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.
Answer key: Number worksheet — GCSE Higher
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- (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.
- (d) 7/20 — Method: write the decimal over the power of ten that matches the number of digits after the point, then divide the numerator and the denominator by their highest common factor. Working: 0.35 has two digits after the point, so it is 35 hundredths and can be written as 35/100; the highest common factor of 35 and 100 is 5, and 35 ÷ 5 = 7 with 100 ÷ 5 = 20. Answer: 7/20. The distractors: 3/10 comes from reading only the first digit after the point and converting 0.3; 7/25 comes from dividing the numerator by 5 but the denominator by 4, using a different factor on the top and on the bottom; 35/10 comes from counting one decimal place instead of two and writing the digits over 10.
- (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.
- (c) £7.25 — Find the total cost of the books first: 3 × 4.25 = 12.75, so the books cost £12.75 in total. Subtract this from the £20 note: 20.00 − 12.75 = 7.25, so the change is £7.25. Stopping after finding the cost and not subtracting it from £20 gives £12.75, which is the amount spent, not the change. Borrowing correctly in the pence column but forgetting to reduce the pounds column by 1 gives £8.25 instead of £7.25. Multiplying 3 × 4.25 as 12.25 instead of 12.75, a multiplication slip, makes the change come out £0.50 too high, at £7.75. So Jack receives £7.25 change.
- (c) 150 m — Method: a fraction acts as an operator, so finding 3/7 of a length means dividing by the denominator and multiplying by the numerator. Working: 350 ÷ 7 = 50, so one seventh of the route is 50 m, and three sevenths is 50 × 3 = 150 m. Answer: 150 m. The distractors: 50 m comes from finding one seventh and stopping there instead of multiplying by 3; 1050 m comes from multiplying by the numerator without dividing by the denominator, giving 350 × 3 = 1050; 200 m comes from working out the stretch of the route the footpath does not run alongside, which is 4/7 of 350 m, instead of the stretch it does.
- (a) 7 — Convert the mixed number to an improper fraction: 5 1/4 = 21/4. Dividing by 3/4 means multiplying by its reciprocal, 4/3: 21/4 × 4/3 gives 84/12, which simplifies to 7. So exactly 7 complete pieces of 3/4 m can be cut. Ignoring the 1/4 m and dividing only the whole number, 5 ÷ 3/4, gives 20/3, which is 6 complete pieces with some wood left over. Multiplying by 3/4 instead of its reciprocal, 21/4 × 3/4, gives 63/16, which is 3 complete pieces. Misreading 5 1/4 as the fraction 5/4, then dividing by 3/4, gives 5/3, which is only 1 complete piece. So 7 complete pieces can be cut from the plank.
- (a) 20 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 588 rounds to 600 (1 s.f.) and 31 rounds to 30 (1 s.f.). 600 ÷ 30 = 20. Answer: 20. 17 comes from cutting 588 down to 500, keeping the leading digit as it stands instead of rounding it up to 1 significant figure, 600, then dividing by the correctly rounded 30. 200 comes from misreading the rounded divisor 30 as 3, giving 600 ÷ 3 instead of 600 ÷ 30. 19 is the exact value of 588 ÷ 31 rounded to the nearest whole number, found without rounding the numbers first.
- (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.
- (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.
- (b) 25 g — Method: substitute the number of years into the model, raise the fraction to that power first, then multiply by the starting mass. Working: with n = 3 the model gives M = 200 × (1/2)³. Since (1/2)³ = 1/8, the mass is 200 ÷ 8 = 25. Answer: 25 g. The distractors: 12.5 g comes from halving four times instead of three, counting the first weighing as a year; 300 g comes from multiplying by 1/2 × 3 = 1.5 instead of raising 1/2 to the power 3; 0.125 g comes from working out (1/2)³ = 0.125 and stopping there, without multiplying by the starting mass.
- (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) An under-estimate, by 8 — Method: work out the exact product, then compare it with the estimate; an estimate that is smaller than the exact value is an under-estimate, and the difference between them is the size of the error. Working: 48 × 21 = 48 × 20 + 48 = 960 + 48 = 1,008, and 1,008 − 1,000 = 8, so the estimate falls short. Answer: an under-estimate, by 8. The distractors: an over-estimate by 8 has the size of the error right but the direction wrong, and comes from assuming that rounding 48 up to 50 must push the estimate above the exact value, without allowing for 21 being rounded down; an over-estimate by 19 comes from working out 48 × 21 as 48 × 20 + 21 = 981, adding a 21 where another 48 belongs; the claim that the estimate is exactly right comes from arguing that one number was rounded up and the other down, so the two changes must cancel.
- (d) 15 — Use √a × √b = √(ab): √20 × √12 = √(20 × 12) = √240. Since 15² = 225 and 16² = 256, and 240 is a little closer to 225 than to 256, √240 is a little under 15.5 — in fact √240 ≈ 15.49, which rounds to 15. Adding the two roots instead of multiplying them, √20 + √12 ≈ 4.47 + 3.46 ≈ 7.94, rounds to 8, but the question asks for the product, not the sum. Multiplying 20 by 12 and stopping there, without ever taking a square root, leaves 240, which is the number under the root, not its value. Rounding each root to the nearest whole number BEFORE multiplying — √20 ≈ 4 and √12 ≈ 3 — gives 4 × 3 = 12, a cruder estimate that loses accuracy by rounding twice instead of once.
- (d) 1.6 × 10⁵ — 5.6 ÷ 3.5 = 1.6, and 7 − 2 = 5, so the average turnover per shop is 1.6 × 10⁵ pounds. Multiplying the exponents instead of subtracting them gives 7 × 2 = 14, so 1.6 × 10¹⁴. Adding the exponents instead of subtracting them gives 7 + 2 = 9, so 1.6 × 10⁹. Subtracting the coefficients instead of dividing them gives 5.6 − 3.5 = 2.1, so 2.1 × 10⁵.
- (a) 2² × 3 — Method: divide repeatedly by the smallest prime that goes in, until 1 is reached, then write the primes used as a product with indices. Working: 12 ÷ 2 = 6, 6 ÷ 2 = 3 and 3 ÷ 3 = 1, so the primes used are 2, 2 and 3, which is written as 2² × 3. Answer: 2² × 3. The distractors: 2 × 6 comes from stopping at the first factor pair without splitting the 6, which is not prime; 2 × 3 comes from listing each prime once and losing the repeat, and it multiplies to 6 rather than 12; 2 × 3² puts the index on the wrong prime and multiplies to 18.
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