Sample Paper 3 Class8 Mathematics power play (Exponents )
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CLASS VIII • MATHEMATICS PRACTICE TEST (2026–27)
Topic: Chapter 2 — Power Play (Exponents)
- All questions are compulsory.
- This question paper contains 20 questions divided into five Sections: A, B, C, D, and E.
- Section A contains 10 Multiple Choice Questions of 1 mark each (Q1 to Q10, including 2 Assertion-Reason questions).
- Section B contains 4 Short Answer questions of 2 marks each (Q11 to Q14).
- Section C contains 3 Short Answer questions of 3 marks each (Q15 to Q17).
- Section D contains 1 Long Answer question of 5 marks (Q18).
- Section E contains 2 Case Study-based questions of 4 marks each (Q19 & Q20).
- Use of calculators is strictly prohibited.
1. What is the value of the numerical expression (4⁻¹ - 5⁻¹)⁻¹?
[1]
✓ Solution (1 Mark):
4⁻¹ - 5⁻¹ = (1/4) - (1/5) = (5 - 4) / 20 = 1/20.
Therefore, (1/20)⁻¹ = 20. (Correct Option: b)
Therefore, (1/20)⁻¹ = 20. (Correct Option: b)
2. The number 0.0000705 expressed in standard scientific notation (m × 10โฟ) is:
[1]
✓ Solution (1 Mark):
Shifting the decimal point 5 places to the right gives 7.05 × 10⁻⁵. (Correct Option: b)
3. If (2/3)หฃ⁺¹ = (9/4)⁻², then the value of x is:
[1]
✓ Solution (1 Mark):
(9/4)⁻² = (4/9)² = [(2/3)²]² = (2/3)⁴.
Equating powers: x + 1 = 4 ⇒ x = 3. (Correct Option: c)
Equating powers: x + 1 = 4 ⇒ x = 3. (Correct Option: c)
4. Find the units digit in the simplified value of (2¹⁰⁰ ÷ 4²⁰).
[1]
✓ Solution (1 Mark):
2¹⁰⁰ ÷ 4²⁰ = 2¹⁰⁰ ÷ (2²)²⁰ = 2¹⁰⁰ ÷ 2⁴⁰ = 2⁶⁰.
Since powers of 2 have a cyclicity of 4 (2, 4, 8, 6) and 60 is a multiple of 4, the units digit is the same as 2⁴, which is 6. (Correct Option: c)
Since powers of 2 have a cyclicity of 4 (2, 4, 8, 6) and 60 is a multiple of 4, the units digit is the same as 2⁴, which is 6. (Correct Option: c)
5. If A = 3⁴ and B = (3²)³, then the quotient B ÷ A is equal to:
[1]
✓ Solution (1 Mark):
B = 3²หฃ³ = 3⁶. A = 3⁴.
B ÷ A = 3⁶ ÷ 3⁴ = 3⁶⁻⁴ = 3² = 9. (Correct Option: c)
B ÷ A = 3⁶ ÷ 3⁴ = 3⁶⁻⁴ = 3² = 9. (Correct Option: c)
6. How many zeros will the evaluated number (100)⁵ have at its end?
[1]
✓ Solution (1 Mark):
100 = 10². Thus (100)⁵ = (10²)⁵ = 10¹⁰, which has exactly 10 zeros at the end. (Correct Option: c)
7. A briefcase lock has 4 independent rotating wheels, each marked with digits 0 to 9. How many unique passcodes are possible?
[1]
✓ Solution (1 Mark):
Each wheel has 10 choices. Total combinations = 10 × 10 × 10 × 10 = 10⁴ = 10,000. (Correct Option: d)
8. The value of the arithmetic expression (5⁰ + 7⁰ - 3⁰) × 8⁰ is:
[1]
✓ Solution (1 Mark):
Since a⁰ = 1 for any non-zero number a:
(1 + 1 - 1) × 1 = 1 × 1 = 1. (Correct Option: b)
(1 + 1 - 1) × 1 = 1 × 1 = 1. (Correct Option: b)
9. Assertion (A): The value of (2/5)⁻³ is equal to 125/8.
Reason (R): For any non-zero rational number (a/b) and integer n, (a/b)⁻โฟ = (b/a)โฟ. [1]
Reason (R): For any non-zero rational number (a/b) and integer n, (a/b)⁻โฟ = (b/a)โฟ. [1]
✓ Solution (1 Mark):
(2/5)⁻³ = (5/2)³ = 125/8. Both Assertion and Reason are true, and the law stated in Reason directly explains the assertion. (Correct Option: a)
