HBSE 8th Class Maths Solutions Chapter 12 Exponents and Powers InText Questions

Haryana State Board HBSE 8th Class Maths Solutions Chapter 12 Exponents and Powers InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 12 Exponents and Powers InText Questions

Try These (Page 194)

Question 1.
Find the multiplicative inverse of the following :
(i) 2-4
(ii) 10-5
(iii) 7-2
(iv) 5-3
(v) 10-100
Solution:
(i) ∵ a-m = \(\frac{1}{a^{m}}\)
where m is a positive integer and a-m is the multiplicative inverse of am.
∴ Multiplicative inverse of 2-4 = \(\frac{1}{2^{4}}\)

(ii) Similarly, multiplicative inverse of 10-5 = \(\frac{1}{10^{5}}\)
(iii) Multiplicative inverse of 7-2 = \(\frac{1}{7^{2}}\)
(iv) Multiplicative inverse of 5-3 = \(\frac{1}{5^{3}}\)
(v) Multiplicative inverse of 10-100 = \(\frac{1}{10^{100}}\)

Question 2.
Expand the following numbers using exponents:
(i) 1025.63
(ii) 1256.249.
Solution:
(i) We learnt how to write numbers like 1025 in expanded form using exponents as 1 × 103 + 0 × 102 + 2 × 101 + 5 × 100.
Similarly we have
1025.63 = 1 × 1000 + 0 × 100 + 2 × 10 + 5 × 1 + \(\frac{6}{10}\) + \(\frac{3}{100}\)
= 1 × 103 + 0 × 102 + 2 × 101 + 5 + 6 × 10-1 + 3 × 10-2.

(ii) 1256.249 = 1 × 1000 + 2 × 100 + 5 × 10 + 6 × 1 + \(\frac{2}{10}\) + \(\frac{4}{100}\) + \(\frac{9}{1000}\)
= 1 × 103 + 2 × 102 + 5 × 101 + 6 × 100 + 2 × 10-1 + 4 × 10-2 + 9 × 10-3.

HBSE 8th Class Maths Solutions Chapter 12 Exponents and Powers InText Questions

Try These (Page 195)

Question 1.
Simplify and write in exponential form :
(i) (-2)-3 × (-2)-4
(ii) p3 × p-10
(iii) 32 × 3-5 × 36.
Solution:
Since, am × an = am+n
(i) (-2)-3 × (-2)-4 = (-2)(-3)+(4)
= (-2)-7

(ii) p3 × p-10 = (p)3+(-10)
= p-7

(iii) 32 × 3-5 × 36 = (3)2+(-5)+6
= (3)8-5 = 33.

Try These (Page 199)

Question 1.
Write the following numbers in standard form :
(ii) 0.000000564
(ii) 0.0000021
(iii) 21600000
(iv) 15240000.
Solution:
(i) 0.000000564
= \(\frac{564}{1000000000}\) = \(\frac{5.64 \times 100}{10^{9}}\)
= 5.64 × 102 × 10-9
= 5.64 × 10-7.

(ii) 0.0000021 = \(\frac{21}{10000000}\) = \(\frac{2.1 \times 10}{10^{7}}\)
= 2.1 × 101 × 10-7
= 2.1 × 10-6

(iii) 21600000 = 2.16 × 107
(iv) 15240000 = 1.524 × 107.

HBSE 8th Class Maths Solutions Chapter 12 Exponents and Powers InText Questions

Question 2.
Write all the facts given in the standard form.
Solution:
(i) Decimal is moved 7 places to the right.
(ii) Decimal is moved 6 places to the right.
(iii) Decimal is moved 7 places to the left.
(iv) Decimal is moved 7 places to the left.
We conclude the above number can be written in the form K × 10n where n is some integer and K is a terminating decimal lying between 1 and 10, i.e.,
1 ≤ K< 10.
That is the fact of standard form.

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