Printable · GCSE Higher · ages 14-16
Number worksheet — GCSE Higher
Fifteen questions across the number statements at Higher 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 Higher
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- (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) 2700 cm² — Convert both lengths to the same unit before multiplying: 0.6 m = 60 cm. The area is 45 × 60 = 2700 cm². Multiplying without converting 0.6 m to centimetres at all, treating it as 0.6 cm, gives 45 × 0.6 = 27 cm². Converting 0.6 m to centimetres by multiplying by 10 instead of 100 gives 6 cm, and 45 × 6 = 270 cm². Converting 0.6 m to centimetres by multiplying by 1000 instead of 100 gives 600 cm, and 45 × 600 = 27000 cm².
- (a) 14 — Multiply both numbers by 10 to clear the decimals: 8.4 becomes 84 and 0.6 becomes 6. Then divide: 84 ÷ 6 = 14, so 14 complete pieces can be cut. Scaling only the divisor by 10 and leaving the dividend as 8.4 gives 8.4 ÷ 6 = 1.4, which rounds down to 1 complete piece — the dividend was never converted. Scaling only the dividend by 10 and leaving the divisor as 0.6 gives 84 ÷ 0.6 = 140. Rounding the divisor from 0.6 to 0.7 before dividing, trading accuracy for a rounder number, gives 8.4 ÷ 0.7 = 12. So 14 complete pieces of ribbon can be cut.
- (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.
- (d) 9 — Method: split into two cases — the soups with no restriction, and the mushroom soup on its own — then add the totals. Working: the 2 soups other than mushroom can be paired with any of the 4 sandwiches: 2 × 4 = 8. The mushroom soup can only be paired with the cheese sandwich: 1 combination. Total = 8 + 1 = 9. Answer: 9. 12 comes from working out 3 × 4 = 12 without applying the restriction at all. 8 comes from correctly finding the 2 unrestricted soups' 8 combinations, but forgetting to add back the 1 allowed mushroom-and-cheese combination. 11 comes from taking the unrestricted total of 12 and removing only 1 mushroom combination instead of all 3 disallowed ones.
- (d) 6 — Method: the square root of a product can be found either by multiplying first and then rooting, or by rooting each factor and multiplying the two roots together. Working: 4 × 9 = 36, and 6 × 6 = 36, so the root is 6; the same value comes from √4 × √9 = 2 × 3. Answer: 6. The distractors: 36 comes from multiplying inside the root and then leaving the root untaken; 5 comes from rooting each factor and adding the results, 2 + 3, instead of multiplying them; 18 comes from rooting the 4 only and leaving the 9 untouched, giving 2 × 9.
- (b) Yes, because 14.8 cm rounds to 15 cm to the nearest cm — Method: a recorded measurement is not an exact length; it stands for every length that rounds to it, so the two records agree if one rod can produce both. Working: Ben's record of 14.8 cm to the nearest 0.1 cm means the rod is between 14.75 cm and 14.85 cm, and 14.8 is nearer to 15 than to 14, so a rod of that length is recorded as 15 cm to the nearest centimetre. Both records can therefore come from the same rod. Answer: Yes, because 14.8 cm rounds to 15 cm to the nearest cm. The distractors: the claim that 14.8 cm rounds to 15.0 cm to 1 decimal place is false, since 14.8 cm is already written to 1 decimal place and stays 14.8 cm; the claim that it rounds to 14 cm is false, because 14.8 is 0.2 away from 15 and 0.8 away from 14; the claim that the two lengths are not the same treats each record as an exact length, when each is only a rounded record of one rod.
- (b) 4 m — Blue fabric is 3 parts and this equals 1.5 m, so one part is 1.5 ÷ 3 = 0.5 m. Yellow fabric is 5 parts, so it is 5 × 0.5 = 2.5 m. The total length is 1.5 + 2.5 = 4 m. 2.5 m is the length of yellow fabric only, without adding the blue fabric back in. 1.5 m is just the given length of blue fabric, with the yellow fabric never worked out. 2.4 m comes from swapping the ratio, treating blue as 5 parts and yellow as 3 parts, giving one part as 1.5 ÷ 5 = 0.3 m and yellow as 3 × 0.3 = 0.9 m, then adding 1.5 + 0.9 = 2.4.
