# Maths Ace Class 8 Solutions Chapter 3

## Maths Ace Class 8 Solutions Chapter 3 Exponents

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Ace Prime Class 8 Math Book, Chapter 3, Exponents. Here students can easily find step by step solutions of all the problems for Exponents, Exercise 3.1, 3.2, 3.3 and 3.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 3 solutions.

Exponents Exercise 3.1 Solution :

Question no – (1)

Solution :

(a) 35

= 35 × 35

= 1225

(b) 120

= (120)2

= (120 × 120)

= 14400

(c) (-214)

= (-214)2

= (-214) × (-214)

= 45796

(d) (-356)

= -356 × (-356)

= 126736

(e) 2/7

= 2/7 × 2/7

= 4/49

(f) (- 9/10)

= (- 9/10) × (- 9/10)

= 21/100

(g) 29/51

= 29/51 × 29/51

= 841/2601

(h) (- 5/13)

= (-5/13) × (-5/13)

= 25/169

Question no – (2)

Solution :

(a) 123

= No, it is not a perfect square

(b) 576

= Yes, it’s a perfect square

(c) 82

= No, it is not a perfect square

(d) 169

= Yes, it’s a perfect square

(e) 64

= Yes, it’s a perfect square

(f) 1127

= No, it is not a perfect square

(g) 5776

= Yes, it’s a perfect square.

(h) 2020

= No, it is not a perfect square.

Question no – (3)

Solution :

(a) 169

= 132 (odd)

(b) 144

= 122 (Even)

(c) 1936

= 44(Even)

(d) 4096

= 642 (Even)

(e) 529

= 232 (Odd)

(f) 64

= 82 (Even)

(g) 81

= 92 (Odd)

(h) 2601

= 512 (Odd)

Exponents Exercise 3.2 Solution :

Question no – (1)

Solution :

(a) 5

= 5 = 1 or 2 (2x – 1, x)

(b) 98

= 3 or 4 (2 × 2 – 1, 2 × 2)

(c) 117

= 5 or 6 (2 × 2 – 1, 2 × 2)

(d) 287

= 5 or 6 (2 × 3 – 1), (2 × 3)

(e) 15

= 3 or 4

Question no – (2)

Solution :

(a) 12

= 12 × 12

= 144 → ‘4’

(b) 52

= 52 × 52

= 270 → ‘4’

(c) 23

= ‘9’

(d) 566

= ‘6’

(e) 179

= 1

Question no – (3)

Solution :

(a) (9)2 and (10)2

= 2 × 9

= 18

(b) (12)2 and (13)2

= 2 × 12

= 24

(c) (24)2 and (25)2

= 2 × 24

= 48

(d) (39)2 and (40)2

= 2 × 39

= 78

Question no – (4)

Solution :

(a) 4th and 5th terms

= 10 + 15

= 25

(b) 2nd and 3rd terms

= 3 + 6

= 9

(c) 9th and 10th terms

= 9 × 9 + ½

= 9 × 10/2 = 45

∴ 45 + 55

= 100

(d) 1st and 2nd terms

= 1 + 3

= 4

Question no – (5)

Solution :

(a) (33, 56, 65)

= 332 + 562 = 1089 + 3136

= 4225

= 652

Yes, It forms Pythagorean triplet

(b) Given, (65, 72, 97)

= 652 + 722 = 4225 + 5184

= 9409

= 972

Yes, it forms Pythagorean triplet

(c) (2, 4, 8)

= 22 + 42 = 4 + 16 = 20

≠ 82

No, it not forms Pythagorean triplet

(d) (7, 24, 25)

= 72 + 242 = 49 + 576

= 625

= 252

Yes, it forms Pythagorean triplet

Question no – (6)

Solution :

(a) Pythagorean triplets using 7

= 2x, x2 – 1, x2 + 1

= 2 × 7, 72 – 1, 7+ 1

= (14, 48, 50)

(b) Pythagorean triplets using 11

= 2 × 11, 112 – 1, 112 + 1

= 22, 120, 122

(c) Pythagorean triplets using 6

= 6 × 2, 62 – 1, 62 + 1

= (12, 35, 37)

(d) Pythagorean triplets using 13

= 2 × 13, 132 – 1, 132 + 1

= (26, 168, 170)

Exponents Exercise 3.3 Solution :

Question no – (1)

Solution :

(a) 256, 361, 225 – by Prime factorisation

√256

= √2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2 × 2 × 2 × 2

= 16

√361

= √19 × 19

= 19

√225

= √5 × 5 × 3 × 3

= 5 × 3

= 15

(b) 36, 81, 121 – by Estimation

(b) √36 = 6

One’s place digit ‘6’

6 × 6 = 36

√81 = 9

√121 = 11

(c) 169, 144, 324 – by Subtraction

Subtraction 169

(1) 169 – 1 = 168

(2) 168 – 3 = 165

(3) 165 – 5 = 160

(4) 160 – 7 = 153

(5) 153 – 9 = 144

(6) 144 – 11 = 133

(7) 133 – 13 = 120

(8) 120 – 15 = 105

(9) 105 – 107 = 88

(10) 88 – 19 = 69

(11) 69 – 21 = 48

(12) 48 – 23 = 25

(13) 25 – 25 = 0

∴ Total No of steps = 13

∴ √169 = 13

= 144

(1) 144 – 1 = 143

(2) 143 – 3 = 140

(3) 140 – 5 = 135

(4) 135 – 7 = 128

(5) 128 – 9 = 118

(6) 118 – 11 = 107

(7) 107 – 13 = 94

(8) 94 – 15 = 79

(9) 79 – 17 = 62

(10) 62 – 19 = 43

(11) 43 – 21 = 22

(12) 22 – 22 = 0

∴ Total steps = 12

∴ √144 = 12

(d) 4761, 7921, 4225 – by Division

Division method :

(i) ∴ √4761 = 69

(ii) ∴ √7921 = 89

(iii) ∴ √4225 = 65

Question no – (2)

Solution :

(a) 248

23 will subtracted to make perfect square

(b) 330

6 will be subtracted to get perfect square

(c) 200

∴ 4 will be subtracted

(d) 175

6 will be subtracted

(e) 230

5 will be subtracted

Question no – (4)

Solution :

(a) 3, 4, 12 and 15

∴ L.C.M = 2 × 2 × 3 × 5 = 60

3, 5 are not pair

So, 60 × 3 × 5 = 900

∴ Required number 900

(b) 4, 6, 8 and 12

= ∴ L.CM is = 2 × 2 × 3 × 2 = 24

Here, 2, 3 are not pair

so, 24 × 2 × 3 = 144

Required number is 144

(c) 6, 15, 18 and 20

= L.C.M is = 2 × 2 × 3 × 3 × 5 = 180

Here, 5 is not pair

180 × 5 = 900 is required number.

Exponents Exercise 3.4 Solution :

Question no – (1)

Solution :

(a) 64/121

= √64/121

= 8/11

(b) 16/25

= √16/25

= √4 × 4/5 × 5

= 4/5

(c) 256/676

= √256/676

= √25 × 25/26 × 26

= 25/26

(d) 361/625

= √361/625

= √19 × 19/25 × 25

= 19/25

(e) 4761/6724

= √4761/6724

= √69 × 69/82 × 82

= 69/82

(f) 196/1681

= √196/1681

= √14 × 14/41 × 41

= 14/41

Question no – (2)

Solution :

(a) 148.84

12.2

(b) 67.24

√67.24

= 8.2

(c) 5.76

√5.76

= 2.4

(d) 27.04

Next Chapter Solution :

Updated: June 16, 2023 — 5:31 am