Rd Sharma Solutions Class 7 Chapter 6 Exponents
Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Rd Sharma Class 7 Mathematics, Chapter 6, Exponents. Here students can easily find Exercise wise solution for chapter 6, Exponents. Students will find proper solutions for Exercise 6.1, 6.2, 6.3 and 6.4. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.
Exponents Exercise 6.1 Solution
Question no – (1)
Solution :
(i) 13²
= 13 × 13
= 169
Thus, the value will be 169
(ii) 7³
= 7 × 7 × 7
= 343
Hence, the value will be 343
(iii) 3⁴
= 3 × 3 × 3 × 3
= 81
Therefore, the value will be 81.
Question no – (2)
Solution :
(i) (-7)2
= (-7) × (-7)
= 49
So, the value will be 49.
(ii) (- 3)4
= -3 × (-3) × (-3) × (-3)
= 81
Hence, the value will be 81.
(iii) (- 5)5
= -5 × (- 5) × (- 5) × (- 5) × (- 5)
= -3125
Therefore, the value will be -3125.
Question no – (3)
Solution :
(i) 3 × 102
= 3 × 10 × 10
= 300…(Simplified)
(ii) 22 × 5³
= 2 × 2 × 5 × 5 × 5
= 4 × 125
= 500…(Simplified)
(iii) 3³ × 52
= 3 × 3 × 3 × 5 × 5
= 27 × 25
= 675…(Simplified)
Question no – (4)
Solution :
(i) 32 × 104
= 3 × 3 × 10 × 10 × 10 × 10
= 9 × 10000
= 90000…(Simplified)
(ii) 24 × 32
= 2 × 2 × 2 × 2 × 3 × 3
= 16 × 9
= 144…(Simplified)
(iii) 52 × 34
= 5 × 5 × 3 × 3 × 3 × 3
= 25 × 81
= 2025…(Simplified)
Question no – (5)
Solution :
(i) (- 2) × (- 3)3
= -2 × 3 × 3 × 3
= -2 × (- 27)
= 54…(Simplified)
(ii) (- 3)2 × (- 5)3
= (-3 × -3 × (-5 × -5 × -5 ×)
= -(9 × 125)
= – 1125…(Simplified)
(iii) (- 2)5 × (- 10)2
= (- 2) × (- 2) × (- 2) × (- 2) × (- 2) × (- 10) × (- 10)
= (- 32) × (100)
= – 3200…(Simplified)
Question no – (6)
Solution :
(i) (3/4)2
= 9/16…(Simplified)
(ii) (-2/3)4
= 16/81…(Simplified)
(iii) (-4/5)5
= – 1024/3125…(Simplified)
Question no – (7)
Solution :
(i) 2⁵ or 5²
= 2⁵ = 32
and 5² = 25
∴ 2⁵ > 5²
Hence, 2⁵ is the greater number.
(ii) 3⁴ or 4³
= 34 = 81
and 4³ = 64
∴ 3⁴ > 43
Thus, 3⁴ is the greater number.
(iii) 3⁵ or 5³
= 3⁵ = 243,
and 5³ = 125
∴ 3⁵ > 5³
Therefore, 3⁵ is the greater number.
