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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- (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.
- (b) 24 — List multiples of 8 and of 12: multiples of 8 are 8, 16, 24, 32; multiples of 12 are 12, 24, 36. The lowest number in both lists is 24, so the lighthouses next flash together after 24 minutes. Multiplying the two numbers together, 8 × 12, gives 96, which double-counts the common factor of 4 shared by 8 and 12. Working out the highest common factor instead of the lowest common multiple gives 4, far too soon a time for both lighthouses to line up again. Adding the two numbers, 8 + 12, gives 20, which is not even a multiple of either 8 or 12. So the lighthouses next flash together after 24 minutes.
- (d) 4/3 — Method: the product of two negative numbers is positive, so work with 2/5 × 10/3 and then simplify. Multiply the numerators together and the denominators together. Working: 2 × 10 = 20 and 5 × 3 = 15, giving 20/15; both 20 and 15 divide by 5, so 20/15 = 4/3. Answer: 4/3. The distractors: −4/3 has the arithmetic right but keeps a minus sign, from treating negative × negative as negative; 3/25 comes from turning the second fraction upside down and multiplying, which divides instead of multiplying and gives 2/5 × 3/10 = 6/50; −56/15 comes from adding the two fractions instead of multiplying them, giving −6/15 − 50/15.
- (a) 1/16 — Method: terms can only be subtracted once they share a denominator, so write every term over the largest denominator, 16, and then subtract the numerators in order from left to right. Working: 1 = 16/16, 1/2 = 8/16, 1/4 = 4/16 and 1/8 = 2/16, so the numerators give 16 − 8 − 4 − 2 − 1 = 1, over a denominator of 16. Answer: 1/16. The distractors: 1/8 comes from stopping one term early, after 16 − 8 − 4 − 2 = 2; 3/16 comes from a sign slip on the last term, adding it instead of subtracting it, which gives 2 + 1 = 3; 15/16 comes from working from the right-hand end as though the last four terms were bracketed together, so that only a single sixteenth is taken away from 1.
- (a) 3.84 × 10⁵ — 384,000 = 3.84 × 100,000 = 3.84 × 10⁵, with the decimal point moved five places and the coefficient kept between 1 and 10. Moving the point six places instead of five gives 3.84 × 10⁶, ten times too large. Leaving the coefficient as 38.4 gives 38.4 × 10⁴, which is not between 1 and 10. Using a negative exponent instead of a positive one gives 3.84 × 10⁻⁵, a number far smaller than 1.
- (b) 30 — Method: 12% of an amount is 12/100 of it, so find 1% by dividing by 100 and then multiply by 12. Working: 1% of 250 is 250 ÷ 100 = 2.5, and 12% is 2.5 × 12 = 30. Answer: 30 seats. The distractors: 3 comes from writing 12% as 0.012 instead of 0.12, giving 0.012 × 250 = 3; 25 comes from finding 10% of the seats and stopping there; 24 comes from counting 12 seats for each whole hundred, 12 + 12 = 24, and ignoring the remaining 50 seats.
- (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.
- (b) 2² × 3 × 5 — Repeatedly divide 60 by prime numbers: 60 ÷ 2 = 30, 30 ÷ 2 = 15, 15 ÷ 3 = 5, and 5 is itself prime. So 60 is 2 × 2 × 3 × 5, which in index notation is 2² × 3 × 5. Stopping the factor tree after only three divisions and writing 2 × 3 × 5 misses that the 2 divides in twice, and gives only 30, not 60. Squaring the 3 as well as the 2 gives 2² × 3² × 5, which comes to 180, far too big. Squaring the 5 instead of the 2 gives 2 × 3 × 5², which comes to 150, also too big. So 60 = 2² × 3 × 5.
- (d) 6 + 2√3 — Multiply √3 by each term in the bracket separately. First term: √3 × 2 = 2√3. Second term: √3 × √12 = √(3 × 12) = √36 = 6. Adding the two results in the order they were found, and writing the whole-number term first, gives 6 + 2√3. Adding the numbers under the root for the second term instead of multiplying them (3 + 12 = 15) gives √15 in place of 6, leading to √15 + 2√3. Multiplying √3 by the 2 but never distributing to the √12 term at all leaves just 2√3. Treating √3 × 2 as if the 3 were multiplied by the 2 inside the root, √3 × 2 → √6, while still getting the second term correct, gives 6 + √6.
- (b) Yes — the actual mass could be as low as 995 g — Method: a mass shown to the nearest 10 g lies within half of 10 g, that is 5 g, of the figure on the display, so compare the smallest mass the bag can have with the checker's limit of 996 g. Working: 1,000 − 5 = 995, so the actual mass of the bag can be as low as 995 g, and 995 g is below the 996 g limit, so a bag showing 1,000 g on the machine can still be rejected. Answer: Yes — the actual mass could be as low as 995 g. The distractors: 990 g comes from going a whole 10 g below the display instead of half of it; 999.5 g comes from treating the display as being to the nearest gram, when it is to the nearest 10 g; the claim that the mass is exactly 1,000 g treats a rounded display as an exact measurement.
- (a) 8 × 10² — Method: to multiply numbers written in standard form, multiply the coefficients and add the indices. Working: 4 × 2 = 8 for the coefficients, and −3 + 5 = 2 for the indices; 8 already lies between 1 and 10, so no adjustment is needed. Answer: 8 × 10². The distractors: 6 × 10² comes from adding the coefficients, 4 + 2, instead of multiplying them; 8 × 10⁸ comes from ignoring the minus sign and adding 3 + 5; 8 × 10⁻¹⁵ comes from multiplying the indices, −3 × 5, instead of adding them.
- (a) 3/5 — The question compares the rock albums directly with the pop albums, so write the rock part over the pop part: 3/5. 5/3 comes from writing the ratio the wrong way round, comparing pop with rock instead of rock with pop. 3/8 comes from comparing the rock albums with the total number of albums, 3 + 5 = 8, instead of with the pop albums. 5/8 comes from comparing the pop albums with the total number of albums instead of the rock albums with the pop albums.
- (d) 26 and 27 — Find the two consecutive perfect squares either side of 700: 26² = 676 and 27² = 729. Since 676 < 700 < 729, √700 lies between 26 and 27. Answering 7 and 8 comes from stripping the two zeros off 700 and using the 7 itself as the size of the root, instead of comparing 700 with the perfect squares around it — dividing the number under the root by 100 divides the root by 10, so the digits do not simply carry across. Answering 25 and 26 comes from checking 25² = 625, seeing that it is less than 700, and stopping there without also checking the square directly above it. Answering 35 and 36 comes from halving 700 to 350 and then treating that halved value as if it were ten times the true root, drifting into the thirties instead of the twenties.
- (c) £8.20 — Method: find the total cost, then subtract from the amount paid. Working: 35 × £1.48 = £51.80. Amount paid = 3 × £20 = £60.00. Change = £60.00 − £51.80 = £8.20. Answer: £8.20. (£58.52 comes from forgetting to multiply the price by the 35 litres and subtracting only £1.48 from £60. £7.50 comes from rounding £1.48 up to £1.50 before multiplying, giving a total of £52.50 instead of £51.80. £9.20 comes from miscarrying in the pence column when subtracting £51.80 from £60.00.)
- (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.
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