Number: Exam Practice

This lesson brings together the whole Number strand of OCR GCSE Mathematics (J560) in a set of exam-style questions. Each question is fully worked, with commentary on where the method marks are earned and on the OCR command words (Work out, Calculate, Show that, Write down, Find, Give a reason). Questions are a mix of Foundation (grades 1–5) and Higher [H] (grades 4–9); marks are shown in brackets and Higher-only items are labelled. Work through each yourself first, then check the solution.

A few exam habits worth carrying through every question below. First, show your method — OCR mark schemes award method (M) marks for a correct approach even when the final answer is wrong, so never write a bare number for a multi-mark question. Second, read the command word: "Write down" expects an instant answer with no working; "Work out" and "Calculate" expect method; "Show that" and "Prove" require every step to a given target; "Give a reason" needs a sentence of explanation. Third, keep full accuracy until the final line and only round at the end, and give the answer in exactly the form requested (index form, standard form, simplest surd form, a mixed number in lowest terms, or to a stated number of significant figures). These habits, more than raw speed, are what separate the grade bands.

Exam-Style Questions

Question 1 (Foundation, grades 1–3) — (2 marks)

Write down the value of the digit $7$ in the number $3{,}472{,}016$ .

Solution. Reading the columns from the right of $3{,}472{,}016$ : units $6$ , tens $1$ , hundreds $0$ , thousands $2$ , ten-thousands $7$ , hundred-thousands $4$ , millions $3$ . The $7$ sits in the ten-thousands column, so its value is $70{,}000$ .

Method-mark note: "Write down" needs no working — just the correct value $70{,}000$ . A common error is to answer "ten thousands" (the column name) instead of the value; OCR wants the number. $[2]$

Question 2 (Foundation, grades 2–4) — (3 marks)

Work out an estimate for $\dfrac{38.7 \times 5.1}{0.21}$ .

Solution. Round each value to 1 significant figure: $38.7 \approx 40$ , $5.1 \approx 5$ and $0.21 \approx 0.2$ . Substitute the rounded values:

$\dfrac{40 \times 5}{0.2} = \dfrac{200}{0.2} = 1000.$

The final division by $0.2$ is the same as multiplying by $5$ , which is why the answer jumps up to $1000$ — a point many candidates miss.

Method-mark note: one mark for each correct rounding line and one for the final value; the " $\approx$ " lines secure the method marks even if the division slips. $[3]$

Question 3 (Foundation, grades 3–4) — (3 marks)

Write $540$ as a product of its prime factors. Give your answer in index form.

Solution. Build a factor tree. Split $540 = 54 \times 10$ . Then $54 = 2 \times 27 = 2 \times 3^{3}$ and $10 = 2 \times 5$ . Collecting all the prime leaves gives $2 \times 3^{3} \times 2 \times 5$ , and grouping the repeated $2$ s into a power:

$540 = 2^{2} \times 3^{3} \times 5.$

Method-mark note: marks for a correct factor tree and for giving the answer in index form. Check by recombining: $4 \times 27 \times 5 = 540$ ✓. A different first split (e.g. $540 = 4 \times 135$ ) gives the same powers — the factorisation is unique. $[3]$

Question 4 (Foundation, grades 3–5) — (3 marks)

Work out $2\dfrac{1}{4} + 1\dfrac{2}{3}$ . Give your answer as a mixed number.

Solution. Convert each mixed number to an improper fraction: $2\dfrac{1}{4} = \dfrac{9}{4}$ and $1\dfrac{2}{3} = \dfrac{5}{3}$ . The LCM of $4$ and $3$ is $12$ , so rewrite with denominator $12$ : $\dfrac{9}{4} = \dfrac{27}{12}$ and $\dfrac{5}{3} = \dfrac{20}{12}$ . Add the numerators: $\dfrac{27}{12} + \dfrac{20}{12} = \dfrac{47}{12}$ . Finally convert back: $47 \div 12 = 3$ remainder $11$ , so $\dfrac{47}{12} = 3\dfrac{11}{12}$ .

Method-mark note: marks for the common denominator, the addition, and converting back to a mixed number in the form the question asks for. $[3]$

Question 5 (Foundation/Higher, grades 4–5) — (3 marks)

A coat costs £75. In a sale it is reduced by $16\%$ . Work out the sale price.

Solution. A $16\%$ decrease means the customer pays $100\% - 16\% = 84\%$ of the original, so the multiplier is $0.84$ . Sale price $= 75 \times 0.84 = 63$ , so £63. (Building it up gives the same: $10\%$ is £7.50 and $6\%$ is £4.50, so $16\%$ is £12, and $75 - 12 = 63$ , i.e. £63.)

Method-mark note: one mark for the multiplier $0.84$ (or for finding that $16\%$ is £12), then marks for the calculation and the final answer. The multiplier method is faster and less error-prone on the calculator papers. $[3]$

Question 6 (Higher [H], grades 4–6) — (2 marks)

Work out the value of $36^{\frac{1}{2}} + 5^{0}$ .

