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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Answer key: Number worksheet — GCSE Foundation
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- (d) £143 — From 09:15 to 12:45 is 3 hours 30 minutes, which is 3.5 hours. The charge for the time is 3.5 × £28 = £98, and the call-out fee is added on once: £98 + £45 = £143. £98 leaves out the call-out fee altogether, £129 charges for 3 hours instead of 3.5 hours, and £255.50 adds the fee to the hourly rate before multiplying, which charges the £45 for every hour.
- (a) 1,850 — Method: find the error interval, then check which value falls outside it. Working: half of 100 is 50, so the actual number of visitors, v, satisfies 1,750 ≤ v < 1,850. 1,850 sits exactly on the excluded upper boundary, since a value of 1,850 would round to 1,900, not 1,800. Answer: 1,850. (1,750 is a genuine possible value — it sits at the included lower boundary. 1,799 is a genuine possible value, just below the upper boundary. 1,760 is a genuine possible value, well inside the interval.)
- (a) 0.5, 0.55, 0.56, 0.6, 0.601 — Compare the decimals by giving them all the same number of decimal places first: 0.600, 0.550, 0.601, 0.500, 0.560. Ordering these from smallest to largest gives 0.500, 0.550, 0.560, 0.600, 0.601, which is 0.5, 0.55, 0.56, 0.6, 0.601. Comparing the digits as though they were whole numbers, reading 0.601 as "601" and 0.5 as "5", without padding to the same number of decimal places, gives the wrong order 0.5, 0.6, 0.55, 0.56, 0.601, because it ignores the place value of each digit. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 0.601, 0.6, 0.56, 0.55, 0.5. Misreading the close values 0.55 and 0.56 and swapping them gives 0.5, 0.56, 0.55, 0.6, 0.601. So the correct order, smallest to largest, is 0.5, 0.55, 0.56, 0.6, 0.601.
- (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) 90 km/h — To convert metres per second to kilometres per hour, multiply by 3.6 (there are 3600 seconds in an hour and 1000 metres in a kilometre, and 3600 ÷ 1000 = 3.6): 25 × 3.6 = 90 km/h. Dividing by 3.6 instead of multiplying gives 25 ÷ 3.6 = 6.9 km/h (to 1 d.p.). Multiplying by 60 instead of 3.6, confusing the conversion from seconds to minutes with the conversion to hours, gives 25 × 60 = 1500 km/h. Multiplying by 3600 to convert seconds to hours but forgetting to convert metres to kilometres gives 25 × 3600 = 90000, which is a speed in metres per hour, not kilometres per hour.
- (c) £384.00 — Adding 20% VAT means multiplying the price by 1.2: £320 × 1.2 = £384.00. Treating the 20% as a flat £20 rather than a percentage of the price, £320 + £20, gives £340.00. Working out the VAT amount alone, £320 × 0.2 = £64.00, and stopping there without adding it back to the original price gives just the VAT, not the total price. Misplacing the decimal point and using 2% instead of 20%, £320 × 1.02, gives £326.40.
- (d) 0.024 cm — Method: significant figures are counted from the first non-zero digit; the zeros in front of it only fix the place value and are not significant. Working: in 0.02384 the first significant figure is 2 and the second is 3, so the rounding is decided by the next digit, 8. As 8 is 5 or more, the second significant figure goes up from 3 to 4, in the same place value. Answer: 0.024 cm. The distractors: 0.023 cm comes from chopping the 84 off instead of rounding it; 0.02 cm comes from counting the leading zeros as significant figures, so the 2 is taken as the second figure and the rounding stops there; 0.0238 cm is 0.02384 correct to 3 significant figures, one figure too many.
- (c) £4.05 — The total cost is 7 × 85p = £5.95. Josh's change is £10.00 − £5.95 = £4.05. Taking off the pounds and then only the 5p digit, £10.00 − £5.00 = £5.00 followed by £5.00 − £0.05, ignoring the 90p altogether, gives £4.95. Rounding the cost up to £6.00 and never accounting for the extra 5p at all gives £4.00. Multiplying 7 × 85 incorrectly as 605 (a slip in 7 × 5) gives a total cost of £6.05, and correctly subtracting that from £10.00 gives £3.95.
- (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) 11 — Method: list every possible total from the smallest to the largest, and count how many different values there are. Working: the smallest total is 1+1=2 and the largest is 6+6=12, and every whole number total from 2 to 12 is possible: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 — that is 11 different totals. Answer: 11. 36 comes from counting the number of possible dice outcomes (6×6) instead of the number of different totals. 10 comes from listing the totals but missing one from the ends of the list, for example starting at 3 instead of 2. 6 comes from counting only the number of different scores on one die, not the totals of both dice together.
- (d) £20 — 1% of £250 = £2.50, so 8% = 8 × £2.50 = £20. A candidate who misplaces the decimal point and finds 0.8% instead gets £2. A candidate who confuses 8% with 80% gets £200. A candidate who rounds 8% up to the nearby 10% gets £25.
- (b) 3.65 ≤ n < 3.75 — Rounding to 1 decimal place means n can be up to half of one decimal place, 0.05, below or above 3.7 before it would round to a different value. This gives a lower bound of 3.7 − 0.05 = 3.65 and an upper bound of 3.7 + 0.05 = 3.75. A value exactly at 3.75 would round up to 3.8, not 3.7, so the upper bound is excluded while the lower bound, 3.65, does still round to 3.7. Writing 3.65 ≤ n ≤ 3.75 wrongly includes 3.75. Writing 3.6 ≤ n < 3.8 uses a whole decimal place, 0.1, either side instead of half of one, 0.05. Writing 3.65 < n < 3.75 wrongly excludes 3.65, which does round to 3.7.
- (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) £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.
- (c) 5/8 — Method: write the decimal over 1000 using its three decimal places, then simplify. Working: 0.625 = 625/1000 = 5/8 (dividing both numerator and denominator by 125). Answer: 5/8. 25/4 comes from writing the decimal over 100 instead of 1000, as if there were only two decimal places. 31/50 comes from rounding 0.625 to 0.62 before converting. 8/5 comes from simplifying correctly to 5/8 and then writing the fraction upside down.
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