Printable · GCSE Foundation · ages 14-16
Number worksheet — GCSE Foundation
Fifteen questions across the number statements at Foundation 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 Foundation
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- 1.Four friends share a restaurant bill of £53.90 equally. A calculator gives each share as 13.475. Work out how much each friend should pay.
- 2.Work out (4 × 10³) × (3 × 10¹). Give your answer in standard form.
- 3.After a price increase of 10%, a laptop costs £330. Work out the original price.
- 4.A water tank holds 80 litres when full. It currently contains 60 litres. Work out what fraction of the tank is empty.
- 5.A recipe states that a cake needs 200 g of flour, correct to the nearest 25 g. Work out the smallest possible mass of flour, in grams, that could actually be used.
- 6.A carton of orange juice holds 1.35 litres. Ruby pours the juice equally into 4 identical glasses. Work out how much juice is in each glass, giving your answer as a fraction of a litre in its simplest form.
- 7.The density of a type of plastic is 0.9 g/cm³. Work out this density in kg/m³.
- 8.A cube-shaped storage box has edges of length 7 cm. A shelf can hold a total volume of 2,000 cm³. Work out the greatest number of these boxes that will fit in that volume.
- 9.Postage on a parcel is calculated as £4.60, correct to the nearest 20p. Which of these could not be the actual cost of the postage?
- 10.The error interval for the mass of a suitcase, m kg, is given as 22.5 ≤ m < 23.5. Write down the mass of the suitcase, correct to the nearest whole number.
- 11.A tap fills a 20 litre bucket in 2 minutes 30 seconds. Work out the rate of flow, in litres per minute.
- 12.Work out (5 × 10⁴) ÷ (2 × 10⁻²). Give your answer in standard form.
- 13.In a science experiment, the temperature of a liquid is recorded as 18.6 °C, correct to the nearest 0.2 °C. Write down the error interval for the actual temperature, T °C.
- 14.A charity raises 8 × 10⁶ pounds. This total is shared equally among 2 × 10² local projects. Work out how much each project receives, in standard form.
- 15.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.
Answer key
- (b) £13.48 — Method: an amount of money is written to the nearest penny, which is 2 decimal places, so the calculator display has to be rounded to 2 decimal places. Working: 53.90 ÷ 4 = 13.475, and the digit in the third decimal place is 5, so the penny digit goes up from 7 to 8. Answer: £13.48. The distractors: £13.47 comes from chopping the third decimal place off instead of rounding with it; £13.50 comes from rounding to the nearest 10p rather than to the nearest penny; £13.40 comes from cutting the display short at 1 decimal place, which is both the wrong degree of accuracy and a truncation rather than a rounding.
- (d) 1.2 × 10⁵ — 4 × 3 = 12, and 3 + 1 = 4, giving 12 × 10⁴ — but 12 is not between 1 and 10, so this must be rewritten as 1.2 × 10⁵. Stopping at 12 × 10⁴ without rewriting it leaves the coefficient out of range. Rewriting 12 as 1.2 but leaving the exponent at 4 instead of increasing it to 5 gives 1.2 × 10⁴, which is ten times too small. Adding the coefficients instead of multiplying them gives 4 + 3 = 7, so 7 × 10⁴.
- (b) £300 — The increased price is 110% of the original, so the original price = £330 ÷ 1.1 = £300. A candidate who finds 10% of £330 and subtracts it, wrongly treating £330 as the original, gets £330 − £33 = £297. A candidate who adds 10% of £330 again instead of reversing the increase gets £330 + £33 = £363. A candidate who divides by 0.1 instead of 1.1 gets £3,300.
- (a) 1/4 — The empty part of the tank is 80 − 60 = 20 litres. As a fraction of the full capacity, this is 20/80, which simplifies to 1/4. Finding the fraction of the tank that is FULL instead of empty, 60/80, simplifies to 3/4 — the wrong quantity for the question asked. Writing the empty amount over the amount remaining instead of over the full capacity, 20/60, simplifies to 1/3. Comparing the empty amount to 100 instead of to the tank's actual capacity of 80, 20/100, gives 1/5.
- (a) 187.5 g — Method: the smallest possible actual mass is half the rounding unit below the given value. Working: half of 25 g is 12.5 g, so the smallest possible mass is 200 − 12.5 = 187.5 g. Answer: 187.5 g. (175 g comes from subtracting the whole rounding unit, 25, instead of half of it. 200 g comes from giving the rounded value itself rather than the lower bound. 212.5 g comes from adding the half unit instead of subtracting it, giving the upper bound.)
