CBSE Class 8 Mathematics Power Play Practice Paper Assignment Set 02 with Answers

Here is Class 8 Maths Ganita Prakash Chapter 2 Power Play Practice Paper 10, a 40-mark paper of 20 questions across Sections A to E with a step-by-step answer to every question. It focuses on converting standard form to ordinary numbers, powers of decimals and fractions, simplifying expressions such as 12⁴ × 9³ × 4 by prime factorisation, and solving exponent equations with fractional bases, while the case studies on a robot factory and treasure tunnels apply powers of 3, 4 and 10 to counting problems.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Simplify and express the result as a power of 2: \( \frac{2^{5} \times 4^{2}}{8^{2}} \)
Answer:
Write 4 and 8 as powers of 2: 4 = 2², 8 = 2³.
4² = (2²)² = 2⁴ and 8² = (2³)² = 2⁶.
So the expression = \( \frac{2^{5} \times 2^{4}}{2^{6}} \) = \( \frac{2^{9}}{2^{6}} \) = 2⁹⁻⁶ = 2³.
In simple words: Change everything to base 2: 2⁵ × 2⁴ ÷ 2⁶ = 2³.

Teacher's Note:
1. A power of a power: multiply the exponents.
2. Check: 32 × 16 ÷ 64 = 8 = 2³.

Question 12. Find the value of x for which \( 5^{x+1} \times 5^{3} = 5^{8} \).
Answer:
Same base, so add exponents: \( 5^{(x+1)+3} = 5^{x+4} \).
So \( 5^{x+4} = 5^{8} \).
Equate the exponents: x + 4 = 8.
x = 4.
In simple words: x + 1 + 3 = 8, so x = 4.

Teacher's Note:
1. Equal bases mean equal exponents.
2. Check: 5⁵ × 5³ = 5⁸. ✓

Question 13. Simplify: 3⁻² + 4⁻²
Answer:
3⁻² = \( \frac{1}{3^{2}} \) = \( \frac{1}{9} \) and 4⁻² = \( \frac{1}{4^{2}} \) = \( \frac{1}{16} \).
LCM of 9 and 16 is 144: \( \frac{1}{9} \) + \( \frac{1}{16} \) = \( \frac{16}{144} \) + \( \frac{9}{144} \) = \( \frac{25}{144} \).
In simple words: 1/9 + 1/16 = \( \frac{25}{144} \).

Teacher's Note:
1. Different bases cannot be combined; find each value first.
2. Note 25/144 = (5/12)², a neat check.

Question 14. The population of a city is 1.25 × 10⁷. If each person consumes 10³ litres of water per year, what is the total water consumption in a year? Express your answer in standard form.
Answer:
Total consumption = population × consumption per person.
= (1.25 × 10⁷) × 10³ = 1.25 × 10⁷⁺³.
= 1.25 × 10¹⁰ litres.
In simple words: Multiply and add the exponents: 1.25 × 10¹⁰ litres.

Teacher's Note:
1. Multiplying by 10³ only changes the power of 10.
2. 1.25 lies between 1 and 10, so the form is standard.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Simplify: \( \frac{12^{4} \times 9^{3} \times 4}{6^{3} \times 8^{2} \times 27} \)
Answer:
Prime factors: 12 = 2² × 3, 9 = 3², 4 = 2², 6 = 2 × 3, 8 = 2³, 27 = 3³.
Numerator: 12⁴ × 9³ × 4 = (2⁸ × 3⁴) × 3⁶ × 2² = 2¹⁰ × 3¹⁰.
Denominator: 6³ × 8² × 27 = (2³ × 3³) × 2⁶ × 3³ = 2⁹ × 3⁶.
Divide: 2¹⁰⁻⁹ × 3¹⁰⁻⁶ = 2¹ × 3⁴ = 2 × 81 = 162.
In simple words: Factorise into powers of 2 and 3 and cancel: the answer is 2 × 3⁴ = 162.

Teacher's Note:
1. Collect powers of 2 and of 3 separately.
2. Write the final answer as a number as well as in power form.

Question 16. Find the value of x for which \( \left(\frac{2}{5}\right)^{x-2} \) = \( \frac{125}{8} \)
Answer:
\( \frac{125}{8} \) = \( \frac{5^{3}}{2^{3}} \) = \( \left(\frac{5}{2}\right)^{3} \) = \( \left(\frac{2}{5}\right)^{-3} \).
So \( \left(\frac{2}{5}\right)^{x-2} \) = \( \left(\frac{2}{5}\right)^{-3} \).
Equate the exponents: x − 2 = −3.
x = −1.
Check: \( \left(\frac{2}{5}\right)^{-3} \) = \( \left(\frac{5}{2}\right)^{3} \) = \( \frac{125}{8} \). ✓
In simple words: 125/8 = (2/5)⁻³, so x − 2 = −3 and x = −1.

Teacher's Note:
1. Turn the right side into a power of the same fraction.
2. Flipping the fraction makes the exponent negative.