10. Assertion (A): The number 4.5 × 10⁻⁶ is greater than 4.5 × 10⁻⁴.
Reason (R): As the negative integer power of 10 decreases, the value of the number becomes smaller. [1]
Reason (R): As the negative integer power of 10 decreases, the value of the number becomes smaller. [1]
✓ Solution (1 Mark):
10⁻⁶ = 0.000001 while 10⁻⁴ = 0.0001, so 4.5 × 10⁻⁶ is smaller than 4.5 × 10⁻⁴. Thus A is false, but Reason R is a true statement. (Correct Option: d)
11. Solve for x: 3²หฃ⁺¹ ÷ 9 = 27
[2]
✓ Solution (2 Marks):
Convert all bases to 3:
3²หฃ⁺¹ ÷ 3² = 3³
3(²หฃ⁺¹⁻²) = 3³ ⇒ 3²หฃ⁻¹ = 3³
Equating powers: 2x - 1 = 3 ⇒ 2x = 4 ⇒ x = 2.
3²หฃ⁺¹ ÷ 3² = 3³
3(²หฃ⁺¹⁻²) = 3³ ⇒ 3²หฃ⁻¹ = 3³
Equating powers: 2x - 1 = 3 ⇒ 2x = 4 ⇒ x = 2.
12. Evaluate the expression: [(1/3)⁻² - (1/2)⁻³] ÷ (1/4)⁻²
[2]
✓ Solution (2 Marks):
(1/3)⁻² = 3² = 9.
(1/2)⁻³ = 2³ = 8.
(1/4)⁻² = 4² = 16.
Substituting: (9 - 8) ÷ 16 = 1 ÷ 16 = 1/16.
(1/2)⁻³ = 2³ = 8.
(1/4)⁻² = 4² = 16.
Substituting: (9 - 8) ÷ 16 = 1 ÷ 16 = 1/16.
13. By what number should (-4)⁻¹ be multiplied so that the resulting product is equal to 10⁻¹?
[2]
✓ Solution (2 Marks):
Let the required multiplier be y.
(-4)⁻¹ × y = 10⁻¹ ⇒ (-1/4) × y = 1/10
y = (1/10) × (-4) = -4/10 = -2/5.
(-4)⁻¹ × y = 10⁻¹ ⇒ (-1/4) × y = 1/10
y = (1/10) × (-4) = -4/10 = -2/5.
14. Express the following measures in standard scientific notation:
(a) 384,000,000,000,000 metres (Cosmic distance)
(b) 0.0000075 metres (Size of a human blood cell) [2]
(a) 384,000,000,000,000 metres (Cosmic distance)
(b) 0.0000075 metres (Size of a human blood cell) [2]
✓ Solution (2 Marks):
(a) 384,000,000,000,000 m = 3.84 × 10¹⁴ m. [1 Mark]
(b) 0.0000075 m = 7.5 × 10⁻⁶ m. [1 Mark]
(b) 0.0000075 m = 7.5 × 10⁻⁶ m. [1 Mark]
15. Simplify and express the result with positive exponents (where t ≠ 0):
[25 × t⁻⁴] ÷ [5⁻³ × 10 × t⁻⁸] [3]
[25 × t⁻⁴] ÷ [5⁻³ × 10 × t⁻⁸] [3]
✓ Solution (3 Marks):
Write base numbers in powers of 5 and 2:
Numerator = 5² × t⁻⁴.
Denominator = 5⁻³ × (2 × 5¹) × t⁻⁸ = 5⁻² × 2 × t⁻⁸.
Fraction = [5² ÷ 5⁻²] × [1/2] × [t⁻⁴ ÷ t⁻⁸]
= 5²⁻⁽⁻²⁾ × (1/2) × t⁻⁴⁻⁽⁻⁸⁾ = 5⁴ × t⁴ / 2 = (625t⁴) / 2.
Numerator = 5² × t⁻⁴.
Denominator = 5⁻³ × (2 × 5¹) × t⁻⁸ = 5⁻² × 2 × t⁻⁸.
Fraction = [5² ÷ 5⁻²] × [1/2] × [t⁻⁴ ÷ t⁻⁸]
= 5²⁻⁽⁻²⁾ × (1/2) × t⁻⁴⁻⁽⁻⁸⁾ = 5⁴ × t⁴ / 2 = (625t⁴) / 2.
