4 3 2− −

= 2. – 4

R. If (23)2 = 4x, then 3. 827 5x =

S. − =−1

64

1 3/

4. 125

(a) P-4, Q-3, R-1, S-2(b) P-3, Q-4, R-2, S-1(c) P-1, Q-4, R-2, S-3(d) P-3, Q-1, R-4, S-2

31Number Systems

13. Column I Column II

P. 1008

634 = 1. 29

90

Q. 2 6 0 82. .− = 2. 2

R. The simplest 3. 34

form of 0 32. is

S. 27 634

6

2

−

= 4. 182

99

(a) P-4, Q-3, R-1, S-2 (b) P-3, Q-4, R-2, S-1(c) P-2, Q-4, R-1, S-3 (d) P-3, Q-2, R-4, S-1

Comprehension TypeDirection (Q. No. 14 to 15) : Read the given passage and answer the following questions.

( )x y x y xy+ = + +2 2 and

( ) ( )x y x y x y+ − = − where x and y are positive real numbers.

14. If x = +3 5 and y = −3 5, thenx4 + y4 =(a) 156 (b) 120(c) 248 (d) 149

15. If x y= + = −3 2 3 2and then,

x4 + y4 + 12x2y2 =(a) 394 (b) 110(c) 120 (d) 256Direction (Q. No. 16 to 17) : Read the given passage and answer the following questions.

For 1

a x b y+, the rationalising factor is

a x b y− .

16. If x =−

1

3 2 2 and y =

+1

3 2 2, then value

of xy3 + x3y is(a) 70 (b) 12 (c) 34 (d) 36

17. An equivalent expression of 57 4 5+

after rationalising the denominator is

(a) 20 5 3531

− (b) 20 5 35129

−

(c) 35 20 531

− (d) 35 20 5121−

Assertion & ReasonEach of the following questions contains two statements : Statement-I (Assertion) and Statement-II (Reason) followed by four options, out of which only one is correct. Select the correct option.(a) Both Statement-I and Statement-II are true.(b) Both Statement-I and Statement-II are false.(c) Statement-I is true, Statement-II is false.(d) Statement-I is false, Statement-II is true.18. Statement-I : Two rational numbers between 27

47

and are 721

821

and .

Statement-II : A rational number between two rational numbers x and y is xy.

19. Statement-I : 1 714285. is a non-terminating repeating decimal and we can express this

number as 127

which is of the form pq

, where p

and q are integers and q ≠ 0.Statement-II : A terminating or non-terminating repeating decimal expansion can be expressed as a rational number.

20. Statement-I : 2 5+ is an irrational number.Statement-II : The sum of a rational number and an irrational number is an irrational number.

21. Statement-I : 0 36. and 2.141414... are examples of irrational numbers.Statement-II : An irrational number can not be

expressed in the form of pq

.

22. Statement-I : If 2 = 1.414, 3 = 1.732, then 2 3 5+ = .Statement-II : Square root of a positive real number always an irrational number.

Integer / Numerical Value

23. If ,x = + +1 2 213

23 then find value of

x3 – 3x2 – 3x – 1.

24. If x = −1 2, then find the value of xx

−

1 3

.

25. Find the value of x, if 2 3 2161 2 5x x− − = .

26. If x = + + −( ) ( ) ,/ /4 15 4 151 3 1 3 then find the value of x3 – 3x.


1. (d)2. (a) : We have,21/3 × 21/4 × (32)1/12 = 21/3 × 21/4 × (25)1/12

= 21/3 × 21/4 × 25/12 = 2(1/3 + 1/4 + 5/12) [Q xm × xn = xm + n]= 2

3. (c) : We have, 3 3 2 2 4 4414 3

13 4

13= = =, and

L.C.M. of 4, 3 and 3 is 12.

