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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- (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) 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.)
- (a) 8,350 — Rounding to the nearest 100 means the true number can be up to half of 100, which is 50, below the recorded figure before it would round down to a lower hundred. The smallest possible number is therefore 8,400 − 50 = 8,350. Adding 50 instead of subtracting it gives 8,450, which is the upper end of the interval rather than the smallest value, and 8,450 is not itself possible because it would round up to 8,500. Subtracting a whole 100 instead of half of it gives 8,300, going too far below the recorded value. Subtracting 10 instead of half of the rounding unit gives 8,390, treating the rounding unit as 100 but the tolerance as only 10.
- (a) 7.5 × 10⁴ — Standard form is A × 10ⁿ, where A is between 1 and 10 (1 ≤ A < 10) and n is an integer — 7.5 × 10⁴ satisfies all of this. 12 × 10³ fails because 12 is not below 10. 0.5 × 10⁴ fails because 0.5 is not at least 1. 6.2 × 4⁵ fails because standard form always uses a power of 10, not a power of 4.
- (c) 36 — Method: set the outcomes out in a grid with one die along the top and the other down the side, so that every cell of the grid is one outcome, and count the cells. Working: the red die can land in 6 ways, so the grid has 6 columns, and the blue die can also land in 6 ways, so the grid has 6 rows; the number of cells is 6 × 6 = 36. Answer: 36. The distractors: 12 comes from adding 6 and 6 instead of multiplying them; 6 comes from counting the outcomes of a single die and forgetting that the second die also has to land; 21 comes from treating the two dice as indistinguishable, so that a red 2 with a blue 3 and a red 3 with a blue 2 are counted as one outcome.
- (d) 9 — Method: split into two cases — the soups with no restriction, and the mushroom soup on its own — then add the totals. Working: the 2 soups other than mushroom can be paired with any of the 4 sandwiches: 2 × 4 = 8. The mushroom soup can only be paired with the cheese sandwich: 1 combination. Total = 8 + 1 = 9. Answer: 9. 12 comes from working out 3 × 4 = 12 without applying the restriction at all. 8 comes from correctly finding the 2 unrestricted soups' 8 combinations, but forgetting to add back the 1 allowed mushroom-and-cheese combination. 11 comes from taking the unrestricted total of 12 and removing only 1 mushroom combination instead of all 3 disallowed ones.
- (a) 1,500 ≤ m < 2,500 — Method: a four-digit figure written to 1 significant figure has been rounded to the nearest 1,000, so the mass lies within half of 1,000, that is 500, of the figure given. Working: 2,000 − 500 = 1,500 and 2,000 + 500 = 2,500. The lower limit is included, because 1,500 kg rounds up to 2,000 kg to 1 significant figure, while 2,500 kg rounds up to 3,000 kg, so the upper limit is not. Answer: 1,500 ≤ m < 2,500. The distractors: 1,950 ≤ m < 2,050 comes from rounding to the nearest 100 instead of to 1 significant figure; 1,000 ≤ m < 3,000 goes a whole 1,000 either side instead of half of it; 1,500 < m ≤ 2,500 has the two limits the wrong way round.
- (a) 33.5 mph — Method: the smallest possible actual value is half the rounding unit below the given value. Working: half of 1 mph is 0.5 mph, so the smallest possible speed is 34 − 0.5 = 33.5 mph. Answer: 33.5 mph. (33 mph comes from subtracting the whole rounding unit, 1, instead of half of it. 34 mph comes from giving the rounded value itself rather than the lower bound. 34.5 mph comes from adding the half unit instead of subtracting it, giving the upper bound.)
- (d) 1050 ml — Convert both amounts to millilitres: 1.8 litres = 1800 ml and 3/4 litre = 750 ml. Subtracting gives 1800 − 750 = 1050 ml. Confusing 3/4 with 75% and converting it as 75 ml instead of 750 ml gives 1800 − 75 = 1725 ml. Adding the amount poured out instead of subtracting it gives 1800 + 750 = 2550 ml. Misreading 1.8 litres as 0.8 litres, losing the whole litre, gives 800 − 750 = 50 ml.
- (b) 6.4 × 10⁸ — There are 5 letter positions, each with 23 choices, and 2 digit positions, each with 10 choices, and every position is independent because repeats are allowed. By the product rule, the total is 23⁵ × 10² = 6,436,343 × 100 = 643,634,300, which is 6.4 × 10⁸ to 2 significant figures. Using all 26 letters instead of the 23 that are actually allowed, ignoring the excluded letters entirely, gives 26⁵ × 10² = 1,188,137,600, which is 1.2 × 10⁹ to 2 significant figures. Adding the seven counts of choices instead of multiplying them, 23 + 23 + 10 + 10 + 23 + 23 + 23, gives 135, which is 1.4 × 10² to 2 significant figures — a total far too small for seven independent positions. Swapping which count of choices belongs to letters and which belongs to digits, working out 23² × 10⁵ instead of 23⁵ × 10², gives 52,900,000, which is 5.3 × 10⁷ to 2 significant figures.
- (a) 11:20 — Method: find the flight time using time = distance ÷ speed, then add this to the departure time. Working: 2340 ÷ 780 = 3 hours; 08:20 + 3 hours = 11:20. Answer: 11:20. 08:40 comes from dividing speed by distance instead of distance by speed, giving a flight time of 1/3 hour (20 minutes) rather than 3 hours. 11:00 comes from adding the 3-hour flight time to the hour of the departure time only, 8 + 3 = 11, and losing the 20 minutes. 03:00 comes from finding the flight time correctly but giving it as a clock time on its own, forgetting to add it to the departure time.
- (a) 33.2 — The radius was rounded to 1 decimal place, so its error interval is 3.15 ≤ r < 3.25. The upper bound for the area uses the upper bound of the radius, squared: area = π × 3.25² ≈ 33.183, which rounds to 33.2 m² (3 s.f.). Using the given value of the radius directly instead of its upper bound, π × 3.2² ≈ 32.2, ignores that the radius itself has a range of possible values. Bounding the radius correctly but forgetting to square it, using area = π × 3.25 ≈ 10.2 instead of π × 3.25², drops the whole squaring step from the area formula. Using the LOWER bound of the radius instead of the upper one, π × 3.15² ≈ 31.2, finds the lower bound of the area, not the upper one.
- (c) 7 × 10⁻⁸ — '100 times smaller' means dividing by 100 = 10². Dividing 7 × 10⁻⁶ by 10² means subtracting 2 from the exponent: −6 − 2 = −8, giving 7 × 10⁻⁸. A candidate who multiplied by 100 instead of dividing added 2 to the exponent, getting 7 × 10⁻⁴. A candidate who divided by 10 instead of 100 subtracted only 1 from the exponent, getting 7 × 10⁻⁵. A candidate who did not apply the scale factor at all left the diameter as 7 × 10⁻⁶, the same as the red blood cell.
- (d) 24.69 — To round to 2 decimal places, look only at the third decimal digit to decide whether the second decimal digit rounds up. In 24.6851 the third decimal digit is 5, and since 5 is 5 or more, the second decimal digit rounds up from 8 to 9, giving 24.69. Rounding to 1 decimal place instead of 2 gives 24.7, one place value too coarse. Keeping the third decimal digit rather than dropping it gives 24.685, which is 3 decimal places. Looking at the fourth decimal digit, 1, instead of the third one, and wrongly deciding that no rounding is needed, leaves the length unrounded at 24.68.
- (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.
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