Number Exam Practice

This capstone lesson walks through a set of Edexcel-style exam questions covering every Number topic in the GCSE Mathematics (1MA1) specification. Each question is presented with a complete mark-scheme-style solution, mark allocation, and an examiner's note highlighting the most common pitfalls. Foundation and Higher content are mixed, with Higher-only items marked [H].

Work through the questions under timed conditions first (aim for roughly 1 minute per mark), then read the model solution. Compare your working against the mark-scheme steps — examiner marks are awarded for showing the right method, not just the right final answer.

How to Use This Lesson

Stage	What to do
1. Attempt	Work each question under non-calculator conditions unless indicated (Paper 2/3 questions are labelled).
2. Mark	Check your working against the mark-scheme breakdown (M marks for method, A marks for accuracy).
3. Diagnose	If you lost marks, read the examiner's note and identify the class of error (sign error, rounding early, missing step).
4. Re-attempt	Re-do any question you scored below full marks on, after reviewing the underlying topic lesson.

Section A — Foundation-accessible questions

Question 1 — Place value and ordering (3 marks)

Here are four numbers: 0.6, 3/5, 58%, 0.58̇. Write them in order from smallest to largest.

Mark-scheme solution:

Convert every number to a decimal to at least 3 decimal places.

Number	Decimal
0.6	0.600
3/5	0.600
58%	0.580
0.58̇	0.5888...

Ordering: 0.580 < 0.5888... < 0.600 = 0.600.

Answer: 58%, 0.58̇, 0.6 = 3/5

Marks: M1 for converting all four to a comparable form · A1 for three correct conversions · A1 for correct order with equal pair identified.

Examiner's note: Candidates routinely lose the final accuracy mark by writing 0.58̇ as 0.58 and placing it on the wrong side of 58%. The dot means the 8 recurs, so the number is larger than 0.58. Always write recurring decimals to at least one digit beyond the first repeat before comparing.

Question 2 — Estimation (3 marks, Paper 1 non-calculator)

Estimate the value of $(29.6 \times 5.12) / 0.48$ . You must show your working.

Mark-scheme solution:

Round each value to 1 significant figure:

$29.6 \approx 30$
$5.12 \approx 5$
$0.48 \approx 0.5$

Calculate: $(30 \times 5) / 0.5 = 150 / 0.5 = 300$ .

Answer: 300

Marks: M1 for each value rounded to 1 s.f. (any two correct) · M1 for a correct division by 0.5 (or equivalent multiplication by 2) · A1 for 300.

Examiner's note: "Estimate" is a technical command word — you must round to 1 s.f. before calculating. Candidates who write the final number only, without showing the three rounded values, receive zero method marks. Dividing by 0.5 also trips students up: remember that $\div 0.5$ is the same as $\times 2$ .

Question 3 — Fractions and percentages of an amount (4 marks, Paper 1)

Liam saves 3/8 of his monthly wage. He spends 20% of his wage on rent. His wage is £2,400. (a) How much does Liam save each month? (b) After paying rent and saving, how much of his wage is left?

Mark-scheme solution:

(a) 1/8 of £ $2,400 = 2400 \div 8 =$ £300. Savings $= 3 \times$ £300 = £900.

(b) Rent = 10% of £2,400 = £240, so 20% = £480. Remainder = £2,400 − £900 − £480 = £1,020.

Marks: (a) M1 for dividing by 8 · A1 for £900. (b) M1 for a correct method to find 20% · M1 for subtracting both savings and rent from £2,400 · A1 for £1,020.

Examiner's note: Part (b) is a two-step question and a surprising number of students only subtract one of the two amounts. Underline what each amount represents before calculating. Rent and savings come out of the same £2,400 — they do not compound.

Question 4 — Prime factorisation and HCF/LCM (5 marks)

(a) Write 420 as a product of its prime factors in index form. (b) Find the HCF and LCM of 420 and 168.

Mark-scheme solution:

(a) Repeated division: $420 \div 2 = 210$ $210 \div 2 = 105$ $105 \div 3 = 35$ $35 \div 5 = 7$ $7 \div 7 = 1$

$420 = 2^{2} \times 3 \times 5 \times 7$

(b) $168 = 2^{3} \times 3 \times 7 ($ from factor tree: $168 \to 84 \to 42 \to 21 \to 7)$ .

HCF (lowest power of each common prime): $2^{2} \times 3 \times 7 = 4 \times 21 =$ 84.
LCM (highest power of each prime): $2^{3} \times 3 \times 5 \times 7 = 8 \times 105 =$ 840.

Check: HCF $\times$ LCM $= 84 \times 840 = 70,560 = 420 \times 168$ ✓.

Marks: (a) M1 for a correct factor tree or repeated division · A1 for $2^{2} \times 3 \times 5 \times 7$ in index form. (b) M1 for both numbers written in prime-factor form · A1 for HCF = 84 · A1 for LCM = 840.