Now, 3 3 3 3 27414

312 3

112

112

1= = = = =( ) ( ) [ ( ) ( ) ]Q x x

mn m n

2 2 2 2 16313

412 4

112

112= = = =( ) ( ) and

4 4 4 4 256313

412 4

112

112= = = =( ) ( )

∵ 16 < 27 < 256

\ ( ) ( ) ( )16 27 2561

121

121

12< <

⇒ 2 3 43 4 3< <

Hence, the ascending order is 2 3 43 4 3, , 4. (b) : We have, g = t2/3 + 4t–1/2 and t = 64\ g = (64)2/3 + 4(64)–1/2

⇒ g = ( )4323 + 4 × ( )82

12−( )

⇒ g x xm n m n= + × =× × −( ) ×( ) ( ) [ ( ) ]4 4 8

3 23

2 12

Q

⇒ g = 42 + 4 × (8)–1

⇒ g = 16 + 4 × 18

1Qx

xn

n− =

⇒ g = 16 + 12

⇒ g = 332

5. (c) : We have, 2 12 1−+

By rationalising, we get

( )( )

( )( )

2 12 1

2 12 1

−+

×−−

= −−

( )( ) ( )

2 12 1

2

2 2

= ( ) ( )2 1

2 12 1

1

2 2−−

=−

= − = −2 1 1 4142 1. [Given, 2 1 4142= . ]= 0.4142

6. (a) : Given, a = −17 12 2 ...(i)

\ 1 1

17 12 2a=

−


1 1

17 12 2

17 12 2

17 12 2a=

−× +

+

1 17 12 2

17 12 2

17 12 2289 288

17 12 212 2a

= +

−= +

−= +

( ) ( ) ...(ii)

On squaring (i) and (ii), we get

aa

2217 12 2 1 17 12 2= − = + and

Now, aa

22

1 17 12 2 17 12 2+ = − + +

⇒ aa

22

1 34+ =

7. (b) : We have, a = +−

7 4 37 4 3


a = + × +− × +

= +−

( ) ( )( ) ( )

( )( ) ( )

7 4 3 7 4 37 4 3 7 4 3

7 4 37 4 3

2

2 2

⇒ a = +−

= +7 4 349 48

7 4 3( )

\ a a2 27 4 3 14 7 4 3 14 4 3 7= + − = + − = −( ) ( ) ( ) and

\ ( ) ( )a − = −14 4 3 72 2

Now, a2(a – 14)2 = + −( ) ( )7 4 3 4 3 72 2

⇒ a2(a – 14)2 = + −( ) ( )4 3 7 4 3 72 2

⇒ a2(a – 14)2 = − = − = −[( ) ( ) ] ( ) ( )4 3 7 48 49 12 2 2 2 2

\ a2(a – 14)2 = 1

8. (c) : We have, 2 7 3 3 55 12 212010 4020−( ) +( )= −( ) + ⋅ +( )2 7 3 3 2 7 2 2 7 3 3 3 3

12010 2 2

14020( ) . ( )

= −( ) +( )2 7 3 3 2 7 3 31

20102

4020

= ( ) − ( )

2 7 3 32 2

12010

= (28 – 27)1/2010 = 1

9. (a) : We have, a = 6 11 6 11− = + and b

⇒ a2 = 6 11− and b2 = 6 11+

\ a2 + b2 = 6 11 6 11− + + = 12

a⋅b = ( ) ( )6 11 6 11 36 11 25− + = − = = 5\ (a + b)2 = a2 + b2 + 2ab= 12 + 2 × 5 = 22

\ a + b = 22

33Number Systems

10. (a) : We have, ( ) ( )12 8 3 25 24

− ++

= ( ) ( )2 3 2 2 3 23 2 2 3 2− ++ +

= 2 3 2 3 2

3 2 2 3 22 2( ) ( )

( ) ( )− +

+ +

= 2 3 23 2

23 22

( )( )

−

+=

+ = 2 3 2

3 2 3 2× −

+ −( )

( ) ( )

= 6 23 2

6 2−−

= −

11. (a) : (P) Let x = 0 57. . Then, x = 0.5777... ...(i)Multiplying (i) by 10 and 100, we get10x = 5.777... ...(ii)100x = 57.777... ...(iii)Subtracting (ii) from (iii), we get(100x – 10x) = (57.777...) – (5.777...)⇒ 90x = 52

⇒ x = =5290

2645

, which is a rational number.

(Q) We have, x = +3 2 2

⇒ 1 13 2 2x

=+

.


3 2 23 2 23 2 2

1 3 2 23 2 22 2x x

=+

× −−

⇒ = −−( ) ( )

[Q (a – b) (a + b) = a2 – b2]

⇒ 1 3 2 29 8

1 3 2 21x x

= −−

⇒ = −

\ xx

+ = + + − =1 3 2 2 3 2 2 6 , which is a rational

number.