- (d) 636 g — Let x = 0.636363... . Since two digits repeat, multiply by 100: 100x = 63.636363... . Subtracting removes the recurring part exactly: 100x − x = 63.636363... − 0.636363... = 63, so 99x = 63, giving x = 63/99 = 7/11 kg. Converting to grams: 7/11 × 1000 = 7000/11 = 636.3636... g, which rounds to 636 g. Treating the decimal as if it terminated, writing 0.63 as 63/100 kg, gives 630 g when multiplied by 1000 — this drops the recurring part entirely. Subtracting 10x instead of x, using 100x − 10x = 90x = 63, is the wrong power of ten for a two-digit block, giving x = 63/90 = 7/10 kg, which is 700 g. A numerator slip in the subtraction, 63 − 1 = 62 instead of 63, gives x = 62/99 kg, which is 62000/99 = 626.26... g, rounding to 626 g.
- (c) 7/30 — Let x = 0.2333... . Because only the 3 recurs, use two multiples of x that line up the recurring part exactly: 10x = 2.333... and 100x = 23.333... . Subtracting removes the recurring tail completely: 100x − 10x = 23.333... − 2.333... = 21, so 90x = 21, giving x = 21/90 = 7/30. Treating the decimal as if it terminated after two places, writing 0.23 as 23/100, ignores that the 3 carries on forever. Misreading which digits recur — treating 0.2333... as if the block '23' repeated, giving 0.232323... — leads to x = 23/99, which is a different, larger recurring decimal from the one given. A numerator slip in the subtraction, computing 22 instead of 21, gives x = 22/90 = 11/45.
- (d) −1.4, −6/5, 0, 5/4, 1.3 — Method: convert the fractions 5/4 and −6/5 to decimals so every number is written the same way, then compare all five decimals. Working: 5/4 = 1.25 and −6/5 = −1.2. Comparing −1.4, −1.2, 0, 1.25 and 1.3 in size gives the order −1.4, −1.2, 0, 1.25, 1.3. Answer: −1.4, −6/5, 0, 5/4, 1.3. −6/5, −1.4, 0, 5/4, 1.3 swaps the two negative numbers, treating −6/5 as more negative than −1.4 even though −1.2 is closer to zero than −1.4. 1.3, 5/4, 0, −6/5, −1.4 lists the numbers from largest to smallest instead of smallest to largest. −1.4, −6/5, 0, 1.3, 5/4 swaps 5/4 and 1.3, comparing the numerator 5 directly with 1.3 instead of converting 5/4 to the decimal 1.25 first.
- (c) 2/3 — Method: multiply the numerators together and the denominators together, then divide both parts of the result by their highest common factor. Working: 3 × 14 = 42 and 7 × 9 = 63, giving 42/63; the highest common factor of 42 and 63 is 21, and 42 ÷ 21 = 2 with 63 ÷ 21 = 3. Answer: 2/3. The distractors: 17/16 comes from adding the numerators and adding the denominators, giving (3 + 14)/(7 + 9); 27/98 comes from turning the second fraction upside down and multiplying, which divides instead of multiplying and gives 3/7 × 9/14; 2/21 comes from cancelling the 7 into the 14 in the numerator but leaving the 7 in the denominator, giving 6/63.
- (a) 200 g — One quarter of 160 g is 40 g. Increasing the amount means adding this on: 160 + 40 = 200 g. Finding the increase, 1/4 of 160 = 40 g, but stopping there without adding it to the original amount leaves just 40 g. Using 4/5 instead of 5/4 as the scaling fraction, 160 × 4/5 = 128 g, actually decreases the amount rather than increasing it. Increasing by a half instead of a quarter, 160 + 80 = 240 g, uses the wrong fraction of 160.
- (c) 5/11 — Let x = 0.45 recurring, so x = 0.454545... . Since two digits repeat, multiply by 100: 100x = 45.454545... . Subtracting the original x removes the recurring part exactly, because it lines up digit for digit: 100x − x = 45.454545... − 0.454545... = 45, so 99x = 45, giving x = 45/99 = 5/11. Treating the decimal as if it terminated at two places gives 45/100 = 9/20, which is only 0.45 and drops the repeating part entirely. Subtracting 10x instead of x — using 100x − 10x = 90x = 45 — is the wrong power of ten for a two-digit repeating block, and gives x = 45/90 = 1/2. Making an arithmetic slip in the numerator, 45 − 1 = 44 instead of 45, gives 44/99 = 4/9.
- (d) 10 — Method: list the pairs systematically, taking each sweet in turn and pairing it only with the sweets that come after it, so that no pair is written down twice. Working: numbering the sweets 1 to 5, the first sweet pairs with 4 others, the second pairs with 3 sweets that come after it, the third with 2 and the fourth with 1, so the total is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 20 comes from working out 5 × 4 and never halving, which counts each pair twice, once in each order; 25 comes from working out 5 × 5, which allows the same sweet to be chosen twice; 9 comes from adding the 5 choices and the 4 remaining choices instead of combining them as a selection of two.
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