16. If 2โฟ⁻⁵ × 5โฟ⁻⁴ = 500, find the value of the integer n.
[3]
✓ Solution (3 Marks):
Rewrite 2โฟ⁻⁵ as 2โฟ⁻⁴ × 2⁻¹:
2โฟ⁻⁴ × 2⁻¹ × 5โฟ⁻⁴ = 500
(2 × 5)โฟ⁻⁴ × (1/2) = 500
10โฟ⁻⁴ = 500 × 2 = 1000 = 10³
Equating exponents: n - 4 = 3 ⇒ n = 7.
Verification: 2² × 5³ = 4 × 125 = 500.
2โฟ⁻⁴ × 2⁻¹ × 5โฟ⁻⁴ = 500
(2 × 5)โฟ⁻⁴ × (1/2) = 500
10โฟ⁻⁴ = 500 × 2 = 1000 = 10³
Equating exponents: n - 4 = 3 ⇒ n = 7.
Verification: 2² × 5³ = 4 × 125 = 500.
17. Find the units digit in each of the following expansions by analyzing power cyclicity:
(a) 3⁶⁵ (b) 8⁴³ (c) 5⁹⁹ × 6⁴⁵ [3]
(a) 3⁶⁵ (b) 8⁴³ (c) 5⁹⁹ × 6⁴⁵ [3]
✓ Solution (3 Marks):
(a) Powers of 3 cycle as (3, 9, 7, 1). 65 ÷ 4 leaves remainder 1 ⇒ Units digit is 3¹ = 3. [1 Mark]
(b) Powers of 8 cycle as (8, 4, 2, 6). 43 ÷ 4 leaves remainder 3 ⇒ Units digit is 8³ → 2. [1 Mark]
(c) Every power of 5 ends in 5; every power of 6 ends in 6. 5 × 6 = 30 ⇒ Units digit is 0. [1 Mark]
(b) Powers of 8 cycle as (8, 4, 2, 6). 43 ÷ 4 leaves remainder 3 ⇒ Units digit is 8³ → 2. [1 Mark]
(c) Every power of 5 ends in 5; every power of 6 ends in 6. 5 × 6 = 30 ⇒ Units digit is 0. [1 Mark]
18. (a) Simplify the following algebraic-exponential expression: [3 Marks]
[16 × 2โฟ⁺¹ - 4 × 2โฟ] ÷ [16 × 2โฟ⁺² - 2 × 2โฟ⁺²]
(b) Find the value of m such that: 9แต⁺¹ × 3⁴ = 27แต⁻¹ ÷ 3⁻² [2 Marks] [5]
[16 × 2โฟ⁺¹ - 4 × 2โฟ] ÷ [16 × 2โฟ⁺² - 2 × 2โฟ⁺²]
(b) Find the value of m such that: 9แต⁺¹ × 3⁴ = 27แต⁻¹ ÷ 3⁻² [2 Marks] [5]
✓ Solution (5 Marks):
(a) Simplification:
Numerator = 2⁴ × 2โฟ⁺¹ - 2² × 2โฟ = 2โฟ⁺⁵ - 2โฟ⁺² = 2โฟ⁺²(2³ - 1) = 2โฟ⁺²(8 - 1) = 7 × 2โฟ⁺².
Denominator = 2โฟ⁺²(16 - 2) = 14 × 2โฟ⁺².
Quotient = [7 × 2โฟ⁺²] / [14 × 2โฟ⁺²] = 7/14 = 1/2. [3 Marks]
(b) Solving for m:
LHS = (3²)แต⁺¹ × 3⁴ = 3²แต⁺² × 3⁴ = 3²แต⁺⁶.
RHS = (3³)แต⁻¹ ÷ 3⁻² = 3³แต⁻³ ÷ 3⁻² = 3³แต⁻³⁻⁽⁻²⁾ = 3³แต⁻¹.
Equating bases: 2m + 6 = 3m - 1 ⇒ 3m - 2m = 6 + 1 ⇒ m = 7. [2 Marks]
Numerator = 2⁴ × 2โฟ⁺¹ - 2² × 2โฟ = 2โฟ⁺⁵ - 2โฟ⁺² = 2โฟ⁺²(2³ - 1) = 2โฟ⁺²(8 - 1) = 7 × 2โฟ⁺².
