· chapter 1 number systems exercise 1.1 write the correct answer in each of the following: q.1)...

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Class 9

NCERT Exemplar Solutions Mathematics Chapter 1 Number Systems

Exercise 1.1 Write the correct answer in each of the following:

Q.1) Every rational number is (A) a natural number (B) an integer (C) a real number (D) a whole number

Sol.1) (c) Since, real numbers are the combination of rational and irrational numbers. Hence, every rational number is a real number.

Q.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

Sol.2) (c) Between two rational numbers, there are infinitely many rational numbers.

e.g., 3

5 and

4

5 are two rational numbers then

31

50,

32

50,

33

50,

34

50,

35

50… are infinite rational

numbers.

Q.3) Decimal representation of a rational number cannot be (A) terminating (B) non-terminating (C) non-terminating repeating (D) non-terminating non-repeating

Sol.3) (d) Decimal representation of a rational number cannot be non-terminating non-repeating because the decimal expansion of rational number is either terminating or non-terminating recurring (repeating).

Q.4) The product of any two irrational numbers is (A) always an irrational number (B) always a rational number (C) always an integer (D) sometimes rational, sometimes irrational

Sol.4) (d) We know that, the product of any two irrational numbers is sometimes rational and sometimes irrational.

e.g., √2 × √2 = 2 (rational) and √2 × √3 = √6 (irrational)

Q.5) The decimal expansion of the number √2 is (A) a finite decimal (B) 1.41421 (C) non-terminating recurring (D) non-terminating non-recurring

Sol.5) The decimal expansion of the number √2 is non-terminating non-recurring. Because √2 is an irrational number. Also, we know that an irrational number is non-terminating non-recurring.

Q.6) Which of the following is irrational?

(a) √4

9 (b)

√12

√3 (c) √7 (d) √81

Sol.6) (c)

√4

9=

2

3 (rational)

√12

√3=

2√3

√3= 2 (rational)

√81 = 9 (rational) But √7 is an irrational number

Q.7) Which of the following is irrational?

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(A) 0.14 (B) 0.1416̅̅̅̅ (C) 0. 1416̅̅ ̅̅ ̅̅ ̅ (D) 0.4014001400014...

Sol.7) (d) An irrational number is non-terminating non-recurring which is 0.4014001400014….

Here, 0.14 is terminating and 0.1416̅̅̅̅ , 1416̅̅ ̅̅ ̅̅ ̅ are non-terminating recurring.

Q.8) A rational number between √2 and √3 is

(A) √2+√3

2 (b)

√2.√3

2 (C) 1.5 (D) 1.8

Sol.8) (c)

A rational number between √2 and √3 i.e., 1.414 and 1.732

a) √2+√3

2, which is an irrational number , so it is not a solution

b) √2.√3

2=

√6

2, which is an irrational number , so it is not a solution

now, 10.5 and 1.8 are the rational numbers but only 1.5 lies between 1.414 and 1.732

Q.9) The value of 1.999... in the form 𝑝

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

(A) 19

10 (B)

1999

1000 (C) 2 (D)

1

9

Sol.9) (c) Let 𝑥 = 1.999 Now, 10𝑥 = 19.999 … On subtraction eq. (i) from Eq. (ii) 10𝑥 − 𝑥 = (19.999 … ) − (1.9999 … ) 9𝑥 = 18

𝑥 =18

9= 2

Q.10) 2√3 + √3 is equal to

(A) 2√6 (B) 6 (C) 3√3 (D) 4√6

Sol.10) 2√3 + √3 = √3 (2 + 1) = 3√3 Q.11) √10 × √15 is equal to

(a) 2√6 (b) 6 (c) 3√3 (d) 4√6

Sol.11) (b) √10 × √15 = √2.5 × √3.5 = √2. √5 × √3. √5 = 5 √6

Q.12) The number obtained on rationalising the denominator of 1

√7−2 is

(A) √7+2

3 (b)

(√7−2)

3 (c)

√7+2

5 (d)

√7+2

45

Sol.12) a) 1

√7−2=

1

√7−2.7+27+2

=√7+2

(√7)2

−(2)2=

√7+2

7−4=

√7+2

3

Q.13) 1

√9−√8 is equal to

(a) 1

2(3 − 2√2) (b)

1

3+2√2 (c) 3 − 2√2 (d) 3 + 2√2

Sol.13) 1

√9−√8=

1

3−2√2=

1

3−2√2.

3+2√2

3+2√2

=3+2√2

9−(2√2)2

=3+2√2

9−8= 3 + 2√2

Q.14) After rationalising the denominator of 7

3√3−2√2 , we get the denominator as

(A) 13 (B) 19 (C) 5 (D) 35

Sol.14) 7

3√3−2√2=

7

3√3−2√2 .

3√3+2√2

3√3+2√2



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=7(3√3+2√2)

(3√3)2

−(2√2)2

=7(3√3+2√2)

27−8=

7(3√3+2√2)

19

Hence, after rationalising the denominator of 7

3√3−2√2 , we get the denominator as 19.

Q.15) The value of √32+√48

√8+√12 is equal to

(A) 2 (B) 2 (C) 4 (D) 8

Sol.15) √32+√48

√8+√12=

√16×2+√16×3

√4×2+√4×3

=4√2+4√3

2√2+2√3=

4(√2+√3)

2(√2+√3)= 2

Q.16) If 2 = 1.4142, then √

√2−1

√2+1 is equal to

(A) 2.4142 (B) 5.8282 (C) 0.4142 (D) 0.1718

Sol.16) √√2−1

√2+1= √√2−1

√2+1.