Examiner's note: Students lose the first accuracy mark for writing " $2 \times 2 \times 3 \times 5 \times 7$ " instead of using index notation; Edexcel mark schemes specifically require indices. For HCF/LCM, a common slip is taking the highest powers for HCF (or the lowest for LCM) — remember: HCF is the floor, LCM is the ceiling.

Question 5 — Standard form without a calculator (4 marks, Paper 1)

(a) Write $6.04 \times 10^{5}$ as an ordinary number. (b) Work out $(3 \times 10^{4}) \times (7 \times 10^{-2})$ . Give your answer in standard form.

Mark-scheme solution:

(a) Move the decimal point 5 places right: 604,000.

(b) Multiply the A values: $3 \times 7 = 21$ . Add the powers: $10^{4} \times 10^{-2} = 10^{2}$ . Combine: $21 \times 10^{2}$ . Adjust to standard form (A must satisfy $1 \leq A < 10)$ : $21 \times 10^{2} =$ $2.1 \times 10^{3}$ .

Marks: (a) B1 for 604,000. (b) M1 for $21 \times 10^{2} ($ or equivalent pre-adjusted form) · M1 for adjustment step · A1 for $2.1 \times 10^{3}$ .

Examiner's note: Part (b) tests whether you recognise that $21 \times 10^{2}$ is not in proper standard form. Leaving it unadjusted is the single most common way to drop the A1 in this type of question. Always finish by checking that your value of A satisfies $1 \leq A < 10$ .

Section B — Higher-tier questions

Question 6 — Fractional and negative indices [H] (4 marks, Paper 1)

Evaluate each of the following, giving each answer as an integer or a fraction. (a) 81^{3/4} (b) 27^{-2/3}

Mark-scheme solution:

(a) Do the root first, then the power. $81^{3/4} = (\sqrt[4]{81})^{3} = 3^{3} =$ 27.

(b) Deal with the fractional index first, then apply the negative. $27^{2/3} = (\sqrt[3]{27})^{2} = 3^{2} = 9$ . $27^{-2/3} = 1/9$ .

Answer: (a) 27 (b) 1/9

Marks: (a) M1 for $\sqrt[4]{81} = 3$ (or $81^{1/4}$ identified) · A1 for 27. (b) M1 for $27^{2/3} = 9$ · A1 for 1/9.

Examiner's note: The Higher-tier staple: when a negative-fractional index appears, split it into two steps — root, power, then reciprocal. Students who try to do all three at once almost always produce 1/27 or −9. Write the intermediate value explicitly.

Question 7 — Surds: simplification and rationalising [H] (5 marks, Paper 1)

(a) Simplify fully $\sqrt{48} + \sqrt{75}$ . (b) Rationalise the denominator of $12 / (3 + \sqrt{5})$ . Write your answer in the form $a + b\sqrt{5}$ where a and b are integers.

Mark-scheme solution:

(a) Simplify each surd by extracting the largest square factor.

$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$

Sum $= 4\sqrt{3} + 5\sqrt{3} =$ $9\sqrt{3}$ .

(b) Multiply top and bottom by the conjugate $(3 - \sqrt{5})$ :

Numerator: $12 \times (3 - \sqrt{5}) = 36 - 12\sqrt{5}$ .

Denominator: $(3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4$ .

$= (36 - 12\sqrt{5})/4 =$ $9 - 3\sqrt{5}$ .

So a = 9, b = −3.

Marks: (a) M1 for one correct simplification $(4\sqrt{3}$ or $5\sqrt{3})$ · A1 for $9\sqrt{3}$ . (b) M1 for multiplying by the conjugate · M1 for correctly expanding both numerator and denominator · A1 for $9 - 3\sqrt{5}$ .

Examiner's note: In (a), splitting $\sqrt{48}$ as $\sqrt{4} \times \sqrt{12} ($ rather than $\sqrt{16} \times \sqrt{3})$ only half-simplifies — expect to lose the A1 if the final answer is $2\sqrt{12} + 5\sqrt{3}$ or similar. Always take the largest square factor. In (b), the denominator expansion is the difference-of-two-squares $(3)^{2} - (\sqrt{5})^{2}$ . Students who expand with FOIL often introduce spurious $\sqrt{5}$ terms that should cancel.

Question 8 — Recurring decimals to fractions [H] (4 marks)

Using algebra, prove that 0.4̇5̇4̇ can be written as the fraction 454/999.

Mark-scheme solution:

Let x = 0.454454454...

Multiply by 1000 (the repeating block has 3 digits):

1000x = 454.454454454...

Subtract: 1000x − x = 454.454454... − 0.454454...

999x = 454

x = 454/999.

