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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Number worksheet — GCSE Higher
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- 1.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.
- 2.A semicircle has a diameter of 8 cm. Work out the exact area of the semicircle, in terms of π.
- 3.Work out (−2)³ + (−3)² − (−4)
- 4.A lorry travels 105 miles in 1 hour 45 minutes. Work out its average speed in miles per hour.
- 5.A supermarket sells apples at £1.85 per kg. Anna buys 3.6 kg of apples. Estimate the cost by rounding each number to 1 significant figure before multiplying. Work out Anna's estimate.
- 6.A paint mix is made from white paint and blue paint in the ratio 5 : 3. Priya uses 1.2 litres of blue paint and the matching amount of white paint. Work out the total volume of the mix.
- 7.A country's population is 8,340,000, and it is estimated that 1,950,000 of them live in the capital city. By rounding each number to 1 significant figure, work out an estimate for the number of people who do not live in the capital city.
- 8.The number 24 can be written as 2³ × 3, and the number 60 can be written as 2² × 3 × 5. Work out the lowest common multiple of 24 and 60.
- 9.A plank of wood is 5 1/4 m long. Pieces of length 3/4 m are cut from it. Work out how many complete pieces of 3/4 m can be cut from the plank.
- 10.Without using a calculator, estimate the value of √70 × ∛65, giving your answer to 1 significant figure.
- 11.Work out the value of (1/3)⁻²
- 12.Ava measures a metal rod and records its length as 15 cm, correct to the nearest centimetre. Ben measures the same rod and records its length as 14.8 cm, correct to the nearest 0.1 cm. Decide whether both records can be correct, and give a reason for your answer.
- 13.Two bags of mixed nuts are combined. Bag A has peanuts and cashews in the ratio 2 : 3. Bag B has peanuts and cashews in the ratio 1 : 4. Both bags contain the same total number of nuts. Work out the fraction of the combined mixture that is peanuts.
- 14.A netball team has 8 players. Two of them are chosen to be captains, and the two captains have equal standing. Work out how many different pairs of captains could be chosen.
- 15.Write 0.875 as a fraction in its simplest form.
Answer key
- (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) 8π cm² — A diameter of 8 cm gives a radius of 4 cm. The area of a full circle would be π × r² = π × 4² = 16π cm², and a semicircle is exactly half of this, giving 16π ÷ 2 = 8π cm². Forgetting to halve the area for the semicircle gives 16π cm², the area of the whole circle. Halving the diameter twice, using a radius of 2 instead of 4, gives π × 2² = 4π cm². Using the diameter itself as the radius, so π × 8² = 64π, and then halving that for the semicircle gives 32π cm².
- (a) 5 — Method: each index is worked out first, and subtracting a negative number is the same as adding the positive. Working: (−2)³ = (−2) × (−2) × (−2) = −8 and (−3)² = (−3) × (−3) = 9, while − (−4) becomes + 4, so the calculation becomes −8 + 9 + 4 = 5. Answer: 5. The distractors: −13 comes from taking (−3)² as −9, giving −8 − 9 + 4 = −13; −3 comes from reading − (−4) as − 4, giving −8 + 9 − 4 = −3; 21 comes from treating every power of a negative number as positive, so that (−2)³ is taken as 8 and the calculation becomes 8 + 9 + 4 = 21.
- (d) 60 mph — Average speed = distance ÷ time, with the time measured in hours. 45 minutes is 45/60 of an hour, which is 0.75 of an hour, so the journey takes 1.75 hours. Speed = 105 ÷ 1.75 = 60 mph. 52.5 mph rounds the time up to 2 hours, 72.4 mph writes 1 hour 45 minutes as 1.45 hours, and 183.75 mph multiplies the distance by the time instead of dividing.
- (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) 3.2 litres — The blue paint is 3 of the 8 equal shares in the mix, that is 3/8 of the total. Three shares are 1.2 litres, so one share is 1.2 ÷ 3 = 0.4 litres. The whole mix is 8 shares: 8 × 0.4 = 3.2 litres. 2 litres is the volume of white paint, 1.92 litres divides the blue paint by the white paint's 5 shares instead of its own 3, and 9.6 litres treats the 1.2 litres as a single share.