(R) We have, xx

= + ⇒ =+

× −−

5 2 6 1 15 2 6

5 2 65 2 6

1 5 2 6

5 2 62 2x= −

( ) − ( )[ (a – b)(a + b) = a2 – b2]

⇒ = −−

⇒ = −1 5 2 625 24

1 5 2 61x x

Now, xx

xx

xx

+

= + + ⋅ ⋅

1 1 2 122

2

⇒ 5 2 6 5 2 6 1 22 2

2+ + −( ) = + +xx

⇒ 10 2 1 1 982 22

22

( ) − = + ⇒ + =xx

xx

(S) p is an irrational number.

12. (d) : (P) We have, 8116

32

32

34 4

4

34 4 3

4

=

=

− − × −

[Q (xm) = xm × n]

= 32

23

3 3

=

=

− −

Q

xy

yx

m m

= 827

(Q) We have,

( ) ( ) [ ( ) ]256 2564 22 32

32−( ) − × −( ) ×

−

= =Q x xm n m n

= = = = = =−−

− × −( ) −−( ) ( ) ( )

( ) /256 256 256 2 2 12

21

2 1 8 8 18 13 3

(R) We have, 2 4 2 43 2 2 3( ) = ⇒ ( ) =x x { (xm)n = (xn)m}

⇒ 43 = 4x

On comparing the power of 4, we get x = 3\ 5x = 53 = 125

(S) −

= −( ) = −( )−1

6464 4

13 3

13

13

( )

= − =× ×( ) [ ( ) ]4

3 13

Q x xm n m n

= – 4

13. (c) : (P) We have, 100863

1127

16 24 4 4 414= = = ( )

= =× ×( ) [ ( ) ]2

4 14

Q x xm n m n

= 2(Q) Let x = 2 6. . Then, x = 2.666... ...(i)Multiplying (i) by 10, we get10x = 26.666... ...(ii)Subtracting (i) from (ii), we get(10x – x) = (26.666 ... )– (2.666...)

⇒ 9x = 24 ⇒ = =x 249

83

Now, let y = 0 82.\ y = 0.828282... ...(iii)Multiplying (iii) by 100, we get100y = 82.8282... ...(iv)Subtracting (iii) from (iv), we get(100y – y) = (82.8282...) – (0.8282...)⇒ 99y = 82

⇒ y = 8299

Now, 2 6 0 82 83

8299

264 8299

18299

. .− = − = − =

(R) Let x = 0.32. Then x = 0.3222 ... ...(i)Multiplying (i) by 10 and 100, we get 10x = 3.2222 ... ...(ii)and 100x = 32.222 ... ...(iii)Subtracting (ii) from (iii), we get100x – 10x = (32.222...) – (3.222...)⇒ 90x = 29

⇒ =x 2990


(S) We have, 27 6 34

27 274

62

216

12

−

= ( ) −

= −

=×

×

×

×3 3

2

3 16

3 12

2 12

2

[ ( ) ]Q x xm n m n

= −

= −

3 3 3

22 3 3 3

2

12

2 2

= −

=3

234

2

14. (c) : We have, x y= + = −3 5 3 5and

\ xy = +( ) −( ) = − = −3 5 3 5 3 5 2Now, x2 + y2 = (x + y)2 – 2xy

⇒ x2 + y2 = + + −( ) − −( )3 5 3 5 2 22

⇒ x2 + y2 = ( ) + ⇒ + = +2 3 4 12 42 2 2x y

⇒ x2 + y2 = 16\ x4 + y4 = (x2 + y2)2 – 2x2y2 = (16)2 – 2 × (–2)2

= 256 – 8 = 248

15. (b) : We have, x y= + = −3 2 3 2and

\ xy = +( ) −( ) = − =3 2 3 2 3 2 1 \ x2 + y2 = (x + y)2 – 2xy

= + + −( ) −3 2 3 2 2 12

( )

