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) to the nearest centimetre — Method: the error interval of a rounded measurement runs from half a unit below the stated value to half a unit above it, so the width of the interval is one whole unit of the accuracy used. Working: the interval runs from 24.5 to 25.5, a width of 25.5 − 24.5 = 1, so the unit of accuracy is 1 cm; the stated value is the midpoint, 25 cm, and 25 correct to the nearest centimetre is exactly what gives 24.5 ≤ L < 25.5. Answer: to the nearest centimetre. To the nearest 0.5 cm comes from reading the half-unit, 0.5, as the accuracy itself instead of doubling it back to the full unit. To 1 decimal place comes from seeing the bounds written with one decimal place and taking that as the accuracy, but the bounds of a value given to 1 decimal place would be only 0.05 either side. To the nearest 10 cm comes from confusing the size of the value, about 25, with the unit it was rounded to; rounding to the nearest 10 cm would give an interval 5 cm either side of the stated value.
- (d) 2 — Method: a fourth root undoes raising to the power 4, so look for the number that gives 16 when it is multiplied by itself four times. Working: 2 × 2 = 4, 4 × 2 = 8 and 8 × 2 = 16, which uses four factors of 2. Answer: 2. The distractors: 4 comes from taking the square root of 16 instead of its fourth root; 8 comes from halving 16, treating any root as a halving; 64 comes from multiplying 16 by 4 instead of taking a fourth root.
- (a) 3.84 × 10⁵ — 384,000 = 3.84 × 100,000 = 3.84 × 10⁵, with the decimal point moved five places and the coefficient kept between 1 and 10. Moving the point six places instead of five gives 3.84 × 10⁶, ten times too large. Leaving the coefficient as 38.4 gives 38.4 × 10⁴, which is not between 1 and 10. Using a negative exponent instead of a positive one gives 3.84 × 10⁻⁵, a number far smaller than 1.
- (b) 17 — Roots and powers are worked out first: √25 = 5 and 4² = 16. Division comes next: 12 ÷ 3 = 4. Then addition and subtraction, left to right: 5 + 16 − 4 = 17. A candidate who treated 4² as 4 × 2 = 8, multiplying the base by the exponent instead of squaring it, worked out 5 + 8 − 4 = 9. A candidate who did not evaluate the root and used 25 itself worked out 25 + 16 − 4 = 37. A candidate who ignored the priority of division and worked through 5 + 16 − 12 ÷ 3 strictly left to right got 5 + 16 = 21, then 21 − 12 = 9, then 9 ÷ 3 = 3.
- (c) 6 — The units digit must be even, so it can be 2 or 8, giving 2 choices. The tens digit can then be any of the remaining 3 digits, since one digit has been used for the units. Multiply: 2 × 3 = 6. 12 comes from working out how many two-digit numbers can be made in total, 4 × 3 = 12, ignoring the requirement that the number is even. 8 comes from choosing the units digit from 2 options and then wrongly allowing any of the 4 digits again for the tens digit, 2 × 4 = 8, which lets a digit repeat. 2 comes from counting only the choices for the units digit and forgetting the tens digit.
- (b) 30 — Method: 12% of an amount is 12/100 of it, so find 1% by dividing by 100 and then multiply by 12. Working: 1% of 250 is 250 ÷ 100 = 2.5, and 12% is 2.5 × 12 = 30. Answer: 30 seats. The distractors: 3 comes from writing 12% as 0.012 instead of 0.12, giving 0.012 × 250 = 3; 25 comes from finding 10% of the seats and stopping there; 24 comes from counting 12 seats for each whole hundred, 12 + 12 = 24, and ignoring the remaining 50 seats.
- (d) 2 hours — Method: the time for a journey is the distance divided by the speed, so round the distance first and then divide by the speed. Working: 95 km rounds to 100 km, and 100 ÷ 50 = 2; the speed is in kilometres per hour, so the answer is a number of hours. Answer: 2 hours. The distractors: 1 hour comes from rounding the distance down to 50 km to match the speed, so that the journey looks like a single hour of driving; 30 minutes comes from dividing the speed by the distance, 50 ÷ 100, instead of the distance by the speed; 1 hour 54 minutes is the exact time, 95 ÷ 50 = 1.9 hours, worked out in full when the question asks for an estimate.
- (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.
- (b) 3/10 — Since each bag's ratio has 5 parts and both bags contain the same total number of nuts, imagine each bag has 5 nuts: Bag A has 2 peanuts and Bag B has 1 peanut, so together there are 2 + 1 = 3 peanuts out of a combined 5 + 5 = 10 nuts, giving 3/10. 1/5 comes from using only Bag A's peanuts, 2 out of 10, without adding Bag B's peanuts. 1/10 comes from using only Bag B's peanut, without adding Bag A's peanuts. 3/5 comes from writing the combined peanuts over the number of parts in one bag instead of the combined total number of nuts.
- (a) 6 — Use √a × √b = √(ab): √3 × √12 = √(3 × 12) = √36 = 6. Adding the numbers under the roots instead of multiplying them, 3 + 12 = 15, gives √15 — that comes from applying the rule for adding surds to a multiplication question. Multiplying the two numbers under the roots but then forgetting to take the square root at the end leaves 36. Simplifying only √12 to 2√3 and then dropping the other √3 factor entirely gives 2√3.
- (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.
- (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.
- (c) 9π cm² — The area of a circle is π × r². With a radius of 3 cm this is π × 3² = 9π cm², and this is exact because π has not been replaced by any approximation. Writing 28.3 cm² replaces π with a rounded decimal value, 3.14, and then rounds the result again, so it is only an approximation. Writing 28.26 cm² uses π ≈ 3.14 without a final rounding step, but this is still only an approximation of 9π, not the exact value. Writing 27 cm² comes from replacing π with the rough approximation 3, which is even further from the true value.
- (c) 0.0479 — Method: round each option to 2 significant figures and check which one gives 0.048. Working: for 0.0479, the first two significant figures are 4 and 7; the next digit is 9, so 7 rounds up to 8, giving 0.048. For 0.0485, the first two significant figures are 4 and 8; the next digit is 5, so 8 rounds up to 9, giving 0.049, not 0.048. 0.052 already has exactly 2 significant figures, 5 and 2, so it stays as 0.052 and does not round to 0.048 at all. 0.04 has only 1 significant figure, so it is already less precise than the 2 significant figures asked for. Answer: 0.0479.
- (b) 750,000,000 — Method: multiplying by 10⁸ moves the decimal point eight places to the right, and every empty place is filled with a zero. Working: 10⁸ = 100,000,000, and moving the decimal point in 7.5 eight places to the right gives 7.5 × 100,000,000. Answer: 750,000,000. The distractors: 7,500,000,000 comes from removing the decimal point first to make 75 and then writing eight zeros after it, which carries the digits one place too far; 600 comes from reading 10⁸ as 10 × 8 = 80 and working out 7.5 × 80; 0.000000075 comes from moving the decimal point eight places to the left, as though the index were negative.
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