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.
Calculator
Answer key: Number worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (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.
- (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.
- (b) 28 — Method: count the ordered choices with the product rule and then correct for the double counting, because the two captains have equal standing and so a pair is the same pair whichever captain is named first. Working: there are 8 players who could be named first and 7 who could be named second, giving 8 × 7 = 56 ordered choices; each pair has been counted twice, once in each order, so the number of pairs is 56 ÷ 2 = 28. Answer: 28. The distractors: 56 comes from stopping at 8 × 7 and never halving, which counts each pair of captains twice; 64 comes from working out 8 × 8, which allows the same player to be chosen as both captains; 16 comes from multiplying the 8 players by the 2 captaincies instead of pairing the players with one another.
- (b) 78 hours — From 10:00 on Monday to 10:00 on Thursday is exactly 3 complete days, which is 3 × 24 = 72 hours. From 10:00 to 16:00 on the Thursday is a further 6 hours, giving a total of 72 + 6 = 78 hours. Counting Monday to Thursday as 4 full calendar days instead of 3 complete 24-hour periods gives 4 × 24 = 96 hours. Undercounting the number of complete days as 2 instead of 3 gives 2 × 24 + 6 = 54 hours. Subtracting the extra 6 hours instead of adding them to the 3 complete days gives 72 − 6 = 66 hours.
- (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.
- (c) 19 — Without any restriction there would be 5 × 4 = 20 different sandwiches. The restriction removes exactly one combination, cheese and mustard on gluten-free bread, so subtract 1: 20 − 1 = 19. 20 comes from ignoring the restriction completely. 15 comes from removing the gluten-free bread altogether, as if none of the fillings were available on it, 5 × 3 = 15. 16 comes from removing the cheese and mustard filling completely, as if it were not available on any bread, 4 × 4 = 16.
- (a) 56 — Method: count the ordered selections with the product rule first, then divide by the number of different orders in which any one committee could have been picked. Working: there are 8 choices for a first pupil, 7 for a second and 6 for a third, giving 8 × 7 × 6 = 336 ordered selections; any particular three pupils could have been picked in 3 × 2 × 1 = 6 orders, so the number of different committees is 336 ÷ 6 = 56. Answer: 56. The distractors: 336 comes from stopping at 8 × 7 × 6 and treating the three places as distinct posts when they are identical; 168 comes from dividing that product by 2 rather than by the 6 orders in which three chosen pupils can be listed; 24 comes from multiplying the 8 volunteers by the 3 places instead of multiplying the choices at each stage.
- (c) 62.35 ≤ r < 62.45 — Method: with a value rounded to 1 decimal place, the error interval reaches half of 0.1 either side. Working: half of 0.1 is 0.05, so the interval runs from 62.4 − 0.05 to 62.4 + 0.05. Answer: 62.35 ≤ r < 62.45. (62 ≤ r < 63 comes from rounding to the nearest whole number instead of 1 decimal place. 62.35 ≤ r ≤ 62.45 comes from including the upper bound with ≤ instead of excluding it with <. 62.3 ≤ r < 62.5 comes from using 0.1 either side instead of half of it.)
- (c) 4 × 10⁻¹ — Multiply the A values: 8 × 5 = 40. Add the powers of 10: −5 + 3 = −2, giving 40 × 10⁻². Since A must satisfy 1 ≤ A < 10, rewrite 40 as 4 × 10¹, so 40 × 10⁻² = 4 × 10¹ × 10⁻² = 4 × 10⁻¹. A candidate who stopped at 40 × 10⁻² did the index arithmetic correctly but left the answer outside standard form, since 40 is not between 1 and 10. A candidate who adjusted the A value to 4 correctly but then took the power of 10 by subtracting the two given powers, −5 − 3 = −8, wrote 4 × 10⁻⁸. A candidate who adjusted the A value to 4 but multiplied the two given powers, −5 × 3 = −15, wrote 4 × 10⁻¹⁵. Both of these forgot that multiplying in standard form means adding the powers.
- (a) 8 × 10³ — Method: divide the capacity of the card by the size of one photograph, dividing the coefficients and subtracting the indices, then bring the coefficient back into the range 1 to 10. Working: 3.2 ÷ 4 = 0.8 and 10 − 6 = 4, which gives 0.8 × 10⁴; a coefficient of 0.8 is smaller than 1, so the decimal point moves one place to the right and the index falls by 1. Answer: 8 × 10³. The distractors: 8 × 10⁴ comes from correcting 0.8 to 8 without reducing the index, which makes the answer ten times too large; 1.28 × 10¹⁷ comes from multiplying the two numbers instead of dividing them, since 3.2 × 4 = 12.8 and 10 + 6 = 16; 8 × 10¹⁵ comes from dividing the coefficients but adding the indices instead of subtracting them.
- (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.
- (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.
- (c) 22 — Without restriction there are 6 × 4 = 24 combinations. Two specific combinations are not available, so subtract 2: 24 − 2 = 22. 24 comes from ignoring the restriction completely. 23 comes from subtracting only 1 of the 2 excluded combinations. 18 comes from removing the whole sport trim level, 6 × 3 = 18, instead of removing just the two excluded combinations.
- (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) 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.)
Build your own mix at the worksheet builder.