OCR GCSE Mathematics: Exact calculations with fractions, pi and surds (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 follow OCR GCSE (9-1) Mathematics, specification Version 2.1 (August 2026), sections 3.03a and 3.03b. Questions 1 to 6 use fractions and multiples of π and are for both tiers. Questions 7 to 12 use surds and are Higher tier only. Work every question without a calculator, as the specification describes for this content, and leave answers exact.

Revise first with the study guide or the revision notes. Tick off topics on the checklist and see the course hub.

Questions

1. Work out 4 1/3 − 2 5/6. Give your answer as a mixed number. [3]

2. A bag of flour has mass 2 1/4 kg. A baker uses 3/8 of the bag. Work out the mass of flour left. Give your answer as a mixed number of kilograms. [4]

3. Work out (2/3 + 3/4) ÷ (5/6 − 1/4). Write your answer as a fraction in its lowest terms. [4]

4. (a) A circle has radius 6 cm. Find its area and its circumference. Give both in terms of π. [2] (b) A semicircle has diameter 10 cm. Find the perimeter of the semicircle (curved edge plus the straight edge) and its area. Give both in terms of π. [3]

5. A sector of a circle has angle 150° and radius 12 cm. Find the area of the sector and the length of its arc. Give both in terms of π. [4]

6. A solid is made from a cylinder of radius 3 cm and height 10 cm, with a hemisphere of radius 3 cm joined to one flat end. The volume of a hemisphere is (2/3)πr³ and the curved surface area of a hemisphere is 2πr².

(a) Show that the volume of the solid is 108π cm³. [3] (b) Find the fraction of the volume of the solid that is in the hemisphere. [2] (c) Find the total surface area of the solid, including the flat circular end. Give your answer in terms of π. [3]

7. (Higher tier only) (a) Simplify √72 − √18 + √8. [3] (b) Write √50 in the form a√2, where a is an integer. [1]

8. (Higher tier only) Expand and simplify. (a) (3 + √5)(2 − √5) [2] (b) (4 − √3)² [2]

9. (Higher tier only) Rationalise the denominator and simplify. (a) 6/√8 [2] (b) 5/(3 − √2) [3]

10. (Higher tier only) A right-angled triangle has shorter sides of length (2 + √3) cm and (2 − √3) cm. (a) Find the exact length of the hypotenuse. [4] (b) Find the exact area of the triangle. [1]

11. (Higher tier only) Show that (√7 + 2)/(√7 − 2) = (11 + 4√7)/3. [4]

12. (Higher tier only) A rectangle has area 6 cm² and width (3 − √3) cm. (a) Show that the length of the rectangle is (3 + √3) cm. [3] (b) Find the perimeter of the rectangle. [2] (c) Find the exact length of a diagonal of the rectangle. Give your answer in the form k√6. [3]

Answers

1. Convert: 4 1/3 = 13/3 and 2 5/6 = 17/6 [1]. Common denominator 6: 26/6 − 17/6 = 9/6 [1]. Simplify: 3/2 = 1 1/2 [1]. [3] Examiner insight: Writing the improper fractions and the common denominator makes your method easy to follow and easy to check. Check that the answer is in the form requested, here a mixed number.

2. Flour used = 9/4 × 3/8 [1] = 27/32 kg [1]. Flour left = 72/32 − 27/32 [1] = 45/32 = 1 13/32 kg [1]. [4] Examiner insight: It is easy to find the amount used and stop there. Underline what the question asks for, which is the amount left, before you start.

3. 2/3 + 3/4 = 8/12 + 9/12 = 17/12 [1]. 5/6 − 1/4 = 10/12 − 3/12 = 7/12 [1]. 17/12 ÷ 7/12 = 17/12 × 12/7 [1]. The twelves cancel, giving 17/7 [1]. [4] Examiner insight: Work out each bracket separately and write the two results down. Then divide, flipping only the second fraction.

4. (a) Area = π × 6² = 36π cm² [1]. Circumference = 2 × π × 6 = 12π cm [1]. [2] (b) Radius 5. Curved edge = 1/2 × π × 10 = 5π [1]. Perimeter = 5π + 10 cm [1]. Area = 1/2 × π × 5² = 25π/2 cm² [1]. [3] Examiner insight: Leaving π as a symbol is what an exact answer needs. A frequent slip in part (b) is forgetting the straight edge of the semicircle, so list the edges before adding.

5. Area = 150/360 × π × 12² [1] = 5/12 × 144π = 60π cm² [1]. Arc = 150/360 × 2 × π × 12 [1] = 5/12 × 24π = 10π cm [1]. [4] Examiner insight: Simplify 150/360 to 5/12 before multiplying. Fewer large numbers means fewer slips, and the working shows the fraction of the circle clearly.