√2−1

√2−1

= √(√2−1)

2

(√2)2

+(1)2

= √(√2−1)2

2−1= √2 − 1 = (1.4142 … ) − 1 = 0.4142 …

Q.17) √√2234 equals

(A) 2−1

6 (B) 2−6 (C) 21

6 (D) 26

Sol.17) (c)

√√2234= [√223

]1

4

= [(22)1

3]

1

4= [2

2

3]

1

4= 2

2

3.1

4 = 21

6

Q.18) The product of √23

. √24

. √3212

equals

(A) √2 (B) 2 (C) √212

(D) √3212

Sol.18) Thinking Process Take the LCM of indices of three irrational numbers. Then, convert all individually in the form whose index. (b) LCM of 3, 4 and 12 = 12

√23

= √2412

√24

= √2312

& √3212

= √2512

Product of √23

. √24

. √3212

= √24. 23. 2512

= 12√24+3+5 = √21212= 212×

1

12 = 2 Q.19) Value of √(81)−24

is

(A) 1

9 (B)

1

3 (C) 9 (D)

1

81

Sol.19) (a) √(81)−24

= √1

(81)2

4=

1

(81)24

=1

(81)12

=1

(92)12

=1

922

=1

9



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Q.20) Value of (256)0.16 × (256)0.09 is (A) 4 (B) 16 (C) 64 (D) 256.25

Sol.20) (a) (256)0.16 × (256)0.09 = (256)16

100 × (256)9

100

= (256)16

100+

9

100

= (256)26

100 = (256)1

4

= (44)1

4 = 4 Q.21) Which of the following is equal to ?

(A) 𝑥12

7 − 𝑥5

7 (B) √(𝑥4)1

3

12

(C) (√𝑥3)2

3 (D) 𝑥12

7 × 𝑥7

12

Sol.21) (c)

(A) 𝑥12

7 − 𝑥5

7

= 𝑥5

7+1 − 𝑥

5

7

= 𝑥5

7. 𝑥 − 𝑥5

7

(B) √(𝑥4)1

3

12

= ((𝑥4)1

3)

1

12

= 𝑥4×1

3×

1

12 = 𝑥1

9 ≠ 𝑥

(C) (√𝑥3)2

3

= (𝑥3

2)

2

3= 𝑥

3

2×

2

3 = 𝑥

(D) 𝑥12

7 × 𝑥7

12

= 𝑥12

7+

7

12

= 𝑥144+49

84 = 𝑥193

84 ≠ 𝑥 Exercise 1.2

Short Answer Type Questions Q.1) Let 𝑥 and 𝑦 be rational and irrational numbers, respectively. Is 𝑥 + 𝑦 necessarily an

irrational number? Give an example in support of your answer

Sol.1) Yes, 𝑥 + 𝑦 is necessarily an irrational number

e.g., Let 𝑥 = 2, 𝑦 = √3

then, 𝑥 + 𝑦 = 2 + √3

if possible, let 𝑥 + 𝑦 = 2 + √3 be a rational number. Consider, On squaring both sides, we get

𝑎2 = (2 + √3)2

𝑎2 = 22 + (√3)2

𝑎2 = 4 + 3 + 4√3

=𝑎2−7

4= √3



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So, 𝑎 is rational 𝑎2−7

4 is rational ,√3 is rational.

Q.2) Let 𝑥 be rational and 𝑦 be irrational. Is 𝑥𝑦 necessarily irrational? Justify your answer by an example.

Sol.2) No, (𝑥𝑦) is necessarily an irrational only when 𝑥 ≠ 0. Let 𝑥 be a non-zero rational and 𝑦 be an irrational. Then, we have to show that 𝑥𝑦 be an irrational. If possible, let 𝑥𝑦 be a rational number.

Since, quotient of two non-zero rational number is a rational number. So,(𝑥𝑦

𝑥) is a

rational number => 𝑦 is a rational number. But, this contradicts the fact that 𝑦 is an irrational number. Thus, our supposition is wrong. Hence, 𝑥𝑦 is an irrational number. But, when 𝑥 = 0, then 𝑥𝑦 = 0, a rational number.

Q.3) State whether the following statements are true or false? Justify your answer.

(i) √2

3 is a rational number.

(ii) There are infinitely many integers between any two integers. (iii) Number of rational numbers between 15 and 18 is finite.

(iv) There are numbers which cannot be written in the form 𝑝

𝑞, 𝑞 ≠ 0 , 𝑝, 𝑞 both

are integers. (v) The square of an irrational number is always rational.

(vi) √12

√3 is not a rational number as 12 and 3 are not integers.

(vii) √15

√3 is written in the form

𝑝

𝑞, 𝑞 ≠ 0 and so it is a rational number

Sol.3) (i) False, here √2 is an irrational number and 3 is a rational number, we know that when we divide irrational number by non-zero rational number it will always give an irrational number. (ii) False, because between two consecutive integers (like 1 and 2), there does not exist any other integer. (iii) False, because between any two rational numbers there exist infinitely many rational numbers. (iv) True, because there are infinitely many numbers which cannot be written in the form 𝑝

𝑞, 𝑞 ≠ 0. 𝑝, 𝑞 both are integers and these numbers are called irrational numbers.

(v) False, e.g., Let an irrational number be √2 and √24

(a) (√2)2

= 2, which is a rational number.

(b) (√24

)2

= √2, which is not a rational number.

Hence, square of an irrational number is not always a rational number.

(vi) False, √12

√13=

√4×3

√3=

(√4×√3)

√3= 2 × 1 = 2, which is a rational number.

(vii) False, √15

√3=

√5×3

√3=

(√5×√3)

√3= √5, which is an irrational number.

Q.4) Classify the following numbers as rational or irrational with justification

(i) √196 (ii) 3√18 (iii) √9

27 (iv)

√28

√343 (v) – √0.4

(vi) √12

√75 (vii) 0.5918 (viii) (1 + √5) − (4 + √5) (ix) 10.124124. ..

(x) 1.010010001. ..

Sol.4) (i) √196 = √(14)2 = 14



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Hence, it is a rational number.

(ii) 3√18 = 3√(3)2 × 2 = 3 × 3√2 = 9√2 Hence, it is an irrational number.

(iii) √9

27= √

9

9×3=

1

√3

Hence, it is an irrational number, because √3 is an irrational number.

(iv) √28

√343=

√2×2×7

√7×7×7=

2√7

7√7=

2

7

Hence, it is a rational number.

