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 minibus can carry 16 passengers. A school is taking 179 pupils on a trip. By rounding 179 to the nearest 10, work out an estimate for the number of minibuses needed, given that the school cannot hire part of a minibus.
- 2.Work out 5/6 × 2/9, giving your answer in its simplest form.
- 3.A company sells 4 × 10³ items per day, each priced at £2.50. It operates for 3 × 10² days a year. Work out the company's total revenue for the year. Give your answer in standard form.
- 4.Simplify (3a²)² × (2a)³
- 5.A supermarket sells apples at £1.85 per kg. Anna buys 3.6 kg of apples. Estimate the cost by rounding each number to 1 significant figure before multiplying. Work out Anna's estimate.
- 6.Which of these numbers rounds to 0.048 when rounded to 2 significant figures?
- 7.Work out 3 + 4 × (−2).
- 8.Work out the value of 6² − 4³.
- 9.A set of kitchen scales displays the mass of a bag of sugar as 0.63 recurring kilograms, meaning 0.636363... kg with the block '63' repeating forever. Convert this mass to a fraction of a kilogram, then work out the mass in grams, giving your answer to the nearest gram.
- 10.A measuring jug shows a volume of 340 ml, correct to the nearest 20 ml. Work out the smallest possible volume in the jug.
- 11.A lorry travels 105 miles in 1 hour 45 minutes. Work out its average speed in miles per hour.
- 12.A 3-character PIN starts with one letter chosen from A, B, C, D and E, followed by two different digits chosen from 1 to 9 (no digit may be used twice in the same PIN). Work out how many different PINs are possible.
- 13.Work out (4 × 10⁻³) × (2 × 10⁵). Give your answer in standard form.
- 14.Work out (4 × 10³) × (3 × 10¹). Give your answer in standard form.
- 15.Insert one pair of brackets into 2 + 3 × 5 − 1 so that the calculation is equal to 24. Which calculation is correct?
Answer key
- (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.
- (b) 5/27 — Method: multiply the numerators together and the denominators together, then simplify. Working: (5 × 2)/(6 × 9) = 10/54 = 5/27. Answer: 5/27. 7/15 comes from adding the fractions instead of multiplying: (5+2)/(6+9) = 7/15. 15/4 comes from flipping the second fraction, as if dividing: (5 × 9)/(6 × 2) = 45/12 = 15/4. 5/3 comes from cancelling the two denominators against each other, dividing both 6 and 9 by 3 to leave 5/2 × 2/3 = 10/6 = 5/3; cancelling is only valid between a numerator and a denominator, never between two denominators.
- (c) 3 × 10⁶ — Daily revenue = (4 × 10³) × 2.5 = 1 × 10⁴ (£10,000). Multiplying by the number of days, (3 × 10²), gives annual revenue = (1 × 10⁴) × (3 × 10²) = 3 × 10⁶ (£3,000,000). A candidate who forgot to multiply by the price and just multiplied the number of items by the number of days worked out (4 × 10³) × (3 × 10²) = 1.2 × 10⁶. A candidate who added the number of days to the daily revenue instead of multiplying worked out 1 × 10⁴ + 3 × 10² = 1.03 × 10⁴. A candidate who misread £2.50 as £25 worked out a daily revenue of (4 × 10³) × 25 = 1 × 10⁵, giving an annual total of (1 × 10⁵) × (3 × 10²) = 3 × 10⁷.
- (d) 72a⁷ — Method: a power outside brackets applies to every factor inside them, and multiplying two powers of the same letter adds their indices. Working: (3a²)² = 3² × a⁴ = 9a⁴, and (2a)³ = 2³ × a³ = 8a³. Multiplying the two results gives 9 × 8 = 72 for the number and 4 + 3 = 7 for the index of a. Answer: 72a⁷. The distractors: 36a⁷ comes from squaring the 2 in (2a)³ instead of cubing it, giving 4a³ and then 9 × 4; 72a¹² comes from multiplying the indices 4 and 3 when the two terms are multiplied, instead of adding them; 17a⁷ comes from adding the coefficients 9 and 8 rather than multiplying them.
