OCR GCSE Mathematics: Powers, roots, laws of indices and standard form (J560) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.01 and 3.02 of the OCR GCSE (9-1) Mathematics specification, Version 2.1 (August 2026): powers, roots, the laws of indices and standard form. Questions on fractional indices, fractional powers and estimating powers and roots are labelled “(Higher tier only)”. All other questions suit both tiers. Learn the method first from the study guide or the revision notes. Course pages: the OCR GCSE Mathematics hub, the checklist and the free 10-minute diagnostics.

Questions

1. Work out the following.

(a) 3⁴ [1] (b) 5⁻², as a fraction [1] (c) √196 + ∛216 [2]

2. Work these by hand.

(a) Write 2⁹ × 2⁻⁴ ÷ 2³ as a single power of 2, then work out its value. [3] (b) Find the value of n when 3ⁿ = 1/81. [2]

3. Simplify fully.

(a) 4a³b⁻² × 5a⁻⁵b⁶ [3] (b) (3x²y⁻¹)³ ÷ (9x⁴y⁻⁵) [4]

4. Work these by hand.

(a) Work out (3/4)⁻². [2] (b) Work out 2⁻³ + 2⁻² as a single fraction. [2]

5. (Higher tier only)

(a) Write √93 to the nearest whole number. Show how you decided. [2] (b) Write ∛30 to the nearest whole number. Show how you decided. [2] (c) Estimate the value of √401 ÷ ∛7.8 by rounding each root to a number you can work out. [3]

6. (Higher tier only) Work out the exact value of each.

(a) 81^(1/4) [1] (b) 125^(2/3) [2] (c) 64^(−1/6) [2] (d) (16/81)^(−3/4) [3]

7. (Higher tier only)

(a) Write 1/∛x in the form xᵖ. [1] (b) Simplify (27x⁹)^(2/3). [2] (c) Solve x^(3/2) = 216. [2]

8. Standard form.

(a) Write 0.0000308 in standard form. [1] (b) Write 5.7 × 10⁻³ as an ordinary number. [1] (c) Write 6.4 × 10⁵ as an ordinary number. [1] (d) Put these in order, smallest first: 6.1 × 10⁻³, 0.0059, 5.8 × 10⁻², 6.0 × 10⁻³. [3]

9. Work these by hand. Give each answer in standard form.

(a) (6 × 10⁴) × (3.5 × 10⁻⁷) [3] (b) (7.2 × 10⁹) ÷ (1.8 × 10³) [2] (c) 3.6 × 10⁵ + 4.8 × 10⁴ [3]

10. One grain of a powder has mass 3.2 × 10⁻⁵ kg. A pot holds 4.8 × 10⁻¹ kg of the powder. You may use a calculator.

(a) Work out the number of grains in the pot. Give your answer in standard form. [3] (b) A warehouse holds 2.5 × 10⁷ grains. Work out their total mass in kg, in standard form. [2]

11. A laboratory counts bacteria in two cultures. Culture A has 4.2 × 10⁸ bacteria and culture B has 7.5 × 10⁷ bacteria.

(a) Work out the total number of bacteria in both cultures, by hand. Give your answer in standard form. [3] (b) Work out how many times as many bacteria there are in culture A as in culture B, by hand. [2] (c) Each bacterium has mass 1.5 × 10⁻¹² g. Work out the total mass of culture A in grams. Give your answer in standard form. [3]

12. (Higher tier only)

(a) Show that 8^(2/3) × 32^(−2/5) = 1. [3] (b) Simplify (16x⁸)^(3/4) ÷ x^(1/2). [4] (c) Given that 4ᵏ = 1/(8√2), find the value of k. [3]

Answers

1. (a) 3⁴ = 3 × 3 × 3 × 3 = 81 [1] (b) 5⁻² = 1/5² = 1/25 [1] (c) √196 = 14 [1] and ∛216 = 6, so the total is 20 [1] [4] Examiner insight: A negative index gives a reciprocal, so a negative answer to part (b) shows the index was read as a sign. Writing each root separately in part (c) means a slip in one root does not hide the other.

