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) 13 °C — Method: subtract the lowest temperature from the highest temperature to find the difference. Working: the highest temperature is 6 °C and the lowest is −7 °C. Difference = 6 − (−7) = 6 + 7 = 13. Answer: 13 °C. 8 °C comes from using −2 °C as the lowest temperature instead of −7 °C: 6 − (−2) = 8. 5 °C comes from finding the difference between the two negative temperatures instead of the highest and lowest: −2 − (−7) = 5. −1 °C comes from adding the highest and lowest temperatures instead of subtracting: 6 + (−7) = −1.
- (b) 5.1 — ∛130 lies between 5 and 6, since 125 < 130 < 216, and closer to 5 because 130 is much nearer 125 than 216. To pin down the first decimal place, test the midpoint of the tenth, 5.05: 5.05³ = 5.05 × 5.05 × 5.05 ≈ 128.79. Since 130 is greater than 128.79, ∛130 lies above 5.05, so it rounds to 5.1 rather than 5.0. Rounding down to 5.0, on the assumption that a value close to the lower bound 125 must round down, ignores that 5.05³ is already less than 130. Estimating 5.2 overshoots the true root: 5.2³ = 140.608, which is well above 130, so ∛130 cannot round to 5.2. Taking 6.0, the upper of the two whole numbers the root lies between, ignores that 130 is far nearer to 5³ = 125 than to 6³ = 216, so the root sits just above 5, not just below 6.
- (d) 10 — Method: list the pairs systematically, taking each sweet in turn and pairing it only with the sweets that come after it, so that no pair is written down twice. Working: numbering the sweets 1 to 5, the first sweet pairs with 4 others, the second pairs with 3 sweets that come after it, the third with 2 and the fourth with 1, so the total is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 20 comes from working out 5 × 4 and never halving, which counts each pair twice, once in each order; 25 comes from working out 5 × 5, which allows the same sweet to be chosen twice; 9 comes from adding the 5 choices and the 4 remaining choices instead of combining them as a selection of two.
- (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.
- (a) 6 × 10² metres — Method: distance = speed × time, so multiply the coefficients and add the indices. Working: 3 × 2 = 6 for the coefficients, and 8 + (−6) = 2 for the indices; 6 lies between 1 and 10, so the coefficient needs no adjustment. Answer: 6 × 10² metres, which is 600 metres. The distractors: 5 × 10² metres comes from adding the coefficients, 3 + 2, instead of multiplying them; 6 × 10¹⁴ metres comes from subtracting the indices, 8 − (−6), which is the rule for dividing rather than for multiplying; 6 × 10⁻⁴⁸ metres comes from multiplying the indices, 8 × (−6), instead of adding them.
- (a) 0.0069 — Leading zeros are not significant, so the significant figures in 0.006852 start at 6: 6, 8, 5, 2. Rounding to 2 significant figures means keeping 6 and 8, and looking at the next digit, 5, to decide whether to round up. Since 5 rounds up, the second significant figure increases from 8 to 9: 0.006852 rounds to 0.0069. A candidate who rounded to 1 significant figure instead of 2 wrote 0.007. A candidate who rounded to 3 significant figures instead of 2 wrote 0.00685. A candidate who did not round up despite the next digit being 5 wrote 0.0068.
- (d) 24 — Method: the three parts of the deal are chosen independently, so every sandwich can be taken with every snack and every one of those pairs with every drink; the product rule multiplies the number of choices in each part. Working: there are 4 choices of sandwich and each can be taken with any of the 3 snacks, giving 4 × 3 = 12 sandwich-and-snack pairs; each of those pairs can be completed with either of the 2 drinks, so the number of meal deals is 12 × 2 = 24. Answer: 24. The distractors: 9 comes from adding the choices, 4 + 3 + 2, instead of multiplying them, and a candidate who adds writes that total down as the count; 12 comes from multiplying the sandwiches by the snacks and never bringing the drink into the count at all; 27 comes from adding the items on the menu to get 9 and then multiplying that by the 3 parts of the deal, which counts the menu rather than the combinations.
- (c) 3⁻³ — Method: when dividing powers of the same base, subtract the index of the number you are dividing by from the index of the number being divided, keeping them in the order the question writes them. Working: 2 − 5 = −3, so 3² ÷ 3⁵ = 3⁻³. It is worth checking this against the numbers: 3² = 9 and 3⁵ = 243, and 9 ÷ 243 = 1/27, which is 3⁻³. 3³ comes from subtracting the other way round, 5 − 2 = 3, which reverses the sign of the index and gives 27 instead of 1/27. 3⁷ comes from working out 2 + 5 = 7, which is the rule for multiplying powers, not dividing them. 3¹⁰ comes from multiplying the indices, 2 × 5 = 10, instead of subtracting them. Answer: 3⁻³.
