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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- (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.)
- (a) 8.35 ≤ L < 8.45 — Method: a length rounded to 1 decimal place lies within half of 0.1 cm, that is 0.05 cm, of the value written down. The lower limit is included because it rounds up to that value, and the upper limit is excluded because it rounds up to the next value instead. Working: 8.4 − 0.05 = 8.35 and 8.4 + 0.05 = 8.45, so a length of 8.35 cm still rounds to 8.4 cm while a length of 8.45 cm rounds to 8.5 cm. Answer: 8.35 ≤ L < 8.45. The distractors: 8.3 ≤ L < 8.5 goes a whole 0.1 cm either side instead of half of it; 8.35 < L ≤ 8.45 has the two limits the wrong way round, excluding the length that does round to 8.4 cm and including the one that does not; 8.4 ≤ L < 8.5 is the interval for a length truncated to 1 decimal place, not one rounded to it.
- (c) 12.3 ≤ t < 12.4 — Method: truncating cuts the later digits off instead of rounding them, so nothing is ever pushed upwards. The displayed value is therefore the smallest the time can be, and the time can run up to, but not reach, the next value the display can show. Working: the display reads 12.3, so the actual time is at least 12.3 seconds; as soon as the time reaches 12.3 + 0.1 = 12.4 seconds the display would read 12.4, so 12.4 is not included. Answer: 12.3 ≤ t < 12.4. The distractors: 12.25 ≤ t < 12.35 is the interval for a time rounded to 1 decimal place, and this display does not round; 12.3 < t ≤ 12.4 excludes the one value the display certainly allows and includes the one it rules out; 12.3 ≤ t ≤ 12.4 treats 12.4 seconds as possible, but at 12.4 seconds the display would no longer read 12.3.
- (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.
- (d) £6 — First find 30% of £200, which is £60, then find 10% of that: £60 × 0.1 = £6. Adding the two percentages together instead of applying them one after the other, 10% + 30% = 40%, and finding 40% of £200 gives £80. Finding 30% of £200 = £60 correctly but stopping before applying the second percentage leaves £60 as the final answer. Finding only 10% of the original £200, ignoring the 30% entirely, gives £20.
- (a) 33.5 mph — Method: the smallest possible actual value is half the rounding unit below the given value. Working: half of 1 mph is 0.5 mph, so the smallest possible speed is 34 − 0.5 = 33.5 mph. Answer: 33.5 mph. (33 mph comes from subtracting the whole rounding unit, 1, instead of half of it. 34 mph comes from giving the rounded value itself rather than the lower bound. 34.5 mph 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) 7.2 × 10⁻⁴ — 0.00072 = 7.2 × 10⁻⁴, moving the decimal point 4 places to the right to reach 7.2, so the exponent is negative. Writing 7.2 × 10⁴ uses a positive exponent, which would give a number far larger than 1, not a small decimal. Writing 7.2 × 10⁻⁵ moves the point one place too many. Writing 0.72 × 10⁻³ leaves the coefficient below 1, which standard form does not allow.
- (c) 4.1 × 10³, 3.2 × 10⁴, 2.9 × 10⁵ — The exponent decides the size first: 10³ is smaller than 10⁴, which is smaller than 10⁵, so the order is 4.1 × 10³, then 3.2 × 10⁴, then 2.9 × 10⁵. Reversing the whole list gives largest to smallest instead of smallest to largest. Comparing 3.2 × 10⁴ and 4.1 × 10³ by their coefficients alone, 3.2 against 4.1, and swapping them ignores that 10⁴ is bigger than 10³ regardless of the coefficient. Comparing 2.9 × 10⁵ and 3.2 × 10⁴ by their coefficients alone and swapping them makes the same mistake at the top of the list.
- (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.
- (b) 0.3, 32%, 7/20 — Method: convert every number to a decimal so they can be compared on the same scale. Working: 7/20 = 0.35, 0.3 stays as 0.3, and 32% = 0.32. Comparing 0.3, 0.32 and 0.35 in size gives the order 0.3, then 0.32, then 0.35. Answer: 0.3, 32%, 7/20. 7/20, 32%, 0.3 lists the numbers from largest to smallest instead of smallest to largest. 0.3, 7/20, 32% swaps 32% and 7/20, treating the fraction 7/20 as smaller even though 7/20 = 0.35 is bigger than 32% = 0.32. 32%, 0.3, 7/20 comes from moving the digits one place too far when converting the percentage, giving 0.032 instead of 0.32, which makes 32% look far smaller than it really is.
- (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.
- (a) £600 — Method: round the number of cakes and the price of each cake to 1 significant figure, then multiply the rounded values. Working: 187 rounds to 200, and £2.95 rounds to £3 (the digit after the first, 9, rounds the 2 up to 3), so the estimate is 200 × £3 = £600. £400 comes from rounding £2.95 down to £2 instead of up to £3, giving 200 × £2 = £400. £561 comes from rounding only the price and using the exact number of cakes, 187 × £3 = £561. £570 comes from rounding 187 to the nearest 10 as 190 instead of to 1 significant figure as 200, giving 190 × £3 = £570. Answer: £600.
- (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.
- (c) 0.0065 — Leading zeros are never significant, so counting from the first non-zero digit, the first two significant figures of 0.006482 are 6 and 4. Look at the next digit along, 8, to decide whether the second figure rounds up: since 8 is 5 or more, the 4 rounds up to 5, giving 0.0065. Rounding to 2 decimal places instead of 2 significant figures gives 0.01, which answers a different question. Wrongly counting one of the leading zeros as a significant figure and stopping one figure short gives 0.006. Keeping an extra digit, as in 0.00648, gives 3 significant figures rather than 2.
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