2. (a) 2⁹ × 2⁻⁴ = 2^(9−4) = 2⁵ [1]; 2⁵ ÷ 2³ = 2² [1]; the value is 4 [1] (b) 1/81 = 1/3⁴ [1]; so 3ⁿ = 3⁻⁴ and n = −4 [1] [5] Examiner insight: Combining the indices in one line (9 − 4 − 3 = 2) is quicker, but writing the single power first shows the law you used. In part (b) the bases must match before you compare indices.

3. (a) Numbers: 4 × 5 = 20 [1]; a: a^(3+(−5)) = a⁻² [1]; b: b^(−2+6) = b⁴ [1]. Answer 20b⁴/a², which is 20a⁻²b⁴ (b) Numerator: (3x²y⁻¹)³ = 27 [1] × x⁶y⁻³ [1]; dividing by 9x⁴y⁻⁵: 27 ÷ 9 = 3 [1]; x^(6−4) = x² and y^(−3−(−5)) = y² [1]. Answer 3x²y² [7] Examiner insight: Cubing the bracket must include the coefficient 3, giving 27. When you divide, subtracting a negative index is an easy step to write as an addition, so write the subtraction out with brackets.

4. (a) (3/4)⁻² = (4/3)² [1] = 16/9 [1] (b) 2⁻³ = 1/8 and 2⁻² = 1/4 [1]; 1/8 + 2/8 = 3/8 [1] [4] Examiner insight: Flip the fraction first and then square it, which keeps the working short. In part (b) a common denominator is a method you can show even if you cannot see the answer at once.

5. (a) 9² = 81 and 10² = 100, so √93 lies between 9 and 10 [1]; 9.5² = 90.25 is below 93, so the answer is 10 [1] (b) 3³ = 27 and 4³ = 64, so ∛30 lies between 3 and 4 [1]; 3.5³ = 42.875 is above 30, so the answer is 3 [1] (c) √401 ≈ √400 = 20 [1]; ∛7.8 ≈ ∛8 = 2 [1]; 20 ÷ 2 = 10 [1] [7] Examiner insight: “To the nearest whole number” needs a reason, so test the midpoint and say which side the number falls. For an estimate, state the rounded values you used, because that is what shows the method.

6. (a) The fourth root of 81 is 3 [1] (b) ∛125 = 5 [1]; 5² = 25 [1] (c) The sixth root of 64 is 2 [1]; the negative index gives the reciprocal, 1/2 [1] (d) Flip the fraction to get (81/16)^(3/4) [1]; the fourth root of 81/16 is 3/2 [1]; (3/2)³ = 27/8 [1] [8] Examiner insight: Work out the root first and then apply the power, which keeps the arithmetic small. In part (d), flipping the fraction first removes the negative sign and makes the rest easy to check.

7. (a) 1/∛x = 1/x^(1/3) = x^(−1/3) [1] (b) 27^(2/3) = (∛27)² = 9 [1]; (x⁹)^(2/3) = x⁶, so the answer is 9x⁶ [1] (c) x^(3/2) = (√x)³ = 216, so √x = 6 [1]; x = 36 [1] [5] Examiner insight: Treat the number and the letter separately inside a bracket with a fractional index. For part (c), checking that 36^(3/2) = 6³ = 216 takes seconds and confirms the answer.

8. (a) The point moves 5 places right: 3.08 × 10⁻⁵ [1] (b) 0.0057 [1] (c) 640 000 [1] (d) 0.0059 = 5.9 × 10⁻³ [1]; the three numbers with 10⁻³ compare as 5.9, 6.0, 6.1 [1]; 5.8 × 10⁻² is the largest, because 10⁻² is bigger than 10⁻³. Order: 0.0059, 6.0 × 10⁻³, 6.1 × 10⁻³, 5.8 × 10⁻² [1] [6] Examiner insight: Converting everything to the same form before comparing is the safest approach. The number with the smallest A value (5.8) is the largest number here, and showing the powers of 10 explains why.