Therefore 0.4̇5̇4̇ = 454/999 ✓.

Marks: M1 for "let x = 0.454454..." (or equivalent variable setup) · M1 for multiplying by 1000 to align the recurring block · M1 for subtraction giving 999x = 454 · A1 for the correct fraction with a concluding statement.

Examiner's note: "Using algebra" and "prove" are both deliberate — writing the fraction by inspection earns zero marks, even if it is correct. The common failure is to multiply by 10 or 100 instead of 1000: remember that the power of 10 must match the length of the repeating block, not an arbitrary choice. Concluding with "Therefore..." is what unlocks the final A1.

Question 9 — Bounds and error intervals [H] (5 marks)

A rectangle has length 12.5 cm and width 8.4 cm, each measured correct to 1 decimal place. (a) Write the error interval for the length. (b) Calculate the upper bound of the area of the rectangle. (c) A student writes the area as " $105 cm^{2} ($ to 3 significant figures)". State, with justification, whether this is a suitable level of accuracy.

Mark-scheme solution:

Length bounds: $12.45 \leq l < 12.55$ . Width bounds: $8.35 \leq w < 8.45$ .

(a) $12.45 \leq$ length < 12.55 cm.

(b) Upper bound of area = upper length $\times$ upper width $= 12.55 \times 8.45 =$ $106.0475 cm^{2}$ .

(c) Lower bound of area $= 12.45 \times 8.35 = 103.9575 cm^{2}$ .

To 3 s.f., the upper bound rounds to 106 and the lower bound rounds to 104. The two bounds do not agree at 3 s.f., so " $105 cm^{2}$ " is not a suitable level of accuracy — the true area could be anywhere between 104 and 106 to 3 s.f.

Trying 2 s.f.: both 103.96 and 106.05 round to 110 — no, both round to 100? Check: 103.96 rounds to 100 (to 2 s.f.), 106.05 rounds to 110 (to 2 s.f.). So 2 s.f. also does not agree.

At 1 s.f., both bounds round to $100 cm^{2}$ . So the only suitable level of accuracy is 1 s.f.: the area is approximately $100 cm^{2}$ .

Marks: (a) B1 for $12.45 \leq l < 12.55$ . (b) M1 for both upper bounds identified · A1 for 106.0475. (c) M1 for calculating the lower bound · A1 for a reasoned conclusion that 3 s.f. is not suitable.

Examiner's note: Part (c) is the discriminator — candidates who merely recalculate without comparing the bounds lose the final mark. The standard technique is: compute both bounds, then round both to progressively finer levels of accuracy until they agree. The coarsest level at which they agree is the suitable one. Also note the strict inequality "<" on the upper bound: 12.55 itself would round up to 12.6.

Question 10 — Standard form with division [H] (4 marks, Paper 2 calculator)

The mass of a proton is $1.673 \times 10^{-27} kg$ . The mass of an electron is $9.109 \times 10^{-31} kg$ . Work out how many times heavier a proton is than an electron. Give your answer to 3 significant figures.

Mark-scheme solution:

Ratio $= (1.673 \times 10^{-27}) / (9.109 \times 10^{-31})$

$= (1.673 / 9.109) \times 10^{-27-(-31)}$

= 0.18366... $\times 10^{4}$

= 1836.6...

To 3 s.f.: 1840 (or equivalently $1.84 \times 10^{3})$ .

Marks: M1 for a correct division setup · M1 for correct handling of the powers (−27 − (−31) = 4) · A1 for 1836 or 1840 · A1 for the answer rounded to 3 s.f. and the correct form stated.

Examiner's note: The power arithmetic is where most marks are dropped: −27 − (−31) = +4, not −58 or −4. A "double negative" slip collapses the whole answer. Enter standard form on the calculator using the dedicated $\times10^{x}$ button — typing " $\times 10 ^ -27$ " without brackets produces the wrong result on many calculators.

Number Exam Practice

Number Exam Practice

How to Use This Lesson

Section A — Foundation-accessible questions

Question 1 — Place value and ordering (3 marks)

Question 2 — Estimation (3 marks, Paper 1 non-calculator)

Question 3 — Fractions and percentages of an amount (4 marks, Paper 1)

Question 4 — Prime factorisation and HCF/LCM (5 marks)

Question 5 — Standard form without a calculator (4 marks, Paper 1)

Section B — Higher-tier questions

Question 6 — Fractional and negative indices [H] (4 marks, Paper 1)

Question 7 — Surds: simplification and rationalising [H] (5 marks, Paper 1)

Question 8 — Recurring decimals to fractions [H] (4 marks)

Question 9 — Bounds and error intervals [H] (5 marks)

Question 10 — Standard form with division [H] (4 marks, Paper 2 calculator)

Question 11 — Multi-topic problem [H] (5 marks, Paper 1)

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