- (d) 27/80 — Method: write the total as a fraction of a litre, then divide by the number of glasses. Working: 1.35 = 27/20, so each glass holds 27/20 ÷ 4 = 27/80 of a litre. Answer: 27/80. 27/20 comes from converting the total correctly to a fraction but forgetting to divide by the number of glasses. 27/5 comes from multiplying the total by 4 instead of dividing. 17/50 comes from rounding 1.35 ÷ 4 to 0.34 before converting to a fraction.
- (a) 900 kg/m³ — Both parts of the compound unit change. There are 100 cm in a metre, so 1 m³ = 100³ = 1 000 000 cm³, and there are 1000 g in a kilogram. So 1 g/cm³ = 1 000 000 ÷ 1000 = 1000 kg/m³, and the plastic is 0.9 × 1000 = 900 kg/m³. 900 000 kg/m³ converts the volume but leaves the mass in grams, 0.0009 kg/m³ converts the mass but leaves the volume in cm³, and 0.09 kg/m³ uses the 100 cm in a metre without cubing it.
- (c) 5 — Method: work out the volume of one box, divide the total volume by it, then round down since a partial box cannot fit. Working: volume of one box = 7³ = 343 cm³. 2000 ÷ 343 = 5.83 (2 d.p.). Since only whole boxes fit, the greatest number is 5. Answer: 5. (6 comes from rounding 5.83 up to the nearest whole number instead of rounding down to the number of boxes that actually fit. 343 comes from giving the volume of one box instead of the number of boxes. 5.8 comes from leaving the division as a decimal instead of rounding down to a whole number of boxes.)
- (d) £4.80 — Method: find the error interval, then check which value falls outside it. Working: half of 20p is 10p, so the actual cost, c, satisfies £4.50 ≤ c < £4.70. £4.80 is above £4.70, so it could not be the actual cost. Answer: £4.80. (£4.50 is a genuine possible cost — it sits at the included lower boundary. £4.65 is a genuine possible cost, below the £4.70 upper boundary. £4.55 is a genuine possible cost, well inside the interval.)
- (b) 23 kg — Method: the rounded value sits exactly in the middle of the error interval. Working: the interval 22.5 ≤ m < 23.5 stretches 0.5 either side of the rounded value, so the rounded value is 23. Answer: 23 kg. (22 kg comes from rounding the lower bound down instead of finding the middle of the interval. 22.5 kg comes from giving the lower bound itself rather than the rounded value. 23.5 kg comes from giving the upper bound itself rather than the rounded value.)
- (a) 8 litres per minute — Method: write the time as a decimal number of minutes, then divide the volume by the time. Working: 30 seconds = 30/60 minute = 0.5 minute, so 2 minutes 30 seconds = 2.5 minutes. Rate = 20 ÷ 2.5 = 8 litres per minute. Answer: 8 litres per minute. (10 litres per minute comes from ignoring the extra 30 seconds and dividing by 2 minutes only. 8.7 litres per minute comes from misreading 2 minutes 30 seconds as 2.3 minutes instead of 2.5 minutes. 0.125 litres per minute comes from dividing the time by the volume instead of the volume by the time.)
- (c) 2.5 × 10⁶ — Divide the A values: 5 ÷ 2 = 2.5. Subtract the powers of 10: 4 − (−2) = 4 + 2 = 6. So the answer is 2.5 × 10⁶. A candidate who worked out 4 − 2 = 2, treating the subtraction of a negative as an ordinary subtraction, wrote 2.5 × 10². A candidate who subtracted in the wrong order, −2 − 4 = −6, wrote 2.5 × 10⁻⁶. A candidate who multiplied the A values instead of dividing, 5 × 2 = 10, and added the powers, 4 + (−2) = 2, then rewrote 10 × 10² in standard form as 1 × 10³.
- (b) 18.5 ≤ T < 18.7 — Method: the error interval reaches half the rounding unit either side of the recorded value. Working: half of 0.2 is 0.1, so the interval runs from 18.6 − 0.1 to 18.6 + 0.1. Answer: 18.5 ≤ T < 18.7. (18.4 ≤ T < 18.8 comes from using the full rounding unit, 0.2, either side instead of half of it. 18.5 ≤ T ≤ 18.7 comes from including the upper bound with ≤ instead of excluding it with <. 18.6 ≤ T < 18.8 comes from treating the recorded value as the start of the interval and adding the whole rounding unit, 0.2, above it.)
- (c) 4 × 10⁴ — 8 ÷ 2 = 4, and 6 − 2 = 4, so each project receives 4 × 10⁴ pounds. Multiplying the exponents instead of subtracting them gives 6 × 2 = 12, so 4 × 10¹². Adding the exponents instead of subtracting them gives 6 + 2 = 8, so 4 × 10⁸. Subtracting the coefficients instead of dividing them gives 8 − 2 = 6, so 6 × 10⁴.
- (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.
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