(v) – √0.4 = −√4

10= −

2

√10

Hence, it is a quotient of a rational and irrational numbers, so it is an irrational number.

(vi) √12

√75=

√4×3

√25×3=

√4√3

√25√3=

2

5

Hence, it is a rational number. (vii) 0.5918

It is a number with terminating decimal, so it can be written in the form of 𝑝

𝑞, where 𝑞 ≠

0, 𝑝 and 𝑞 are integers. Hence, it is a rational number.

(viii) (1 + √5) − (4 + √5)

= 1 − 4 + √5 − √5 = −3 Hence, it is a rational number. (ix) 10.124124. .. It is a number with non-terminating recurring decimal expansion. Hence, is a rational number. (x) 1.010010001. .. It is a number with non-terminating non-recurring decimal expansion. Hence, it is a rational number.

Exercise 1.3 Short Answer Type Questions

Q.1) Find which of the variables 𝑥, 𝑦, 𝑧 and 𝑢 represent rational numbers and which irrational numbers:

(i) 𝑥2 = 5 (ii) 𝑦2 = 9 (iii) 𝑧2 = .04 (iv) 𝑢2 =17

4

Sol.1) i) Given, 𝑥2 = 5 On taking square root both sides we get

𝑥 = ±√5 ii) Given, 𝑦2 = 9 On taking square root both sides, we get

𝑦 = ±√9 = ±3



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iii) Given, 𝑧2 = 0.04 On taking square root both sides, we get

𝑧 = √0.04 = √4

100=

2

10 (rational number)

iv) Given, 𝑢2 =17

4

On taking square root of both sides, we get

𝑢 = ±√17

4= ±

√17

2

Q.2) Find three rational numbers between

(i) –1 and –2 (ii) 0.1 and 0.11 (iii) 5

7 and

6

7 (iv)

1

4 and

1

5

Sol.2) i) Let 𝑦 = −1 and 𝑥 = −2 Here, 𝑥 < 𝑦 and we have to find three rational numbers, so 𝑛 = 3

𝑑 =𝑦−𝑥

𝑛+1=

−1+2

3+1=

1

4

Since, the three rational numbers between 𝑥 and 𝑦 are 𝑥 + 𝑑, 𝑥 + 2𝑑 and 𝑥 + 3𝑑

Now, 𝑥 + 𝑑 = −2 +1

4=

−8+1

4= −

7

4

𝑥 + 2𝑑 = −2 +2

4=

−8+2

4= −

6

4= −

3

2

& 𝑥 + 3𝑑 = −2 +3

4=

−8+3

4= −

5

4

Hence, three rational numbers between -1 and -2 are −7

4 , −

3

2 and −

5

4

Alternate method: Let 𝑥 = −1 and 𝑦 = −2

We know, a rational number between 𝑥 and 𝑦 =𝑥+𝑦

2

A rational number between -1 and -2 =−1−2

2= −

3

2

& a rational number between -1 and −3

2=

−1−3

2

2=

−2−3

4= −

5

4

Similarly, −7

4 is a rational number between -1 and -2.

Hence, required solution = −3

2 , −

5

4 , −

7

4

ii) Let 𝑥 = 0.1 and 𝑦 = 0.11 here, 𝑥 < 𝑦 and we have to find three rational numbers, so consider 𝑛 = 3

𝑑 =𝑦−𝑥

𝑛+1=

0.11−0.1

3+1=

0.01

4

Since, the three rational numbers between 𝑥 and 𝑦 are (𝑥 + 𝑑), (𝑥 + 2𝑑) and (𝑥 + 3𝑑)

Now, 𝑥 + 𝑑 = 0.1 +0.01

4=

0.4+0.01

4=

0.41

4= 0.1025

𝑥 + 2𝑑 = 0.1 +0.02

4=

0.4+0.02

4=

0.42

4= 0.105

& 𝑥 + 3𝑑 = 0.1 +0.03

4=

0.4+0.03

4=

0.43

4= 0.1075

Hence, three rational numbers between 0.1 and 0.11 are 0.1025, 0.105, 0.1075. Also, without using above formula the three rational numbers between 0.1 and 0.11 are 0.101, 0.102, 0.103.

iii) Let 𝑥 =5

7 and 𝑦 =

6

7

Here, 𝑥 < 𝑦 Here, we have to find three rational numbers. Consider, 𝑛 = 3

𝑑 =𝑦−𝑥

𝑛+1



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𝑑 =6

7−

5

7

4=

1

7

4=

1

28

Since, the three rational numbers between 𝑥 and 𝑦 are (𝑥 + 𝑑), (𝑥 + 2𝑑) and (𝑥 + 3𝑑)

Now, 𝑥 + 𝑑 =5

7+

1

28=

20+1

28=

21

28

𝑥 + 2𝑑 =5

7+

2

28=

20+2

28=

22

28

& 𝑥 + 3𝑑 =5

7+

3

28=

20+3

28=

23

28

Hence, three rational numbers between 5

7 and

6

7 are

21

28,

22

28 ,

23

28

Also, without using above formula, the three rational numbers between 5

7 and

6

7 are

51

70 ,

52

70 ,

53

70

iv) Let 𝑥 =1

5 and 𝑦 =

1

4

Here, 𝑥 < 𝑦 Consider, 𝑛 = 3

𝑑 =𝑦−𝑥

𝑛+1=

1

4−

1

5

3+1=

5−4

20

4=

1

80

Since, the three rational numbers between 𝑥 and 𝑦 are 𝑥 + 𝑑, 𝑥 + 2𝑑 and 𝑥 + 3𝑑

Now, 𝑥 + 𝑑 =1

5+

1

80=

16+1

80=

17

80

𝑥 + 2𝑑 =1

5+

2

80=

16+2

80=

18

80=

9

40

& 𝑥 + 3𝑑 =1

5+

3

80=

16+3

80=

19

80

Hence, three rational numbers between 1

4 and

1

5 are

17

80 ,

9

40 ,

19

80

Q.3) Insert a rational number and an irrational number between the following :