- (b) £8 — Rounding to 1 significant figure: £1.85 rounds to £2, and 3.6 kg rounds to 4 kg. The estimate is £2 × 4 = £8. A candidate who used the unrounded values instead of estimating worked out 1.85 × 3.6 = £6.66. A candidate who rounded only the mass and used the exact price worked out 1.85 × 4 = £7.40. A candidate who rounded the price to the nearest 10p instead of 1 significant figure worked out 1.9 × 4 = £7.60.
- (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.
- (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) −28 — 6² = 36 and 4³ = 64. Work out 36 − 64 = −28. A candidate who subtracts in the wrong order gets 64 − 36 = 28. A candidate who adds instead of subtracting gets 36 + 64 = 100. A candidate who multiplies the base by the exponent instead of raising the power (6 × 2 − 4 × 3 = 12 − 12) gets 0.
- (d) 636 g — Let x = 0.636363... . Since two digits repeat, multiply by 100: 100x = 63.636363... . Subtracting removes the recurring part exactly: 100x − x = 63.636363... − 0.636363... = 63, so 99x = 63, giving x = 63/99 = 7/11 kg. Converting to grams: 7/11 × 1000 = 7000/11 = 636.3636... g, which rounds to 636 g. Treating the decimal as if it terminated, writing 0.63 as 63/100 kg, gives 630 g when multiplied by 1000 — this drops the recurring part entirely. Subtracting 10x instead of x, using 100x − 10x = 90x = 63, is the wrong power of ten for a two-digit block, giving x = 63/90 = 7/10 kg, which is 700 g. A numerator slip in the subtraction, 63 − 1 = 62 instead of 63, gives x = 62/99 kg, which is 62000/99 = 626.26... g, rounding to 626 g.
- (a) 330 ml — Correct to the nearest 20 ml means the true volume could be up to 10 ml (half of 20) either side of 340 ml. The smallest possible volume is 340 − 10 = 330 ml. A candidate who subtracted the full 20 ml instead of half of it worked out 340 − 20 = 320 ml. A candidate who added instead of subtracted, finding the largest possible volume instead of the smallest, worked out 340 + 10 = 350 ml. A candidate who halved the interval again by mistake, using 5 ml instead of 10 ml, worked out 340 − 5 = 335 ml.
- (d) 60 mph — Average speed = distance ÷ time, with the time measured in hours. 45 minutes is 45/60 of an hour, which is 0.75 of an hour, so the journey takes 1.75 hours. Speed = 105 ÷ 1.75 = 60 mph. 52.5 mph rounds the time up to 2 hours, 72.4 mph writes 1 hour 45 minutes as 1.45 hours, and 183.75 mph multiplies the distance by the time instead of dividing.
- (c) 360 — There are 5 choices for the letter. The first digit can be any of the 9 digits from 1 to 9, giving 9 choices, and the second digit must differ from the first, leaving 8 choices. By the product rule, the number of PINs is 5 × 9 × 8 = 360. Allowing the second digit to repeat the first, ignoring the 'no digit twice' rule, gives 5 × 9 × 9 = 405. Adding the numbers of choices instead of multiplying them, 5 + 9 + 8, gives 22. Treating the pair of digits as an unordered choice, rather than as a first digit followed by a second digit in a fixed order, halves the digit count: 5 × (9 × 8 ÷ 2) = 180.
- (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.
- (d) 1.2 × 10⁵ — 4 × 3 = 12, and 3 + 1 = 4, giving 12 × 10⁴ — but 12 is not between 1 and 10, so this must be rewritten as 1.2 × 10⁵. Stopping at 12 × 10⁴ without rewriting it leaves the coefficient out of range. Rewriting 12 as 1.2 but leaving the exponent at 4 instead of increasing it to 5 gives 1.2 × 10⁴, which is ten times too small. Adding the coefficients instead of multiplying them gives 4 + 3 = 7, so 7 × 10⁴.
- (b) (2 + 3) × 5 − 1 — 2 + 3 = 5, then 5 × 5 = 25, then 25 − 1 = 24, so the brackets belong around 2 + 3. Placing them around 5 − 1 instead gives 5 − 1 = 4, then 3 × 4 = 12, then 2 + 12 = 14. Leaving the multiplication bracketed instead changes nothing, because it already had priority: 3 × 5 = 15, then 2 + 15 = 17, then 17 − 1 = 16. Bracketing both 2 + 3 and 5 − 1 uses two pairs instead of the one asked for: 2 + 3 = 5, 5 − 1 = 4, then 5 × 4 = 20.
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