Exponents and Powers For Class 7 Maths Important Queastions
Q 1 – Verify the following:
Q 2 – Simplify and write in exponential form:
Q 3 – Express each of the following as a product of prime factors is the exponential form:
(i) 729 × 125
(ii) 384 × 147
(i) 729 × 125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5 = 36 × 53
Thus, 729 × 125 = 36 × 53
(ii) 384 × 147 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 = 27 × 32 × 72
Thus, 384 × 147 = 27 × 32 × 72
Q 4 – Write the following in expanded form:
(i) 70,824
(ii) 1,69,835
(i) 70,824
= 7 × 10000 + 0 × 1000 + 8 × 100 + 2 × 10 + 4 × 100
= 7 × 104 + 8 × 102 + 2 × 101 + 4 × 100
(ii) 1,69,835
= 1 × 100000 + 6 × 10000 + 9 × 1000 + 8 × 100 + 3 × 10 + 5 × 100
= 1 × 105 + 6 × 104 + 9 × 103 + 8 × 102 + 3 × 101 + 5 × 100
Q 5 – Find the number from each of the expanded forms:
(i) 7 × 108 + 3 × 105 + 7 × 102 + 6 × 101 + 9
(ii) 4 × 107 + 6 × 103 + 5
(i) 7 × 108 + 3 × 105 + 7 × 102 + 6 × 101 + 9
= 7 × 100000000 + 3 × 100000 + 7 × 100 + 6 × 10 + 9
= 700000000 + 300000 + 700 + 60 + 9
= 700300769
(ii) 4 × 107 + 6 × 103 + 5
= 4 × 10000000 + 6 × 1000 + 5
= 40000000 + 6000 + 5
= 40006005
Q 6 – Find the value of k in each of the following:
Q 7 – Find the value of
(a) 30 ÷ 40
(b) (80 – 20) ÷ (80 + 20)
(c) (20 + 30 + 40) – (40 – 30 – 20)
(a) We have 30 ÷ 40 = 1 ÷ 1 = 1 [∵ a0 = 1]
(6) (80 – 20) ÷ (80 + 20) = (1 – 1) ÷ (1 + 1) = 0 ÷ 2 = 0
(c) (20 + 30 + 40) – (40 – 30 – 20)
= (1 + 1 + 1) – (1 – 1 – 1) [∵ a0 = 1]
= 3 – 1
= 2
Q 8 – Evaluate:
Q 9 – Find the value of x, if
Q 10 – Express the following in exponential form:
(i) 5 × 5 × 5 × 5 × 5
(ii) 4 × 4 × 4 × 5 × 5 × 5
(iii) (–1) × (–1) × (–1) × (–1) × (–1)
(iv) a × a × a × b × c × c × c × d × d
(i) 5 × 5 × 5 × 5 × 5 = (5)5
(ii) 4 × 4 × 4 × 5 × 5 × 5 = 43 × 53
(iii) (–1) × (–1) × (–1) × (–1) × (–1) = (–1)5
(iv) a × a × a × b × c × c × c × d × d = a3b1c3d2
Q 11 – If a = 2 and b= 3, the 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
(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
Q 12 – Using laws of exponents, simplify and write the answer in exponential form
(i) 23 × 24 × 25
(ii) 512 ÷ 53
(iii) (72)3
(iv) (32)5 ÷ 34
(v) 37 × 27
(vi) (521 ÷ 513) × 57
(i) Given 23 × 24 × 25
We know that first law of exponents states that am × an × ap = a (m+n+p)
Therefore above equation can be written as 23 x 24 x 25 = 2 (3 + 4 + 5)
= 212
(ii) Given 512 ÷ 53
According to the law of exponents we have am ÷ an = am-n
Therefore given question can be written as 512 ÷ 53 = 512 – 3 = 59
(iii) Given (72)3
According to the law of exponents we have (am)n = amn
Therefore given question can be written as (72)3 = 76
(iv) Given (32)5 ÷ 34
According to the law of exponents we have (am)n = amn
Therefore (32)5 ÷ 34 = 3 10 ÷ 34
According to the law of exponents we have am ÷ an = am-n
3 10 ÷ 34 = 3(10 – 4) = 36
(v) Given 37 × 27
We know that law of exponents states that am x bm = (a x b)m
37 × 27 = (3 x 2)7 = 67
(vi) Given (521 ÷ 513) × 57
According to the law of exponents we have am ÷ an = am-n
= 5(21 -13) x 57
= 58 x 57
According to the law of exponents we have am x an = a(m +n)
= 5(8+7) = 515
Q 13 – The exponential form of 64 is
(a) 25
(b) 26
(c) 27
(d) 28
(b) 26
Q 14 – The value of (– 2) 4 is
(a) 8
(b) -8
(c) 16
(d) -16
(c) 16
Q 15 – (–1)even number =
(a) -1
(b) 1
(c) 0
(d) none of these
(b) 1
Q 16 – If 23 × 24 = 2?, then ? =
(a) 3
(b) 4
(c) 1
(d) 7
(d) 7
Q 17 – 1353000000 in standard form is
(a) 1.353 × 109
(b) 1.353 × 106
(c) 1.353 × 103
(d) 1.353 × 1012
(a) 1.353 × 109
Q 18 – 100000000000 in standard form is
(a) 1 × 108
(b) 1 × 109
(c) 1 × 1010
(d) 1 × 1011
(d) 1 × 1011
Q 19 – Which of the following is true?
(a) 20 = (100)0
(b) 102 × 108 = 1016
(c) 22 × 33 = 65
(d) 23 > 32
(a) 20 = (100)0