Get the free PDF of RD Sharma Solutions for Class 7 Maths Exercise 6.1 of Chapter 6 Exponents from the given links. Students who aim to score high marks in the annual examination can download the PDFs available here. The BYJU’S subject expert team has solved the questions in such a way that learners can understand easily. Students who aim to score high in the Maths of Class 7 are advised to practise all the questions present in the RD Sharma Solutions for Class 7 as many times as possible. This exercise includes definitions, the meaning of exponents, along with how to read exponents. Here, students will be thorough with how to express a given number in terms of exponents and why it is introduced and so on.
RD Sharma Solutions for Class 7 Maths Chapter 6 – Exponents Exercise 6.1
Access answers to Maths RD Sharma Solutions for Class 7 Chapter 6 – Exponents Exercise 6.1
1. Find the values of each of the following:
(i) 132
(ii) 73
(iii) 34
Solution:
(i) Given 132
132 = 13 × 13 =169
(ii) Given 73
73 = 7 × 7 × 7 = 343
(iii) Given 34
34 = 3 × 3 × 3 × 3
= 81
2. Find the value of each of the following:
(i) (-7)2
(ii) (-3)4
(iii) (-5)5
Solution:
(i) Given (-7)2
We know that (-a) even number= positive number
(-a) odd number = negative number
We have, (-7)2 = (-7) × (-7)
= 49
(ii) Given (-3)4
We know that (-a) even number= positive number
(-a) odd number = negative number
We have, (-3)4 = (-3) × (-3) × (-3) × (-3)
= 81
(iii) Given (-5)5
We know that (-a) even number= positive number
(-a) odd number = negative number
We have, (-5)5 = (-5) × (-5) × (-5) × (-5) × (-5)
= -3125
3. Simplify:
(i) 3 × 102
(ii) 22 × 53
(iii) 33 × 52
Solution:
(i) Given 3 × 102
3 × 102 = 3 × 10 × 10
= 3 × 100
= 300
(ii) Given 22 × 53
22 × 53 = 2 × 2 × 5 × 5 × 5
= 4 × 125
= 500
(iii) Given 33 × 52
33 × 52 = 3 × 3 × 3 × 5 × 5
= 27 × 25
= 675
4. Simply:
(i) 32 × 104
(ii) 24 × 32
(iii) 52 × 34
Solution:
(i) Given 32 × 104
32 × 104 = 3 × 3 × 10 × 10 × 10 × 10
= 9 × 10000
= 90000
(ii) Given24 × 32
24 × 32 = 2 × 2 × 2 × 2 × 3 × 3
= 16 × 9
= 144
(iii) Given 52 × 34
52 × 34 = 5 × 5 × 3 × 3 × 3 × 3
= 25 × 81
= 2025
5. Simplify:
(i) (-2) × (-3)3
(ii) (-3)2 × (-5)3
(iii) (-2)5 × (-10)2
Solution:
(i) Given (-2) × (-3)3
(-2) × (-3)3 = (-2) × (-3) × (-3) × (-3)
= (-2) × (-27)
= 54
(ii) Given (-3)2 × (-5)3
(-3)2 × (-5)3 = (-3) × (-3) × (-5) × (-5) × (-5)
= 9 × (-125)
= -1125
(iii) Given (-2)5 × (-10)2
(-2)5 × (-10)2 = (-2) × (-2) × (-2) × (-2) × (-2) × (-10) × (-10)
= (-32) × 100
= -3200
6. Simplify:
(i) (3/4)2
(ii) (-2/3)4
(iii) (-4/5)5
Solution:
(i) Given (3/4)2
(3/4)2 = (3/4) × (3/4)
= (9/16)
(ii) Given (-2/3)4
(-2/3)4 = (-2/3) × (-2/3) × (-2/3) × (-2/3)
= (16/81)
(iii) Given (-4/5)5
(-4/5)5 = (-4/5) × (-4/5) × (-4/5) × (-4/5) × (-4/5)
= (-1024/3125)
7. Identify the greater number in each of the following:
(i) 25 or 52
(ii) 34 or 43
(iii) 35 or 53
Solution:
(i) Given 25 or 52
25 = 2 × 2 × 2 × 2 × 2
= 32
52 = 5 × 5
= 25
Therefore, 25 > 52
(ii) Given 34 or 43
34 = 3 × 3 × 3 × 3
= 81
43 = 4 × 4 × 4
= 64
Therefore, 34 > 43
(iii) Given 35 or 53
35 = 3 × 3 × 3 × 3 × 3
= 243
53 = 5 × 5 × 5
= 125
Therefore, 35 > 53
8. Express each of the following in exponential form:
(i) (-5) × (-5) × (-5)
(ii) (-5/7) × (-5/7) × (-5/7) × (-5/7)
(iii) (4/3) × (4/3) × (4/3) × (4/3) × (4/3)
Solution:
(i) Given (-5) × (-5) × (-5)