Question 17. By what number should \( \left(\frac{4}{5}\right)^{-3} \) be divided so that the quotient is 4⁻²?
Answer:
Let the number be x. Then \( \left(\frac{4}{5}\right)^{-3} \) ÷ x = 4⁻², so x = \( \left(\frac{4}{5}\right)^{-3} \) ÷ 4⁻².
\( \left(\frac{4}{5}\right)^{-3} \) = \( \left(\frac{5}{4}\right)^{3} \) = \( \frac{125}{64} \).
4⁻² = \( \frac{1}{16} \).
x = \( \frac{125}{64} \) ÷ \( \frac{1}{16} \) = \( \frac{125}{64} \) × 16 = \( \frac{125}{4} \) (= 31.25).
In simple words: The number is (5/4)³ × 4² = \( \frac{125}{4} \).

Teacher's Note:
1. Dividend ÷ divisor = quotient, so divisor = dividend ÷ quotient.
2. Dividing by 1/16 is the same as multiplying by 16.

Section D: Long Answer Question (5 Marks)

Question 18. Find the value of x such that \( \frac{3^{2x} \times 9^{-2x}}{3^{-2}} = 27 \).
Answer:
Write every term with base 3: 9⁻²ˣ = (3²)⁻²ˣ = 3⁻⁴ˣ and 27 = 3³.
Numerator: \( 3^{2x} \times 3^{-4x} = 3^{2x - 4x} = 3^{-2x} \).
Dividing by 3⁻²: \( 3^{-2x} \div 3^{-2} = 3^{-2x + 2} \).
So \( 3^{2 - 2x} = 3^{3} \), which gives 2 − 2x = 3.
−2x = 1, so x = −1/2.
Check: with x = \( -\frac{1}{2} \), 3²ˣ = 3⁻¹ and 9⁻²ˣ = 9¹ = 9; (3⁻¹ × 9) ÷ 3⁻² = 3 × 9 = 27. ✓
In simple words: Changing all terms to base 3 gives 3²⁻²ˣ = 3³, so x = −1/2.

Teacher's Note:
1. Use (aᵐ)ⁿ = aᵐⁿ to change 9 into base 3.
2. Dividing by 3⁻² adds 2 to the exponent.
3. A fractional value of x is allowed; always check by substituting.

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

Question 19. Case Study 1: The Robot Army. A robotics engineer is building an army of robots. The number of robots built each day is a power of 3. On the first day, 3 robots are built. On the second day, 9 robots are built, and on the third day, 27 are built. The factory has a capacity to produce 3⁹ robots in a single day.
(i) Write an expression for the number of robots built on the 5th day.
(ii) On which day will the number of robots built be 243?
(iii) If the production rate tripled every 2 days instead of every day, starting with 3 robots on Day 0, how many robots would be built on the 6th day?
(iv) What is the ratio of the number of robots built on the 7th day to the number of robots built on the 3rd day?
Answer:
Day 1 has 3¹, day 2 has 3², day 3 has 3³ robots, so day n has 3ⁿ robots.
(i) 5th day: 3⁵ robots (= 243).
(ii) 243 = 3⁵, so it is the 5th day.
(iii) From Day 0 to Day 6 there are 6 ÷ 2 = 3 tripling periods. Starting with 3 robots: 3 × 3³ = 3⁴ = 81 robots.
(iv) Ratio = 3⁷ : 3³ = 3⁷⁻³ = 3⁴ = 81, that is 81 : 1.
In simple words: Day n has 3ⁿ robots, so day 5 has 3⁵ = 243 and day 7 has 3⁴ = 81 times as many as day 3.

Teacher's Note:
1. Match the exponent with the day number.
2. In part (iii) count the tripling periods, not the days.

Question 20. Case Study 2: The Treasure Tunnels. Two treasure hunters, Ali and Zora, find a map with a riddle. The riddle leads them to a sequence of tunnels. The number of paths from the entrance of each tunnel is determined by a power of 4. From the first tunnel, there are 4 paths. From the second tunnel, there are 16 paths, and so on.
(i) How many paths are there from the 4th tunnel?
(ii) If a tunnel has 1024 paths, which tunnel number is it?
(iii) To find a hidden treasure, they must follow a tunnel that has 4³ paths from its entrance. Each of these paths leads to a treasure of 4² gold coins. What is the total number of gold coins they could find from all such paths?
(iv) The treasure is protected by a lock that requires a 6-digit code. Each digit can be any number from 0 to 9. How many possible combinations are there for the lock? Express your answer in exponential form.
Answer:
Tunnel 1 has 4¹ paths, tunnel 2 has 4² paths, so tunnel n has 4ⁿ paths.
(i) 4th tunnel: 4⁴ = 256 paths.
(ii) 1024 = 4⁵ (4 × 4 × 4 × 4 × 4), so it is the 5th tunnel.
(iii) Total coins = number of paths × coins per path = 4³ × 4² = 4³⁺² = 4⁵ = 1024 gold coins.
(iv) Each of the 6 digits has 10 choices, so the number of codes = 10 × 10 × 10 × 10 × 10 × 10 = 10⁶.
In simple words: Tunnel n has 4ⁿ paths; 4³ paths with 4² coins each give 4⁵ coins, and a 6-digit lock has 10⁶ codes.

Teacher's Note:
1. Multiplying powers of the same base: add the exponents.
2. For a code, multiply the choices for each position.