Printable · GCSE Foundation · ages 14-16
Number worksheet — GCSE Foundation
Fifteen questions across the number statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Answer key: Number worksheet — GCSE Foundation
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- (a) 4.5 × 10⁴ — 45,000 = 4.5 × 10,000 = 4.5 × 10⁴, with the coefficient between 1 and 10 as standard form requires. Writing 45 × 10³ keeps the coefficient too large — 45 is not between 1 and 10. Writing 4.5 × 10³ undercounts the places moved, giving only 4,500. Writing 4.5 × 10⁵ overcounts the places moved, giving 450,000.
- (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) £8.20 — Method: find the total cost, then subtract from the amount paid. Working: 35 × £1.48 = £51.80. Amount paid = 3 × £20 = £60.00. Change = £60.00 − £51.80 = £8.20. Answer: £8.20. (£58.52 comes from forgetting to multiply the price by the 35 litres and subtracting only £1.48 from £60. £7.50 comes from rounding £1.48 up to £1.50 before multiplying, giving a total of £52.50 instead of £51.80. £9.20 comes from miscarrying in the pence column when subtracting £51.80 from £60.00.)
- (a) 40 — Multiply the number of choices for each item: 4 × 5 × 2 = 40. 11 comes from adding the three numbers instead of multiplying them. 20 comes from multiplying only the crisps and chocolate bars, 4 × 5, and forgetting the drink. 10 comes from multiplying only the chocolate bars and drinks, 5 × 2, and forgetting the crisps.
- (b) 3.65 ≤ n < 3.75 — Rounding to 1 decimal place means n can be up to half of one decimal place, 0.05, below or above 3.7 before it would round to a different value. This gives a lower bound of 3.7 − 0.05 = 3.65 and an upper bound of 3.7 + 0.05 = 3.75. A value exactly at 3.75 would round up to 3.8, not 3.7, so the upper bound is excluded while the lower bound, 3.65, does still round to 3.7. Writing 3.65 ≤ n ≤ 3.75 wrongly includes 3.75. Writing 3.6 ≤ n < 3.8 uses a whole decimal place, 0.1, either side instead of half of one, 0.05. Writing 3.65 < n < 3.75 wrongly excludes 3.65, which does round to 3.7.
- (c) 2/3 — Method: fractions with the same denominator are added by adding the numerators and leaving the denominator alone, because the parts are already the same size. Working: 1/3 + 1/3 has numerators 1 + 1 = 2 and the denominator stays as 3, giving 2/3. Answer: 2/3. The distractors: 2/6 comes from adding the denominators as well as the numerators, 1 + 1 over 3 + 3; 2/9 comes from adding the numerators but multiplying the denominators, 1 + 1 over 3 × 3; 1/9 comes from multiplying throughout instead of adding, 1 × 1 over 3 × 3.
- (d) 1,000 — Method: round each number to 1 significant figure, then multiply the rounded values. Working: 49 rounds to 50 and 21 rounds to 20, and 50 × 20 = 1,000 because 5 × 2 = 10 and the two rounded numbers carry one zero each. Answer: 1,000. The distractors: 800 comes from rounding 49 down to 40 instead of to the nearest ten; 1,500 comes from rounding 21 up to 30 rather than down to 20; 1,029 is the exact product 49 × 21, worked out in full when the question asks for an estimate.
- (c) 24 — Method: round each number to the nearest whole number, then multiply the rounded values. Working: 6.4 rounds to 6 (nearest whole number) and 3.9 rounds to 4 (nearest whole number). 6 × 4 = 24. Answer: 24. 18 comes from rounding 3.9 down to 3 instead of up to the nearest whole number, 4, giving 6 × 3. 28 comes from rounding 6.4 up to 7 instead of down to the nearest whole number, 6, giving 7 × 4. 25 is the exact value of 6.4 × 3.9, which is 24.96, rounded to the nearest whole number after multiplying, rather than estimated by rounding first.
- (a) 10π cm — The circumference of a circle is found from C = 2 × π × r. With a radius of 5 cm, this gives C = 2 × π × 5 = 10π cm. Using the radius directly in the formula without doubling it gives 5π cm, missing the factor of 2. Using the formula for area, π × r², instead of circumference gives 25π cm, which is also the wrong units for a length. Doubling the radius to get a diameter of 10 and then applying the circumference formula a second time gives 20π cm, doubling the answer that is already correct.
- (d) 0.25 — Dividing by 1000 moves every digit three place-value columns, so 250 ÷ 1000 = 0.25. A candidate who divides by 100 instead of 1000 gets 2.5. A candidate who divides by 10,000 instead of 1000 gets 0.025. A candidate who divides by 10 instead of 1000 gets 25.
- (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.)
- (a) 0.35, 3/8, 0.4 — Method: convert the fraction to a decimal so all three values can be compared directly. Working: 3/8 = 0.375, so the three values are 0.35, 0.375 and 0.4. In order from smallest to largest: 0.35, 3/8 (0.375), 0.4. Answer: 0.35, 3/8, 0.4. "3/8, 0.35, 0.4" comes from assuming a fraction must be smaller than any decimal, without converting it first. "0.4, 3/8, 0.35" comes from writing the numbers in reverse order. "0.35, 0.4, 3/8" comes from converting 3/8 by flipping it to 8/3, making it seem larger than both decimals.
- (b) 1 — Any non-zero number raised to the power 0 equals 1, so 7⁰ = 1. Writing 0 confuses this rule with multiplying by zero. Writing 7 mistakes the power of 0 for leaving the base unchanged, as if nothing happened. Writing 1/7 confuses the zero index with a negative one — it is 7⁻¹ that equals 1/7, while 7⁰ equals 1.
- (b) 12 — Method: round the number of pupils to the nearest 10, divide by the number of passengers each minibus can carry, then round up because a part-full minibus still needs a whole vehicle. Working: 179 rounds to 180 (nearest 10); 180 ÷ 16 = 11.25; 11 minibuses only carry 176 passengers, so a 12th minibus is needed for the rest. Answer: 12. 11 comes from rounding 11.25 to the nearest whole number in the usual way, without checking that the leftover pupils still need transporting. 10 comes from rounding 179 down to 170 instead of to the nearest 10, 180. 180 comes from stopping after rounding the number of pupils, without dividing by the number of passengers each minibus carries at all.
- (d) 30 kg — Round 0.485 kg to 1 significant figure: 0.5 kg. Multiply by the 60 cakes: 0.5 × 60 = 30 kg. A candidate who rounded to 2 significant figures instead of 1 used 0.49 kg, giving 0.49 × 60 = 29.4 kg. A candidate who used the unrounded amount instead of the estimate worked out 0.485 × 60 = 29.1 kg. A candidate who rounded 0.485 down to 0.4 kg instead of up to 0.5 kg worked out 0.4 × 60 = 24 kg.
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