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) 2.00 kg — Convert the tin's mass to kilograms first: 650 g = 0.65 kg. Adding this to the bag's mass gives 0.65 + 1.35 = 2.00 kg. Converting 650 g to kilograms by dividing by 100 instead of 1000 gives 6.5 kg, and adding this to 1.35 kg gives 7.85 kg. Adding the two masses without converting grams to kilograms at all — treating 650 as if it were already measured in kilograms — gives 651.35 kg. Subtracting the tin's mass from the bag's mass instead of adding the two together, 1.35 − 0.65, gives 0.70 kg.
- (d) 10 — Method: picking 3 flowers from 5 leaves 2 flowers behind, so counting the different pairs that could be left out counts the bunches, and those pairs can be listed systematically. Working: number the flowers 1 to 5; the first flower can be left out alongside any of the 4 flowers after it, the second alongside any of the 3 after it, the third alongside any of the 2 after it and the fourth alongside the last one, so the number of pairs left out is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 60 comes from working out 5 × 4 × 3 and treating the three picks as an ordered selection when the order does not matter; 30 comes from dividing that product by 2 instead of by the 6 orders in which three chosen flowers could have been picked; 15 comes from multiplying the 5 flowers by the 3 flowers picked instead of counting the selections.
- (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.
- (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.
- (d) 64 — Method: each time the power increases by 1, the value doubles, since one more 2 is multiplied in. Working: 2⁶ = 2⁵ × 2 = 32 × 2 = 64. Answer: 64. (37 comes from adding the power to the value, 32 + 5, instead of multiplying by the base. 192 comes from multiplying 32 by the new power, 6, instead of by the base, 2. 34 comes from adding the base to the value, 32 + 2, instead of multiplying.)
- (b) 245 ≤ m < 255 — Method: a mass given to the nearest 10 g lies within half of 10 g, that is 5 g, of the figure written down. The lower limit is included because it rounds up to that figure, and the upper limit is excluded because it rounds up to the next multiple of 10. Working: 250 − 5 = 245 and 250 + 5 = 255, and a mass of 245 g rounds to 250 g while a mass of 255 g rounds to 260 g. Answer: 245 ≤ m < 255. The distractors: 240 ≤ m < 260 goes a whole 10 g either side instead of half of it; 245 < m ≤ 255 has the two limits the wrong way round; 240 ≤ m < 250 treats the stated 250 g as the largest mass possible, as though rounding were always upwards.
- (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.
- (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.
- (d) 30 kg — Round 0.485 kg to 1 significant figure: 0.5 kg. Multiply by the 60 cakes: 0.5 × 60 = 30 kg. A candidate who rounded to 2 significant figures instead of 1 used 0.49 kg, giving 0.49 × 60 = 29.4 kg. A candidate who used the unrounded amount instead of the estimate worked out 0.485 × 60 = 29.1 kg. A candidate who rounded 0.485 down to 0.4 kg instead of up to 0.5 kg worked out 0.4 × 60 = 24 kg.
- (c) 25 cm — Method: for a cube, the edge length is the cube root of the volume. Working: 25 × 25 × 25 = 15,625, so the edge length is 25 cm. 5 cm comes from cube-rooting 125 instead of 15,625, misreading the number of digits. 50 cm comes from working out 25 × 2 = 50, doubling the correct edge length. 125 cm comes from taking the square root of the volume instead of the cube root, since 125 × 125 = 15,625 — that would be the side of a SQUARE of area 15,625, not the edge of a cube of that volume. Answer: 25 cm.
- (a) 17 — Method: work out each power separately, then combine them as the question asks. Working: $2^3 = 8$ and $3^2 = 9$, and 8 + 9 = 17. 72 comes from working out 8 × 9 = 72, multiplying the two powers instead of adding them. 12 comes from misreading the powers as repeated multiplication of the base by the index, 2 × 3 + 3 × 2 = 6 + 6 = 12. −1 comes from working out 8 − 9 = −1, subtracting the powers instead of adding them. Answer: 17.
- (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) 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.
- (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.
- (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.
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