(i) 2 and 3 (ii) 0 and 0.1 (iii) 1

3 and

1

2 (iv) −

2

5 and

1

2

(v) 0.15 and 0.16 (vi) √2 and √3 (vii) 2.357 and 3.121 (viii) .0001 and .001 (ix) 3.623623 and 0.484848 (x) 6.375289 and 6.375738

Sol.3) We know that, there are infinitely many rational and irrational values between any two numbers. (i) A rational number between 2 and 3 is 2.1. To find an irrational number between 2 and 3. Find a number which is non-terminating non-recurring lying between them. Such number will be 2.040040004…….. (ii) A rational number between 0 and 0.1 is 0.03. An irrational number between 0 and 0.1 is 0.007000700007………. (iii) A rational number between 1/3 and 1/2 is 5/12. An irrational number between 1/3 and 1/2 i.e., between 0-3 and 0.5 is 0.4141141114…………. (iv) A rational number between -2/5 and 1/2 is 0. An irrational number between -2/5 and 1/2 i.e., between – 0.4 and 0.5 is 0.151151115……….. (v) A rational number between 0.15 and 0.16 is 0.151. An irrational number between 0.15 and 0.16 is 0.1515515551…….



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(vi) A rational number between √2 and √3 i.e.,, between 1.4142…… and 1.7320…… is 1.5. An irrational number between √2 and √3 is 1.585585558………. (vii) A rational number between 2.357 and 3.121 is 3. An irrational number between 2.357 and 3.121 is 3.101101110…….. (viii) A rational number between 0.0001 and 0.001 is 0.00011. An irrational number between 0.0001 and 0.001 is 0.0001131331333……….. (ix) A rational number between 3.623623 and 0.484848 is 1. An irrational number between 3.623623 and 0.484848 is 1.909009000………. (x) A rational number between 6.375289 and 6.375738 is 6.3753. An irrational number between 6.375289 and 6.375738 is 6.375414114111………

Q.4) Represent the following numbers on the number line :

7, 7.2, −3

2 , −

12

5

Sol.4) Firstly, we draw a number line whose mid-point is 0. Marks a positive numbers on right hand side of 0 and negative numbers on left hand side of 0. (i) Number 7 is a positive number. So we mark a number 7 on the right hand side of 0, which is a 7 units distance from zero.

(ii) Number 7.2 is a positive number. So, we mark a number 7.2 on the right hand side of 0, which is a 7.2 units distance from zero. (iii) Number -3/2 or -1.5 is a negative number. So, we mark a number 1.5 on the left hand side of 0, which is a 1.5 units distance from zero. (iv) Number – 12/5 or -2.4 is a negative number. So, we mark a number 2.4 on the left hand side of 0, which is a 2.4 units distance from zero.

Q.5) Locate √5 , √10 and √17 on the number line.

Sol.5) i) Here, 5 = 22 + 12 So, draw a right angles ∆𝑂𝐴𝐵, in which 𝑂𝐴 = 2 𝑢𝑛𝑖𝑡𝑠 and 𝐴𝐵 = 1 𝑢𝑛𝑖𝑡 and ∠𝑂𝐴𝐵 =

90° By using Pythagoras theorem, we get

𝑂𝐵 = √𝑂𝐴2 + 𝐴𝐵2

= √22 + 12 = √4 + 1 = √5

Taking 𝑂𝐵 = √5 as radius & point O as centre, draw an arc, which meets the number line at point P on the positive side of it.



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Hence, it is clear that point P represents √5 on the number line. ii) Here, 10 = 32 + 1 So, draw a right angled ∆𝑂𝐴𝐵, in which 𝑂𝐴 = 3 𝑢𝑛𝑖𝑡𝑠 and 𝐴𝐵 = 1 𝑢𝑛𝑖𝑡 and ∠𝑂𝐴𝐵 =

90° By using Pythagoras theorem, we get


= √32 + 12 = √9 + 1 = √10

Taking 𝑂𝐵 = √10 as radius & point O as centre, draw an arc, which meets the number line at the point P on positive side.

The point P represents √10 on the number line. iii) Here, 17 = 42 + 1 So, draw a right angled ∆𝑂𝐴𝐵, in which 𝑂𝐴 = 4 𝑢𝑛𝑖𝑡𝑠 and 𝐴𝐵 = 1 𝑢𝑛𝑖𝑡 and ∠𝑂𝐴𝐵 =

90°. By using Pythagoras theorem, we get


= √42 + 12 = √16 + 1 = √17

Taking 𝑂𝐵 = √17 as radius & point O as centre, draw an arc, which meets the number

line at the point P on positive side. The point P represents √17 on the number line.

Q.6) Represent geometrically the following numbers on the number line :

(i) √4.5 (ii) √5.6 (iii) √8.1 (iv) √2.3

Sol.6) (i) Firstly, we draw a line segment 𝐴𝐵 = 4.5 𝑢𝑛𝑖𝑡𝑠 and extend it to C such that BC = 1 unit. Let O be the mid-point of AC. Now, draw a semi-circle with centre O and radius OA Let us draw BD perpendicular to AC passing through point B and intersecting the semi-circle at point D.

Hence, the distance BD is √4.5 𝑢𝑛𝑖𝑡𝑠. Draw an arc with centre B and radius BD, meeting

AC produced at E, then 𝐵𝐸 = 𝐵𝐷 = √4.5 𝑢𝑛𝑖𝑡𝑠. (ii) Firstly, we draw a line segment AB = 5.6 units and extend it to C such that BC = 1 unit. Let O be the mid-point of AC. Now, draw a semi-circle with centre O and radius OA. Let



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us draw BD perpendicular to AC passing through point B and intersecting the semi-circle at point D.

