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) 63 — 10% of 180 = 18, so 5% = 9. 35% = (3 × 18) + 9 = 54 + 9 = 63. A candidate who uses 25% instead of 35% gets 45. A candidate who doubles 35% to get 70% by mistake gets 126. A candidate who subtracts 35 from 180 instead of finding a percentage gets 145.
- (a) 300 N/m² — Pressure is force ÷ area, and the unit N/m² says so: newtons divided by square metres. The calculation is 240 ÷ 0.8. Multiplying both numbers by 10 clears the decimal: 2400 ÷ 8 = 300 N/m². 192 N/m² multiplies the force by the area instead of dividing, 30 N/m² divides by 8 and so loses the decimal point, and 3000 N/m² multiplies only the force by 10 when clearing the decimal.
- (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.
- (d) £800 — Rounding 38.7 to 1 significant figure gives 40, and rounding 21.40 to 1 significant figure gives 20. Multiplying the rounded values gives an estimate of 40 × 20 = £800. Rounding 21.40 to the nearest whole number instead of to 1 significant figure gives 21, and 40 × 21 = £840, one place value too fine for the price. Adding the rounded values instead of multiplying them gives 40 + 20 = £60. Rounding both numbers to 2 significant figures instead of 1, giving 39 and 21, produces 39 × 21 = £819.
- (a) £177.60 — Total raised = 240 × £1.85 = £444.00. The hospital receives 40% of this: £444.00 × 0.4 = £177.60. A candidate who works out the remaining 60% instead of the 40% given away gets £266.40. A candidate who forgets to find the percentage and gives the full total gets £444.00. A candidate who halves 40% by mistake and uses 20% gets £88.80.
- (a) 40 — Method: a number is odd exactly when its units digit is odd, so the restricted position is filled first and the two free positions are then filled from the digits that are left, multiplying the number of choices at each stage. Working: of the six digits only 3 and 9 are odd, so there are 2 choices for the units digit; once that digit has been used, 5 digits remain for the hundreds position and then 4 remain for the tens position, so the count is 2 × 5 × 4 = 40. Answer: 40. The distractors: 120 comes from ignoring the word odd altogether and counting every three-digit number that can be made from the six digits, 6 × 5 × 4; 60 comes from filling the hundreds and tens positions first, 6 then 5, and only then allowing 2 odd digits for the units position, which overcounts because one of 3 and 9 may already have been used, giving 6 × 5 × 2; 72 comes from restricting the units digit to 3 or 9 correctly but overlooking the condition that no digit may be used twice, so all six digits are still counted as available for each of the other two positions, giving 2 × 6 × 6.
- (a) 1,850 — Method: find the error interval, then check which value falls outside it. Working: half of 100 is 50, so the actual number of visitors, v, satisfies 1,750 ≤ v < 1,850. 1,850 sits exactly on the excluded upper boundary, since a value of 1,850 would round to 1,900, not 1,800. Answer: 1,850. (1,750 is a genuine possible value — it sits at the included lower boundary. 1,799 is a genuine possible value, just below the upper boundary. 1,760 is a genuine possible value, well inside the interval.)
- (c) 0.0065 — Leading zeros are never significant, so counting from the first non-zero digit, the first two significant figures of 0.006482 are 6 and 4. Look at the next digit along, 8, to decide whether the second figure rounds up: since 8 is 5 or more, the 4 rounds up to 5, giving 0.0065. Rounding to 2 decimal places instead of 2 significant figures gives 0.01, which answers a different question. Wrongly counting one of the leading zeros as a significant figure and stopping one figure short gives 0.006. Keeping an extra digit, as in 0.00648, gives 3 significant figures rather than 2.
- (a) 17 — Method: work out each power separately, then combine them as the question asks. Working: $2^3 = 8$ and $3^2 = 9$, and 8 + 9 = 17. 72 comes from working out 8 × 9 = 72, multiplying the two powers instead of adding them. 12 comes from misreading the powers as repeated multiplication of the base by the index, 2 × 3 + 3 × 2 = 6 + 6 = 12. −1 comes from working out 8 − 9 = −1, subtracting the powers instead of adding them. Answer: 17.
- (c) 4.1 × 10³, 3.2 × 10⁴, 2.9 × 10⁵ — The exponent decides the size first: 10³ is smaller than 10⁴, which is smaller than 10⁵, so the order is 4.1 × 10³, then 3.2 × 10⁴, then 2.9 × 10⁵. Reversing the whole list gives largest to smallest instead of smallest to largest. Comparing 3.2 × 10⁴ and 4.1 × 10³ by their coefficients alone, 3.2 against 4.1, and swapping them ignores that 10⁴ is bigger than 10³ regardless of the coefficient. Comparing 2.9 × 10⁵ and 3.2 × 10⁴ by their coefficients alone and swapping them makes the same mistake at the top of the list.
- (a) the nearest whole number — The error intervals are 23.5 ≤ p < 24.5 and 5.5 ≤ q < 6.5. The minimum of p ÷ q is 23.5 ÷ 6.5 ≈ 3.615, and the maximum is 24.5 ÷ 5.5 ≈ 4.455. Both of these round to 4 at the nearest whole number, so the answer is guaranteed correct to the nearest whole number — but not to the nearest 0.1, since 3.615 rounds to 3.6 while 4.455 rounds to 4.5, which do not agree. Claiming the nearest 0.1 assumes every figure a calculator shows is trustworthy, without checking whether the bounds actually agree that far. Claiming only the nearest 10 badly understates how much can be guaranteed here, since both bounds already round to 4, not merely to 0. Saying no degree of accuracy can be guaranteed gives up before checking whether the bounds agree at any level at all.
- (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.
- (d) £120 — Total parts = 5 + 3 = 8, so one part is worth 320 ÷ 8 = 40 pounds. The cake sale is 3 parts, so it raised 3 × 40 = 120 pounds. £200 comes from working out the raffle's share, 5 × 40, instead of the cake sale's share. £40 comes from finding the value of one part but forgetting to multiply by 3. £192 comes from dividing the total by 5 instead of 8 to find the value of one part, 320 ÷ 5 = 64, then multiplying by 3, 3 × 64 = 192.
- (d) £70 — Method: the discount is a percentage of the original price only, so work it out, subtract it, and add the fixed delivery charge afterwards. Working: 25% of £80 is £80 ÷ 4 = £20, so the discounted price is £80 − £20 = £60, and the total is £60 + £10 = £70. Answer: £70. The distractors: £60 comes from working out the discounted price and stopping there, leaving the delivery charge out of the total; £65 comes from taking £25 off the price instead of 25% of it, giving £80 − £25 = £55 and then £55 + £10 = £65; £67.50 comes from adding the delivery charge before the discount and reducing the whole amount, giving 75% of £90 = £67.50.
- (b) 120 — Method: fill the positions on the shelf one at a time; each book placed leaves one fewer book available for the next position, and the product rule multiplies the choices. Working: there are 5 books for the first position, 4 for the second, 3 for the third, 2 for the fourth and 1 for the last, so the number of orders is 5 × 4 × 3 × 2 × 1 = 120. Answer: 120. The distractors: 25 comes from multiplying the 5 books by the 5 positions rather than multiplying the shrinking number of choices at each position; 60 comes from halving the correct product, as though each order had been counted twice in the way that pairs are; 720 comes from carrying the product one factor too far and working out 6 × 5 × 4 × 3 × 2 × 1, as though there were six books.
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