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.
Number worksheet — GCSE Higher
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- 1.A van has a mass of 2,000 kg, correct to 1 significant figure. Using m for the mass of the van in kilograms, write down the error interval for m.
- 2.Write 60 as a product of its prime factors, using index notation.
- 3.A carpenter has a plank of wood 4.8 m long. She cuts off 3 pieces, each 0.9 m long, to make shelves. Work out the length of wood remaining.
- 4.A student writes 0.08 as the fraction 8/10, reading the 8 as if it stood in the tenths column and ignoring the zero. Work out the correct fraction that 0.08 is equal to, giving your answer in its simplest form.
- 5.3/5 of a 5/6 litre bottle of juice is poured out. Work out the exact volume poured out, in litres.
- 6.A recipe uses 160 g of flour. Sam wants to increase the amount by 1/4. Work out the new amount of flour.
- 7.A gardener has 42 tulip bulbs and 56 daffodil bulbs. She plants them in rows, with every row containing the same number of tulip bulbs and the same number of daffodil bulbs, and no bulbs left over. Work out the greatest number of rows she can plant.
- 8.A driving instructor is working out how many current-style UK number plates are possible. The format is 2 letters, then 2 digits, then a space, then 3 letters (for example AB12 CDE). The letters I, O and Q are never used in any letter position, leaving 23 allowed letters, but every letter and every digit may repeat anywhere on the plate. Work out how many different number plates are possible, giving your answer in standard form to 2 significant figures.
- 9.A tank contains 120 litres of water. Water is drained out at a rate of 8 litres per minute for 6 minutes, and then a hose adds 15 litres. Work out how much water is left in the tank.
- 10.Expand and simplify (2 + √3)², giving your answer in the form a + b√3.
- 11.An allotment plot is divided into vegetables and flowers in the ratio 9 : 4. A third of the vegetable section is used for potatoes. What fraction of the whole plot is potatoes?
- 12.Find the missing number: 17 × ▢ = 391
- 13.A charity shop buys a coat for £24 and sells it for a profit that is 3/8 of the buying price. Work out the selling price.
- 14.Decide which of 2³⁰ and 3²⁰ is the larger number, and write down the correct statement.
- 15.At a bus station one bus leaves every 6 minutes and another leaves every 9 minutes. The two buses leave together at the start of the day. Work out how long it is until they next leave together.
Answer key
- (a) 1,500 ≤ m < 2,500 — Method: a four-digit figure written to 1 significant figure has been rounded to the nearest 1,000, so the mass lies within half of 1,000, that is 500, of the figure given. Working: 2,000 − 500 = 1,500 and 2,000 + 500 = 2,500. The lower limit is included, because 1,500 kg rounds up to 2,000 kg to 1 significant figure, while 2,500 kg rounds up to 3,000 kg, so the upper limit is not. Answer: 1,500 ≤ m < 2,500. The distractors: 1,950 ≤ m < 2,050 comes from rounding to the nearest 100 instead of to 1 significant figure; 1,000 ≤ m < 3,000 goes a whole 1,000 either side instead of half of it; 1,500 < m ≤ 2,500 has the two limits the wrong way round.
- (b) 2² × 3 × 5 — Repeatedly divide 60 by prime numbers: 60 ÷ 2 = 30, 30 ÷ 2 = 15, 15 ÷ 3 = 5, and 5 is itself prime. So 60 is 2 × 2 × 3 × 5, which in index notation is 2² × 3 × 5. Stopping the factor tree after only three divisions and writing 2 × 3 × 5 misses that the 2 divides in twice, and gives only 30, not 60. Squaring the 3 as well as the 2 gives 2² × 3² × 5, which comes to 180, far too big. Squaring the 5 instead of the 2 gives 2 × 3 × 5², which comes to 150, also too big. So 60 = 2² × 3 × 5.
- (b) 2.1 m — The three pieces use 3 × 0.9 = 2.7 m of wood. Remaining wood = 4.8 − 2.7 = 2.1 m. A candidate who miscounts and only subtracts 2 pieces instead of 3 gets 4.8 − 1.8 = 3.0 m. A candidate who adds instead of subtracting gets 4.8 + 2.7 = 7.5 m. A candidate who gives the length used instead of the length remaining gets 2.7 m.
- (d) 2/25 — Method: write the decimal over 100 using its two decimal places, then simplify. Working: 0.08 = 8/100 = 2/25 (dividing both numerator and denominator by 4). Answer: 2/25. The student's fraction, 8/10, comes from ignoring the zero in the tenths column and reading 0.08 as though it were 0.8; it simplifies to 4/5. 1/125 comes from writing the decimal over 1000 instead of 100, as if there were three decimal places. 25/2 comes from flipping the correct fraction upside down.
- (a) 1/2 — To find a fraction of an amount, multiply the fractions together: 3/5 × 5/6 = 15/30, which simplifies to 1/2 litre. Adding the fractions instead of multiplying them, using a common denominator of 30, gives 18/30 + 25/30 = 43/30, a value greater than the whole bottle. Dividing by 5/6 instead of multiplying by it, using its reciprocal 6/5, gives 3/5 × 6/5 = 18/25. Multiplying 5/6 by itself instead of by 3/5 gives 25/36.
