CBSE Class 8 Mathematics Power Play Short Answer Questions with Answers

Short Answer Questions

Question 1. Express 1500 as a product of powers of its prime factors, writing each prime factor in exponential form. Show the repeated-division steps you used.
Answer:
Divide by the smallest possible prime each time:
1500 ÷ 2 = 750, 750 ÷ 2 = 375, 375 ÷ 3 = 125, 125 ÷ 5 = 25, 25 ÷ 5 = 5, 5 ÷ 5 = 1.
So 1500 = 2 × 2 × 3 × 5 × 5 × 5.
In exponential form, 1500 = 2² × 3 × 5³.
Check: 4 × 3 × 125 = 1500.
In simple words: Keep dividing 1500 by primes, then group equal primes as powers: 2² × 3 × 5³.

Teacher's Note:
1. Stop only when the quotient is 1.
2. A prime used once is written without an exponent.

Question 2. For the power 7⁵:
(i) Name the base and the exponent, and read the power aloud in words.
(ii) Write it as a repeated product and find its value.
Answer:
(i) The base is 7 and the exponent (power) is 5. It is read as “7 raised to the power 5” or “the 5th power of 7”.
(ii) 7⁵ = 7 × 7 × 7 × 7 × 7 = 16807.
In simple words: 7⁵ means five 7s multiplied together, which gives 16807.

Teacher's Note:
1. Multiply step by step: 49, 343, 2401, 16807.
2. Reading powers aloud correctly helps in oral tests.

Question 3. Find the value of each of the following and state in one line how the sign of the answer is decided:
(i) (−1)⁷
(ii) (−1)²⁰
(iii) (−3)⁴
Answer:
(i) (−1)⁷ = −1
(ii) (−1)²⁰ = 1
(iii) (−3)⁴ = 81
The sign depends only on the exponent: an odd number of negative factors gives a negative answer, while an even number gives a positive answer.
In simple words: Minus signs cancel in pairs, so even exponents give positive answers and odd exponents give negative ones.

Teacher's Note:
1. The size comes from the number; the sign comes from the exponent.
2. (−1) to any even power is 1.

Question 4. Answer the following about negative exponents:
(i) Write 2⁻⁵ as a fraction in its simplest form.
(ii) Write \( \frac{1}{3^{4}} \) as a power of 3 with a negative exponent.
Answer:
(i) 2⁻⁵ = \( \frac{1}{2^{5}} \) = \( \frac{1}{32} \)
(ii) \( \frac{1}{3^{4}} \) = 3⁻⁴
Both use the rule \( n^{-a} = \frac{1}{n^{a}} \), where n ≠ 0.
In simple words: A negative exponent simply means 'one over' the same power.

Teacher's Note:
1. The rule works in both directions.
2. The value stays positive.

Question 5. Find the value of 15⁰ + 7⁰. Then explain, using the division law of exponents, why n⁰ = 1 and why n cannot be 0.
Answer:
15⁰ + 7⁰ = 1 + 1 = 2.
By the division law, \( n^{a} \div n^{a} = n^{a-a} = n^{0} \).
But any non-zero number divided by itself is 1, so n⁰ = 1.
If n = 0, then n^a ÷ n^a becomes \( \frac{0}{0} \), which is not defined. That is why n ≠ 0.
In simple words: Dividing a power by itself gives both n⁰ and 1, so they are equal; with 0 we would divide by zero.

Teacher's Note:
1. The answer is 2, not 0.
2. Always mention the condition n ≠ 0.

Question 6. Write 3085 in expanded form using powers of 10. Which power of 10 has the digit 0 as its coefficient?
Answer:
3085 = (3 × 10³) + (0 × 10²) + (8 × 10¹) + (5 × 10⁰).
Here 10⁰ = 1, so the last term is 5.
The power 10² (the hundreds place) has the digit 0 as its coefficient.
In simple words: Each digit is multiplied by the power of 10 for its place, and the hundreds place holds 0.

Teacher's Note:
1. The units place is 10⁰.
2. Zero digits still keep their place.

Question 7. Express each number in standard form (scientific notation) x × 10^y, where 1 ≤ x < 10:
(i) 59,40,000
(ii) 0.00032
Answer:
(i) 59,40,000 = 5.94 × 10⁶ (the decimal point moves 6 places to the left).
(ii) 0.00032 = 3.2 × 10⁻⁴ (the decimal point moves 4 places to the right, so the exponent is negative).
In each case the coefficient lies between 1 and 10.
In simple words: Place the decimal point after the first non-zero digit and count how far it moved.

Teacher's Note:
1. Left moves give positive exponents; right moves give negative exponents.
2. Drop the trailing zeros in the coefficient.

Question 8. A tissue sheet is 0.008 cm thick. It is folded over and over, so its thickness doubles with every fold. Write an expression for its thickness after 6 folds using exponential notation, and find that thickness.
Answer:
Each fold multiplies the thickness by 2, so after 6 folds it is multiplied by 2 six times.
Thickness after 6 folds = 0.008 × 2⁶ cm.
2⁶ = 64, so thickness = 0.008 × 64 = 0.512 cm.
Just six folds make the tissue 64 times thicker.
In simple words: Doubling six times multiplies 0.008 cm by 64, giving 0.512 cm.

