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.
Number worksheet — GCSE Higher
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- 1.A restaurant adds a service charge of 15% to a bill of £120. Work out the service charge.
- 2.Work out 6² − 4².
- 3.A café buys 18 boxes of teabags at £3.45 each, and sells all the teabags for £108 in total. Work out the café's profit.
- 4.Work out (−36) ÷ (−6) × 2
- 5.Work out (4 × 10⁻³) × (2 × 10⁵). Give your answer in standard form.
- 6.Write 5/6 as a decimal, showing clearly which digit recurs.
- 7.Work out (−6) + (−4) × 3
- 8.Work out an estimate for 588 ÷ 31, by rounding each number to 1 significant figure.
- 9.Write these three numbers in order, starting with the smallest: 3/5, 0.55, 58%
- 10.The error interval for the mass of a suitcase, m kg, is given as 22.5 ≤ m < 23.5. Write down the mass of the suitcase, correct to the nearest whole number.
- 11.Write these three numbers in order, starting with the smallest: 7/20, 0.3, 32%
- 12.During one night on a mountain, four hikers each recorded the temperature they felt: Ben −6 °C, Priya −2 °C, Sam −9 °C, Alex 1 °C. Work out the difference between the coldest and the warmest of these four temperatures.
- 13.Work out how many factors 100 has.
- 14.A car travels 100 km, correct to the nearest km, in a time of 2 hours, correct to the nearest 0.1 hour. Work out the average speed, in km/h, to the greatest degree of accuracy the bounds can guarantee.
- 15.A garden path is measured by Jon as 12 m, correct to the nearest metre, and by Mia as 12.6 m, correct to the nearest 0.1 m. Which statement about the two measurements is correct?
Answer key
- (a) £18 — Method: the percentage acts as an operator on the bill, and 15% can be built from 10% and 5%, where 5% is half of 10%. Working: 10% of £120 is £120 ÷ 10 = £12, and 5% is half of that, £6, so the charge is £12 + £6 = £18. Answer: £18. The distractors: £8 comes from reading 15% as one fifteenth and working out £120 ÷ 15 = £8; £12 comes from finding 10% of the bill and stopping there; £138 comes from finding the charge correctly and then giving the new total, £120 + £18, rather than the charge the question asks for.
- (c) 20 — Method: work out each power separately before subtracting. Working: 6² = 36 and 4² = 16, so 6² − 4² = 36 − 16 = 20. Answer: 20. (4 comes from subtracting first, 6 − 4 = 2, and then squaring that result, instead of squaring each number first. 52 comes from adding the two squares, 36 + 16, instead of subtracting them. 2 comes from subtracting the two numbers, 6 − 4, and forgetting to square at all.)
- (b) £45.90 — The cost is 18 × £3.45 = £62.10. Profit = £108 − £62.10 = £45.90. A candidate who does not borrow in the tenths column, doing 1 − 0 = 1 instead of borrowing to make 10 − 1 = 9 and so leaving the units as 8 − 2 = 6, gets £46.10. A candidate who adds the cost to the selling price instead of subtracting gets £108 + £62.10 = £170.10. A candidate who gives the cost instead of the profit gets £62.10.
- (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.
- (a) 8 × 10² — Method: to multiply numbers written in standard form, multiply the coefficients and add the indices. Working: 4 × 2 = 8 for the coefficients, and −3 + 5 = 2 for the indices; 8 already lies between 1 and 10, so no adjustment is needed. Answer: 8 × 10². The distractors: 6 × 10² comes from adding the coefficients, 4 + 2, instead of multiplying them; 8 × 10⁸ comes from ignoring the minus sign and adding 3 + 5; 8 × 10⁻¹⁵ comes from multiplying the indices, −3 × 5, instead of adding them.
- (b) 0.83333... — Divide 5 by 6 using long division. 5.000... ÷ 6: 50 ÷ 6 = 8 remainder 2, giving the first decimal digit 8. Bring down a 0 to make 20, and 20 ÷ 6 = 3 remainder 2 — the remainder 2 has reappeared, so from here the digit 3 repeats forever. This gives 5/6 = 0.83333... . Stopping after two decimal places and writing 0.83 treats the division as if it terminated, when the remainder never reaches zero. Shifting the decimal point one place too far to the left gives 0.083333..., the same digits divided by an extra power of ten. A slip in the long division itself, misreading a remainder, can produce the wrong repeating digit, 0.85555... .
- (b) −18 — Method: the multiplication is carried out before the addition, and a negative multiplied by a positive is negative. Working: (−4) × 3 = −12, so the calculation becomes (−6) + (−12) = −18. Answer: −18. The distractors: −30 comes from adding first and multiplying afterwards, giving (−6 + −4) × 3 = −10 × 3 = −30; 6 comes from treating (−4) × 3 as +12 on the grounds that a minus sign makes a product positive, giving −6 + 12 = 6; 18 comes from ignoring both minus signs and working out 6 + 4 × 3 = 18.