Hence, the distance BD is √5.6 𝑢𝑛𝑖𝑡𝑠. Draw an arc with centre B and radius BD, meeting

AC produced at E, then 𝐵𝐸 = 𝐵𝐷 = √5.6 𝑢𝑛𝑖𝑡𝑠. (iii) Firstly, we draw a line segment 𝐴𝐵 = 8.1 𝑢𝑛𝑖𝑡𝑠 and extend it to C such that SC = 1 unit. Let O be the mid-point of AC. Now, draw a semi-circle with centre 0 and radius OA Let us draw BD perpendicular to AC passing through point 6 intersecting the semi-circle at point D:

Hence, the distance BD is √8.1 𝑢𝑛𝑖𝑡𝑠. Draw an arc with centre Sand radius BD, meeting

AC produced at E, then 𝐵𝐸 = 𝐵𝐷 = √8.1 𝑢𝑛𝑖𝑡𝑠. (iv) Firstly, we draw a line segment 𝐴𝑆 = 2.3 𝑢𝑛𝑖𝑡𝑠 and extend it to C such that 𝑆𝐶 = 1 𝑢𝑛𝑖𝑡. Let O be the mid-point of AC. Now, draw a semi-circle with centre 0 and radius OA. Let us draw BD perpendicular to AC passing through point 6 and intersecting the semi-circle at point D.

Hence, the distance BD is √2.3 𝑢𝑛𝑖𝑡𝑠. Draw an arc with centre 6 and radius BD, meeting AC produced at E, then 𝐵𝐸 = 𝐵𝐷 =

√2.3 𝑢𝑛𝑖𝑡𝑠 Q.7) Express the following in the form

𝑝

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

(i) 0.2 (ii) 0.888... (iii) 5. 2̅ (iv) 0. 001̅̅ ̅̅ ̅ (v) 0.2555...

(vi) 0.134̅̅̅̅ (vii) .00323232... (viii) .404040...

Sol.7) i) 0.2

Let 𝑥 = 0.2 =2

10=

1

5

(ii) 0.888... Let 𝑥 = 0.888 … …(i) On multiplying both sides of eq. (i) by 10, we get 10𝑥 = 8.888 … …(ii) On subtracting eq. (i) from eq. (ii), we get 10𝑥 − 𝑥 = (8.88) − (0.888) 9𝑥 = 8

𝑥 =8

9

(iii) 5. 2̅

𝑥 = 5. 2̅ = 5.222 … …(i) On multiplying both sides of Eq. (i) by 10, we get



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10𝑥 = 52.222 … …(ii) On subtracting Eq. (i) from Eq.(ii), we get 10𝑥 − 𝑥 = (52.222 … ) − (5.222 … ) 9𝑥 = 47

𝑥 =47

9

(iv) 0. 001̅̅ ̅̅ ̅

Let 𝑥 = 0. 001̅̅ ̅̅ ̅ …(i) On multiplying both sides of Eq. (i) by 1000, we get 1000𝑥 = 001.001 … …(ii) On subtracting Eq. (i) from eq. (ii), we get 1000𝑥 − 𝑥 = 001.001 − (0.001001 … ) 999𝑥 = 001

𝑥 =1

999

(v) 0.2555... Let 𝑥 = 0.2555 … …(i) On multiplying both sides of eq. (i) by 10, we get 10𝑥 = 25.555 … …(ii) On multiplying both sides of eq. (i) from eq. (ii), we get 100𝑥 − 10𝑥 = 25.555 … − (2.555 … ) 90𝑥 = 23

𝑥 =23

90

(vi) 0.134̅̅̅̅

Let 𝑥 = 0.134̅̅̅̅

𝑥 = 0.134̅̅̅̅ = 0.13434 … …(i) On multiplying both sides of eq.(i) by 10, we get 10𝑥 = 1.3434 … …(ii) On multiplying both sides of eq. (ii) by 100, we get 1000𝑥 = 134.3434 … …(iii) On subtracting eq. (ii) from eq. (iii), we get 1000𝑥 − 10𝑥 = 134.34 … − (1.3434 … . ) 990𝑥 = 133

𝑥 =133

990

(vii) .00323232... Let 𝑥 = 0.00323232 … …(i) On multiplying both sides of eq. (i) by 100, we get 100𝑥 = 0.3232 … …(ii) On multiplying both sides of eq. (ii) by 100, we get 10000𝑥 = 32.3232 … On subtracting eq. (ii) from eq. (iii), we get 10000𝑥 − 100𝑥 = 32.32 … − 3.3232 … 9900𝑥 = 32

𝑥 =32

9900=

8

2475



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(viii) .404040... Let 𝑥 = 0.404040 … …(i) On multiplying both sides by 100, we get 100𝑥 = 40.4040 …(ii) On subtracting eq. (i) from eq.(ii), we get 100𝑥 − 𝑥 = 40.4040 … − 0.404040 … 99𝑥 = 40

𝑥 =40

99

Q.8) Show that 0.142857142857 … =1

7

Sol.8) Let 𝑥 = 0.142857142857 …………………..(i) On multiplying both sides of Eq. (i) by 1000000, we get 1000000 𝑥 = 142857.142857…………………………(ii) On subtracting Eq. (i) from Eq. (ii), we get 1000000 𝑥 – 𝑥 = (142857.142857 … ) – (0.142857. . ) ⇒ 999999 𝑥 = 142857

∴ 𝑥 =142857

999999=

1

7 Hence proved.

Q.9) Simplify the following:

(i) √45 − 3√20 + 4√5 (ii) √24

8+

√54

9 (iii) √12

4× √6

7

(iv) √284

÷ 3√7 ÷ √73

(v) 3√3 + 2√27 +7

√3 (vi) (√3 − √2)

2

(vii) √814

− 8√2163

+ 15√325

+ √225 (viii) 3

√8+

1

√2

(ix) 2√3

3−

√3

6

Sol.9) (i) √45 − 3√20 + 4√5

= √3 × 3 × 5 − 3√2 × 2 × 2 + 4√5

= 3√5 − 3 × 2√5 + 4√5

= 3√5 − 6√5 + 4√5 = √5

(ii) √24

8+

√54

9

=√2×2×2×3

8+

√3×3×3×2

9

=2√6

8+

3√6

9=

√6

4+

√6

3

=3√6+4√6

12=

7√6

12

(iii) √124

× √67

= (12)1

4 × (6)1

7

= (2 × 2 × 3)1

4 × (2 × 3)1

7 = 21

4. 21

4. 31

4. 21

7. 31

7

= 21

4+

1

4+

1

7 × 31

4+

1

7

= 29

14 × 311

28 = √2914√31128

= √21828√331128

= √218 × 31128

(iv) √284

÷ 3√7 ÷ √73

= (4 √4 × 7 ÷ 3√7) ÷ √73

= (8√7

3√7) ÷ (7)