Teacher's Note:
1. The starting thickness is multiplied by a power of 2.
2. 8 × 64 = 512, then place the decimal point.

Question 9. Simplify, writing each answer in exponential form and then as a number:
(i) 6⁹ × 6⁻⁴ ÷ 6³
(ii) (4³)² ÷ 4⁵
Answer:
(i) 6⁹ × 6⁻⁴ ÷ 6³ = 6^{9 + (−4) − 3} = 6² = 36.
(ii) (4³)² = 4⁶, so (4³)² ÷ 4⁵ = 4^{6 − 5} = 4¹ = 4.
Laws used: \( n^{a} \times n^{b} = n^{a+b} \), \( n^{a} \div n^{b} = n^{a-b} \) and \( (n^{a})^{b} = n^{ab} \).
In simple words: Combine all the exponents of the same base first, then find the small final value.

Teacher's Note:
1. Work out the exponent before computing any number.
2. Name the law used at each step.

Question 10. Write 5¹² as a power of a power in two different ways, and also as a product of two powers of 5 in one way. State the law of exponents you used each time.
Answer:
5¹² = 244140625.
As a power of a power: 5¹² = (5²)⁶ and 5¹² = (5³)⁴ (also (5⁴)³ or (5⁶)²), using \( (n^{a})^{b} = n^{ab} \), since 2 × 6 = 3 × 4 = 12.
As a product of two powers: 5¹² = 5⁵ × 5⁷, using \( n^{a} \times n^{b} = n^{a+b} \), since 5 + 7 = 12.
In simple words: Split 12 as a product for a power of a power, and as a sum for a product of powers.

Teacher's Note:
1. Product of exponents: power of a power. Sum of exponents: product of powers.
2. Many correct answers are possible.

Question 11. Using the rule \( m^{a} \times n^{a} = (mn)^{a} \):
(i) Show that 3⁴ × 5⁴ = 15⁴ by finding both values.
(ii) Hence find the value of 2⁵ × 5⁵ in one step.
Answer:
(i) 3⁴ = 81 and 5⁴ = 625, so 3⁴ × 5⁴ = 81 × 625 = 50625. Also 15⁴ = 50625. Hence 3⁴ × 5⁴ = 15⁴, because the factors can be regrouped as (3 × 5) taken four times.
(ii) 2⁵ × 5⁵ = (2 × 5)⁵ = 10⁵ = 1,00,000.
In simple words: When the exponents match, multiply the bases first and then take the power.

Teacher's Note:
1. Pairing 2 with 5 to make 10 is a useful trick.
2. 15² = 225, so 15⁴ = 225² = 50625.

Question 12. A cycle lock uses a code of 4 slots, and every slot can carry any one of the 26 letters A to Z.
(i) How many different codes are possible?
(ii) If a thief could try one code every second, roughly how many days would be needed to try them all?
Answer:
(i) Each of the 4 slots has 26 choices, so the number of codes = 26 × 26 × 26 × 26 = 26⁴ = 4,56,976.
(ii) At one code per second this takes 4,56,976 seconds. One day has 86400 seconds, and 456976 ÷ 86400 ≈ 5.3, so about 5 days are needed.
In simple words: There are 26⁴ = 4,56,976 codes, which takes about 5 days at one per second.

Teacher's Note:
1. Seconds in a day = 24 × 60 × 60 = 86400.
2. Round sensibly for 'roughly'.

Question 13. Arrange 2², 2⁻³, 2⁰ and 2⁻¹ in ascending order by first writing the value of each. Then find 2² ÷ 2⁻³ in exponential form and as a number.
Answer:
Values: 2⁻³ = \( \frac{1}{8} \), 2⁻¹ = \( \frac{1}{2} \), 2⁰ = 1, 2² = 4.
Ascending order: 2⁻³ < 2⁻¹ < 2⁰ < 2².
2² ÷ 2⁻³ = 2^{2 − (−3)} = 2⁵ = 32.
Check: 4 ÷ \( \frac{1}{8} \) = 4 × 8 = 32.
In simple words: For base 2, a smaller exponent means a smaller value, and dividing by 2⁻³ is multiplying by 8.

Teacher's Note:
1. For a base greater than 1, order powers by their exponents.
2. Subtracting −3 means adding 3.

Question 14. Work out the following, giving each answer in standard form:
(i) (3 × 10⁸) × (2 × 10⁵)
(ii) (8 × 10⁹) ÷ (4 × 10⁴)
Answer:
(i) (3 × 10⁸) × (2 × 10⁵) = (3 × 2) × 10⁸⁺⁵ = 6 × 10¹³.
(ii) (8 × 10⁹) ÷ (4 × 10⁴) = (8 ÷ 4) × 10⁹⁻⁴ = 2 × 10⁵.
The coefficients are multiplied or divided, and the exponents are added or subtracted.
In simple words: Handle the coefficients and the powers of 10 separately.