Denominator = 2โฟ⁺²(16 - 2) = 14 × 2โฟ⁺².
Quotient = [7 × 2โฟ⁺²] / [14 × 2โฟ⁺²] = 7/14 = 1/2. [3 Marks]
(b) Solving for m:
LHS = (3²)แต⁺¹ × 3⁴ = 3²แต⁺² × 3⁴ = 3²แต⁺⁶.
RHS = (3³)แต⁻¹ ÷ 3⁻² = 3³แต⁻³ ÷ 3⁻² = 3³แต⁻³⁻⁽⁻²⁾ = 3³แต⁻¹.
Equating bases: 2m + 6 = 3m - 1 ⇒ 3m - 2m = 6 + 1 ⇒ m = 7. [2 Marks]
19. Case Study 1: Technological Depreciation Model
[4]
A high-tech aerial mapping company purchases an industrial drone fleet for ₹2,56,000. Due to rapid technological advances, the market value of the equipment depreciates by half every single year. The value V of the drone equipment after n years is modeled by the exponential formula V = 2,56,000 / 2โฟ rupees.
(a) Calculate the monetary value of the drone after 3 years. [1 Mark]
(b) What will be the value of the drone equipment after 7 years? [1 Mark]
(c) After how many complete years will the equipment's value reduce to exactly ₹1,000? [1 Mark]
(d) Express the ratio of the equipment value at year 5 to its initial purchase value in negative exponential form. [1 Mark]
✓ Solution (4 Marks):
(a) V₃ = 256000 / 2³ = 256000 / 8 = ₹32,000. [1 Mark]
(b) V₇ = 256000 / 2⁷ = 256000 / 128 = ₹2,000. [1 Mark]
(c) 256000 / 2โฟ = 1000 ⇒ 2โฟ = 256000 / 1000 = 256 = 2⁸ ⇒ n = 8 years. [1 Mark]
(d) Ratio = V₅ / V₀ = (256000 / 2⁵) / 256000 = 1 / 2⁵ = 2⁻⁵. [1 Mark]
(b) V₇ = 256000 / 2⁷ = 256000 / 128 = ₹2,000. [1 Mark]
(c) 256000 / 2โฟ = 1000 ⇒ 2โฟ = 256000 / 1000 = 256 = 2⁸ ⇒ n = 8 years. [1 Mark]
(d) Ratio = V₅ / V₀ = (256000 / 2⁵) / 256000 = 1 / 2⁵ = 2⁻⁵. [1 Mark]
20. Case Study 2: Microbial Proliferation in Dairy Incubators
[4]
In a bio-dairy laboratory, a culture of probiotic bacteria is incubated under ideal temperature conditions. Initially, there are 250 bacteria in the sample. The bacterial population triples every hour. The population P after t hours is represented by the exponential growth relation P = 250 × 3แต.
(a) Calculate the population of bacteria after 4 hours. [1 Mark]
(b) By what factor does the population multiply after 6 hours compared to the initial count? Express your answer as a power of 3. [1 Mark]
(c) If the maximum holding capacity of the culture tube is 6,750 bacteria, after how many hours will this limit be reached? [1 Mark]
(d) If a single bacterium has an average length of 2 × 10⁻⁶ metres, calculate the total length if 5 × 10⁷ bacteria were placed end-to-end in a straight line (express in standard scientific notation). [1 Mark]
✓ Solution (4 Marks):
(a) P = 250 × 3⁴ = 250 × 81 = 20,250 bacteria. [1 Mark]
(b) Multiplication factor = P₆ / P₀ = (250 × 3⁶) / 250 = 3⁶ times (or 729 times). [1 Mark]
(c) 250 × 3แต = 6750 ⇒ 3แต = 6750 / 250 = 27 = 3³ ⇒ t = 3 hours. [1 Mark]
(d) Total length = (2 × 10⁻⁶ m) × (5 × 10⁷) = 10 × 10¹ = 100 m = 1.0 × 10² m. [1 Mark]
(b) Multiplication factor = P₆ / P₀ = (250 × 3⁶) / 250 = 3⁶ times (or 729 times). [1 Mark]
(c) 250 × 3แต = 6750 ⇒ 3แต = 6750 / 250 = 27 = 3³ ⇒ t = 3 hours. [1 Mark]
(d) Total length = (2 × 10⁻⁶ m) × (5 × 10⁷) = 10 × 10¹ = 100 m = 1.0 × 10² m. [1 Mark]