- (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.
- (b) 14 — Method: the greatest number of identical rows is the highest common factor of the two bulb totals, found by taking every prime factor the two totals share. Working: 42 = 2 × 3 × 7 and 56 = 2 × 2 × 2 × 7, so the prime factors common to both are 2 and 7, giving a highest common factor of 2 × 7 = 14. 2 comes from taking only the common factor 2 and forgetting the common factor 7. 7 comes from taking only the common factor 7 and forgetting the common factor 2. 168 is the lowest common multiple of 42 and 56, not their highest common factor. Answer: 14.
- (b) 6.4 × 10⁸ — There are 5 letter positions, each with 23 choices, and 2 digit positions, each with 10 choices, and every position is independent because repeats are allowed. By the product rule, the total is 23⁵ × 10² = 6,436,343 × 100 = 643,634,300, which is 6.4 × 10⁸ to 2 significant figures. Using all 26 letters instead of the 23 that are actually allowed, ignoring the excluded letters entirely, gives 26⁵ × 10² = 1,188,137,600, which is 1.2 × 10⁹ to 2 significant figures. Adding the seven counts of choices instead of multiplying them, 23 + 23 + 10 + 10 + 23 + 23 + 23, gives 135, which is 1.4 × 10² to 2 significant figures — a total far too small for seven independent positions. Swapping which count of choices belongs to letters and which belongs to digits, working out 23² × 10⁵ instead of 23⁵ × 10², gives 52,900,000, which is 5.3 × 10⁷ to 2 significant figures.
- (c) 87 litres — Work out how much water is drained: 8 × 6 = 48 litres. Subtract this from the starting amount: 120 − 48 = 72 litres. Then add the 15 litres from the hose: 72 + 15 = 87 litres. Subtracting the 15 litres instead of adding it, as though the hose also removed water, gives 120 − 48 − 15 = 57 litres. Stopping after the drain step, without adding the hose water back in, leaves the working at 72 litres. Adding the rate and the time instead of multiplying them, 8 + 6 = 14 litres drained, and then working from there gives 120 − 14 + 15 = 121 litres. So 87 litres of water is left in the tank.
- (c) 7 + 4√3 — Expand the brackets fully: (2 + √3)² = 2² + 2 × 2 × √3 + (√3)² = 4 + 4√3 + 3. Adding the two whole-number terms, 4 + 3 = 7, gives 7 + 4√3. Using (a + b)² = a² + b² and skipping the middle cross term entirely gives just 4 + 3 = 7, with no surd term at all. Treating (√3)² as if it stayed √3 rather than becoming 3, then merging it with the existing surd term, gives 4 + 5√3. Squaring only the surd term correctly but carrying the whole-number term as 2 instead of squaring it to 4 gives 2 + 3 + 4√3 = 5 + 4√3.
- (a) 3/13 — Vegetables are 9 of the 9 + 4 = 13 parts, so vegetables are 9/13 of the plot. Potatoes are a third of the vegetable section, so potatoes are 1/3 of 9/13, which is 9/39, simplifying to 3/13, of the whole plot. 9/13 comes from stopping after finding the fraction of the plot that is vegetables, without taking the further third for potatoes. 1/3 gives the fraction of the vegetable section that is potatoes, not the fraction of the whole plot. 4/39 comes from taking a third of the flowers' fraction, 4/13, instead of the vegetables' fraction.
- (a) 23 — Division undoes multiplication, so the missing number is 391 ÷ 17 = 23. Writing down 17 repeats the number already given instead of solving for the missing one. Subtracting instead of dividing gives 391 − 17 = 374. Multiplying instead of dividing gives 391 × 17 = 6647.
- (d) £33.00 — The profit is 3/8 of £24 = (£24 ÷ 8) × 3 = £3 × 3 = £9.00. Selling price = £24 + £9.00 = £33.00. A candidate who gives the profit instead of the selling price gets £9.00. A candidate who subtracts the profit instead of adding it gets £24 − £9 = £15.00. A candidate who works out one eighth of £24 and adds that on, forgetting to multiply by the numerator 3, gets £24 + £3 = £27.00.
- (b) 3²⁰ is larger — Method: two powers with different bases and different indices can be compared once they are rewritten with a common index, which is possible whenever the indices share a factor. Working: 30 and 20 have a highest common factor of 10, so 2³⁰ = (2³)¹⁰ = 8¹⁰ and 3²⁰ = (3²)¹⁰ = 9¹⁰. Both are now tenth powers, and since 9 is larger than 8, 9¹⁰ is larger than 8¹⁰. Answer: 3²⁰ is larger. The distractors: 2³⁰ is larger comes from comparing only the indices and choosing the power with the bigger index; They are equal comes from multiplying base by index, 2 × 30 and 3 × 20, and finding 60 each time; They cannot be compared without a calculator comes from assuming that powers this large can only be ranked by evaluating them in full.
- (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.
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