CBSE Class 8 Mathematics Power Play Practice Paper Worksheet Set 01 with Answers

Practice Paper 09 for Class 8 Maths Ganita Prakash Chapter 2 Power Play has 20 questions worth 40 marks, arranged in Sections A to E, and each one is solved step by step below. The questions practise negative exponents of fractions, finding reciprocals, equations like 2ˣ × 2ˣ⁺¹ = 32, adding and converting numbers in standard form, and real-life doubling such as dividing cells and file storage, with case studies on a king's guard formation and an architect's block design.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Simplify: \( \left(\frac{2}{3}\right)^{2} \) × \( \left(\frac{3}{2}\right)^{-3} \)
Answer:
Flip the second fraction to make the exponent positive: \( \left(\frac{3}{2}\right)^{-3} \) = \( \left(\frac{2}{3}\right)^{3} \).
Now the bases are equal: \( \left(\frac{2}{3}\right)^{2} \) × \( \left(\frac{2}{3}\right)^{3} \) = \( \left(\frac{2}{3}\right)^{2+3} \) = \( \left(\frac{2}{3}\right)^{5} \).
\( \left(\frac{2}{3}\right)^{5} \) = \( \frac{2^{5}}{3^{5}} \) = \( \frac{32}{243} \).
In simple words: Turn (3/2)⁻³ into (2/3)³ and add exponents to get (2/3)⁵ = \( \frac{32}{243} \).

Teacher's Note:
1. Make the bases the same before using the product rule.
2. Raise both numerator and denominator to the power.

Question 12. If 5ˣ = 25 and 3ʸ = 9², find the value of 2x − y.
Answer:
25 = 5², so 5ˣ = 5² and x = 2.
9² = (3²)² = 3⁴, so 3ʸ = 3⁴ and y = 4.
2x − y = 2 × 2 − 4 = 4 − 4 = 0.
In simple words: x = 2 and y = 4, so 2x − y = 0.

Teacher's Note:
1. Write 9 as 3² before comparing exponents.
2. Substitute carefully: 2x means 2 × x.

Question 13. Find the reciprocal of \( \left(\frac{2}{3}\right)^{-2} \) + \( \left(\frac{1}{2}\right)^{-3} \).
Answer:
\( \left(\frac{2}{3}\right)^{-2} \) = \( \left(\frac{3}{2}\right)^{2} \) = \( \frac{9}{4} \).
\( \left(\frac{1}{2}\right)^{-3} \) = 2³ = 8.
Sum = \( \frac{9}{4} \) + 8 = \( \frac{9}{4} \) + \( \frac{32}{4} \) = \( \frac{41}{4} \).
The reciprocal of \( \frac{41}{4} \) is \( \frac{4}{41} \).
In simple words: The sum is 9/4 + 8 = 41/4, so its reciprocal is \( \frac{4}{41} \).

Teacher's Note:
1. Find the sum first, then flip it.
2. The reciprocal of a sum is not the sum of reciprocals.

Question 14. A particle has a mass of 5.2 × 10⁻¹² kg. Another particle has a mass of 3.5 × 10⁻¹² kg. What is the total mass of the two particles?
Answer:
Total mass = 5.2 × 10⁻¹² + 3.5 × 10⁻¹².
Both terms have the same power of 10, so take it out as a common factor: (5.2 + 3.5) × 10⁻¹².
Total mass = 8.7 × 10⁻¹² kg.
In simple words: Same power of 10, so just add 5.2 and 3.5: 8.7 × 10⁻¹² kg.

Teacher's Note:
1. Add numbers in standard form only when the powers of 10 match.
2. 8.7 is between 1 and 10, so the answer is already in standard form.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Simplify: \( \frac{(2^{3})^{2} \times 3^{-2} \times 4^{0}}{12^{-1}} \)
Answer:
(2³)² = 2⁶ = 64.
3⁻² = \( \frac{1}{9} \) and 4⁰ = 1.
Dividing by 12⁻¹ is the same as multiplying by 12.
So the value = 64 × \( \frac{1}{9} \) × 1 × 12 = \( \frac{768}{9} \) = \( \frac{256}{3} \).
Using powers: 2⁶ × 12 = 2⁶ × 2² × 3 = 2⁸ × 3, and 2⁸ × 3 ÷ 3² = \( \frac{2^{8}}{3} \) = \( \frac{256}{3} \) (= 85⅓).
In simple words: 64 × 1/9 × 1 × 12 = \( \frac{256}{3} \).

Teacher's Note:
1. Dividing by a negative power is the same as multiplying by the positive power.
2. Remember that 4⁰ = 1.

Question 16. Find the value of x for which \( \left(\frac{1}{5}\right)^{x-1} \) = 125
Answer:
Write both sides with base 5: \( \frac{1}{5} \) = 5⁻¹ and 125 = 5³.
So \( (5^{-1})^{x-1} = 5^{3} \), that is \( 5^{-(x-1)} = 5^{3} \).
Equate exponents: −(x − 1) = 3, so −x + 1 = 3.
−x = 2, so x = −2.
Check: \( \left(\frac{1}{5}\right)^{-3} \) = 5³ = 125. ✓
In simple words: Write 1/5 as 5⁻¹; then 1 − x = 3, so x = −2.

Teacher's Note:
1. Turn 1/5 into 5⁻¹ to get a common base.
2. Watch the minus sign when opening the bracket.