Exponential form of (-5) × (-5) × (-5) = (-5)3
(ii) Given (-5/7) × (-5/7) × (-5/7) × (-5/7)
Exponential form of (-5/7) × (-5/7) × (-5/7) × (-5/7) = (-5/7)4
(iii) Given (4/3) × (4/3) × (4/3) × (4/3) × (4/3)
Exponential form of (4/3) × (4/3) × (4/3) × (4/3) × (4/3) = (4/3)5
9. Express each of the following in exponential form:
(i) x × x × x × x × a × a × b × b × b
(ii) (-2) × (-2) × (-2) × (-2) × a × a × a
(iii) (-2/3) × (-2/3) × x × x × x
Solution:
(i) Given x × x × x × x × a × a × b × b × b
Exponential form of x × x × x × x × a × a × b × b × b = x4a2b3
(ii) Given (-2) × (-2) × (-2) × (-2) × a × a × a
Exponential form of (-2) × (-2) × (-2) × (-2) × a × a × a = (-2)4a3
(iii) Given (-2/3) × (-2/3) × x × x × x
Exponential form of (-2/3) × (-2/3) × x × x × x = (-2/3)2 x3
10. Express each of the following numbers in exponential form:
(i) 512
(ii) 625
(iii) 729
Solution:
(i) Given 512
Prime factorization of 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 29
(ii) Given 625
Prime factorization of 625 = 5 x 5 x 5 x 5
= 54
(iii) Given 729
Prime factorization of 729 = 3 x 3 x 3 x 3 x 3 x 3
= 36
11. Express each of the following numbers as a product of powers of their prime factors:
(i) 36
(ii) 675
(iii) 392
Solution:
(i) Given 36
Prime factorization of 36 = 2 x 2 x 3 x 3
= 22 x 32
(ii) Given 675
Prime factorization of 675 = 3 x 3 x 3 x 5 x 5
= 33 x 52
(iii) Given 392
Prime factorization of 392 = 2 x 2 x 2 x 7 x 7
= 23 x 72
12. Express each of the following numbers as a product of powers of their prime factors:
(i) 450
(ii) 2800
(iii) 24000
Solution:
(i) Given 450
Prime factorization of 450 = 2 x 3 x 3 x 5 x 5
= 2 x 32 x 52
(ii) Given 2800
Prime factorization of 2800 = 2 x 2 x 2 x 2 x 5 x 5 x 7
= 24 x 52 x 7
(iii) Given 24000
Prime factorization of 24000 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 5 x 5 x 5
= 26 x 3 x 53
13. Express each of the following as a rational number of the form (p/q):
(i) (3/7)2
(ii) (7/9)3
(iii) (-2/3)4
Solution:
(i) Given (3/7)2
(3/7)2 = (3/7) x (3/7)
= (9/49)
(ii) Given (7/9)3
(7/9)3 = (7/9) x (7/9) x (7/9)
= (343/729)
(iii) Given (-2/3)4
(-2/3)4 = (-2/3) x (-2/3) x (-2/3) x (-2/3)
= ((16/81)
14. Express each of the following rational numbers in power notation:
(i) (49/64)
(ii) (- 64/125)
(iii) (-12/16)
Solution:
(i) Given (49/64)
We know that 72 = 49 and 82= 64
Therefore (49/64) = (7/8)2
(ii) Given (- 64/125)
We know that 43 = 64 and 53 = 125
Therefore (- 64/125) = (- 4/5)3
(iii) Given (-1/216)
We know that 13 = 1 and 63 = 216
Therefore -1/216) = – (1/6)3
15. Find the value of the following:
(i) (-1/2)2 × 23 × (3/4)2
(ii) (-3/5)4 × (4/9)4 × (-15/18)2
Solution:
(i) Given (-1/2)2 × 23 × (3/4)2
(-1/2)2 × 23 × (3/4)2 = 1/4 × 8 × 9/16
= 9/8
(ii) Given (-3/5)4 × (4/9)4 × (-15/18)2
(-3/5)4 × (4/9)4 × (-15/18)2 = (81/625) × (256/6561) × (225/324)
= (64/18225)
16. If a = 2 and b= 3, then find the values of each of the following:
(i) (a + b)a
(ii) (a b)b
(iii) (b/a)b
(iv) ((a/b) + (b/a))a
Solution:
(i) Consider (a + b)a
Given a = 2 and b= 3
(a + b)a = (2 + 3)2
= (5)2
= 25
(ii) Given a = 2 and b = 3
Consider, (a b)b = (2 × 3)3
= (6)3
= 216
(iii) Given a =2 and b = 3
Consider, (b/a)b = (3/2)3
= 27/8
(iv) Given a = 2 and b = 3
Consider, ((a/b) + (b/a))a = ((2/3) + (3/2))2
= (4/9) + (9/4)
LCM of 9 and 6 is 36
= 169/36
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