CBSE Class 8 Mathematics Exponents and Powers Practice Paper Assignment Set 13 with Answers (Old NCERT Syllabus)

This Class 8 Maths Exponents and Powers practice paper has 17 questions worth 40 marks in Sections A to E, with step-by-step answers throughout. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it is a good fit for state-board students and for revising this chapter. The questions test negative exponents, the laws of exponents, solving for an unknown exponent, multiplicative inverses, and writing very large and very small numbers in standard form, including a case study on planetary distances.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. Find the value of x for which 2ˣ ÷ 2⁻⁴ = 4⁵.
Answer:
Left side: 2ˣ ÷ 2⁻⁴ = 2ˣ⁻⁽⁻⁴⁾ = 2ˣ⁺⁴.
Right side: 4⁵ = (2²)⁵ = 2¹⁰.
So x + 4 = 10, giving x = 6.
In simple words: Write both sides as powers of 2, then match the exponents.

Teacher's Note:
1. aᵐ ÷ aⁿ = aᵐ⁻ⁿ; subtracting −4 means adding 4.
2. Convert 4⁵ to base 2 before comparing exponents.

Question 8. Find the value of x for which \( \left(\frac{5}{7}\right)^{-3} \times \left(\frac{5}{7}\right)^{-11} = \left(\frac{5}{7}\right)^{7x} \)
Answer:
Same base, so add the exponents: \( \left(\frac{5}{7}\right)^{-3 + (-11)} = \left(\frac{5}{7}\right)^{-14} \)
So \( \left(\frac{5}{7}\right)^{-14} = \left(\frac{5}{7}\right)^{7x} \), which gives 7x = −14.
x = −2
In simple words: Add the powers on the left to get −14, then 7x = −14 gives x = −2.

Teacher's Note:
1. aᵐ × aⁿ = aᵐ⁺ⁿ, even when m and n are negative.
2. Equal bases (other than 0, 1 and −1) mean equal exponents.

Question 9. Find the value of m for which 5ᵐ ÷ 5⁻³ = 5⁵.
Answer:
5ᵐ ÷ 5⁻³ = 5ᵐ⁻⁽⁻³⁾ = 5ᵐ⁺³.
So 5ᵐ⁺³ = 5⁵, and m + 3 = 5.
m = 2
In simple words: Dividing by 5⁻³ adds 3 to the power, so m + 3 = 5 and m = 2.

Teacher's Note:
1. Dividing by a negative power is the same as multiplying by the positive power.
2. Check: 5² ÷ 5⁻³ = 5² × 5³ = 5⁵.

Section C: Short Answer Questions (3 Marks Each)

Question 10. Simplify: (a) \( \left[\left(\frac{1}{2}\right)^{2} - \left(\frac{1}{4}\right)^{3}\right] \times 2^{3} \) (b) \( \left(3^2 - 2^2\right) \div \left(\frac{1}{5}\right)^{2} \)
Answer:
(a) \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \) and \( \left(\frac{1}{4}\right)^3 = \frac{1}{64} \)
\( \frac{1}{4} - \frac{1}{64} = \frac{16 - 1}{64} = \frac{15}{64} \)
\( \frac{15}{64} \times 8 = \frac{15}{8} \), so the value is \( \frac{15}{8} \).
(b) 3² − 2² = 9 − 4 = 5 and \( \left(\frac{1}{5}\right)^2 = \frac{1}{25} \)
\( 5 \div \frac{1}{25} = 5 \times 25 \) = 125.
In simple words: Work out each power first, then do the bracket, then multiply or divide.

Teacher's Note:
1. Follow the order: brackets, powers, then multiplication or division.
2. Dividing by 1/25 is the same as multiplying by 25.

Question 11. Find the value of a⁶ − (b + 1)³ when a = \( \frac{1}{2} \), b = \( \frac{-3}{4} \).
Answer:
a⁶ = \( \left(\frac{1}{2}\right)^6 = \frac{1}{64} \)
b + 1 = \( \frac{-3}{4} + 1 = \frac{1}{4} \), so (b + 1)³ = \( \left(\frac{1}{4}\right)^3 = \frac{1}{64} \)
a⁶ − (b + 1)³ = \( \frac{1}{64} - \frac{1}{64} \) = 0
In simple words: Both parts work out to 1/64, so their difference is 0.

Teacher's Note:
1. Find b + 1 first, then cube it.
2. Note (1/2)⁶ = ((1/2)²)³ = (1/4)³, which is why the two parts are equal.

Question 12. By what number should \( \left(\frac{5}{4}\right)^{-3} \) be divided so that the quotient may be \( \left(\frac{15}{16}\right)^{-2} \)?
Answer:
Let the number be y. Then \( \left(\frac{5}{4}\right)^{-3} \div y = \left(\frac{15}{16}\right)^{-2} \).
So y = \( \left(\frac{5}{4}\right)^{-3} \div \left(\frac{15}{16}\right)^{-2} = \left(\frac{4}{5}\right)^{3} \div \left(\frac{16}{15}\right)^{2} \)
= \( \frac{64}{125} \div \frac{256}{225} = \frac{64}{125} \times \frac{225}{256} \)
= \( \frac{14400}{32000} = \frac{9}{20} \)
The required number is \( \frac{9}{20} \).
In simple words: Divide the first number by the quotient you want: (4/5)³ ÷ (16/15)² = 9/20.