Question no – (8)
Solution :
(i) (- 5) × (- 5) × (- 5)
= (- 5) × (- 5) × (- 5)
= (-5)3
Thus, Exponential form will be = (-5)3
(ii) – 5/7 × – 5/7 × – 5/7 × – 5/7
= (- 5/7) × (- 5/7) × (- 5/7) × (- 5/7)
= (- 5/7)4
Hence, Exponential form will be = (- 5/7)4
(iii) 4/3 × 4/3 × 4/3 × 4/3 × 4/3
= 4/3 × 4/3 × 4/3 × 4/3 × 4/3
= (4/3)5
Therefore, Exponential form will be = (4/3)5
Question no – (9)
Solution :
(i) x × x × x × x × a × a × b × b × b × b
= x3a2b3
Thus, the exponential form will be x3a2b3
(ii) (- 2) × (- 2) × (- 2) × (- 2) × a × a × a
= (- 2)4 × a3
Hence, the exponential form will be (- 2)4 × a3
Question no – (10)
Solution :
(i) 512
= 29
Thus, the exponential form of 512 is 29
(ii) 625
= 54
Therefore, the exponential form of 625 is 54
(iii) 729
= 36
Thus, the exponential form of 729 is 36
Question no – (11)
Solution :
(i) 36
= 4 × 9
= 22 × 32
(ii) 675
= 25 × 27
= 52 × 33
(iii) 392
= 8 × 49
= 23 × 72
Question no – (12)
Solution :
(i) 450
= 2 × 9 × 25
= 2 × 32 × 52
(ii) 2800
= 7 × 400
= 7 × 22 × 102
Question no – (13)
Solution :
(i) (3/7)2
= 3 × 3 × 7 × 7
= 9/49
(ii) (7/9)3
= 7 × 7 × 7 × 9 × 9 × 9
= 343/729
(iii) (-2/3)4
= -2 × -2 × -2 × -2 × 3 × 3 × 3 × 3
= 16/81
Question no – (14)
Solution :
(i) 49/64
= (7/8)2
(ii) -64/125
= (- 4/5)3
(iii) -1/216
= (-1/6)3
Question no – (15)
Solution :
(i) (- 1/2)2 × 23 × (3/4)2
= 1/4 × 8 × 9/16
= 9/8
Thus, the value will be 9/8
(ii) (- 3/5)4 × (4/9)4 × (- 15/18)2
= 81/625 × 256/6561 × 225/18 × 18
= 64/18225
Therefore, the value will be 64/18225.
Exponents Exercise 6.2 Solution
Question no – (1)
Solution :
(i) Given, 23 × 24 × 25
= 2 (3 + 4 + 5)
= 212
∴ The answer in exponential form = 212
(ii) Given in the question, 512 ÷ 53
∴ 512 ÷ 53
= 512-3
= 59
∴ The answer in exponential form = 59
(iii) Given, (72)3
= 72 × 3
= 76
∴ The answer in exponential form = 76
(iv) Given, (32)5 ÷ 34
= 310 – 4
= 36
∴ The answer in exponential form = 36
(v) Given, 37 × 27
= (3 × 2)7
= 67
∴ The answer in exponential form = 67
(vi) (521 ÷ 613) × 57
= 521 – 13 × 57
= 58+7
= 515
∴ The answer in exponential form = 515
Question no – (2)
Solution :
(i) {(23)4 × 28} ÷ 212
= {212 +8} ÷ 212
= 220-12
= 28…(Simplified)
(ii) (82 × 84) ÷ 83
= 86 – 3
= 83…(Simplified)
(iii) (57/52) × 53
= 5 7-2+3
= 58…(Simplified)
(iv) 54 × x10 y5/54 × x7 y4
= x3 y…(Simplified)
Question no – (3)
Solution :
(i) Given, {(32)3 × 26} × 56
= {36 × 26} × 56
= 66 × 56
= 306…(Simplified)
(ii) Given, (x/y)12 × y24 × (23)4
= 212 x12 y12…(Simplified)
(iii) Given, (5/2)6 × (5/2)2
= (5/2)8…(Simplified)
Question no – (4)
Solution :
In the given question, 9 × 9 × 9 × 9 × 9
∴ 9 × 9 × 9 × 9 × 9
= 32 × 32 × 32 × 32 × 32
= 32+2+2+2+2+2
= 310
Therefore, the answer will be 310
Question no – (5)
Solution :
(i) (25)3 ÷ 53
= 56 ÷ 53
= 56-3
= 53…(Simplified)
(ii) (81)5 ÷ (32)5
= 320 ÷ 3 10
= 320 – 10
= 310…(Simplified)
(iii) 98 × (x2)5/ (27)4 × (x3)2
= 31 × x10/ 312 × x6
= 34 × x4
= (3x)4 …(Simplified)
Question no – (6)
Solution :
(i) In the given question, (35)11 × (315)4 – (35)18 × (35)5
∴ (35)11 × (315)4 – (35)18 × (35)5
= 355 × 360 – 390 × 325
= 3115 – 3115
= 0…(Simplified)
(ii) Given in the question, 16 × 2n+1 – 4 × 2n/16 × 2n +2 – 2 × 2n + 2
= 16 × 2n + 21 – 2 × 2n+2/ 16 × 2n 22 – 2 × 2n 22
= 2n [16.2 -4]/2 × 2n [16 × 2 – 22]
= 1/2…(Simplified)
(iii) In the given question, 10 × 5n+1 + 25 × 5n/3 × 5n+2 + 10 × 5 n+1
∴ 10 × 5n × 5 + 25 × 5n/3 × 5n × 52 + 10 × 5n × 5
= 5n (10 × 5 + 25)/5n (3 × 25 + 50)
= 75/125
= 3/5
Question no – (7)
Solution :
(i) In the given question, 52n × 53 = 511
= 5 2n+3
= 511
So, 2n + 3 = 11
= 2n = 11 – 3
= n = 8/2
= 4
Thus, the value of n will be 4.
