AQA GCSE Mathematics 8300: Structure and calculation – 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 section 3.1.1, Structure and calculation (N1 to N9), of the AQA GCSE Mathematics (8300) specification, for teaching from September 2015 with exams from May/June 2017 (version 1.0). Parts on the product rule (N5), estimating roots (N6), fractional indices (N7) and surds (N8) are Higher tier only and are labelled “(Higher)”; everything else suits both tiers. Each question is labelled “(non-calculator)” in the style of Paper 1 or “(calculator)” in the style of Papers 2 and 3. These questions are different from the ones in the Number practice questions, which cover the whole of Topic 1.

Learn the content first in the study guide and the revision notes. The course hub is AQA GCSE Mathematics and the printable checklist lists every statement.

Questions

1. (non-calculator)

(a) Write these numbers in order, smallest first: 0.7, 2/3, −0.75, −4/5, 0.67 [2] (b) Write < or > between −3/8 and −0.4 [1]

2. (non-calculator) You are told that 63 × 48 = 3024. Use this fact to write down the value of:

(a) 6.3 × 0.48 [1] (b) 302.4 ÷ 4.8 [1] (c) 0.3024 ÷ 0.063 [1]

3. (non-calculator) Work out:

(a) 3 2/5 − 1 3/4, giving your answer as a mixed number [2] (b) −2 1/4 ÷ 1 1/2 [2] (c) (−0.3) × (−0.4) − 0.5 [1]

4. (non-calculator)

(a) Work out 20 − 4 × 3² ÷ 6 [2] (b) Insert one pair of brackets to make this correct: 8 + 12 ÷ 4 − 2 = 14 [1] (c) Work out 2/5 of the reciprocal of 0.25 [2]

5. (non-calculator)

(a) Write 252 as a product of its prime factors in index form. [2] (b) A = 2³ × 3 × 5² and B = 2 × 3² × 5 × 7. Find the HCF and the LCM of A and B, leaving your answers in index form or as ordinary numbers. [2] (c) Two warning lights are switched on together. One flashes every 40 seconds and the other every 75 seconds. After how many minutes do they next flash together? [2]

6. (non-calculator)

(a) Ravi chooses two different fruits from apple (A), banana (B), cherry (C) and date (D). Order does not matter. List all the possible pairs. [2] (b) (Higher) A password is three letters from A, B, C, D, E (repeats allowed) followed by two odd digits (repeats allowed). How many passwords are possible? [2] (c) (Higher) How many four-digit numbers greater than 5000 have every digit even? [2]

7. (non-calculator)

(a) Write down the value of 13² and of ∛125 [2] (b) (Higher) Show that √90 lies between 9.4 and 9.5 [2] (c) Write 7⁵ × 7⁻² ÷ 7⁶ as a single power of 7, then find its value as a fraction. [2] (d) 2ⁿ = 1/32. Find n. [1] (e) (Higher) Work out 125^(−2/3) [2] (f) (Higher) Work out (16/81)^(3/4) [2]

8. (non-calculator) A shape is made from a square of side 8 cm with a semicircle drawn outwards on one side, so the diameter of the semicircle is that side.

(a) Find the area of the shape. Give your answer in terms of π. [2] (b) Find the perimeter of the shape. Give your answer in terms of π. [2]

9. (non-calculator) (Higher) A rectangle has length √75 cm and width √12 cm.

(a) Show that its area is 30 cm². [2] (b) Find its perimeter in the form k√3. [2] (c) Write 30/√12 in the form a√3, and explain what your answer represents. [2]

10. (non-calculator) Give each answer in standard form.

(a) (6 × 10⁴) × (4 × 10⁻⁹) [2] (b) (9 × 10⁷) ÷ (1.5 × 10³) [1] (c) 5.2 × 10⁶ − 8 × 10⁵ [2]

11. (calculator)

(a) A calculator shows 4.56E−03. Write this as an ordinary number. [1] (b) A country spends £9.38 × 10¹¹ in a year and has 6.7 × 10⁷ people. Work out the spending per person. Give your answer in standard form. [2] (c) A virus particle has a diameter of 1.2 × 10⁻⁷ m. How many particles placed end to end would make a line 3 cm long? Give your answer in standard form. [3]

12. (calculator)

(a) Leah’s bank account has a balance of £245.60. She then has debits of £82.35 and £129.99, a credit of £310.00 and a debit of £412.50. Work out her new balance and say what it means. [3] (b) A trader buys 24 cushions at a cost price of £3.85 each. He sells 15 of them at a selling price of £5.20 each. Has he made a profit or a loss so far, and how much? [2] (c) What is the least number of extra cushions he must sell at £5.20 to make a profit on all 24? [3]

Answers

1. (a) Convert: −4/5 = −0.8, 2/3 = 0.666… [1]; −4/5, −0.75, 2/3, 0.67, 0.7 [1] (b) −3/8 = −0.375 so −3/8 > −0.4 [1] Examiner insight: The final order must use the original values; a list of decimals only loses the accuracy mark, and 2/3 must sit below 0.67 because 0.666… is less than 0.670.

2. (a) 3.024 [1] (b) 63 [1] (c) 4.8 [1] Examiner insight: Each part is a single accuracy mark with no method credit, so check every answer with an estimate (300 ÷ 5 ≈ 60 for part (b)).

3. (a) 17/5 − 7/4 = 68/20 − 35/20 [1] = 33/20 = 1 13/20 [1] (b) −9/4 × 2/3 [1] = −18/12 = −1 1/2 [1] (c) 0.12 − 0.5 = −0.38 [1] Examiner insight: Method marks are for converting to improper fractions with a common denominator or a reciprocal; an unsimplified 33/20 when a mixed number is asked for loses the final mark.

