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.Using only 20p coins and 10p coins, and at least one of each, work out how many different ways there are to make exactly 60p. List the possibilities systematically.
- 2.A shelf is 60 cm long, correct to the nearest 10 cm. Write down the upper bound for the length of the shelf, in centimetres.
- 3.Work out (5 × 10⁴) ÷ (2 × 10⁻²). Give your answer in standard form.
- 4.A charity raises money from a raffle and a cake sale in the ratio 5 : 3. Altogether the charity raises £320. Work out how much money the cake sale raised.
- 5.A train journey takes 45 minutes, correct to the nearest 5 minutes. Using t for the actual time of the journey in minutes, write down the error interval for t.
- 6.Round 6.283 to 1 significant figure.
- 7.A circle has a radius of 3 cm. Which of these is the exact area of the circle?
- 8.A photography studio offers 4 backdrops and 3 outfits for a portrait session. The plain grey backdrop, one of the 4 backdrops, cannot be used with the formal suit outfit, one of the 3 outfits. Work out how many different backdrop-and-outfit combinations are possible.
- 9.A car travels 180 km using 6 litres of fuel. Work out the car's fuel consumption in kilometres per litre, then work out how many kilometres it can travel on a full tank of 12 litres at this rate.
- 10.A cinema has 250 seats. 12% of the seats are reserved. Work out how many of the seats are reserved.
- 11.A jacket normally costs £65. In a sale it is reduced by 20%, and the shop then takes a further £5 off at the till. Work out the final price.
- 12.A courier's van has a weight limit of 850 kg for its parcels. The driver's display shows the total mass of the parcels loaded as 850 kg, correct to the nearest 5 kg. Decide whether the parcels are definitely within the weight limit.
- 13.Bella's electricity supplier charges 28.5p per unit (kWh). Last month she used 340 units. Work out the total cost, giving your answer in pounds.
- 14.A box stands on a table. The box exerts a force of 240 N on the table, and the base of the box has an area of 0.8 m². Work out the pressure the box exerts on the table, in N/m².
- 15.Work out (8 × 10⁻⁵) × (5 × 10³). Give your answer in standard form.
Answer key
- (a) 2 — Method: systematically try each possible number of 20p coins, starting from one, and check whether the amount left over can be made exactly using whole 10p coins. Working: one 20p coin leaves 40p, made from four 10p coins — valid. Two 20p coins leave 20p, made from two 10p coins — valid. Three 20p coins leave 0p, which needs zero 10p coins — not valid, since at least one 10p coin is required. So there are 2 different ways. Answer: 2. 3 comes from counting the case of three 20p coins and no 10p coins as if it were allowed, even though at least one 10p coin is required. 4 comes from ignoring the 'at least one of each' condition altogether and counting every way of making 60p, including three 20p coins with no 10p coins and six 10p coins with no 20p coins. 1 comes from finding only one of the two valid combinations and stopping the systematic list too early.
- (c) 65 cm — Method: the upper bound is half the rounding unit above the given value. Working: half of 10 cm is 5 cm, so the upper bound is 60 + 5 = 65 cm. Answer: 65 cm. (70 cm comes from adding the whole rounding unit, 10, instead of half of it. 60 cm comes from giving the rounded value itself rather than the upper bound. 55 cm comes from subtracting the half unit instead of adding it, giving the lower bound.)
- (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.
- (d) £120 — Total parts = 5 + 3 = 8, so one part is worth 320 ÷ 8 = 40 pounds. The cake sale is 3 parts, so it raised 3 × 40 = 120 pounds. £200 comes from working out the raffle's share, 5 × 40, instead of the cake sale's share. £40 comes from finding the value of one part but forgetting to multiply by 3. £192 comes from dividing the total by 5 instead of 8 to find the value of one part, 320 ÷ 5 = 64, then multiplying by 3, 3 × 64 = 192.
- (a) 42.5 ≤ t < 47.5 — Rounding to the nearest 5 minutes means the actual time can be up to half of 5 minutes, 2.5 minutes, below or above 45 before it would round to a different multiple of 5. The lower bound is 45 − 2.5 = 42.5 and the upper bound is 45 + 2.5 = 47.5. A time of exactly 47.5 minutes would round up to 50, not 45, so 47.5 is excluded while 42.5 does still round to 45. Writing 42.5 ≤ t ≤ 47.5 wrongly includes 47.5. Writing 40 ≤ t < 50 uses a whole rounding unit, 5, either side instead of half of it. Writing 44.5 ≤ t < 45.5 treats the rounding unit as 1 minute instead of 5 minutes.
- (c) 6 — Method: the first significant figure is the first non-zero digit; round using the digit after it to decide whether to round up or down. Working: the first significant figure of 6.283 is the 6; the next digit is 2, which rounds down, so 6.283 rounds to 6. 6.3 comes from rounding to 2 significant figures instead of 1. 10 comes from rounding up to the nearest 10 instead of finding 1 significant figure of the number itself. 0.6 comes from misplacing the decimal point after rounding. Answer: 6.