- (a) 20 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 588 rounds to 600 (1 s.f.) and 31 rounds to 30 (1 s.f.). 600 ÷ 30 = 20. Answer: 20. 17 comes from cutting 588 down to 500, keeping the leading digit as it stands instead of rounding it up to 1 significant figure, 600, then dividing by the correctly rounded 30. 200 comes from misreading the rounded divisor 30 as 3, giving 600 ÷ 3 instead of 600 ÷ 30. 19 is the exact value of 588 ÷ 31 rounded to the nearest whole number, found without rounding the numbers first.
- (a) 0.55, 58%, 3/5 — Converting all three to decimals: 3/5 = 0.6, 0.55 stays as 0.55, and 58% = 0.58. In order from smallest to largest, this is 0.55, then 58%, then 3/5. Writing the numbers in the reverse order, largest to smallest, gives 3/5, 58%, 0.55. Misconverting 3/5 as 0.5 instead of 0.6 makes it appear smaller than both other values, giving the order 3/5, 0.55, 58%. Misconverting 58% as 0.058 instead of 0.58, by moving the decimal point two extra places, makes it appear smallest of the three, giving the order 58%, 0.55, 3/5.
- (b) 23 kg — Method: the rounded value sits exactly in the middle of the error interval. Working: the interval 22.5 ≤ m < 23.5 stretches 0.5 either side of the rounded value, so the rounded value is 23. Answer: 23 kg. (22 kg comes from rounding the lower bound down instead of finding the middle of the interval. 22.5 kg comes from giving the lower bound itself rather than the rounded value. 23.5 kg comes from giving the upper bound itself rather than the rounded value.)
- (b) 0.3, 32%, 7/20 — Method: convert every number to a decimal so they can be compared on the same scale. Working: 7/20 = 0.35, 0.3 stays as 0.3, and 32% = 0.32. Comparing 0.3, 0.32 and 0.35 in size gives the order 0.3, then 0.32, then 0.35. Answer: 0.3, 32%, 7/20. 7/20, 32%, 0.3 lists the numbers from largest to smallest instead of smallest to largest. 0.3, 7/20, 32% swaps 32% and 7/20, treating the fraction 7/20 as smaller even though 7/20 = 0.35 is bigger than 32% = 0.32. 32%, 0.3, 7/20 comes from moving the digits one place too far when converting the percentage, giving 0.032 instead of 0.32, which makes 32% look far smaller than it really is.
- (a) 10 °C — Method: work out the coldest and warmest of the four temperatures, then subtract to find the difference. Working: the coldest temperature is Sam's, −9 °C, and the warmest is Alex's, 1 °C. The difference is 1 − (−9) = 1 + 9 = 10. Answer: 10 °C. 8 °C comes from working out 1 − 9 = −8 and reporting 8, dropping the negative sign on −9 instead of turning the subtraction into an addition. 3 °C comes from comparing the wrong pair, Sam's −9 °C and Ben's −6 °C, instead of the coldest and the warmest: −6 − (−9) = 3. 7 °C comes from comparing Ben's −6 °C with Alex's 1 °C, mistakenly treating Ben's reading as the coldest instead of Sam's.
- (c) 9 — Method: factors come in pairs that multiply to give the number, so work through the pairs in order; a factor paired with itself is counted only once. Working: the pairs are 1 × 100, 2 × 50, 4 × 25, 5 × 20 and 10 × 10. The first four pairs give eight different factors, and the last pair adds only one more, so the factors are 1, 2, 4, 5, 10, 20, 25, 50 and 100. Answer: 9. The distractors: 10 comes from counting the pair 10 × 10 as two separate factors; 8 comes from leaving 1 out of the list, on the view that 1 is not a proper factor; 4 comes from writing 100 = 2² × 5² and multiplying the two indices together instead of adding 1 to each index first.
- (d) 50 km/h — The error intervals are 99.5 ≤ distance < 100.5 and 1.95 ≤ time < 2.05. The minimum speed is 99.5 ÷ 2.05 ≈ 48.54 km/h, and the maximum speed is 100.5 ÷ 1.95 ≈ 51.54 km/h. These two bounds round to different whole numbers, 49 and 52, so the speed cannot be guaranteed to the nearest whole number — but every value between them rounds to 50 at the nearest 10, so 50 km/h is the value that can safely be guaranteed. Quoting 49 km/h uses only the minimum bound's rounding, without checking that the maximum bound rounds to something different. Quoting 52 km/h makes the same mistake using only the maximum bound instead. Quoting 48.54 km/h states one bound to the full accuracy a calculator shows, as if the smallest possible speed were the answer, when the true speed could be anything up to 51.54 km/h.
- (a) They cannot both be describing the same path — Jon's measurement means the true length, l, satisfies 11.5 m ≤ l < 12.5 m. Mia's measurement means the true length satisfies 12.55 m ≤ l < 12.65 m. These two ranges do not overlap, so the two measurements cannot both be describing the same path. 'They must both be describing the same path' ignores that the two ranges do not overlap at all. 'Jon's measurement must be wrong' wrongly assumes Jon is the one at fault, when the mismatch does not show which measurement, if either, is wrong. 'Mia's measurement must be wrong' makes the same unjustified assumption in the other direction.
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