- (b) 5 — Method: find the square root first, then divide. Working: √225 = 15, and 15 ÷ 3 = 5. Answer: 5. (75 comes from dividing 225 by 3 first and forgetting to take the square root at all. 8.7 comes from dividing 225 by 3 inside the root, √(225 ÷ 3) ≈ 8.7, instead of taking the root first. 45 comes from misreading the divisor as 5 instead of 3, working out 225 ÷ 5 = 45.)
- (a) 37.5 — The error intervals are 45 ≤ c < 55 and 17.5 ≤ d < 18.5. The maximum possible value of a difference comes from the largest possible value being reduced by the smallest amount: use the upper bound of c together with the LOWER bound of d, since subtracting less gives a bigger result: 55 − 17.5 = 37.5. Using the upper bound for both quantities, 55 − 18.5 = 36.5, forgets that subtracting a bigger number gives a smaller answer, not a bigger one. Using the lower bounds for both, 45 − 17.5 = 27.5, gives the lower bound of the difference instead of the upper one. Using the lower bound of c with the upper bound of d, 45 − 18.5 = 26.5, combines the two bounds the wrong way round entirely.
- (c) 23.375 — The error intervals are 7.5 ≤ base < 8.5 and 4.5 ≤ height < 5.5. The upper bound of the area uses the upper bound of both the base and the height, then halves the product: 8.5 × 5.5 ÷ 2 = 23.375 cm². Using the lower bound of both dimensions instead, 7.5 × 4.5 ÷ 2 = 16.875, gives the lower bound of the area rather than the upper one. Multiplying the two upper bounds together but forgetting to halve for the triangle formula, 8.5 × 5.5 = 46.750, treats the triangle as if it were a rectangle. Using the given values directly without applying any bound at all, 8 × 5 ÷ 2 = 20.000, ignores that each rounded measurement has its own range of possible values.
- (b) £13.48 — Method: an amount of money is written to the nearest penny, which is 2 decimal places, so the calculator display has to be rounded to 2 decimal places. Working: 53.90 ÷ 4 = 13.475, and the digit in the third decimal place is 5, so the penny digit goes up from 7 to 8. Answer: £13.48. The distractors: £13.47 comes from chopping the third decimal place off instead of rounding with it; £13.50 comes from rounding to the nearest 10p rather than to the nearest penny; £13.40 comes from cutting the display short at 1 decimal place, which is both the wrong degree of accuracy and a truncation rather than a rounding.
- (c) 3.2 litres — The blue paint is 3 of the 8 equal shares in the mix, that is 3/8 of the total. Three shares are 1.2 litres, so one share is 1.2 ÷ 3 = 0.4 litres. The whole mix is 8 shares: 8 × 0.4 = 3.2 litres. 2 litres is the volume of white paint, 1.92 litres divides the blue paint by the white paint's 5 shares instead of its own 3, and 9.6 litres treats the 1.2 litres as a single share.
- (b) 1/9 — Method: let x stand for the recurring decimal, multiply by the power of ten that moves exactly one repeating block past the decimal point, subtract the original equation so that the recurring tail cancels, and solve the equation that is left. Working: let x = 0.111...; the repeating block is one digit long, so multiply by 10 to give 10x = 1.111...; subtracting the first equation from the second gives 10x − x = 1.111... − 0.111..., that is 9x = 1; dividing both sides by 9 gives x = 1/9. Answer: 1/9. The distractors: 1/10 comes from dividing by the multiplier 10 at the last step instead of by the 9 that is left in front of x; 1/11 comes from recalling the elevenths family instead of the ninths, although 1/11 = 0.0909... has a two-digit repeating block rather than a one-digit one; 11/100 comes from stopping the decimal after two digits and converting 0.11 into hundredths.
- (b) 0.3, 32%, 7/20 — Method: convert every number to a decimal so they can be compared on the same scale. Working: 7/20 = 0.35, 0.3 stays as 0.3, and 32% = 0.32. Comparing 0.3, 0.32 and 0.35 in size gives the order 0.3, then 0.32, then 0.35. Answer: 0.3, 32%, 7/20. 7/20, 32%, 0.3 lists the numbers from largest to smallest instead of smallest to largest. 0.3, 7/20, 32% swaps 32% and 7/20, treating the fraction 7/20 as smaller even though 7/20 = 0.35 is bigger than 32% = 0.32. 32%, 0.3, 7/20 comes from moving the digits one place too far when converting the percentage, giving 0.032 instead of 0.32, which makes 32% look far smaller than it really is.
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