HBSE 7th Class Maths Solutions Chapter 13 Exponents and Powers Ex 13.1

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.