Printable · GCSE Foundation · ages 14-16
Number worksheet — GCSE Foundation
Fifteen questions across the number statements at Foundation 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 Foundation
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- (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.
- (c) 3.05 kg — There are 1000 g in 1 kg, so 450 g = 450 ÷ 1000 = 0.45 kg. The total mass is 0.45 + 2.6 = 3.05 kg. 7.1 kg divides the grams by 100 instead of 1000, 2.645 kg divides them by 10 000, and 452.6 kg adds the two numbers without converting the grams at all.
- (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.
- (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.
- (a) 0.55, 58%, 3/5 — Converting all three to decimals: 3/5 = 0.6, 0.55 stays as 0.55, and 58% = 0.58. In order from smallest to largest, this is 0.55, then 58%, then 3/5. Writing the numbers in the reverse order, largest to smallest, gives 3/5, 58%, 0.55. Misconverting 3/5 as 0.5 instead of 0.6 makes it appear smaller than both other values, giving the order 3/5, 0.55, 58%. Misconverting 58% as 0.058 instead of 0.58, by moving the decimal point two extra places, makes it appear smallest of the three, giving the order 58%, 0.55, 3/5.
- (c) 6 — Method: the first significant figure is the first non-zero digit; round using the digit after it to decide whether to round up or down. Working: the first significant figure of 6.283 is the 6; the next digit is 2, which rounds down, so 6.283 rounds to 6. 6.3 comes from rounding to 2 significant figures instead of 1. 10 comes from rounding up to the nearest 10 instead of finding 1 significant figure of the number itself. 0.6 comes from misplacing the decimal point after rounding. Answer: 6.
- (c) 6.3 × 10⁷ — To add numbers in standard form, first write them with the same power of 10. 6 × 10⁷ = 60 × 10⁶, so the sum is 60 × 10⁶ + 3 × 10⁶ = 63 × 10⁶ = 6.3 × 10⁷. A candidate who added the A values without adjusting for the different powers worked out 6 + 3 = 9 and kept the larger power, writing 9 × 10⁷. A candidate who added the powers of 10 as if multiplying wrote 9 × 10¹³. A candidate who added the A values but used the smaller power wrote 9 × 10⁶.
- (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) £70 — Method: 20% of an amount is 20/100 of it; a reliable route is to find 10% by dividing by 10 and then double it. Working: 10% of £350 is £350 ÷ 10 = £35, and 20% is twice as much, £35 × 2 = £70. Answer: £70. The distractors: £35 comes from finding 10% and stopping there; £17.50 comes from reading 20% as one twentieth and working out £350 ÷ 20 = £17.50; £280 comes from taking 20% off the money raised rather than finding 20% of it, giving £350 ÷ 5 = £70 and then £350 − £70 = £280.
- (d) 0.024 cm — Method: significant figures are counted from the first non-zero digit; the zeros in front of it only fix the place value and are not significant. Working: in 0.02384 the first significant figure is 2 and the second is 3, so the rounding is decided by the next digit, 8. As 8 is 5 or more, the second significant figure goes up from 3 to 4, in the same place value. Answer: 0.024 cm. The distractors: 0.023 cm comes from chopping the 84 off instead of rounding it; 0.02 cm comes from counting the leading zeros as significant figures, so the 2 is taken as the second figure and the rounding stops there; 0.0238 cm is 0.02384 correct to 3 significant figures, one figure too many.
- (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.
- (d) 6 × 10⁷ — 3 × 2 = 6, and 2 + 5 = 7, so (3 × 10²) × (2 × 10⁵) = 6 × 10⁷. Multiplying the exponents instead of adding them gives 2 × 5 = 10, so 6 × 10¹⁰. Adding the coefficients instead of multiplying them gives 3 + 2 = 5, so 5 × 10⁷. Subtracting the exponents instead of adding them gives 5 − 2 = 3, so 6 × 10³.
- (b) 1.45 ≤ m < 1.55 — The flour's mass is labelled 1.5 kg, correct to the nearest 0.1 kg, so half of 0.1 kg is added to and subtracted from 1.5 kg to find the interval: 1.5 − 0.05 = 1.45 and 1.5 + 0.05 = 1.55, giving 1.45 ≤ m < 1.55. '1.4 ≤ m < 1.6' comes from taking the whole 0.1 kg as the margin either side, instead of half of it. '1.45 < m ≤ 1.55' comes from writing the inequality signs the wrong way round — the lower bound should be included and the upper bound excluded, not the other way round. '1.45 ≤ m ≤ 1.55' comes from including the upper bound, when the convention is that the upper bound is never actually reached.
- (c) Sam is correct — The leading zeros in 0.070268 are not significant, so the first three significant figures are 7, 0 and 2. The next digit along is 6, and since 6 is 5 or more, the third significant figure rounds up from 2 to 3, giving 0.0703. This means Sam's answer is correct. Writing 0.070 keeps only 2 significant figures, one short of what was asked. Writing 0.0702 ignores the digit 6 that follows and leaves the third figure unrounded. Writing 0.0704 rounds the third figure up twice, as if a later digit had also pushed it up.
- (d) 10 — Method: picking 3 flowers from 5 leaves 2 flowers behind, so counting the different pairs that could be left out counts the bunches, and those pairs can be listed systematically. Working: number the flowers 1 to 5; the first flower can be left out alongside any of the 4 flowers after it, the second alongside any of the 3 after it, the third alongside any of the 2 after it and the fourth alongside the last one, so the number of pairs left out is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 60 comes from working out 5 × 4 × 3 and treating the three picks as an ordered selection when the order does not matter; 30 comes from dividing that product by 2 instead of by the 6 orders in which three chosen flowers could have been picked; 15 comes from multiplying the 5 flowers by the 3 flowers picked instead of counting the selections.
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