= ( ) − = − =2 3 2 12 2 102

Now, x4 + y4 + 12x2y2 = (x2 + y2)2 – 2x2y2 + 12x2y2

= (10)2 + 10x2y2 = 100 + 10 × (1)2 = 100 + 10 = 110

16. (c) : We have, x =−

13 2 2

Rationalising the denominator, we get

x =−

× ++

= +−

13 2 2

3 2 23 2 2

3 2 23 2 22 2( ) ( )

[ ( )( ) ]Q a b a b a b+ − = −2 2

= +3 2 2

Now, y =+

13 2 2

Rationalising the denominator, we get

y =+

× −−

= −−

= −13 2 2

3 2 23 2 2

3 2 23 2 2

3 2 22 2( ) ( )

\ x + y = 3 + 2 2 + 3 – 2 2 = 6

and xy = + −( )( )3 2 2 3 2 2 = − = − =( ) ( )3 2 2 9 8 12 2

(x + y)2 = x2 + y2 + 2xy

⇒ (6)2 = x2 + y2 + 2(1)

⇒ x2 + y2 = 36 – 2

⇒ x2 + y2 = 34

Then, xy3 + x3y = xy(x2 + y2) = 1(34) = 34.

17. (a) : We have, 57 4 5

57 4 5

7 4 57 4 5+

=+

×−−

= 35 20 549 16 5

35 20 531

20 5 3531

−− ×

=−−

=−

18. (c) : Given two rational numbers 27

47

and .

i.e., 27

2 37 3

621

47

4 37 3

1221

= and =××

= ××

=

Two rational numbers between 621

1221

721

821

and are and .

A rational number between x and y is x y+2

.

19. (a) : Let x = 1 714285. , then x = 1.714285714285... ...(i)Multiplying (i) 1000000, we get1000000x = 1714285.714285... ...(ii)Subtracting (i) from (ii), we get(1000000x – x) = (1714285.714285...) – (1.714285...)⇒ 999999x = 1714284

⇒ x = 1714284999999

127

=

20. (a) : 2 5+ is an irrational number.[Q Sum of a rational and an irrational number is irrational]21. (d) : Let x = 0 36. . Then x = 0.363636... ...(i)Multiplying (i) by 100, we get100x = 36.3636 ... ...(ii)Subtracting (i) from (ii), we get(100x – x) = (36.3636...) – (0.3636 ...)⇒ 99x = 36

⇒ x = =3699

411


Similarly, 2.141414 ... = 2 14.Let y = 2.1414... ...(iii)Multiplying (iii) by 100, we get100y = 214.1414 ... ...(iv)Subtracting (iii) from (iv), we get100y – y = (214.1414...) – (2.1414...)⇒ 99y = 212

⇒ y = 21299


22. (b) : 2 3 3 146 5+ = ≠.

23. (0) : Given, x = + +1 2 213

23

\ x − = +1 2 213

23 ...(i)

On cubing both sides, we get

( )x − = +( )1 2 2313

23

3

35Number Systems

⇒ x3 – 3x2 + 3x – 1

= ( ) + ( ) + +( )2 2 3 2 2 2 213

3 23

3 13

23

13

23. .

[∵ (a + b)3 = a3 + b3 + 3ab (a + b); (a – b)3 = a3 + b3 – 3ab (a – b)]= 2 + 22 + 3 ⋅ 2 (x – 1) [Using (i)]= 6 + 6 (x – 1) = 6x\ x3 – 3x2 + 3x – 6x – 1 = 0⇒ x3 – 3x2 – 3x – 1 = 024. (8) : We have, x = −1 2

\ 1 11 2x

=−

On rationalising, we get 1 11 2

1 21 2x

=−

× ++

⇒ 1 1 2

1 22 2x= +

( ) − ( ) [ (a + b)(a – b) = a2 – b2]

⇒ 1 1 21 2

1 1 2x x= +( )

−⇒ = − +( )

\ xx

− = −( )+ +( ) =1 1 2 1 2 2

So, xx

−

= =1 2 8

33

25. (4) : Given, 2x – 1 . 32x – 5 = 216⇒ 2x – 1 . 32x – 5 = 63

⇒ 2x – 1 . 32x – 5

= (2 × 3)3 = 23 × 33 [ (x × y)m = xm × ym]On comparing the power of 2 and 3, we getx – 1 = 3, 2x – 5 = 3⇒ x = 426. Given, x = + + −( ) ( )/ /4 15 4 151 3 1 3