Teacher's Note:
1. Check the coefficient is still between 1 and 10.
2. Here no adjustment is needed.

Question 15. Write 40.506 in expanded form using powers of 10, including the negative powers for the digits after the decimal point.
Answer:
40.506 = (4 × 10¹) + (0 × 10⁰) + (5 × 10⁻¹) + (0 × 10⁻²) + (6 × 10⁻³).
The places after the decimal point use 10⁻¹ = \( \frac{1}{10} \), 10⁻² = \( \frac{1}{100} \) and 10⁻³ = \( \frac{1}{1000} \).
Check: 40 + 0.5 + 0.006 = 40.506.
In simple words: Tenths, hundredths and thousandths are the negative powers of 10.

Teacher's Note:
1. The exponents fall by 1 at each place to the right.
2. Keep the zero terms to show every place.

Question 16. Write one lakh, one crore, one arab and one kharab as powers of 10. Hence find how many lakhs make one kharab, giving your answer as a power of 10.
Answer:
One lakh = 10⁵, one crore = 10⁷, one arab = 10⁹, one kharab = 10¹¹.
Each named step is 100 = 10² times the one before it.
Number of lakhs in a kharab = 10¹¹ ÷ 10⁵ = 10¹¹⁻⁵ = 10⁶.
So one kharab is ten lakh lakhs.
In simple words: Divide the two powers of 10 by subtracting exponents: 10¹¹ ÷ 10⁵ = 10⁶.

Teacher's Note:
1. The Indian names go up in steps of 10².
2. 'How many lakhs make' means divide by a lakh.

Question 17. Find the value of 3¹⁸ ÷ 9⁷ as a power of 3, and hence write the units digit of this value. (Hint: first write 9 as a power of 3.)
Answer:
9 = 3², so 9⁷ = (3²)⁷ = 3¹⁴.
3¹⁸ ÷ 9⁷ = 3¹⁸ ÷ 3¹⁴ = 3¹⁸⁻¹⁴ = 3⁴.
3⁴ = 81, so the units digit is 1.
In simple words: Turn 9⁷ into 3¹⁴, subtract the exponents and get 81.

Teacher's Note:
1. Division law needs the same base.
2. Convert the larger base into the smaller one.

Question 18. Given that 15² = 225, write down the values of (1.5)², (0.15)² and (150)² without multiplying again. Explain the pattern you used in terms of powers of 10.
Answer:
Write each number as 15 times a power of 10 and square it.
1.5 = 15 × 10⁻¹, so (1.5)² = 225 × 10⁻² = 2.25.
0.15 = 15 × 10⁻², so (0.15)² = 225 × 10⁻⁴ = 0.0225.
150 = 15 × 10¹, so (150)² = 225 × 10² = 22500.
The digits stay 225; only the power of 10 changes, and squaring doubles its exponent.
In simple words: Squaring 15 × 10^k gives 225 × 10^2k, so only the decimal point moves.

Teacher's Note:
1. (ab)² = a²b² is the rule behind this.
2. Count decimal places to check: 0.15 has 2, so its square has 4.

Question 19. Decide whether each statement is Always True, Only Sometimes True or Never True, and justify your answer using algebra:
(i) The product of two square numbers is a square number.
(ii) A fourth power is also a cube number.
Answer:
(i) Always True. If the two squares are a² and b², then a² × b² = (a × b)², which is again a square. For example, 9 × 16 = 144 = 12².
(ii) Only Sometimes True. n⁴ is a cube only when n is itself a cube: if n = k³, then n⁴ = k¹² = (k⁴)³. For example, 8⁴ = 4096 = 16³ is a cube, but 2⁴ = 16 is not a cube.
In simple words: Squares multiply to squares every time, but a fourth power is a cube only for some numbers.

Teacher's Note:
1. One counterexample is enough to show 'not always'.
2. One example where it works shows 'not never'.

Question 20. A Class 8 student claims, “I am about 10⁹ seconds old.” Using the chapter’s idea of estimating with powers of 10, check the claim: find roughly how many seconds a 13-year-old has lived, write it in standard form, and state what age 10⁹ seconds actually corresponds to.
Answer:
The claim is not correct.
Seconds in 13 years ≈ 13 × 365 × 24 × 3600 = 40,99,68,000 ≈ 4.1 × 10⁸ seconds.
So a 13-year-old is of the order of 10⁸ seconds old, not 10⁹.
10⁹ seconds ≈ 31.7 years, the age of an adult, about 2.4 times the student’s age.
One extra power of 10 changes a quantity ten times over.
In simple words: A 13-year-old has lived about 4 × 10⁸ seconds; 10⁹ seconds is nearly 32 years.

Teacher's Note:
1. Seconds in a year ≈ 3.15 × 10⁷.
2. Compare exponents to judge such claims quickly.