1

3

=8

3÷ 7

1

3 =8

3 √73



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(v) 3√3 + 2√27 +7

√3

= 3√3 + 2√3 × 3 × 3 +7

√3×

√3

√3

= 3√3 + 6√3 +7√3

3+ 9√3 +

7√3

3

=(27√3+7√3)

3=

34√3

3

(vi) (√3 − √2)2

= (√3)2

+ (√2)2

− 2√3 × √2

= 3 + 2 − 2√3 × 2 = 5 − 2√6

(vii) √814

− 8√2163

+ 15√325

+ √225

= (81)1

4 − 8 × (216)1

3 + 15 × (32)1

5 + √(15)2

= (34)1

4 − 8 × (63)1

3 + 15 × (25)1

5 + 15

= 34×1

4 − 8 × 63×1

3 + 15 × 25×1

5 + 15 = 31 − 8 × 61 + 15 × 21 + 15 = 3 − 48 + 30 + 15 = 48 − 48 = 0

(viii) 3

√8+

1

√2

=3

√2×2×2+

1

√2

=3

2√2=

5

2√2×

√2

√2

=5√2

2×2=

5√2

4

(ix) 2√3

3−

√3

6

=(4√3−√3)

6=

3√3

6=

√3

2

Q.10) Rationalise the denominator of the following:

(i) 2

3√3 (ii)

√40

√3 (iii)

3+√2

4√2 (iv)

16

√41−5 (v)

2+√3

2−√3

(vi) √6

√2+√3 (vii)

√3+√2

√3−√2 (viii)

3√5+√3

√5−√3 (ix)

4√3+5√2

√48+√18

Sol.10) (i) 2

3√3

Let 𝐸 =2

3√3

For rationalising the denominator, multiplying numerator & denominator by √3.

𝐸 =2

3√3×

√3

√3

=2√3

3×3=

2√3

9

(ii) √40

√3

Let 𝐸 =√40

√3




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𝐸 =√40

√3×

√3

√3=

√40×3

(√3)2

=√120

3

=√2×2×2×5×3

3=

2

3√30

(iii) 3+√2

4√2

Let 𝐸 =(3+√2)

4√2


𝐸 =3+√2

4√2×

√2

√2=

3√2+(√2)2

4(√2)2

=3√2+2

4×2=

3√2+2

8

(iv) 16

√41−5

Let 𝐸 =16

√41−5

For rationalising the denominator, multiplying numerator & denominator by √41 + 5

𝐸 =16

√41−5×

√41+5

√41+5

= (16(√41+5)

(√41)2

−(5)2

=(16(√41+5))

41−25

=16(√41+5)

16= √41 + 5

(v) 2+√3

2−√3

For rationalising the denominator, multiplying numerator & denominator by 2 + √3

𝐸 =(2+√3)

2−√3×

(2+√3)

2+√3=

((2+√3)2

)

(2)2−(√3)2

=22+(√3)

2+2×2×√3

4−3

= 4 + 3 + 4√3 = 7 + 4√3

(vi) √6

√2+√3

For rationalising the denominator, multiplying numerator & denominator by √2 − √3

𝐸 =√6

√2+√3×

√2−√3

√2−√3=

√6(√2−√3)

(√2)2

−(√3)2

=√6(√2−√3)

2−3=

√6(√2−√3)

−1= √6(√3 − √2)

= √18 − √12 = √9 × 2 − √4 × 3 = 3√2 − 2√3

(vii) √3+√2

√3−√2

For rationalising the denominator, multiplying numerator & denominator by √3 + √2



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𝐸 =√3+√2

√3−√2×

√3+√2

√3+√2=

(√3+√2)2

(√3)2

+(√2)2

=(√3)

2+(√2)

2+2√3√2

3−2

= 3 + 2 + 2√6 = 5 + 2√6

(viii) 3√5+√3

√5−√3

For rationalising the denominator, multiplying numerator & denominator by √5 + √3

𝐸 =3√5+√3

√5−√3×

√5+√3

√5+√3

=3√5(√5+√3)+√3(√5+√3)

(√5)2

−(√3)2

=15+3√15+√15+3

5−3=

(18+4√15)

2= 9 + 2√15

(ix) 4√3+5√2

√48+√18

For rationalising the denominator, multiplying numerator & denominator by 4√3 − 3√2

=4√3+5√2

4√3+3√2×

(4√3−3√2)

(4√3−3√2)

=4√3(4√3−3√2)+5√2(4√3−3√2)

(4√3)2

−(3√2)2

=48−12√6+20√6−30

30

=18+8√6

30=

9+4√6

15

Q.11) Find the values of a and b in each of the following

(i) 5+2√3

7+4√3= 𝑎 − 6√3 (ii)

3−√5

3+2√5= 𝑎√5 −

19

11 (iii)

√2+√3

3√2−2√3= 2 − 𝑏√6

(iv) 7+√5

7−√5−

7−√5

7+√5= 𝑎 +

7

11√5𝑏

Sol.11) (i) 5+2√3

7+4√3= 𝑎 − 6√3

For rationalising the above equation, we multiply numerator & denominator of LHS by

7 − 4√3 , we get

5+2√3

7+4√3×

(7−4√3)

7−4√3= 𝑎 − 6√3

5(7−4√3)+2√3(7−4√3)

72−(4√3)2 = 𝑎 − 6√3

35−20√3+14√3−24

49−48= 𝑎 − 6√3

11 − 6√3 = 𝑎 − 6√3 = 𝑎 = 11

(ii) 3−√5

3+2√5= 𝑎√5 −

19

11


3 − 2√5, we get

(3−√5)