Solution. Deal with each term separately. The index $\frac{1}{2}$ means a square root, so $36^{\frac{1}{2}} = \sqrt{36} = 6$ . Any non-zero number to the power $0$ is $1$ , so $5^{0} = 1$ . Adding: $6 + 1 = 7$ .

Method-mark note: one mark for each correct evaluation. The trap is $5^{0}$ — it is $1$ , not $0$ and not $5$ ; this follows from the division law since $5^{n} \div 5^{n} = 5^{0} = 1$ . $[2]$

Question 7 (Higher [H], grades 5–7) — (2 marks)

Write $0.000\,406$ in standard form.

Solution. Standard form needs a coefficient $A$ with $1 \le A < 10$ , so place the point after the first non-zero digit: $A = 4.06$ . To turn $4.06$ back into $0.000\,406$ the point must move $4$ places to the left, and since the number is less than $1$ the power is negative: $4.06 \times 10^{-4}$ .

Method-mark note: one mark for $4.06$ , one for $10^{-4}$ . Common wrong answers ( $406 \times 10^{-6}$ or $4.06 \times 10^{4}$ ) come from an out-of-range coefficient or the wrong sign on the power. $[2]$

Question 8 (Higher [H], grades 5–7) — (3 marks)

Simplify fully $\sqrt{48} + \sqrt{27}$ .

Solution. $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$ and $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$ . These are like surds: $4\sqrt{3} + 3\sqrt{3} = 7\sqrt{3}$ .

Method-mark note: a mark for each simplified surd and one for combining; do not write $\sqrt{75}$ . $[3]$

Question 9 (Higher [H], grades 6–8) — (3 marks)

Show that the recurring decimal $0.\dot{3}\dot{6}$ is equal to $\dfrac{4}{11}$ .

Solution. Let $x = 0.\dot{3}\dot{6}$ , which written out is $x = 0.363636\ldots$ The repeating block " $36$ " has two digits, so multiply both sides by $100$ to shift exactly one block across the point:

$100x = 36.363636\ldots$

Now subtract the original $x$ — the recurring tails are identical, so they cancel:

$100x - x = 36.363636\ldots - 0.363636\ldots \;\Rightarrow\; 99x = 36.$

So $x = \dfrac{36}{99}$ , and dividing top and bottom by the HCF $9$ gives $x = \dfrac{4}{11}$ , the required result.

Method-mark note: "Show that" means every line must be present — "let $x =$ ", the $\times 100$ , the subtraction, and the simplification to the given answer. Skipping straight to $\dfrac{36}{99}$ would lose method marks even though the value is right. $[3]$

Question 10 (Higher [H], grades 6–8) — (4 marks)

$a = 7.5$ and $b = 2.5$ , both correct to $1$ decimal place. Work out the upper bound of $\dfrac{a}{b}$ . Give your answer to $3$ significant figures.

Solution. Each value is to 1 d.p., so half the accuracy is $0.05$ . The bounds are $7.45 \le a < 7.55$ and $2.45 \le b < 2.55$ . A quotient $\dfrac{a}{b}$ is largest when the numerator is as big as possible and the denominator as small as possible, so use the upper bound of $a$ and the lower bound of $b$ :

$\dfrac{\text{UB}_a}{\text{LB}_b} = \dfrac{7.55}{2.45} = 3.08163\ldots$

Keeping full accuracy and rounding only now: $3.08163\ldots \approx 3.08$ (3 s.f.).

Method-mark note: marks for both bounds, for the correct choice $\text{UB} \div \text{LB}$ , and for rounding correctly to 3 s.f. at the end. Using $\dfrac{\text{UB}}{\text{UB}}$ is the common error — it understates the maximum. $[4]$

Question 11 (Foundation/Higher, grades 4–6) — (4 marks)

A jumper costs £40. In a sale its price is reduced by $25\%$ . After the sale, the reduced price is increased by $25\%$ . Work out the final price, and state whether it is more than, less than, or equal to £40.

Number: Exam Practice

Number: Exam Practice

Exam-Style Questions

Question 1 (Foundation, grades 1–3) — (2 marks)

Question 2 (Foundation, grades 2–4) — (3 marks)

Question 3 (Foundation, grades 3–4) — (3 marks)

Question 4 (Foundation, grades 3–5) — (3 marks)

Question 5 (Foundation/Higher, grades 4–5) — (3 marks)

Question 6 (Higher [H], grades 4–6) — (2 marks)

Question 7 (Higher [H], grades 5–7) — (2 marks)

Question 8 (Higher [H], grades 5–7) — (3 marks)

Question 9 (Higher [H], grades 6–8) — (3 marks)

Question 10 (Higher [H], grades 6–8) — (4 marks)

Question 11 (Foundation/Higher, grades 4–6) — (4 marks)

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