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.
Answer key: Number worksheet — GCSE Higher
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- (c) 693 — Method: multiplying a bracket by a number multiplies every term inside it, so 7 × (100 − 1) = 7 × 100 − 7 × 1. Working: 7 × 100 = 700 and 7 × 1 = 7, so the calculation becomes 700 − 7 = 693. Answer: 693. The distractors: 699 comes from subtracting the 1 itself rather than 7 lots of it, giving 700 − 1 = 699; 707 comes from adding the second product instead of subtracting it, giving 700 + 7 = 707; 700 comes from rounding 99 up to 100 and then offering the estimate 7 × 100 as an exact value.
- (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.
- (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.
- (a) 1/3 — To subtract these fractions, first write 3/4 with a denominator of 12: 3/4 = 9/12. Then 9/12 − 5/12 = 4/12, which simplifies to 1/3. Subtracting the numerators and the denominators separately, (3 − 5)/(4 − 12), gives −2/−8, which simplifies to 1/4. Changing 3/4 to twelfths by only changing the denominator, without scaling the numerator to match, gives 3/12 − 5/12 = −2/12, which simplifies to −1/6. Adding the fractions instead of subtracting them, 9/12 + 5/12, gives 14/12, which simplifies to 7/6.
- (b) £300 — The increased price is 110% of the original, so the original price = £330 ÷ 1.1 = £300. A candidate who finds 10% of £330 and subtracts it, wrongly treating £330 as the original, gets £330 − £33 = £297. A candidate who adds 10% of £330 again instead of reversing the increase gets £330 + £33 = £363. A candidate who divides by 0.1 instead of 1.1 gets £3,300.
- (a) 30 cm — The tile's side length must be a common factor of 90 and 120. The factors of 90 include 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90; the factors of 120 include 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. The highest number common to both lists is 30, so the largest square tile has a side length of 30 cm. Picking 15 cm, a common factor but not the largest, gives tiles that are smaller than necessary. Picking 10 cm, also a common factor but smaller still, wastes even more of the possible tile size. Working out the lowest common multiple instead of the highest common factor gives 360 cm, a length far bigger than either side of the patio. So the largest square tile Ben can use has a side length of 30 cm.
- (a) 20 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 588 rounds to 600 (1 s.f.) and 31 rounds to 30 (1 s.f.). 600 ÷ 30 = 20. Answer: 20. 17 comes from cutting 588 down to 500, keeping the leading digit as it stands instead of rounding it up to 1 significant figure, 600, then dividing by the correctly rounded 30. 200 comes from misreading the rounded divisor 30 as 3, giving 600 ÷ 3 instead of 600 ÷ 30. 19 is the exact value of 588 ÷ 31 rounded to the nearest whole number, found without rounding the numbers first.
- (a) −22 — Method: both multiplications are carried out before the addition, and a positive multiplied by a negative is negative. Working: (−3) × 4 = −12 and 2 × (−5) = −10, so the calculation becomes −12 + (−10) = −22. Answer: −22. The distractors: 22 comes from ignoring the minus signs and working out 3 × 4 + 2 × 5 = 22; 50 comes from working from left to right with no priority at all, giving −12 + 2 = −10 and then −10 × (−5) = 50; −2 comes from taking 2 × (−5) as +10, so that −12 + 10 = −2.
- (b) −0.7 < −0.25 — Method: compare the two negative decimals by their distance from zero on a number line. Working: −0.7 is 0.7 away from zero and −0.25 is 0.25 away from zero, so −0.7 is further from zero in the negative direction, making it the smaller number. Answer: −0.7 < −0.25 is true. "−0.7 > −0.25" comes from comparing 0.7 and 0.25 as if both numbers were positive, ignoring the negative signs. "−0.7 = −0.25" comes from assuming the two numbers are equal because they are both negative decimals. "−0.7 ≥ −0.25" combines the false statement "−0.7 > −0.25" with the false statement "−0.7 = −0.25".
- (b) 120 — Method: fill the positions on the shelf one at a time; each book placed leaves one fewer book available for the next position, and the product rule multiplies the choices. Working: there are 5 books for the first position, 4 for the second, 3 for the third, 2 for the fourth and 1 for the last, so the number of orders is 5 × 4 × 3 × 2 × 1 = 120. Answer: 120. The distractors: 25 comes from multiplying the 5 books by the 5 positions rather than multiplying the shrinking number of choices at each position; 60 comes from halving the correct product, as though each order had been counted twice in the way that pairs are; 720 comes from carrying the product one factor too far and working out 6 × 5 × 4 × 3 × 2 × 1, as though there were six books.
- (d) £800 — Rounding 38.7 to 1 significant figure gives 40, and rounding 21.40 to 1 significant figure gives 20. Multiplying the rounded values gives an estimate of 40 × 20 = £800. Rounding 21.40 to the nearest whole number instead of to 1 significant figure gives 21, and 40 × 21 = £840, one place value too fine for the price. Adding the rounded values instead of multiplying them gives 40 + 20 = £60. Rounding both numbers to 2 significant figures instead of 1, giving 39 and 21, produces 39 × 21 = £819.
- (b) 1/9 — Method: let x stand for the recurring decimal, multiply by the power of ten that moves exactly one repeating block past the decimal point, subtract the original equation so that the recurring tail cancels, and solve the equation that is left. Working: let x = 0.111...; the repeating block is one digit long, so multiply by 10 to give 10x = 1.111...; subtracting the first equation from the second gives 10x − x = 1.111... − 0.111..., that is 9x = 1; dividing both sides by 9 gives x = 1/9. Answer: 1/9. The distractors: 1/10 comes from dividing by the multiplier 10 at the last step instead of by the 9 that is left in front of x; 1/11 comes from recalling the elevenths family instead of the ninths, although 1/11 = 0.0909... has a two-digit repeating block rather than a one-digit one; 11/100 comes from stopping the decimal after two digits and converting 0.11 into hundredths.
- (a) 28.8 km/h — A compound unit is converted one part at a time. There are 3600 seconds in an hour, so in one hour the cyclist travels 8 × 3600 = 28 800 metres. There are 1000 metres in a kilometre, so 28 800 m = 28 800 ÷ 1000 = 28.8 km/h. 28 800 km/h leaves the distance in metres, 0.48 km/h converts the seconds to minutes rather than to hours, and 2.22 km/h divides by 3.6 instead of multiplying.
- (a) 18 — Method: round each number to the nearest whole number, then square each rounded number and add the results. Working: 2.9 rounds to 3 and 3.1 rounds to 3, so the estimate is 3² + 3² = 9 + 9. Answer: 18. The distractors: 36 comes from adding before squaring, working out (3 + 3)² instead of 3² + 3²; 12 comes from doubling each rounded number instead of squaring it, adding 6 and 6; 6 comes from adding the two rounded numbers and forgetting to square them at all.
- (b) 7 — Method: trap the number between the two square numbers on either side of it, then decide which of them it is nearer to. Working: 6² = 36 and 7² = 49, so √45 lies between 6 and 7; 49 − 45 = 4 while 45 − 36 = 9, so 45 is nearer to 49. Answer: 7. The distractors: 6 comes from taking the square number below 45 and stopping there, without checking which of 36 and 49 is nearer; 22.5 comes from halving 45 instead of looking for the number that multiplies by itself to give 45; 2,025 comes from squaring 45 instead of taking its square root.
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