Printable · GCSE Higher · ages 14-16
Number worksheet — GCSE Higher
Fifteen questions across the number statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Answer key: Number worksheet — GCSE Higher
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- (c) 0.0479 — Method: round each option to 2 significant figures and check which one gives 0.048. Working: for 0.0479, the first two significant figures are 4 and 7; the next digit is 9, so 7 rounds up to 8, giving 0.048. For 0.0485, the first two significant figures are 4 and 8; the next digit is 5, so 8 rounds up to 9, giving 0.049, not 0.048. 0.052 already has exactly 2 significant figures, 5 and 2, so it stays as 0.052 and does not round to 0.048 at all. 0.04 has only 1 significant figure, so it is already less precise than the 2 significant figures asked for. Answer: 0.0479.
- (d) 11 mm — Method: for a square, the side length is the square root of the area. Working: 11 × 11 = 121, so the side length is 11 mm. 60.5 mm comes from working out 121 ÷ 2 = 60.5, halving the area instead of finding its square root. 242 mm comes from working out 121 × 2 = 242, doubling the area instead of finding its square root. 22 mm comes from working out 11 × 2 = 22, doubling the correct side length. Answer: 11 mm.
- (c) (12 + 3π) cm — The perimeter of a quarter-circle is made up of two straight radii plus a quarter of the circumference. The two radii give 2 × 6 = 12 cm, and a quarter of the circumference is (1/4) × 2 × π × 6 = 3π cm, so the total perimeter is (12 + 3π) cm. Giving only the curved part, 3π cm, forgets the two straight edges entirely. Using the full circumference, 2 × π × 6 = 12π, instead of a quarter of it gives (12 + 12π) cm. Including only one radius instead of two gives (6 + 3π) cm.
- (a) 14 — Multiply both numbers by 10 to clear the decimals: 8.4 becomes 84 and 0.6 becomes 6. Then divide: 84 ÷ 6 = 14, so 14 complete pieces can be cut. Scaling only the divisor by 10 and leaving the dividend as 8.4 gives 8.4 ÷ 6 = 1.4, which rounds down to 1 complete piece — the dividend was never converted. Scaling only the dividend by 10 and leaving the divisor as 0.6 gives 84 ÷ 0.6 = 140. Rounding the divisor from 0.6 to 0.7 before dividing, trading accuracy for a rounder number, gives 8.4 ÷ 0.7 = 12. So 14 complete pieces of ribbon can be cut.
- (a) 5 — 16 + 9 = 25, then √25 = 5. Splitting the root over the addition instead gives √16 = 4 and √9 = 3, then 4 + 3 = 7 — but a root does not split over a sum like this. Multiplying those two roots instead of adding them gives 4 × 3 = 12. Taking the negative square root instead of the positive one gives −5.
- (d) £4.80 — Method: find the error interval, then check which value falls outside it. Working: half of 20p is 10p, so the actual cost, c, satisfies £4.50 ≤ c < £4.70. £4.80 is above £4.70, so it could not be the actual cost. Answer: £4.80. (£4.50 is a genuine possible cost — it sits at the included lower boundary. £4.65 is a genuine possible cost, below the £4.70 upper boundary. £4.55 is a genuine possible cost, well inside the interval.)
- (c) −5 — Method: the bracket is worked out first, then the multiplication and the division, which stand as separate parts, and the subtraction that joins them is carried out last; subtracting a negative is the same as adding. Working: (−3) + 5 = 2, so the product is 2 × (−4) = −8; the division gives (−6) ÷ 2 = −3; joining them gives −8 − (−3) = −8 + 3 = −5. Answer: −5. The distractors: −11 comes from taking away 3 instead of taking away −3, giving −8 − 3 = −11; −1 comes from working from left to right once the bracket is done, giving −8 − (−6) = −2 and then −2 ÷ 2 = −1; 11 comes from treating the first product as positive because it was worked out from a bracket, giving 8 − (−3) = 11.
- (a) 5 — Method: each index is worked out first, and subtracting a negative number is the same as adding the positive. Working: (−2)³ = (−2) × (−2) × (−2) = −8 and (−3)² = (−3) × (−3) = 9, while − (−4) becomes + 4, so the calculation becomes −8 + 9 + 4 = 5. Answer: 5. The distractors: −13 comes from taking (−3)² as −9, giving −8 − 9 + 4 = −13; −3 comes from reading − (−4) as − 4, giving −8 + 9 − 4 = −3; 21 comes from treating every power of a negative number as positive, so that (−2)³ is taken as 8 and the calculation becomes 8 + 9 + 4 = 21.