So, x313

3 13

34 15 4 15= + + −

× ×( ) ( )

+ + −3 4 15 4 151 3 1 3{( ) ( ) }/ / [( ) ( ) ]/ /4 15 4 151 3 1 3+ + − [∵ (a + b)3 = a3 + b3 + 3ab (a + b)]= + + − + + −( ) ( ) [( )( )] /4 15 4 15 3 4 15 4 15 1 3

[( ) ( ) ]/ /4 15 4 151 3 1 3+ + −

⇒ x3 = 8 + 3x [ ( ) ( ) ]/ /Qx = + + −4 15 4 151 3 1 3

⇒ x3 – 3x = 8


SECTION - A

Multiple Choice Type Questions1. Which of the following decimal expansion

represents the fraction 78

.

(a) 0.785 (b) 0.875 (c) 0.778 (d) 0.582. Find the value of (81)0.16 × (81)0.09.(a) 3–2 (b) 0.3 (c) 3 (d) 81

3. The irrational number 11 lies between(a) 3 and 3.5 (b) 3.5 and 4(c) 4.5 and 5 (d) 4 and 4.5

4. If 22

16m n

n m

+

− = , then value of m is

(a) 2 (b) 14

(c) 9 (d) 18

5. Find the value of 3 6

2 3

3 2

2 1 3

×× / .

(a) 28 2 (b) 7 2

(c) 81 93 (d) 25 2

Fill in the Blanks

6. If x =−

1

2 1, then the rationalising factor

of x is equal to ______ .

7. 7.010 in the form of pq

is ______ .

VSA Type Questions8. What is the nature of the number( ) ?− + −5 2 5 5

9. If 3 1 732= . , find the approximate value of 2

3.

10. Simplify : 1

3 5+.

11. Find the value of (13 + 23 + 33)–3/2.

12. Identify 45 as rational number or irrational number.13. Find the quotient when 1500 is divided by 2 15.

SECTION - B14. Find the value of

1 9 9 10 11 5 5 16+ + + + .

15. If x = +8 3 7, then find the value of xx

+ 1 .

16. Simplify : 2 50 3 32 4 18× × .

17. Find the value of 2 1

5

− to three

places of decimals. It is given that 5 2 236 10 3 162= =. . . and

18. Show that ( ) ( ) ( ) ( )2 3 2 3 5 2 5 2+ − + − is a rational number.19. Find two rational numbers between 5 and 6.20. If 2m + 21+m = 96, then find the value of m which satisfy the given equation.

21. Express 2 113. in the form of pq

, where p and q are integers and q ≠ 0.

22. If 22

16m n

n m

+

− = and a = 21/10, then find the value

of a3m.

37Number Systems

Time: 2 hrs. 30 mins. Max. Marks: 75

(i) All the questions are compulsory.(ii) The question paper consists of 35 questions.(iii) Section A comprises of 13 questions of 1 mark each. Section B comprises of 9 questions

of 2 marks each. Section C comprises of 8 questions of 3 marks each. Section D comprises of 5 questions of 4 marks each.

(iv) Use of calculators is not permitted.

SECTION - C23. Represent 10 on number line.24. If x y= + + = + −2 3 5 3 3 5and , then find the value of x2 + y2 – 4x – 6y – 3.

25. If and a ba

= + =5 2 61 then what will be

the value of a3 + b3?

26. If 13 10 8 5+ = +a , then find the value

of aa−+

11

.

27. If x x x+ − − =1 1, then find the value of x.

28. If 5 11

3 2 1111

+−

= +x y , then find the values

of x and y.

29. If x = 7 + 40 , then find the value of xx

+ 1.

30. If a1/3 + b1/3 + c1/3 = 0, then find the value of (a + b + c)3.

SECTION - D31. Find the sum of 0.0333... and 0.444... .

32. Represent geometrically the number 8 1. on the number line.

33. If x aa

= +

12

1, then find the value of

x

x x

2

2

1

1

−

− −.

34. Visualise 3.2345 on the number line, using successive magnification.

35. If and x y= +−

= −+

5 2

5 2

5 2

5 2, find the

value of 9x2 + 2xy + 9y2.

For EXAM DRILL Solutions visit http://bit.ly/mtg-100-mt-examdrill-9


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