- (d) 6,000,000 — Method: round each number to 1 significant figure, then subtract. Working: 8,340,000 rounds to 8,000,000 (1 s.f.); 1,950,000 rounds to 2,000,000 (1 s.f.); 8,000,000 − 2,000,000 = 6,000,000. Answer: 6,000,000. 6,390,000 is the exact difference, found without rounding the numbers first. 8,000,000 comes from rounding the population correctly but forgetting to subtract the capital's population at all. 6,300,000 comes from rounding 8,340,000 to the nearest hundred thousand, 8,300,000, instead of to 1 significant figure, then subtracting the correctly rounded 2,000,000.
- (a) 120 — For the lowest common multiple, take each prime that appears in either factorisation, raised to the higher power. In 2³ × 3 and 2² × 3 × 5, the prime 2 appears with power 3 in one and power 2 in the other — take the higher, 2³; the prime 3 appears with the same power in both, 3¹; and the prime 5 appears only in the second factorisation, so use 5¹. Multiplying these, 2³ × 3 × 5, gives 120. Taking the lower power of 2 instead of the higher, and leaving out 5 altogether, gives the highest common factor, 12, instead. Multiplying the two original numbers together, 24 × 60, gives 1440, which double-counts every shared prime factor. Assuming the lowest common multiple is simply the larger of the two numbers gives 60, but 60 is not a multiple of 24 — 60 ÷ 24 does not divide exactly. So the lowest common multiple of 24 and 60 is 120.
- (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.
- (b) 30 — 70 is close to the perfect square 64, so √70 ≈ 8. 65 is close to the perfect cube 64, so ∛65 ≈ 4. Multiplying these estimates: 8 × 4 = 32, which rounds to 30 to 1 significant figure. Estimating ∛65 as 5 instead of 4, perhaps by confusing it with the nearby cube 125 = 5³ rather than the much closer 64 = 4³, and then multiplying by 8, gives 8 × 5 = 40. Adding the two estimates instead of multiplying them, 8 + 4 = 12, rounds to 10 to 1 significant figure. Rounding both estimates up to the next whole number using the wrong nearby power for each, taking √70 as 9 and ∛65 as 5, gives 9 × 5 = 45, which rounds to 50 to 1 significant figure.
- (b) 9 — Method: a negative index means take the reciprocal of the base first and then apply the positive power. Working: the reciprocal of 1/3 is 3, so (1/3)⁻² = 3² = 3 × 3 = 9. Answer: 9. The distractors: 1/9 comes from ignoring the minus sign and squaring 1/3 as it stands; −9 comes from reading the negative index as a minus sign on the result; 6 comes from multiplying the denominator by the index, 3 × 2, instead of squaring the reciprocal.
- (b) Yes, because 14.8 cm rounds to 15 cm to the nearest cm — Method: a recorded measurement is not an exact length; it stands for every length that rounds to it, so the two records agree if one rod can produce both. Working: Ben's record of 14.8 cm to the nearest 0.1 cm means the rod is between 14.75 cm and 14.85 cm, and 14.8 is nearer to 15 than to 14, so a rod of that length is recorded as 15 cm to the nearest centimetre. Both records can therefore come from the same rod. Answer: Yes, because 14.8 cm rounds to 15 cm to the nearest cm. The distractors: the claim that 14.8 cm rounds to 15.0 cm to 1 decimal place is false, since 14.8 cm is already written to 1 decimal place and stays 14.8 cm; the claim that it rounds to 14 cm is false, because 14.8 is 0.2 away from 15 and 0.8 away from 14; the claim that the two lengths are not the same treats each record as an exact length, when each is only a rounded record of one rod.
- (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.
- (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.
- (a) 7/8 — 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: 0.875 has three digits after the point, so it is 875 thousandths and can be written as 875/1000; the highest common factor of 875 and 1000 is 125, and 875 ÷ 125 = 7 with 1000 ÷ 125 = 8. Answer: 7/8. The distractors: 8/7 comes from cancelling correctly but writing the two parts the wrong way round; 9/10 comes from rounding 0.875 to one decimal place as 0.9 before converting; 7/80 comes from counting four decimal places instead of three and using a denominator of 10000, giving 875/10000.
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