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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
= 442 (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, 72 + 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 :
👉 Chapter 4 👈
