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) 3/50 — Method: write the decimal over the power of ten that matches the number of digits after the point, counting every digit including a zero, then divide the numerator and the denominator by their highest common factor. Working: 0.06 has two digits after the point, so it is 6 hundredths and can be written as 6/100; the highest common factor of 6 and 100 is 2, and 6 ÷ 2 = 3 with 100 ÷ 2 = 50. Answer: 3/50. The distractors: 3/5 comes from ignoring the zero straight after the point and converting 0.6 instead, giving 6/10, which cancels to 3/5; 3/500 comes from counting three decimal places instead of two and writing 6/1000, which cancels to 3/500; 1/6 comes from putting 1 over the digits after the point, as though 0.06 meant one sixth.
- (b) 1 — Method: BIDMAS deals with the index before the subtraction, and a negative number multiplied by itself gives a positive result. Working: (−2)² = (−2) × (−2) = 4, so the calculation becomes 4 − 3 = 1. Answer: 1. The distractors: −7 comes from squaring only the 2 and leaving the minus sign outside the index, giving −(2²) − 3 = −4 − 3 = −7; −1 comes from subtracting the square from 3 instead of 3 from the square, giving 3 − 4 = −1; 25 comes from carrying out the subtraction before the index, giving (−2 − 3)² = (−5)² = 25.
- (d) 6 × 10⁷ — Method: the coefficients and the powers of ten are handled separately — multiply the coefficients, and add the indices because the powers share the base 10. Working: 2 × 3 = 6 for the coefficients, and 10³ × 10⁴ = 10⁷ for the powers; 6 lies between 1 and 10, so the coefficient needs no adjustment. Answer: 6 × 10⁷. The distractors: 5 × 10⁷ comes from adding the coefficients, 2 + 3, instead of multiplying them; 6 × 10¹² comes from multiplying the indices, 3 × 4, instead of adding them; 6 × 10¹ comes from subtracting the indices, 4 − 3, which is the rule for dividing rather than for multiplying.
- (b) 20 — Each length has its own error interval: 11.5 ≤ a < 12.5 and 6.5 ≤ b < 7.5. The upper bound of a sum is found by adding the upper bounds of both quantities: 12.5 + 7.5 = 20. Bounding only one of the two lengths and adding the other quantity's given value unbounded, 12.5 + 7 = 19.5, misses that both measurements carry their own uncertainty. Adding the lower bounds instead of the upper bounds, 11.5 + 6.5 = 18, gives the lower bound of the sum, not the upper one. Using a whole centimetre of error either side instead of half a centimetre, (12 + 1) + (7 + 1) = 21, comes from forgetting the error is half the rounding unit.
- (b) £72 — One fifth of £60 = £12. New price = £60 + £12 = £72. A candidate who gives the increase instead of the new price gets £12. A candidate who subtracts the increase instead of adding it gets £60 − £12 = £48. A candidate who uses 1/4 instead of 1/5 gets £60 + £15 = £75.
- (a) 330 ml — Correct to the nearest 20 ml means the true volume could be up to 10 ml (half of 20) either side of 340 ml. The smallest possible volume is 340 − 10 = 330 ml. A candidate who subtracted the full 20 ml instead of half of it worked out 340 − 20 = 320 ml. A candidate who added instead of subtracted, finding the largest possible volume instead of the smallest, worked out 340 + 10 = 350 ml. A candidate who halved the interval again by mistake, using 5 ml instead of 10 ml, worked out 340 − 5 = 335 ml.
- (c) 8,000,000,000 — Method: a power of ten is 1 followed by that many zeros, so 8 × 10⁹ is 8 multiplied by 1 followed by nine zeros. Working: 10⁹ = 1,000,000,000, and 8 × 1,000,000,000 puts an 8 in front of those nine zeros. Answer: 8,000,000,000. The distractors: 800,000,000 comes from writing nine digits altogether instead of nine zeros after the 8; 720 comes from reading 10⁹ as 10 × 9 = 90 and then working out 8 × 90; 0.000000008 comes from treating the index as negative and carrying the decimal point nine places to the left.
- (a) 18 minutes — Method: the buses leave together again after a number of minutes that is a multiple of both intervals, and the first such time is the lowest common multiple. Working: the multiples of 6 are 6, 12, 18, 24 … and the multiples of 9 are 9, 18, 27 … The first value in both lists is 18, which is 6 × 3 and 9 × 2. Answer: 18 minutes. The distractors: 54 minutes comes from multiplying 6 by 9, which does give a common multiple but not the lowest one; 3 minutes is the highest common factor of 6 and 9 rather than their lowest common multiple; 15 minutes comes from adding the two intervals together.
- (a) 200 g — One quarter of 160 g is 40 g. Increasing the amount means adding this on: 160 + 40 = 200 g. Finding the increase, 1/4 of 160 = 40 g, but stopping there without adding it to the original amount leaves just 40 g. Using 4/5 instead of 5/4 as the scaling fraction, 160 × 4/5 = 128 g, actually decreases the amount rather than increasing it. Increasing by a half instead of a quarter, 160 + 80 = 240 g, uses the wrong fraction of 160.
- (d) £70 — Method: the discount is a percentage of the original price only, so work it out, subtract it, and add the fixed delivery charge afterwards. Working: 25% of £80 is £80 ÷ 4 = £20, so the discounted price is £80 − £20 = £60, and the total is £60 + £10 = £70. Answer: £70. The distractors: £60 comes from working out the discounted price and stopping there, leaving the delivery charge out of the total; £65 comes from taking £25 off the price instead of 25% of it, giving £80 − £25 = £55 and then £55 + £10 = £65; £67.50 comes from adding the delivery charge before the discount and reducing the whole amount, giving 75% of £90 = £67.50.
- (d) √13 — Method: everything under a root sign is worked out first, because a square root cannot be taken term by term across an addition. Working: 2² = 4 and 3² = 9, so the expression under the root is 4 + 9 = 13. As 13 is not a square number, the exact value is left in root form as √13. Answer: √13. The distractors: 5 comes from rooting each square separately and adding, 2 + 3, which treats the root of a sum of squares as the sum of the numbers; 13 comes from working out the sum under the root correctly and then forgetting to take the root; √5 comes from subtracting the two squares, 9 − 4, instead of adding them.
- (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.
- (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.
- (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.
- (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.
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