- (c) 9π cm² — The area of a circle is π × r². With a radius of 3 cm this is π × 3² = 9π cm², and this is exact because π has not been replaced by any approximation. Writing 28.3 cm² replaces π with a rounded decimal value, 3.14, and then rounds the result again, so it is only an approximation. Writing 28.26 cm² uses π ≈ 3.14 without a final rounding step, but this is still only an approximation of 9π, not the exact value. Writing 27 cm² comes from replacing π with the rough approximation 3, which is even further from the true value.
- (c) 11 — Method: work out the total number of combinations as if there were no restriction, then subtract the one combination that is not allowed. Working: without any restriction there are 4 backdrops × 3 outfits = 12 combinations. The grey backdrop with the formal suit is not allowed, removing 1 combination: 12 − 1 = 11. Answer: 11. 12 comes from forgetting to remove the combination that is not allowed. 8 comes from removing the entire formal suit outfit from the count instead of just the one combination with the grey backdrop. 10 comes from removing two combinations instead of just the one that is not allowed.
- (d) 360 km — Method: first find the kilometres per litre by dividing distance by fuel used, then multiply this rate by the new tank size. Working: 180 ÷ 6 = 30 km per litre; 30 × 12 = 360 km. Answer: 360 km. 30 km comes from finding the correct fuel consumption but stopping there, without scaling it up to the full tank. 2160 km comes from multiplying the original distance (180) by the tank size (12) directly, skipping the unit rate. 90 km comes from pairing the numbers the wrong way round: dividing the distance by the new tank size, 180 ÷ 12 = 15, and then multiplying by the original 6 litres, 15 × 6 = 90.
- (b) 30 — Method: 12% of an amount is 12/100 of it, so find 1% by dividing by 100 and then multiply by 12. Working: 1% of 250 is 250 ÷ 100 = 2.5, and 12% is 2.5 × 12 = 30. Answer: 30 seats. The distractors: 3 comes from writing 12% as 0.012 instead of 0.12, giving 0.012 × 250 = 3; 25 comes from finding 10% of the seats and stopping there; 24 comes from counting 12 seats for each whole hundred, 12 + 12 = 24, and ignoring the remaining 50 seats.
- (c) £47.00 — First apply the 20% reduction: £65 × 0.8 = £52.00. Then take off the further £5: £52.00 − £5 = £47.00. Treating the 20% as a flat £20 rather than a percentage of the price, £65 − £20 − £5, gives £40.00. Applying the 20% reduction correctly but forgetting to take off the extra £5 leaves £52.00. Taking off the £5 first and then applying the 20% reduction to the smaller amount, (£65 − £5) × 0.8, gives £48.00.
- (b) No, the true mass could be as high as 852.5 kg — Method: find the upper bound of the true mass and compare it with the weight limit. Working: the display is correct to the nearest 5 kg, so half of 5 kg is 2.5 kg, and the true mass, m kg, satisfies 847.5 ≤ m < 852.5. Part of that interval lies above 850 kg, so the parcels are not definitely within the limit. Answer: the true mass could be as high as 852.5 kg, which is above the limit. ("Yes, the display reads 850 kg, which is not above the limit" compares the limit with the displayed value instead of with the largest value the true mass could take. "Yes, the true mass is at least 847.5 kg and at most 850 kg" uses the correct half unit below but caps the interval at the limit instead of at 852.5 kg. "No, 850 kg on the display rounds up to 855 kg" wrongly treats the displayed value as if it rounds again.)
- (b) £96.90 — Method: multiply the number of units by the cost per unit, then convert the result from pence to pounds. Working: 340 × 28.5p = 9690p; converting to pounds, 9690p ÷ 100 = £96.90. Answer: £96.90. £969.00 comes from misplacing the decimal point when converting pence to pounds, dividing by 10 instead of 100. £9.69 comes from misplacing the decimal point the other way, dividing by 1000 instead of 100. £102.00 comes from rounding the rate to 30p per unit before multiplying.
- (a) 300 N/m² — Pressure is force ÷ area, and the unit N/m² says so: newtons divided by square metres. The calculation is 240 ÷ 0.8. Multiplying both numbers by 10 clears the decimal: 2400 ÷ 8 = 300 N/m². 192 N/m² multiplies the force by the area instead of dividing, 30 N/m² divides by 8 and so loses the decimal point, and 3000 N/m² multiplies only the force by 10 when clearing the decimal.
- (c) 4 × 10⁻¹ — Multiply the A values: 8 × 5 = 40. Add the powers of 10: −5 + 3 = −2, giving 40 × 10⁻². Since A must satisfy 1 ≤ A < 10, rewrite 40 as 4 × 10¹, so 40 × 10⁻² = 4 × 10¹ × 10⁻² = 4 × 10⁻¹. A candidate who stopped at 40 × 10⁻² did the index arithmetic correctly but left the answer outside standard form, since 40 is not between 1 and 10. A candidate who adjusted the A value to 4 correctly but then took the power of 10 by subtracting the two given powers, −5 − 3 = −8, wrote 4 × 10⁻⁸. A candidate who adjusted the A value to 4 but multiplied the two given powers, −5 × 3 = −15, wrote 4 × 10⁻¹⁵. Both of these forgot that multiplying in standard form means adding the powers.
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