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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- (d) 50 km/h — The error intervals are 99.5 ≤ distance < 100.5 and 1.95 ≤ time < 2.05. The minimum speed is 99.5 ÷ 2.05 ≈ 48.54 km/h, and the maximum speed is 100.5 ÷ 1.95 ≈ 51.54 km/h. These two bounds round to different whole numbers, 49 and 52, so the speed cannot be guaranteed to the nearest whole number — but every value between them rounds to 50 at the nearest 10, so 50 km/h is the value that can safely be guaranteed. Quoting 49 km/h uses only the minimum bound's rounding, without checking that the maximum bound rounds to something different. Quoting 52 km/h makes the same mistake using only the maximum bound instead. Quoting 48.54 km/h states one bound to the full accuracy a calculator shows, as if the smallest possible speed were the answer, when the true speed could be anything up to 51.54 km/h.
- (a) 0.5, 0.55, 0.56, 0.6, 0.601 — Compare the decimals by giving them all the same number of decimal places first: 0.600, 0.550, 0.601, 0.500, 0.560. Ordering these from smallest to largest gives 0.500, 0.550, 0.560, 0.600, 0.601, which is 0.5, 0.55, 0.56, 0.6, 0.601. Comparing the digits as though they were whole numbers, reading 0.601 as "601" and 0.5 as "5", without padding to the same number of decimal places, gives the wrong order 0.5, 0.6, 0.55, 0.56, 0.601, because it ignores the place value of each digit. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 0.601, 0.6, 0.56, 0.55, 0.5. Misreading the close values 0.55 and 0.56 and swapping them gives 0.5, 0.56, 0.55, 0.6, 0.601. So the correct order, smallest to largest, is 0.5, 0.55, 0.56, 0.6, 0.601.
- (d) £70 — Method: the discount is a percentage of the original price only, so work it out, subtract it, and add the fixed delivery charge afterwards. Working: 25% of £80 is £80 ÷ 4 = £20, so the discounted price is £80 − £20 = £60, and the total is £60 + £10 = £70. Answer: £70. The distractors: £60 comes from working out the discounted price and stopping there, leaving the delivery charge out of the total; £65 comes from taking £25 off the price instead of 25% of it, giving £80 − £25 = £55 and then £55 + £10 = £65; £67.50 comes from adding the delivery charge before the discount and reducing the whole amount, giving 75% of £90 = £67.50.
- (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) £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.
- (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) £96.90 — Method: multiply the number of units by the cost per unit, then convert the result from pence to pounds. Working: 340 × 28.5p = 9690p; converting to pounds, 9690p ÷ 100 = £96.90. Answer: £96.90. £969.00 comes from misplacing the decimal point when converting pence to pounds, dividing by 10 instead of 100. £9.69 comes from misplacing the decimal point the other way, dividing by 1000 instead of 100. £102.00 comes from rounding the rate to 30p per unit before multiplying.
- (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.
- (a) 1,850 — Method: find the error interval, then check which value falls outside it. Working: half of 100 is 50, so the actual number of visitors, v, satisfies 1,750 ≤ v < 1,850. 1,850 sits exactly on the excluded upper boundary, since a value of 1,850 would round to 1,900, not 1,800. Answer: 1,850. (1,750 is a genuine possible value — it sits at the included lower boundary. 1,799 is a genuine possible value, just below the upper boundary. 1,760 is a genuine possible value, well inside the interval.)
- (c) 2.5 × 10⁶ — Divide the A values: 5 ÷ 2 = 2.5. Subtract the powers of 10: 4 − (−2) = 4 + 2 = 6. So the answer is 2.5 × 10⁶. A candidate who worked out 4 − 2 = 2, treating the subtraction of a negative as an ordinary subtraction, wrote 2.5 × 10². A candidate who subtracted in the wrong order, −2 − 4 = −6, wrote 2.5 × 10⁻⁶. A candidate who multiplied the A values instead of dividing, 5 × 2 = 10, and added the powers, 4 + (−2) = 2, then rewrote 10 × 10² in standard form as 1 × 10³.
- (b) 23 kg — Method: the rounded value sits exactly in the middle of the error interval. Working: the interval 22.5 ≤ m < 23.5 stretches 0.5 either side of the rounded value, so the rounded value is 23. Answer: 23 kg. (22 kg comes from rounding the lower bound down instead of finding the middle of the interval. 22.5 kg comes from giving the lower bound itself rather than the rounded value. 23.5 kg comes from giving the upper bound itself rather than the rounded value.)
- (d) 40 miles — Method: round each number to 1 significant figure first, then divide to estimate the daily distance. Working: 830 rounds to 800, and 19 rounds to 20, and 800 ÷ 20 = 40, so the estimate is 40 miles per day. 41.5 miles comes from rounding only the number of days and working out 830 ÷ 20 = 41.5, without rounding the distance too. 830 miles is the total distance for the whole trek, given as the answer without dividing by the number of days at all. 4 miles comes from working out 80 ÷ 20 = 4, misplacing a digit in the rounded distance. Answer: 40 miles.
- (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) 6 — Method: the two scoops sit in different places on the cone, so a cone is an ordered choice; the possibilities can be listed systematically or counted by multiplying the choices available at each stage. Working: there are 3 flavours for the bottom scoop, and once that flavour is used only 2 flavours remain for the top scoop, so there are 3 × 2 = 6 cones; listing them confirms this, since vanilla on the bottom allows mango or pistachio on top, mango on the bottom allows vanilla or pistachio, and pistachio on the bottom allows vanilla or mango. Answer: 6. The distractors: 3 comes from treating the two scoops as interchangeable, so that vanilla under mango and mango under vanilla are counted as one cone; 9 comes from allowing the same flavour to be used for both scoops, giving 3 × 3; 5 comes from adding the 3 choices for the bottom scoop to the 2 choices left for the top scoop instead of multiplying them.
- (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.
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