- (b) −5 — Using the order of operations, work out the multiplication first: 4 × (−2) = −8. Then 3 + (−8) = −5. A candidate who adds before multiplying gets (3 + 4) × (−2) = −14. A candidate who drops the negative sign on the multiplication gets 3 + 4 × 2 = 11. A candidate who works out the multiplication correctly but gives that as the final answer, forgetting to combine it with the 3, gets −8.
- (c) 12 — Method: division and multiplication have equal priority, so they are carried out in the order they are written, from left to right; a negative divided by a negative is positive. Working: (−36) ÷ (−6) = 6, and then 6 × 2 = 12. Answer: 12. The distractors: 3 comes from carrying out the multiplication first, (−6) × 2 = −12 followed by (−36) ÷ (−12) = 3; −12 comes from treating a negative divided by a negative as negative, giving −6 and then −6 × 2 = −12; 6 comes from stopping at the division and never carrying out the multiplication by 2.
- (c) 5 × 10⁻³ — Method: in standard form the coefficient must be at least 1 and less than 10, and for a number smaller than 1 the index is negative and counts the places the decimal point moves to the right. Working: the only significant digit is 5, so the coefficient is 5; moving the decimal point in 0.005 three places to the right gives 5, so the index is −3. Answer: 5 × 10⁻³. The distractors: 0.5 × 10⁻² is the same value written the wrong way, because 0.5 is smaller than 1 and so is not an allowed coefficient; 5 × 10⁻² comes from counting the two zeros after the decimal point instead of the three places the point moves; 5 × 10³ comes from taking the index as positive, which describes a number in the thousands rather than one smaller than 1.
- (d) 2³ × 5² — Method: divide repeatedly by the smallest prime number, then write any repeated prime using a power. Working: 200 ÷ 2 = 100, 100 ÷ 2 = 50, 50 ÷ 2 = 25, 25 ÷ 5 = 5, and 5 is prime, so 200 = 2 × 2 × 2 × 5 × 5, written as 2³ × 5². 2² × 5³ swaps the two powers, giving 4 × 125 = 500, not 200. 2³ × 5 leaves out one of the two 5s, giving 8 × 5 = 40, not 200. 2 × 5³ leaves out two of the three 2s, giving 2 × 125 = 250, not 200. Answer: 2³ × 5².
- (c) Sam is correct — The leading zeros in 0.070268 are not significant, so the first three significant figures are 7, 0 and 2. The next digit along is 6, and since 6 is 5 or more, the third significant figure rounds up from 2 to 3, giving 0.0703. This means Sam's answer is correct. Writing 0.070 keeps only 2 significant figures, one short of what was asked. Writing 0.0702 ignores the digit 6 that follows and leaves the third figure unrounded. Writing 0.0704 rounds the third figure up twice, as if a later digit had also pushed it up.
- (b) 4.05 — Method: line up the decimal points (or place value columns) before adding. Working: 3.60 + 0.45 = 4.05. Answer: 4.05. 3.65 is Priya's answer, from adding the digits without lining up the place value columns, which effectively treats 0.45 as 0.05. 4.5 comes from rounding both numbers up first, 3.6 to 4 and 0.45 to 0.5, and adding those instead of adding the exact values. 0.81 comes from adding the digits 36 and 45 together to get 81, then placing the decimal point in the wrong position.
- (a) The tape can only give the length to the nearest centimetre — Method: a measurement should never be written to a finer degree of accuracy than the instrument used can read. Working: the tape is marked in centimetres, so the smallest division Leah can read is 1 cm, which is 0.01 m and two decimal places in metres; writing 7.3157 m claims the length to the nearest tenth of a millimetre, four decimal places, which the markings cannot support. A record of 7.32 m, to the nearest centimetre, is what this tape justifies. Answer: The tape can only give the length to the nearest centimetre. The distractors: the nearest millimetre contradicts the markings described in the question, which are centimetres, and would still claim more accuracy than the tape offers; the rule that a length in metres must be written to 2 decimal places borrows the habit of writing money to the penny, when the accuracy of a length depends on the instrument; rounding to the nearest metre would throw away accuracy the tape genuinely provides.
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