Question 17. By what number should \( \left(\frac{1}{3}\right)^{-2} \) be divided to obtain the result 27?
Answer:
Let the number be x. Then \( \left(\frac{1}{3}\right)^{-2} \) ÷ x = 27.
\( \left(\frac{1}{3}\right)^{-2} \) = 3² = 9.
So 9 ÷ x = 27, which gives x = 9 ÷ 27.
x = \( \frac{9}{27} \) = \( \frac{1}{3} \) (= 3⁻¹).
Check: 9 ÷ \( \frac{1}{3} \) = 9 × 3 = 27. ✓
In simple words: (1/3)⁻² = 9, and 9 ÷ \( \frac{1}{3} \) = 27, so the number is \( \frac{1}{3} \).

Teacher's Note:
1. Dividing by a fraction less than 1 gives a bigger answer.
2. Set up the equation in the order the question states.

Section D: Long Answer Question (5 Marks)

Question 18. Find the value of x such that \( \left(\frac{3}{4}\right)^{-2x} \) ÷ \( \left(\frac{4}{3}\right)^{x} \) = \( \left(\frac{9}{16}\right)^{3} \).
Answer:
Make all bases \( \frac{4}{3} \).
\( \left(\frac{3}{4}\right)^{-2x} \) = \( \left(\frac{4}{3}\right)^{2x} \).
\( \left(\frac{9}{16}\right)^{3} \) = \( \left(\left(\frac{3}{4}\right)^{2}\right)^{3} \) = \( \left(\frac{3}{4}\right)^{6} \) = \( \left(\frac{4}{3}\right)^{-6} \).
Left side: \( \left(\frac{4}{3}\right)^{2x} \) ÷ \( \left(\frac{4}{3}\right)^{x} \) = \( \left(\frac{4}{3}\right)^{2x-x} \) = \( \left(\frac{4}{3}\right)^{x} \).
So \( \left(\frac{4}{3}\right)^{x} \) = \( \left(\frac{4}{3}\right)^{-6} \), which gives x = −6.
Check: \( \left(\frac{3}{4}\right)^{12} \) ÷ \( \left(\frac{4}{3}\right)^{-6} \) = \( \left(\frac{3}{4}\right)^{12} \) ÷ \( \left(\frac{3}{4}\right)^{6} \) = \( \left(\frac{3}{4}\right)^{6} \) = \( \left(\frac{9}{16}\right)^{3} \). ✓
In simple words: Bring every term to base 4/3; then x = −6.

Teacher's Note:
1. 9/16 is (3/4)², so the right side is (3/4)⁶.
2. Flipping a fraction changes the sign of its exponent.
3. Always substitute back to check.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1: The King's Guard. In a kingdom, a king's guard is deployed in a special formation. The number of guards in each row is determined by a power of 2. In the first row, there are 2 guards. In the second row, there are 4, in the third, there are 8, and so on. The king needs to know the total number of guards for different formations.
(i) How many guards will be in the 6th row?
(ii) If a new formation requires 2¹⁰ guards in a single row, which row number is this?
(iii) The number of guards in a different formation doubles every 2 days, starting with 4 guards. How many guards will there be after 6 days?
(iv) What is the ratio of the number of guards in the 8th row to the number of guards in the 4th row?
Answer:
Row 1 has 2¹, row 2 has 2², row 3 has 2³ guards, so row n has 2ⁿ guards.
(i) 6th row: 2⁶ = 64 guards.
(ii) 2ⁿ = 2¹⁰ gives n = 10, so it is the 10th row.
(iii) In 6 days there are 6 ÷ 2 = 3 doublings. Starting with 4 = 2² guards: 2² × 2³ = 2⁵ = 32 guards.
(iv) Ratio = 2⁸ : 2⁴ = 2⁸⁻⁴ = 2⁴ = 16, that is 16 : 1.
In simple words: Row n has 2ⁿ guards, so row 6 has 64, 2¹⁰ is row 10, and row 8 has 16 times as many as row 4.

Teacher's Note:
1. Spot the pattern first: the exponent equals the row number.
2. Count doubling periods, not days, in part (iii).

Question 20. Case Study 2: The Architect's Blueprint. An architect is designing a large-scale housing project. The design is based on a pattern of blocks. The number of blocks in each layer increases exponentially. The first layer has 5 blocks. Each subsequent layer has 5 times the number of blocks as the previous layer. The final (top) layer of the building will have 5¹⁰ blocks.
(i) Write an expression for the number of blocks in the 4th layer.
(ii) The total number of blocks in a building is calculated as the sum of blocks in all layers. The architect wants to build a 3-layer building. What is the total number of blocks?
(iii) If another building has a design where the number of blocks triples each layer, starting with 3 blocks, how many blocks will be in the 4th layer?
(iv) The architect decides to use smaller blocks, with a volume of 5⁻² m³. How many such smaller blocks are required to construct a structure with a total volume of 5³ m³?
Answer:
Layer 1 has 5¹ blocks, layer 2 has 5², layer 3 has 5³, so layer n has 5ⁿ blocks.
(i) 4th layer: 5⁴ blocks (= 625).
(ii) Total = 5¹ + 5² + 5³ = 5 + 25 + 125 = 155 blocks.
(iii) Layer n has 3ⁿ blocks, so the 4th layer has 3⁴ = 81 blocks.
(iv) Number of blocks = total volume ÷ volume of one block = 5³ ÷ 5⁻² = 5³⁻⁽⁻²⁾ = 5⁵ = 3125 blocks.
In simple words: Layer n has 5ⁿ blocks; three layers make 155 blocks, and 5³ ÷ 5⁻² = 5⁵ = 3125 small blocks.

Teacher's Note:
1. Powers are added only after working out each value (part ii).
2. Dividing by a negative power adds to the exponent.