4. (a) 3² = 9, then 4 × 9 ÷ 6 = 6 [1]; 20 − 6 = 14 [1] (b) 8 + 12 ÷ (4 − 2) = 14 [1] (c) Reciprocal of 0.25 is 4 [1]; 2/5 × 4 = 8/5 = 1.6 [1] Examiner insight: In (a), working left to right gives (20 − 4) × 9 ÷ 6 = 24 and scores nothing, because the method mark is for applying priority correctly.

5. (a) 252 = 2 × 2 × 3 × 3 × 7 [1] = 2² × 3² × 7 [1] (b) HCF = 2 × 3 × 5 = 30 [1]; LCM = 2³ × 3² × 5² × 7 = 12 600 [1] (c) 40 = 2³ × 5 and 75 = 3 × 5², LCM = 2³ × 3 × 5² = 600 seconds [1]; 10 minutes [1] Examiner insight: A correct product without index form earns only the method mark in (a), and in (c) an answer of 600 without converting to minutes loses the final mark.

6. (a) AB, AC, AD, BC, BD, CD: at least four correct with no repeats [1]; all 6 pairs and no extras [1] (b) 5³ × 5² (five letters, five odd digits) [1] = 3125 [1] (c) First digit 6 or 8, then 5 choices for each other digit: 2 × 5 × 5 × 5 [1] = 250 [1] Examiner insight: Listing BA as well as AB counts as a repeat and loses the second mark in (a); in (b) and (c) adding the choices instead of multiplying earns no method mark.

7. (a) 169 [1]; 5 [1] (b) 9.4² = 88.36 and 9.5² = 90.25 [1]; 88.36 < 90 < 90.25, so 9.4 < √90 < 9.5 [1] (c) 7^(5 − 2 − 6) = 7⁻³ [1]; 1/343 [1] (d) 32 = 2⁵, so n = −5 [1] (e) Cube root of 125 = 5, 5² = 25 [1]; reciprocal gives 1/25 [1] (f) Fourth root of 16/81 = 2/3 [1]; (2/3)³ = 8/27 [1] Examiner insight: For a “show that” in (b), both squares must be worked out and compared with 90; stating “√90 ≈ 9.49” from memory earns no marks on a non-calculator paper.

8. (a) Square 64; semicircle radius 4, area ½ × π × 4² = 8π [1]; total 64 + 8π cm² [1] (b) Three sides 3 × 8 = 24; curved edge ½ × 2 × π × 4 = 4π [1]; total 24 + 4π cm [1] Examiner insight: The side covered by the semicircle is not part of the perimeter, so 32 + 4π loses the accuracy mark; a decimal answer loses it too when “in terms of π” is asked.

9. (a) √75 × √12 = √900 [1] = 30, so the area is 30 cm² [1] (b) √75 = 5√3 and √12 = 2√3 [1]; 2(5√3 + 2√3) = 14√3 cm [1] (c) 30/(2√3) = 15/√3 = 15√3/3 = 5√3 [1]; it is the length of the rectangle (area ÷ width) [1] Examiner insight: In (a) “show that” needs the √900 step written down; in (c) leaving 15√3/3 uncancelled loses the accuracy mark.

10. (a) 24 × 10⁻⁵ [1] = 2.4 × 10⁻⁴ [1] (b) 6 × 10⁴ [1] (c) 5.2 × 10⁶ − 0.8 × 10⁶ [1] = 4.4 × 10⁶ [1] Examiner insight: 24 × 10⁻⁵ is correct in value but not in standard form, so it earns the method mark only.

11. (a) 0.00456 [1] (b) (9.38 × 10¹¹) ÷ (6.7 × 10⁷) = 14 000 [1] = £1.4 × 10⁴ [1] (c) 3 cm = 0.03 m (or 3 × 10⁻² m) [1]; 0.03 ÷ (1.2 × 10⁻⁷) [1] = 2.5 × 10⁵ [1] Examiner insight: In (c) the units must match before dividing; using 3 instead of 0.03 gives 2.5 × 10⁷ and keeps only the division method mark.

12. (a) 245.60 − 82.35 − 129.99 + 310.00 − 412.50 [1] = −69.24 [1]; −£69.24, so she is overdrawn by £69.24 [1] (b) Cost 24 × 3.85 = £92.40 and takings 15 × 5.20 = £78.00 [1]; a loss of £14.40 [1] (c) 92.40 ÷ 5.20 = 17.769… [1]; so 18 must be sold in total (18 × 5.20 = £93.60 > £92.40) [1]; 3 more cushions [1] Examiner insight: In (c) rounding 17.769… down to 17 gives £88.40, still a loss; the final mark needs the answer as extra cushions (3), not the total (18).

Where marks are usually lost

• Ordering negative decimals by the size of their digits, so −0.75 is placed before −0.8.
• Moving the decimal point the wrong number of places when using a given product.
• Working left to right instead of using priority of operations.
• Using the HCF when the context needs the LCM, or giving seconds when minutes were asked for.
• Counting AB and BA as different when order does not matter.
• (Higher) Adding instead of multiplying with the product rule.
• (Higher) Ignoring the minus sign in a negative fractional index, so 125^(−2/3) is given as 25 instead of 1/25.
• Including the hidden side in the perimeter of a compound shape.
• Leaving standard form with A outside 1 ≤ A < 10.
• Rounding down in a “how many must be sold” problem.

Next steps

• Structure and calculation revision notes
• Structure and calculation study guide
• Number practice questions
• AQA GCSE Mathematics course hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA GCSE Mathematics (8300) specification, for teaching from September 2015, exams from May/June 2017, version 1.0, published by AQA – section 3.1.1 Structure and calculation (N1 to N9).