Selina Concise Class 7 Math Chapter 5 Exponents Exercise 5B Solution
EXERCISE 5B
(1) Fill in the blanks:
(i) In 52 = 25, base = 5 and index = 2
(ii) If index = 3x and base = 2y, the number = (2y)3x
(2) Evaluate:
(i) 28 ÷ 23 = (2)8-3 = 25 = 32
(ii) 23 ÷ 28 =(2)3 – 8 = (2)– 5 = ½5 = 1/32
(iii) (26)0 = (2)6 × 0 = 20 = 1
(iv) (30)6 = (3)0 × 6 = 30 = 1
(v) 83 × 8 – 5 × 84 = (8) 3 – 5 + 4 = 8 7 – 5 = 82 = 64
(vi) 54 × 53 ÷ 55 = (5) 4 + 3 – 5 = (5)7 – 5 = 52 = 25
(vii) 54 ÷ 53 × 55 = (5) 4 – 3 + 5 = (5) 1+5 = 56 = 15625
(viii) 44 ÷ 43 × 40 = (4) 4 – 3 + 0 = 41 = 4
(ix) (35 × 47 × 58)0 = 1
(3) Simplify, giving answers with positive index:
(i) (2 × 5) b(6+3+4) = 10b13
(ii) (6 × 9) x(2+5+3) y(3+1+4) = 54x10y8
(iii) (- a5) (a2) = – a5 + 2 = – a7
(iv) (- y2) (- y3) = y2+3 = y5
(v) (- 3)2 (3)3 = – 32+3 = – 35 = – 243
(vi) (- 4x) (- 5x2) = (4 × 5) x1+2 = 20x3
(vii) (5a2b) (2ab2) (a3b)
= (5 × 2) a(2+1+3) b(1+2+1)
= 10 a6b4
(viii) x2a + 7 + 2a – 8 = x4a – 1
(ix) 3(y + 2 – 4) = 3y – 2
(x) 2(4a + 3a – a) = 27a – a = 26a
(xii) (10) 2 × 3 (x)8 × 12
= 106 x96
= 1000000x96
(xiii) (a)10 × 10 (1)6 × 10
= a100 160
= a100
(xiv) (n)2 × 2 (- n)2 × 3
= – n 4+6 = – n10
(xv) – 9a2b2 × 25a4b2c8
= – (9 × 25) a2+4 b2+2 c8
= – 225 a6b4c8
(xvi) 4 × 0 × 27
= 0
(xvii) (2 × 2 × 2 × 2) a12 × (4 × 4) a4
= (16 × 16) a12+4
= 256 a16
(4) Simplify and express the answer in the positive exponents form:
(5) Evaluate:
(iii) 53 × 32 + (17)0 × 73
= (125 × 9) + (1 × 343)
= 1125 + 343
= 1468
(vi) 5n × 25n – 1 ÷ (5n – 1 × 25n – 1)
= 5(n – n + 1) × 25(n – 1 – n + 1)
= 5 × 250
= 5 × 1 = 5
(6) If m = – 2 and n = 2; find the value of:
(i) m2 + n2 – 2mn
= (- 2)2 + (2)2 – [2 × (- 2) × 2]
= 4 + 4 – [- 8]
= 8 + 8 = 16
(iv) 2n3 – 3m
= 2 × (2)3 – [3 × (- 2)]
= 16 + 6 = 22
Tanks very very much
Thara bhai
Thank God