(iii) In the question, 8 × 2n+2 = 32
∴ 8 × 2n+2 = 32
= 23 × 2n +2 = 32
= 2 n+2+3 = 25
So, n + 2 + 3 = 5
= n = 0
Therefore, the value of n will be 0.
(iv) In the given question, 72n+1 ÷ 49 = 73
= 72n+1/72 = 73
= 72n+1 = 7 3+2
So, 2n+1 = 5
= n = 5 -1/2
= 2
Hence, the value of n will be 2.
(v) Given, (3/2)4 × (3/2)5 = (3/2)2n+1
= (3/2)9 = (3/2)2n+1
So, 9 = 2n + 1
= n = 9 – 1/2
= 4
Hence, the value of n will be 4.
Question no – (8)
Solution :
As per the question,
= 9n × 32 × 3n – (27)n/(33)5 × 23 = 1/27
∴ 32n × 32 × 3x – 33x/315 × 23 = 1/27
= 33n+2 – 33n/315 × 23 = 1/33
= 33n+2 – 33n = 315 -3 × 23
= 32n [32 – 1] = 312 × 23
= 32n × 8 = 312 × 23
∴ 2n = 12
= 6
Therefore, the value of n will be 6.
Exponents Exercise 6.3 Solution
Question no – (1)
Solution :
(i) 3908.78 in standard form,
= 3.90878 × 1000
= 3.90878 × 103
(ii) 5, 00, 00, 000 in standard form,
= 5 × 107
(iii) 3, 18, 65, 00, 000 in standard form,
= 31865 × 105
(iv) 846 × 107 in standard form,
= 8, 46 × 109
(v) 723 × 109 in standard form,
= 7.23 × 1011
Question no – (2)
Solution :
(i) 4.83 × 107 in usual form,
= 48300000
(ii) 3.21 × 105 in usual form,
= 3, 21, 000
(iii) 3.5 × 103 in usual form,
= 35,00
Question no – (3)
Solution :
(i) From the question we get,
The distance between the Earth and the moon is 384,000,000 metres.
= 384,000,000 in standard form,
= 384 × 106 m
(ii) In the question we get,
Diameter of the Earth is 1, 27, 56, 000 metres.
= 1, 27, 56, 000 in standard form,
= 1, 2756 × 107 m
(iii) In the question,
Diameter of the Sun is 1, 400, 000, 000 metres.
= 1, 400, 000, 000 in standard form,
= 14 × 109 m
(iv) As per the question,
= The universe is estimated to be about 12, 000, 000, 000 years old.
= 12, 000, 000, 000 in standard form,
= 12 × 109 m.
Exponents Exercise 6.4 Solution
Question no – (1)
Solution :
(i) 20068
= 20000 + 68
= 2 × 104 + 6 × 101 + 8…(Expanded exponential form)
(ii) 420719
= 420000 + 700 + 10 + 9
= 42 × 105 + 2 × 104 + 7 × 102 + 1 × 101 + 9…(Expanded exponential form)
(iii) 7805192
= 7 × 106 + 8 × 105 + 5 × 103 + 1 × 102 + 9 × 10 + 2…(Expanded exponential form)
(iv) 5004132
= 5 × 106 + 4 × 103 + 1 × 102 + 3 × 10 + 2…(Expanded exponential form)
(v) 927303
= 9 × 105 + 2 × 104 + 7 × 103 + 3 × 102 +0 × 101 + 3 × 100…(Expanded exponential form)
Question no – (2)
Solution :
(i) 7 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
= 76045
Therefore, the numbers will be 76045
(ii) 5 × 105 + 4 × 104 + 2 × 103 + 3 × 100
= 542003
Hence, the numbers will be 542003
(iii) 9 × 105 + 5 × 102 + 3 × 101
= 900530
Thus, the numbers will be 900530
(iv) 3 × 104 + 4 × 102 + 5 × 100
= 30405
So, the numbers will be 30405
Next Chapter Solution :
👉 Chapter 7 👈