9. (a) 6 × 3.5 = 21 [1]; 10⁴ × 10⁻⁷ = 10⁻³ [1]; 21 × 10⁻³ = 2.1 × 10⁻² [1] (b) 7.2 ÷ 1.8 = 4 [1]; 10⁹ ÷ 10³ = 10⁶, so 4 × 10⁶ [1] (c) 4.8 × 10⁴ = 0.48 × 10⁵ [1]; 3.6 + 0.48 = 4.08 [1]; 4.08 × 10⁵ [1] [8] Examiner insight: An answer of 21 × 10⁻³ is not in standard form, so finish the tidy-up. For addition, writing both numbers with the same power of 10 is the step to show.

10. (a) (4.8 × 10⁻¹) ÷ (3.2 × 10⁻⁵) [1]; = 15 000 [1]; 1.5 × 10⁴ grains [1] (b) (2.5 × 10⁷) × (3.2 × 10⁻⁵) [1]; = 800, so 8 × 10² kg [1] [5] Examiner insight: Write the calculation you typed before the answer, so an examiner can see the method. Convert the final number to standard form, as the question asks, rather than leaving 15 000 or 800.

11. (a) 7.5 × 10⁷ = 0.75 × 10⁸ [1]; 4.2 + 0.75 = 4.95 [1]; 4.95 × 10⁸ [1] (b) (4.2 × 10⁸) ÷ (7.5 × 10⁷) [1]; 4.2 ÷ 7.5 = 0.56 and 10⁸ ÷ 10⁷ = 10, so 5.6 times [1] (c) (4.2 × 10⁸) × (1.5 × 10⁻¹²) [1]; 4.2 × 1.5 = 6.3 [1]; 10⁸ × 10⁻¹² = 10⁻⁴, so 6.3 × 10⁻⁴ g [1] [8] Examiner insight: The question for part (b) asks “how many times”, which signals a division. Part (a) needs the powers matched, and part (c) multiplies, so the indices are added. Read the operation before you start.

12. (a) 8^(2/3) = (∛8)² = 4 [1]; 32^(−2/5) = 1/(⁵√32)² = 1/4 [1]; 4 × 1/4 = 1 [1] (b) 16^(3/4) = (⁴√16)³ = 8 [1]; (x⁸)^(3/4) = x⁶ [1]; x⁶ ÷ x^(1/2) = x^(6−1/2) [1]; 8x^(11/2) [1] (c) 8√2 = 2³ × 2^(1/2) = 2^(7/2) [1]; 1/(8√2) = 2^(−7/2) and 4ᵏ = 2^(2k) [1]; 2k = −7/2, so k = −7/4 [1] [10] Examiner insight: A “show that” answer needs each stage written down, not just the final 1. In part (c), turning every number into a power of the same base is the idea that makes the equation solvable.

Where marks are usually lost

• Giving −25 for 5⁻², when the negative index means a reciprocal.
• Forgetting to raise the coefficient to the power when a bracket is cubed or squared.
• Subtracting a negative index as if it were positive, so y^(−3−(−5)) becomes y⁻⁸.
• Applying index laws to different bases, such as 2³ × 3² = 6⁵.
• Taking the power before the root with a fractional index, then losing track of large numbers.
• Skipping the midpoint check when a question says “to the nearest whole number”.
• Leaving an answer in a form such as 21 × 10⁻³ instead of tidying it into standard form.
• Adding standard form numbers without first matching the powers of 10.
• Comparing standard form numbers by their A values only, not their powers of 10.

Next steps

• Re-read the revision notes for any method you lost marks on.
• Go back to the study guide for the full worked examples.
• Check coverage on the OCR GCSE Mathematics hub and the printable checklist.
• Take the free 10-minute diagnostics to see what to revise next.
• Book a free trial class if you would like help with a topic.

Official syllabus

OCR GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017, published by OCR. Sections 3.01 and 3.02.