Teacher's Note:
1. If A ÷ y = Q, then y = A ÷ Q.
2. Turn negative exponents into positive ones by flipping the fractions first.

Question 13. In a stack, there are 5 books, each having a thickness of 20 mm, and 5 paper sheets, each having a thickness of 0.016 mm. What is the total thickness of the stack?
Answer:
Thickness of 5 books = 5 × 20 = 100 mm.
Thickness of 5 sheets = 5 × 0.016 = 0.08 mm.
Total = 100 + 0.08 = 100.08 mm.
In standard form: 1.0008 × 10² mm.
In simple words: The books make 100 mm and the sheets only 0.08 mm, so the stack is 100.08 mm thick.

Teacher's Note:
1. Add quantities only when they are in the same unit (both are in mm here).
2. Standard form: move the decimal point so one non-zero digit is before it.

Section D: Long Answer Questions (4 Marks Each)

Question 14. Simplify: (i) \( \frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \) (ii) \( \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}} \)
Answer:
(i) \( \frac{5^2 \times t^{-4}}{5^{-3} \times 2 \times 5 \times t^{-8}} = \frac{5^{2+3-1} \times t^{-4+8}}{2} = \frac{5^4 t^4}{2} \)
= \( \frac{625\,t^4}{2} \)
(ii) Write 10⁻⁵ = 2⁻⁵ × 5⁻⁵, 125 = 5³ and 6⁻⁵ = 2⁻⁵ × 3⁻⁵.
\( \frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^{3}}{5^{-7} \times 2^{-5} \times 3^{-5}} = 5^{-5+3+7} = 5^5 \)
= 3125
In simple words: Break every number into prime powers, then add or subtract exponents of the same base.

Teacher's Note:
1. Write composite bases like 10 and 6 as products of primes so like bases can be combined.
2. Dividing powers of the same base: subtract the exponents (a negative minus a negative may be positive).

Question 15. Simplify: (i) \( \left[\left(\frac{2}{5}\right)^{-3}\right]^{4} \) (ii) \( \left[\left(\frac{-6}{11}\right)^{-5}\right]^{-3} \) (iii) \( \left(-\frac{2}{3}\right)^{-4} \times \left(\frac{1}{8}\right)^{-4} \) (iv) \( \left(\frac{5}{7}\right)^{-1} \times \left(\frac{7}{3}\right)^{-1} \)
Answer:
(i) \( \left(\frac{2}{5}\right)^{-3 \times 4} = \left(\frac{2}{5}\right)^{-12} \) = \( \left(\frac{5}{2}\right)^{12} \)
(ii) \( \left(\frac{-6}{11}\right)^{(-5)(-3)} = \left(\frac{-6}{11}\right)^{15} \) = \( -\frac{6^{15}}{11^{15}} \) (odd power of a negative number is negative)
(iii) \( \left(-\frac{3}{2}\right)^{4} \times 8^{4} = \frac{81}{16} \times 4096 = 81 \times 256 \) = 20736
(iv) \( \frac{7}{5} \times \frac{3}{7} = \frac{3}{5} \), so the value is \( \frac{3}{5} \)
In simple words: Use (aᵐ)ⁿ = aᵐⁿ, flip fractions with negative powers, then multiply.

Teacher's Note:
1. Power of a power: multiply the exponents.
2. In (iii) you can also combine first: (−2/3 × 1/8)⁻⁴ = (−1/12)⁻⁴ = 12⁴ = 20736.
3. An even power of a negative number is positive; an odd power stays negative.

Question 16. Express the following numbers in standard form.
(i) 0.0000000000085 (ii) 0.00000000000942 (iii) 6020000000000000 (iv) 0.00000000837
Answer:
(i) 0.0000000000085: move the point 12 places right, so 8.5 × 10⁻¹²
(ii) 0.00000000000942: move the point 12 places right, so 9.42 × 10⁻¹²
(iii) 6020000000000000: move the point 15 places left, so 6.02 × 10¹⁵
(iv) 0.00000000837: move the point 9 places right, so 8.37 × 10⁻⁹
In simple words: Place the decimal point after the first non-zero digit and count how many places it moved.

Teacher's Note:
1. Moving the point to the right (small numbers) gives a negative power of 10.
2. Moving it to the left (large numbers) gives a positive power of 10.

Section E: Case Study Based Question (4 Marks)

Teacher's Note:
1. Mean of two values = their sum ÷ 2.
2. Drop trailing zeros after the decimal in standard form: 1.6610 × 10⁸ is written 1.661 × 10⁸.