Haryana State Board HBSE 7th Class Maths Solutions Chapter 13 Exponents and Powers Ex 13.1 Textbook Exercise Questions and Answers.
Haryana Board 7th Class Maths Solutions Chapter 13 Exponents and Powers Exercise 13.1
Question 1.
Find the value of:
(i) 26
(ii) 93
(iii) 112
(iv) 54
Solution:
(i) 28 = 2 × 2 × 2 × 2 × 2 × 2
= 64. Ans.
(ii) 93 = 9 × 9 × 9 = 729. Ans.
(iii) 112 = 11 × 11 = 121. Ans.
(iv) 54 = 5 × 5 × 5 × 5 = 625. Ans.
Question 2.
Express the following in
exponential from:
(i) 6 × 6 × 6 × 6
(ii) t × t
(iii) b x b × 6 × 6
(iv) 5 × 5 x 7 x 7 x 7
(v) 2 × 2 × a × a
(vi) axaxaxcxcxcxcxd
Solutoon:
(i) 6 × 6 × 6 × 6 = 64
(ii) t × t = t2
(iii) b × b × b × b = 64
(iv) 5 × 5 × 7 × 7 × 7= 52 × 73
(v) 2 × 2 × a × a = 22 × a2
(vi)a xaxaxcxcxcxcxd
= a3 × c4 × d1 .
Question 3.
Express each of the following numbers using exponential notation:
512 (ii) 343 (iii) 729 (iv) 3125
Solution:
(i) 512 = 8 × 8 × 8 = 83 Ans.
(ii) 343 = 7 × 7 × 7 = 73 Ans.
(iii) 729 = 9 × 9 x9 = 729. Ans.
(iv) 3125 = 5 × 5 × 5 × 5 × 5 = 55 Ans.
Question 4.
Identify the greater number, whever possible, in each of the following ?
(i) 43 or 34
(ii) 53 or 35 (iii) 210 or 82
(iv) 1002 or 2100 (iv) 210 or 102
Sol.
(i) 43 = 4 × 4 × 4 = 64 ,
34= 3x3x3x3 = 81 34 > 43. Ans.
(ii) 53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
.•. 35 > 53. Ans.
(iii) 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 256
82 = 8 × 8 = 64
82 < 28. Ans.
(iv) 1002 or 2100 i.e., 1002
= 100×100 = 10000
2100 = 2 × 2 × 2 × 2 x 2 × 2
= 2100
2100 > 1002. Ans.
(v) 210 or 102 i.e.,210
= 2 × 2 × ………………… x 2 = 210
102 = 10 × 10 = 100
.-. 210 > 102. Ans.
Question 5.
Express each of the following as product of powers of their prime factors :
(i) 648 (ii) 405 (Hi) 540 (iv) 3600
Sol.
(i) 648 = 2 × 324 = 2 × r. × 162
= 2 × 2 × 2 × 81
= 2 × 2 × 2 × 9 × 9
= 2 × 2 × 2 × 3 × 3 × 3 × 3
= 23 × 34. Ans.
405 = 3 × 135
= 3 × 3 × 45
= 3 × 3 × 3 × 3 × 15
= 3 × 3 × 3 × 3 × 5
= 34 × 51. Ans.
540 = 2 × 270
= 2 × 2 × 135
= 2 x 2 x 3 x 45
= 2 x 2 x 3 x 3 x 15
= 2 × 2 × 3 × 3 × 3 x 5
= 22 × 33 × 51. Ans.
3600 = 2 × 1800
= 2 × 2 × 900
= 2 × 2 × 2 × 450
= 2 × 2 × 2 × 2 × 225
= 2 × 2 × 2 × 2 × 3 × 75
= 2 × 2 × 2 × 2 × 3 × 3 × 25
= 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 24 × 32 × 54. Ans.
Question 6.
Simplify :
(i) 2 × 103
(ii) 72 × 22
(iii) 23 × 5
(iv) 3 × 44
(v) 0 × 102
(vi) 52 × 33
(vii) 24 × 32
(viii) 32 × 104
Solution:
(i) 2 × 103 = 2 × 10 × 10 × 10
= 2 × 1000 = 2000 Ans.
(ii) 72 × 22 = 7 × 7 × 2 × 2
= 49 × 4 = 196. Ans.
(iii) 23 × 5 = 2 × 2 × 2 × 5
= 8 × 5 = 40. Ans.
(iv) 3 × 44 = 3 × 4 × 4 × 4 × 4
= 768. Ans.
(v) 0 × 102 = 0 × 10 × 10 = 0. Ans.
(vi) 52 × 33 = 5 × 5 × 3 × 3 × 3
= 25 × 27 = 675 Ans.
(vii) 24 × 32 = 2 × 2 × 2 × 2 × 3 × 3
= 16 × 9 = 144. Ans.
(viii) 32 × 104 = 3 × 3 × 10 × 10 × 10 × 10
= 9 × 10000 = 90000. Ans.
Question 7.
Simplify:
(i) (-4)3
(ii) (-3) × (-2)3
(iii) (- 3)2 × (- 5)2
(iv) (- 2)3 × (- 10)4
Solution:
(i) (- 4)3 = – 4 × – 4 × – 4 = – 64. Ans.
(ii)(-3) x(-2)3 = – 3 × -2 × -2 × -2 = – 3 × – 8 = 24. Ans.
(iii) (- 3)2 × (- 5)2 = – 3 x – 3 x-5x – 5
= 9 × 25 = 225. Ans.
(iv) (- 2)3 × (- 10)4 = – 2 x – 2x – 2x – 10 × – 10 × – 10 × – 10
= -8 x 10000 = -80000 Ans.
Question 8.
Compare the following numbers:
(i) 2.7 ’ × 1012; 1.5 x10s
(ii) 4 x : 10u; 3 × 1017.
Sol. (i) 27
2.7 × 1012 = \(\frac{27}{10}\) × 10 × 1011
= 27 × 1011
= 3 × 3 × 3 × 1011
= 33 × 1011
1.5 × 108 = \(\frac{15}{10}\) × 10 × 107
= 15 × 107
= 3 × 5 × 107
33 × 1011 > 3 × 5 × 107
or 2.7 × 1012 > 1.5 × 108
(ii) 4 x 1014; 3 × 1017
4 × 1014 = 4 × 1014
and 3 × 1017 = 3 × 1017
3 × 1017 > 4 × 1014. Ans.