6. (a) Cylinder = π × 3² × 10 = 90π [1]. Hemisphere = 2/3 × π × 27 = 18π [1]. Total = 90π + 18π = 108π cm³ [1]. [3] (b) Fraction = 18π/108π [1] = 1/6 [1]. [2] (c) Curved surface of cylinder = 2π × 3 × 10 = 60π and hemisphere = 2π × 9 = 18π [1]. Flat circular end = π × 9 = 9π [1]. Total = 60π + 18π + 9π = 87π cm² [1]. [3] Examiner insight: On a “show that” question, write each volume separately and then the sum, so that the printed result is reached by visible steps. In part (c), sketch the solid and name each surface that is on the outside.

7. (a) √72 = 6√2, √18 = 3√2, √8 = 2√2 [2]. 6√2 − 3√2 + 2√2 = 5√2 [1]. [3] (b) √50 = √(25 × 2) = 5√2 [1]. [1] Examiner insight: Simplify every surd before you combine. Using the largest square factor each time avoids a second stage of simplifying.

8. (a) 6 − 3√5 + 2√5 − 5 [1] = 1 − √5 [1]. [2] (b) (4 − √3)(4 − √3) = 16 − 4√3 − 4√3 + 3 [1] = 19 − 8√3 [1]. [2] Examiner insight: Squaring a bracket means writing it twice and expanding. The term (−√3) × (−√3) is +3, so check the sign of each product.

9. (a) √8 = 2√2, so 6/√8 = 6/(2√2) = 3/√2 [1]. Multiply by √2/√2 to get 3√2/2 [1]. [2] (b) Multiply top and bottom by (3 + √2) [1]. Denominator: 9 − 2 = 7 [1]. Result: (15 + 5√2)/7 [1]. [3] Examiner insight: A two-term denominator needs its conjugate, with the sign in the middle changed. Write the new denominator as a difference of two squares so you can see it is rational.

10. (a) Hypotenuse² = (2 + √3)² + (2 − √3)² [1]. (2 + √3)² = 7 + 4√3 [1]. (2 − √3)² = 7 − 4√3 [1]. Sum = 14, so the hypotenuse is √14 cm [1]. [4] (b) Area = 1/2 × (2 + √3)(2 − √3) = 1/2 × (4 − 3) = 1/2 cm² [1]. [1] Examiner insight: Pythagoras with surds works well in exact form because the √3 terms cancel. State the hypotenuse squared first, then take the root last.

11. Multiply top and bottom by (√7 + 2) [1]. Denominator: (√7 − 2)(√7 + 2) = 7 − 4 = 3 [1]. Numerator: (√7 + 2)² = 7 + 4√7 + 4 [1]. This equals 11 + 4√7, so the fraction is (11 + 4√7)/3 [1]. [4] Examiner insight: In a “show that” question, work from the left-hand side to the printed result and write every line. A conclusion that is only stated without the expansion leaves the argument incomplete.

12. (a) Length = 6/(3 − √3). Multiply top and bottom by (3 + √3) [1]. Denominator = 9 − 3 = 6 [1]. Length = 6(3 + √3)/6 = 3 + √3 cm [1]. [3] (b) Perimeter = 2 × [(3 + √3) + (3 − √3)] [1] = 2 × 6 = 12 cm [1]. [2] (c) Diagonal² = (3 + √3)² + (3 − √3)² [1] = (12 + 6√3) + (12 − 6√3) = 24 [1]. Diagonal = √24 = 2√6 cm [1]. [3] Examiner insight: Part (b) is much quicker if you notice that the surd terms cancel. For part (c), simplify √24 to the requested form k√6.

Where marks are usually lost

• Combining fractions before the denominators match.
• Not reading what the question asks for, such as the amount left instead of the amount used.
• Replacing π with a decimal when the question says “in terms of π”.
• Forgetting a straight edge in a perimeter or a flat face in a surface area.
• Using diameter where the radius is needed.
• Leaving a surd such as √72 unsimplified.
• Treating (a + √b)² as a² + b.
• Rationalising a two-term denominator with only the surd instead of the conjugate.
• Dropping the negative sign when expanding with a minus inside a bracket.
• Skipping lines in a “show that” question.

Next steps

• Review the revision notes and the study guide
• Visit the OCR GCSE Mathematics hub and the checklist
• Take one of the free 10-minute diagnostics
• Book a free trial class

Official syllabus

OCR GCSE (9-1) Mathematics (J560), Version 2.1 (August 2026), for first assessment in 2017, published by OCR. These questions cover sections 3.03a and 3.03b.