3+2√5×

3−2√5

3−2√5= 𝑎√5 −

19

11

3(3−2√5)−√5(3−2√5)

(3)2−(2√5)2 = 𝑎√5 −

19

11

9−6√5−3√5+10

9−4×5= 𝑎√5 −

19

11



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19−9√5

9−20= 𝑎√5 −

19

11

19−9√5

−11= 𝑎√5 −

19

11

9√5

11−

19

11= 𝑎√5 −

19

11

9√5

11= 𝑎√5

𝑎 =9

11

(iii) √2+√3

3√2−2√3= 2 − 𝑏√6


3√2 + 2√3 , we get

√2+√3

3√2−2√3×

3√2+2√3

3√2+2√3= 2 − 𝑏√6

(√2(3√2+2√3)+√3(3√2+2√3))

(3√2)2

−(2√3)2 = 2 − 𝑏√6

6+2√6+3√6+6

18−12= 2 − 𝑏√6

12+5√6

6= 2 − 𝑏√6

2 +5√6

6= 2 − 𝑏√6

𝑏√6 = −5√6

6

𝑏 = −5

6

(iv) 7+√5

7−√5−

7−√5

7+√5= 𝑎 +

7

11√5𝑏

(7+√5)

2−(7−√5)

2

(7−√5)(7+√5)= 𝑎 +

7

11√5𝑏

[72+(√5)

2+2×7×√5]−[72+(√5)

2−2×7×√5]

72−(√5)2 = 𝑎 +

7

11√5𝑏

49+5+14√5−49−5+14√5

49−5= 𝑎 +

7

11√5𝑏

28√5

44= 𝑎 +

7

11√5𝑏

7

11√5 = 𝑎 +

7

11√5𝑏

0 +7

11√5 = 𝑎 +

7

11√5𝑏

On comparing both sides, we get 𝑎 = 0 and 𝑏 = 1

Q.12) If 𝑎 = 2 + √3 , then find the value of 𝑎 −1

𝑎.

Sol.12) We have, 𝑎 = 2 + √3

Then, 1

𝑎=

1

2+√3×

2−√3

2−√3

=2−√3

(2)2−(√3)2 =

2−√3

4−3= 2 − √3

=1

𝑎= 2 − √3

𝑎 −1

𝑎= (2 + √3) − (2 − √3) = 2 + √3 − 2 + √3 = 2√3



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Q.13) Rationalise the denominator in each of the following and hence evaluate by taking √2 =

1.414 , √3 = 1.732 and √5 = 2.236 , upto three places of decimal

(i) 4

√3 (ii)

6

√6 (iii)

√10−√5

2 (iv)

√2

2+√2 (v)

1

√3+√2

Sol.13) (i) 4

√3

Let 𝐸 =4

√3


𝐸 =4

√3×

√3

√3=

4√3

3

=4

3× 1.732 =

6.928

3= 2.309

ii) 6

√6

For rationalising the denominator, multiplying numerator & denominator by √6, we get

𝐸 =6

√6×

√6

√6=

6√6

6= √2 × √3

= 1.414 × 1.732 = 2.449

(iii) √10−√5

2

Let 𝐸 =(√10−√5)

2

=√5 √2−√5

2=

√5(√2−1)

2

=2.236(1.414−1)

2= 1.118 × 0.414 = 0.46285 = 0.463

(iv) √2

2+√2

For rationalising the denominator, multiplying numerator & denominator by 2 − √2, we get

=√2

2+√2×

2−√2

2−√2=

√2(2−√2)

(2)2−(√2)2

=√2×√2(√2−1)

2=

2(√2−1)

2

= √2 − 1 = 1.414 − 1 = 0.414

(v) 1

√3+√2

For rationalising the denominator, multiplying numerator & denominator by √3 − √2, we get

1

√3+√2×

√3−√2

√3−√2=

√3−√2

(√3)2

−(√2)2

=√3−√2

3−2= √3 − √2

= 1.732 − 1.414 = 0.318 Q.14) Simplify:

(i) (13 + 23 + 33)1

2 (ii) 3

5

4

8

5

−12

32

5

6 (iii)

1

27

−2

3 (iv) (625)(−1

2)

−14

2

(v) 9

13×27

−12

316×3

−23

(vi) 64−1

3 641

3 − 642

3 (vii) 8

13×16

13

32−

13



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Sol.14) (i) (13 + 23 + 33)1

2

= (1 + 8 + 27)1

2

= (36)1

2 = (62)1

2 = 62×1

2 = 6

(ii) 3

5

4

8

5

−12

32

5

6

=34

54 × (5

23)12

× (25

5)

6

=34

54 ×512

236 ×230

56

=34×512−4−6

236−30

=34

26 × 52 =81×25

64=

2025

64

(iii) 1

27

−2

3

= (1

33)−

2

3= (3−3 )−

2

3

= 3−3×−2

3 = 32 = 9

(iv) (625)(−1

2)

−14

2

= [((252)−1

2)−

1

4]

2

= (25−1)−1

4×2 = [(52)−1](−

1

4×2)

= 5−2×−1

4×2 = 51 = 5

(v) 9

13×27

−12

316×3

−23

=(32)

13×(33)

−12

316×3

−23

=3

23×3

−32

316×3

−23

=3

23

−32

316

−23

=3

4−96

31−4

6

=3

−56

3−

36

= 3−5

6+

3

6 = 3−2

6 = 3−1

3

(vi) 64−1

3 [641

3 − 642

3]

= (43)−1

3 [(43)1

3 − (43)2

3]

= 43×−1

3 (43×1

3 − 43×2

3) = 4−1(4 − 42)

=1

4(4 − 16) = −

12

4= −3

(vii) 8

13×16

13

32−

13

=(23)

13×(24)

