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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- 1.2/3 of an amount of money is £30. Work out the amount.
- 2.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.
- 3.A lorry travels 105 miles in 1 hour 45 minutes. Work out its average speed in miles per hour.
- 4.A number, y, is equal to 8.2 when rounded to 1 decimal place. Write down the error interval for y.
- 5.Work out 5⁰ + 5¹ + 5²
- 6.Find the missing number: ▢ ÷ 15 = 24
- 7.Which of these is written correctly in standard form?
- 8.Work out the value of .
- 9.Jack buys 3 books, each costing £4.25, and pays with a £20 note. Work out how much change he receives.
- 10.A crowd of 8,400 people is recorded correct to the nearest 100. Work out the smallest possible number of people in the crowd.
- 11.Work out the value of √(16 + 9)
- 12.A photography studio offers 4 backdrops and 3 outfits for a portrait session. The plain grey backdrop, one of the 4 backdrops, cannot be used with the formal suit outfit, one of the 3 outfits. Work out how many different backdrop-and-outfit combinations are possible.
- 13.A plank has length a = 12 cm and a second plank has length b = 7 cm, each correct to the nearest centimetre. Work out the upper bound of a + b.
- 14.Which one of these statements is true?
- 15.Two allotments are combined. Allotment A grows vegetables and flowers in the ratio 3 : 2, and has 20 plots in total. Allotment B grows vegetables and flowers in the ratio 1 : 4, and has 20 plots in total. Work out the fraction of all the plots, in both allotments combined, that grow vegetables.
Answer key
- (b) £45 — Since 2/3 of the amount is £30, one third is £30 ÷ 2 = £15, and the whole amount is three thirds: £15 × 3 = £45. Applying the fraction forwards to £30 instead of reversing it, £30 × 2/3 = £20, treats the given amount as the whole rather than as two thirds of it. Finding one third correctly as £15 but forgetting to multiply by 3 to get the whole amount leaves £15 as the final answer. Reading £30 as one third of the amount rather than as two thirds, and so multiplying straight by 3, gives £30 × 3 = £90.
- (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.
- (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.
- (a) 8.15 ≤ y < 8.25 — Rounding to 1 decimal place means the error interval spans half of 0.1, so 0.05, either side of 8.2: 8.2 − 0.05 = 8.15 and 8.2 + 0.05 = 8.25. The lower bound uses ≤ because 8.15 itself rounds to 8.2, but the upper bound uses < because 8.25 would round up to 8.3. So the error interval is 8.15 ≤ y < 8.25. A candidate who used the wrong rounding band gave 8.1 ≤ y < 8.2. A candidate who used a strict inequality at both ends wrote 8.15 < y < 8.25, wrongly excluding 8.15 itself. A candidate who added the full 0.1 instead of half of it wrote 8.2 ≤ y < 8.3.
- (c) 31 — Method: work out each power separately, remembering that any non-zero base raised to the power 0 is 1 and a base raised to the power 1 is itself, then add the three values. Working: 5⁰ = 1, 5¹ = 5 and 5² = 25, so the total is 1 + 5 + 25 = 31. Answer: 31. The distractors: 30 comes from taking 5⁰ as 0 instead of 1; 35 comes from taking 5⁰ as 5, treating a zero index as leaving the base unchanged; 125 comes from adding the indices first, as though the three terms were being multiplied, and working out 5³.
- (a) 360 — The inverse of ÷ 15 is × 15, so the missing number is 24 × 15 = 360. Subtracting instead of multiplying gives 24 − 15 = 9. Dividing by 15 again instead of multiplying gives 24 ÷ 15 = 1.6. Adding instead of multiplying gives 24 + 15 = 39.
