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.
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Answer key: Number worksheet — GCSE Higher
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- (b) £300 — The increased price is 110% of the original, so the original price = £330 ÷ 1.1 = £300. A candidate who finds 10% of £330 and subtracts it, wrongly treating £330 as the original, gets £330 − £33 = £297. A candidate who adds 10% of £330 again instead of reversing the increase gets £330 + £33 = £363. A candidate who divides by 0.1 instead of 1.1 gets £3,300.
- (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.
- (d) 64 — Method: each time the power increases by 1, the value doubles, since one more 2 is multiplied in. Working: 2⁶ = 2⁵ × 2 = 32 × 2 = 64. Answer: 64. (37 comes from adding the power to the value, 32 + 5, instead of multiplying by the base. 192 comes from multiplying 32 by the new power, 6, instead of by the base, 2. 34 comes from adding the base to the value, 32 + 2, instead of multiplying.)
- (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.)
- (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.)
- (a) 200 g — One quarter of 160 g is 40 g. Increasing the amount means adding this on: 160 + 40 = 200 g. Finding the increase, 1/4 of 160 = 40 g, but stopping there without adding it to the original amount leaves just 40 g. Using 4/5 instead of 5/4 as the scaling fraction, 160 × 4/5 = 128 g, actually decreases the amount rather than increasing it. Increasing by a half instead of a quarter, 160 + 80 = 240 g, uses the wrong fraction of 160.
- (d) £6 — First find 30% of £200, which is £60, then find 10% of that: £60 × 0.1 = £6. Adding the two percentages together instead of applying them one after the other, 10% + 30% = 40%, and finding 40% of £200 gives £80. Finding 30% of £200 = £60 correctly but stopping before applying the second percentage leaves £60 as the final answer. Finding only 10% of the original £200, ignoring the 30% entirely, gives £20.
- (b) 30 — Method: 12% of an amount is 12/100 of it, so find 1% by dividing by 100 and then multiply by 12. Working: 1% of 250 is 250 ÷ 100 = 2.5, and 12% is 2.5 × 12 = 30. Answer: 30 seats. The distractors: 3 comes from writing 12% as 0.012 instead of 0.12, giving 0.012 × 250 = 3; 25 comes from finding 10% of the seats and stopping there; 24 comes from counting 12 seats for each whole hundred, 12 + 12 = 24, and ignoring the remaining 50 seats.
- (b) 3 — Method: list all valid two-digit numbers that can be made without starting with 0, then keep only the ones that are multiples of 5. Working: the two-digit numbers possible are 30, 35, 50 and 53. A number is a multiple of 5 only if it ends in 0 or 5: 30 ends in 0, 35 ends in 5, 50 ends in 0, but 53 ends in 3. So there are 3 multiples of 5. Answer: 3. 4 comes from including 53 as a multiple of 5 without checking that its last digit is not 0 or 5. 2 comes from leaving out 50, wrongly assuming 0 cannot be used as the second digit either. 6 comes from listing every two-digit arrangement of the three digits, including ones that start with 0, without applying either restriction.
- (d) £50.00 — The sale price is 85% of the original, so the original price = £42.50 ÷ 0.85 = £50.00. 15% of £42.50 is £6.375. A candidate who finds 15% of £42.50 and subtracts it from the sale price gets £42.50 − £6.375 = £36.125, which is £36.13 to the nearest penny. A candidate who adds 15% of £42.50 instead of reversing the decrease gets £42.50 + £6.375 = £48.875, which is £48.88 to the nearest penny. A candidate who divides by 0.15 instead of 0.85 gets £283.33.
- (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) 1/2, 3/5, 2/3, 5/6 — Convert all four fractions to a common denominator of 30: 2/3 is 20/30, 3/5 is 18/30, 5/6 is 25/30, and 1/2 is 15/30. Ordering by these numerators, smallest to largest, gives 15/30, 18/30, 20/30, 25/30, which is 1/2, 3/5, 2/3, 5/6. Ordering by the size of the numerator in the original fractions, 1, 2, 3, 5, rather than converting to a common denominator, gives the wrong order 1/2, 2/3, 3/5, 5/6, because it ignores that the denominators are different. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 5/6, 2/3, 3/5, 1/2. Using the rule "the bigger the denominator, the smaller the fraction" to place the last two, so that 5/6 is put below 2/3 because 6 is bigger than 3, gives 1/2, 3/5, 5/6, 2/3 — that rule only holds when the numerators are the same, and here 20/30 really is smaller than 25/30. So the correct order, smallest to largest, is 1/2, 3/5, 2/3, 5/6.
- (a) £4.80 — The total cost of the melons is 8 × £1.35 = £10.80. Profit is the selling total minus the cost, so £15.60 − £10.80 = £4.80. Subtracting the price of just one melon from the selling total instead of the total cost of all eight, £15.60 − £1.35, gives £14.25. Miscalculating the cost of the melons as 8 × £1.30 = £10.40, dropping the 5p from each price, gives a profit of £15.60 − £10.40 = £5.20. Adding the selling total and the cost together instead of subtracting them, £15.60 + £10.80, gives £26.40.
- (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.
- (c) 12.44 — The error intervals are 155.5 ≤ mass < 156.5 and 11.5 ≤ volume < 12.5. To make a quotient as small as possible, use the SMALLEST possible numerator together with the LARGEST possible denominator: 155.5 ÷ 12.5 = 12.44 g/cm³. Using the lower bound for both mass and volume, 155.5 ÷ 11.5 ≈ 13.52, forgets that dividing by a smaller number makes the result bigger, not smaller — that pairing does not give a minimum at all. Dividing the two given rounded values directly, 156 ÷ 12 = 13, ignores that both measurements have their own error interval. Using the upper bound of mass with the upper bound of volume, 156.5 ÷ 12.5 = 12.52, takes both bounds the same way round; it is neither the minimum nor the maximum, since the maximum needs the largest mass with the smallest volume, 156.5 ÷ 11.5 ≈ 13.61.
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