13

(25)−13

=2

3×13×2

4×13

25×−

13



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=2

33

+43

2−

53

=2

73

2−

53

= 27

3+

5

3 = 212

3 = 24 = 16

EXERCISE 1.4 Long Answer Type Questions

Q.1) Express 0.6 + 0. 7̅ + 0.47̅ in the form 𝑝

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

Sol.1) Let 𝑥 = 0. 7̅ = 0.777 … (i) On multiplying both sides of eq. (i) by, we get 10𝑥 = 7.77 …. … (ii) On subtracting eq. (i) from eq. (ii), we get 10𝑥 − 𝑥 = (7.77 … ) − (0.77 … ) 9𝑥 = 7

𝑥 =7

9

Now, let 𝑦 = 0.47̅ = 0.4777 … …(iii) On multiplying both sides of eq. (iii) by 10, we get 10𝑦 = 4.777 … … (iv) On multiplying both sides eq. (iv) by 10, we get 100𝑦 = 47.777 … On subtracting eq. (iv) from eq. (v), we get (100𝑦 − 10𝑦) = (47.777 … ) − (4.777 … )

90𝑦 = 43 =43

90

0.6 + 0. 7̅ + 0.47̅ =6

10+

7

9+

43

90

=54+70+43

90=

167

90

Q.2) Simplify: 7√3

√10+√3−

2√5

√6+√5−

3√2

√15+3√2

Sol.2) We have, 7√3

√10+√3−

2√5

√6+√5−

3√2

√15+3√2 … (i)

Now, 7√3

√10+√3×

√10−√3

√10−√3=

7√30−21

(√10)2

−(√3)2

=7√30−21

10−3=

7(√30−3)

7= √30 − 3

2√5

√6+√5=

2√5

√6+√5×

√6−√5

√6−√5

=2√30−10

(√6)2

+(√5)2

=2√30−10

6−5= 2√30 − 10

and 3√2

√15+3√2=

3√2

√15+3√2×

√15−3√2

√15−3√2

=3√30−18

(√15)2

−(3√2)2

=3(√30−6)

15−18=

3(√30−6)

−3= (−√30 + 6)

= 6 − √30

From eq. (i), 7√3

√10+√3−

2√5

√6+√5−

3√2

√15+3√2

= (√30 − 3) − (2√30 − 10) − (6 − √30)

= √30 − 3 − 2√30 + 10 − 6√30

= 2√30 − 2√30 + 10 − 9 = 1



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Q.3) If √2 = 1.414, √3 = 1.732 , then find the value of 4

3√3−2√2+

3

3√3+2√2

Sol.3) We have, 4

3√3−2√2+

3

3√3+2√2

=4(3√3+2√2)+3(3√3−2√2)

(3√3−2√2)(3√3+2√2)

=12√3+8√2+9√3−6√2

(3√3)2

−(2√2)2

=21√3+2√2

27−8=

21√3+2√2

19

=21×1.732+2×1.414

19=

36.372+2.828

19=

39.2

19= 2.06316 = 2.063

Q.4) If 𝑎 =3+√5

2, then find the value of 𝑎2 −

1

𝑎2.

Sol.4) Given, 𝑎 =3+√5

2

Now, 1

𝑎=

2

3+√5=

2

3+√5×

3−√5

3−√5 … (i)

=6−2√5

32−(√5)2

=6−2√5

9−5=

6−2√5

4

1

𝑎=

2(3−√5)

4=

3−√5

2

𝑎2 +1

𝑎2 = 𝑎2 +1

𝑎2 + 2 − 2 = (𝑎 +1

𝑎)

2− 2

= (3+√5

2+

3−√5

2)

2

− 2

= (6

2)

2− 2 = (3)2 − 2

= 9 − 2 = 7 Q.5) If 𝑥 =

√3+√2

√3−√2 and 𝑦 =

√3−√2

√3+√2, the find the value of 𝑥2 + 𝑦2.

Sol.5) Now, 𝑥 =√3+√2

√3−√2×

√3+√2

√3+√2

=(√3+√2)

2

(√3)2

−(√2)2

=(√3)

2+(√2)

2+2.√3.√2

3−2

=3+2+2√6

1= 3 + 2 + 2√6

𝑥 = 5 + 2√6 … (i) On squaring both sides, we get

𝑥2 = (5 + 2√6)2

= (5)2 + (2√6)2

+ 2.5.2√6

𝑥2 = 25 + 24 + 20√6 = 49 + 20√6

𝑥2 = 49 + 20√6 … (ii)

∴ 𝑦 =√3−√2

√3+√2=

1

𝑥=

1

5+2√6 [from eq. (i)]

=1

5+2√6×

5−2√6

5−2√6

=5−2√6

(5)2−(2√6)2 =

5−2√6

25−24=

5−2√6

1

On squaring both sides, we get

𝑦2 = (5 − 2√6)2



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𝑦2 = (5)2 + (2√6)2

− 2 × 5 × 2√6

𝑦2 = 25 + 24 − 20√6

𝑦2 = 49 − 20√6 … (iii) On adding eq. (ii) and (iii), we get

𝑥2 + 𝑦2 = 49 + 20√6 + 49 − 20√6 = 98 Q.6)

Simplify: (256)−(4

−32)

Sol.6) (256)

−(4−

32)

= (256)−(4)−32 = (256)−(22)

−32

= (256)−(2

2×−32)

= (256)−(2−3)

= (28)−(1

23)= (28)−

1

8

= 28×−1

8 = 2−1 =1

2

Q.7) Find the value of 4

(216)−23

+1

(256)−34

+2

(243)−

15

Sol.7) We have, 4

(216)−23

+1

(256)−34

+2

(243)−

15

=4

(63)−23

+1

(162)−34

+2

(35)−

15

=4

63×(−

23

)+

1

(16)2×(−

34

)+

2

(3)5×(−

15

)

=4

6−2 +1

16−

32

+2

3−1

= 4 × 62 + 163

2 + 2 × 31

= 4 × 36 + ((4)2)3

2 + 2 × 31 = 4 × 36 + 43 + 6 = 144 + 64 + 6 = 214



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· chapter 1 number systems exercise 1.1 write the correct answer in each of the following: q.1)...

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