- (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) 0.001 — Method: a negative index means 'one over' the positive power, so $10^{-3}$ means one over $10^{3}$. Working: ten cubed is 1000, and one over 1000 is 0.001. 1000 comes from ignoring the negative sign and working out ten cubed instead of its reciprocal. −1000 comes from ignoring what the negative index does to the size, while still writing a negative sign on the large value. −0.001 comes from correctly finding the size, 0.001, but wrongly keeping a negative sign because the index was negative. Answer: 0.001.
- (c) £7.25 — Find the total cost of the books first: 3 × 4.25 = 12.75, so the books cost £12.75 in total. Subtract this from the £20 note: 20.00 − 12.75 = 7.25, so the change is £7.25. Stopping after finding the cost and not subtracting it from £20 gives £12.75, which is the amount spent, not the change. Borrowing correctly in the pence column but forgetting to reduce the pounds column by 1 gives £8.25 instead of £7.25. Multiplying 3 × 4.25 as 12.25 instead of 12.75, a multiplication slip, makes the change come out £0.50 too high, at £7.75. So Jack receives £7.25 change.
- (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) 5 — 16 + 9 = 25, then √25 = 5. Splitting the root over the addition instead gives √16 = 4 and √9 = 3, then 4 + 3 = 7 — but a root does not split over a sum like this. Multiplying those two roots instead of adding them gives 4 × 3 = 12. Taking the negative square root instead of the positive one gives −5.
- (c) 11 — Method: work out the total number of combinations as if there were no restriction, then subtract the one combination that is not allowed. Working: without any restriction there are 4 backdrops × 3 outfits = 12 combinations. The grey backdrop with the formal suit is not allowed, removing 1 combination: 12 − 1 = 11. Answer: 11. 12 comes from forgetting to remove the combination that is not allowed. 8 comes from removing the entire formal suit outfit from the count instead of just the one combination with the grey backdrop. 10 comes from removing two combinations instead of just the one that is not allowed.
- (b) 20 — Each length has its own error interval: 11.5 ≤ a < 12.5 and 6.5 ≤ b < 7.5. The upper bound of a sum is found by adding the upper bounds of both quantities: 12.5 + 7.5 = 20. Bounding only one of the two lengths and adding the other quantity's given value unbounded, 12.5 + 7 = 19.5, misses that both measurements carry their own uncertainty. Adding the lower bounds instead of the upper bounds, 11.5 + 6.5 = 18, gives the lower bound of the sum, not the upper one. Using a whole centimetre of error either side instead of half a centimetre, (12 + 1) + (7 + 1) = 21, comes from forgetting the error is half the rounding unit.
- (c) −5 ≤ −5 — The symbol ≤ means 'less than or equal to', and −5 is equal to −5, so this statement is true. −3 ≥ −1 is false: a candidate who ignores the negative signs and compares 3 with 1 would wrongly think −3 is the bigger number, but on the number line −3 is smaller than −1. 0.4 < 2/5 is false because 2/5 converts to exactly 0.4, so the two values are equal, not one strictly less than the other — a candidate who assumes a fraction is automatically bigger than a similar-looking decimal without converting it would miss this. 7/10 ≤ 0.6 is false because 7/10 converts to 0.7, which is bigger than 0.6; a candidate who misplaces the decimal point and converts 7/10 as 0.07 would wrongly believe this statement is true.
- (a) 2/5 — Method: find the number of vegetable plots in each allotment, add them, then divide by the total number of plots in both allotments. Working: Allotment A: 3/5 × 20 = 12 vegetable plots. Allotment B: 1/5 × 20 = 4 vegetable plots. Total vegetable plots = 12 + 4 = 16. Total plots = 20 + 20 = 40. Fraction = 16/40 = 2/5. Answer: 2/5. 3/10 comes from using only Allotment A's 12 vegetable plots over the combined total of 40 plots, forgetting to add Allotment B's vegetable plots. 1/5 comes from using only Allotment B's ratio (1:4) as the fraction of vegetables, ignoring Allotment A altogether. 3/5 comes from working out the fraction of the combined plots that grow flowers instead of vegetables.
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