OCR GCSE Mathematics: Integer arithmetic, number theory, priority of operations and inverse operations (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 1.01 to 1.04 of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026): integer arithmetic, whole number theory, priority of operations and inverse operations. All of it is Foundation and Higher content, so every question is suitable for both tiers. Do the arithmetic by hand and show your working.

Revise first with the study guide or the revision notes. Course pages: OCR GCSE Mathematics hub and the printable checklist.

Questions

1. Work out, showing your method where there is more than one step.

(a) -9 - (-17) + 4 [1] (b) (-8) x (-5) ÷ (-10) [1] (c) 2 856 ÷ 12 [2]

2.

(a) Write down the value of the digit 3 in the number 538. [1] (b) Work out 538 x 47. [3]

3. Here is a list of numbers: 27, 36, 51, 59, 91, 97.

(a) Write down the prime numbers in the list. [2] (b) Show that 91 is not a prime number. [1] (c) Write down a cube number and a square number from the list. [2]

4.

(a) Write 540 as a product of its prime factors. Use power notation. [3] (b) Use your answer to (a) to write 1 080 as a product of its prime factors. [1]

5. A = 2³ x 3 x 5² and B = 2² x 3² x 5 x 7. Write each answer as a product of prime factors in power notation.

(a) Find the highest common factor of A and B. [2] (b) Find the lowest common multiple of A and B. [2]

6. By listing factors and multiples, find

(a) the highest common factor of 28 and 40, [2] (b) the lowest common multiple of 28 and 40. [2]

7. Work out the value of

(a) 36 - 4 x (3² - 5) ÷ 2 [3] (b) (5² - 1) / (√49 - 4) [2]

8. Put one pair of brackets in the calculation below so that it is correct.

6 + 4 x 3 - 1 = 29 [2]

9. Use inverse operations to make each calculation easier. Show your steps.

(a) 1 003 - 698 [2] (b) 45 x 24 [2]

10. I think of a number. I multiply it by 4, subtract 6, then halve the result. My answer is 15. Work out my number, and show how you could check it. [4]

11. Each letter stands for a missing digit.

(a) 4a7 + 28b = 703. Find a and b. [3] (b) 8c x 6 = 5d2. Find c and d. [3]

12. A shop sells pens in packs of 15 and pencils in packs of 25.

(a) Chen wants exactly the same number of pens as pencils. What is the smallest number of each kind of pack he can buy? [3]

A florist has 90 roses and 135 lilies. She makes identical bunches using all the flowers, with no flowers left over, and wants as many bunches as possible.

(b) Write 90 and 135 as products of prime factors. [2] (c) How many bunches can she make? [1] (d) How many roses and how many lilies are in each bunch? [2]

Answers

1. (a) -9 - (-17) + 4 = -9 + 17 + 4 [1] = 12 (b) (-8) x (-5) = 40, then 40 ÷ (-10) [1] = -4 (c) 12 x 100 = 1 200, leaving 1 656; 12 x 100 = 1 200 again, leaving 456; 12 x 30 = 360, leaving 96; 12 x 8 = 96 [1]. Total 100 + 100 + 30 + 8 = 238 [1]. [4] Examiner insight: Show the chunks or the bus stop layout in part (c); a correct method with one slip still shows what you know. A negative answer in (b) comes from counting the negative signs.

2. (a) The 3 is in the tens column, so its value is 30 [1] (b) 538 x 40 = 21 520 [1]; 538 x 7 = 3 766 [1]; 21 520 + 3 766 = 25 286 [1]. [4] Examiner insight: “Value” means the amount the digit stands for (30), not the digit itself. In long multiplication, writing the two partial products separately makes an addition slip easy to spot.

3. (a) 59 [1] and 97 [1] Check: 59 is not divisible by 2, 3, 5 or 7, and 97 is not divisible by 2, 3, 5 or 7, so both are prime. (b) 91 = 7 x 13, so it has factors other than 1 and itself [1] (c) Cube: 27 (3³) [1]. Square: 36 (6²) [1]. [5] Examiner insight: A “show that” answer needs the actual factor pair, not just a statement. Take care with odd numbers that look prime, such as 51 and 91.

4. (a) 540 ÷ 2 = 270, 270 ÷ 2 = 135 [1]; 135 ÷ 3 = 45, 45 ÷ 3 = 15, 15 ÷ 3 = 5 [1]; 540 = 2² x 3³ x 5 [1] Check: 4 x 27 x 5 = 540. (b) 1 080 = 2 x 540 = 2³ x 3³ x 5 [1]. [4] Examiner insight: Use the power notation the question asks for, and multiply back to check. In (b), using your answer to (a) is quicker than starting again.

5. (a) Lowest power of each shared prime: 2², 3, 5 [1]. HCF = 2² x 3 x 5 (= 60) [1] (b) Highest power of every prime: 2³, 3², 5², 7 [1]. LCM = 2³ x 3² x 5² x 7 (= 12 600) [1]. [4] Examiner insight: Keep the answer in the form requested. A prime that appears in only one number, like 7, is left out of the HCF and included in the LCM.

6. (a) Factors of 28: 1, 2, 4, 7, 14, 28. Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40 [1]. Common factors 1, 2, 4, so HCF = 4 [1] (b) Multiples of 40: 40, 80, 120, 160, 200, 240, 280 [1]. Multiples of 28 include 252 and 280 (28 x 10), so LCM = 280 [1]. [4] Examiner insight: Write out both lists, as the question tells you to. Check the LCM from both sides: 280 ÷ 28 = 10 and 280 ÷ 40 = 7.

7. (a) Brackets: 3² - 5 = 9 - 5 = 4 [1]. Then 4 x 4 ÷ 2 = 16 ÷ 2 = 8 [1]. 36 - 8 = 28 [1] (b) Numerator 25 - 1 = 24 and denominator 7 - 4 = 3 [1]. 24 ÷ 3 = 8 [1]. [5] Examiner insight: Work out the top and bottom of a fraction separately, as if each were in brackets. Writing each stage on a new line keeps the order of operations visible.

8. Try (6 + 4) x 3 - 1 = 10 x 3 - 1 = 30 - 1 = 29 [1]. The answer is (6 + 4) x 3 - 1 [1]. [2] Examiner insight: Test one placement at a time and show that your choice gives 29. Other placements, such as 6 + 4 x (3 - 1), give a different value.

9. (a) 1 003 - 698 = 1 003 + 2 - 700 [1] = 305 [1] (b) 45 x 24 = 90 x 12 = 180 x 6 [1] = 1 080 [1]. [4] Examiner insight: Show the adjustment step. Doubling one number and halving the other keeps the product the same. In a subtraction, if you add 2 to the number you take away, put the 2 back at the end.

10. Reverse the steps. Halving is undone by doubling: 15 x 2 = 30 [1]. Subtracting 6 is undone by adding 6: 30 + 6 = 36 [1]. Multiplying by 4 is undone by dividing by 4: 36 ÷ 4 = 9 [1]. Check: 9 x 4 = 36, 36 - 6 = 30, 30 ÷ 2 = 15 [1]. [4] Examiner insight: Undo the steps in the opposite order. The check runs the original steps forward and shows the answer is right.

11. (a) Units: 7 + b ends in 3, so b = 6, carry 1 [1]. Tens: a + 8 + 1 ends in 0, so a = 1, carry 1 [1]. Hundreds: 4 + 2 + 1 = 7, so it fits: a = 1, b = 6 (417 + 286 = 703) [1] (b) Units: 6 x (units digit) ends in 2, so c = 2 or c = 7 [1]. 82 x 6 = 492, which does not start with 5 [1]. 87 x 6 = 522, so c = 7, d = 2 [1]. [6] Examiner insight: Work from the units column and say what you are carrying. When two digits are possible, test both and say why one is rejected.

12. (a) Find the LCM of 15 and 25. Multiples of 15: 15, 30, 45, 60, 75; multiples of 25: 25, 50, 75 [1]. LCM = 75 [1]. Pens: 75 ÷ 15 = 5 packs; pencils: 75 ÷ 25 = 3 packs [1] (b) 90 = 2 x 3² x 5 [1]; 135 = 3³ x 5 [1] (c) HCF = 3² x 5 = 45, so 45 bunches [1] (d) Roses: 90 ÷ 45 = 2 [1]. Lilies: 135 ÷ 45 = 3 [1]. [8] Examiner insight: “Smallest number that both divide into” points to the LCM, and sharing out with nothing left over points to the HCF. Say which one you are finding and state the final answer in context.

Where marks are usually lost

• Treating 1 as prime, or calling an odd number prime without testing it.
• Leaving a prime factorisation without powers when power notation is requested.
• Giving 2 x 3 x 5 when a repeated prime needs a power, such as 3³.
• Muddling HCF and LCM in word problems.
• Doing multiplications before divisions that stand to their left.
• Forgetting that a subtraction of a negative number adds.
• Reversing a number puzzle in the order given rather than the opposite order.
• Not checking a missing-digit answer by adding or multiplying it out.
• Writing only a final answer where the question asks you to show a method.

Next steps

• Revise with the revision notes or the study guide.
• Back to the OCR GCSE Mathematics hub and the printable checklist.
• Find your weak spots with the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), published by OCR (Cambridge OCR): sections 1.01a, 1.02a to 1.02c, 1.03a and 1.04a.