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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- (a) 8 × 10³ — Method: divide the capacity of the card by the size of one photograph, dividing the coefficients and subtracting the indices, then bring the coefficient back into the range 1 to 10. Working: 3.2 ÷ 4 = 0.8 and 10 − 6 = 4, which gives 0.8 × 10⁴; a coefficient of 0.8 is smaller than 1, so the decimal point moves one place to the right and the index falls by 1. Answer: 8 × 10³. The distractors: 8 × 10⁴ comes from correcting 0.8 to 8 without reducing the index, which makes the answer ten times too large; 1.28 × 10¹⁷ comes from multiplying the two numbers instead of dividing them, since 3.2 × 4 = 12.8 and 10 + 6 = 16; 8 × 10¹⁵ comes from dividing the coefficients but adding the indices instead of subtracting them.
- (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.
- (a) 33.5 mph — Method: the smallest possible actual value is half the rounding unit below the given value. Working: half of 1 mph is 0.5 mph, so the smallest possible speed is 34 − 0.5 = 33.5 mph. Answer: 33.5 mph. (33 mph comes from subtracting the whole rounding unit, 1, instead of half of it. 34 mph comes from giving the rounded value itself rather than the lower bound. 34.5 mph comes from adding the half unit instead of subtracting it, giving the upper bound.)
- (b) 78 hours — From 10:00 on Monday to 10:00 on Thursday is exactly 3 complete days, which is 3 × 24 = 72 hours. From 10:00 to 16:00 on the Thursday is a further 6 hours, giving a total of 72 + 6 = 78 hours. Counting Monday to Thursday as 4 full calendar days instead of 3 complete 24-hour periods gives 4 × 24 = 96 hours. Undercounting the number of complete days as 2 instead of 3 gives 2 × 24 + 6 = 54 hours. Subtracting the extra 6 hours instead of adding them to the 3 complete days gives 72 − 6 = 66 hours.
- (c) 3.05 kg — There are 1000 g in 1 kg, so 450 g = 450 ÷ 1000 = 0.45 kg. The total mass is 0.45 + 2.6 = 3.05 kg. 7.1 kg divides the grams by 100 instead of 1000, 2.645 kg divides them by 10 000, and 452.6 kg adds the two numbers without converting the grams at all.
- (b) £72 — One fifth of £60 = £12. New price = £60 + £12 = £72. A candidate who gives the increase instead of the new price gets £12. A candidate who subtracts the increase instead of adding it gets £60 − £12 = £48. A candidate who uses 1/4 instead of 1/5 gets £60 + £15 = £75.
- (a) 0.55, 58%, 3/5 — Converting all three to decimals: 3/5 = 0.6, 0.55 stays as 0.55, and 58% = 0.58. In order from smallest to largest, this is 0.55, then 58%, then 3/5. Writing the numbers in the reverse order, largest to smallest, gives 3/5, 58%, 0.55. Misconverting 3/5 as 0.5 instead of 0.6 makes it appear smaller than both other values, giving the order 3/5, 0.55, 58%. Misconverting 58% as 0.058 instead of 0.58, by moving the decimal point two extra places, makes it appear smallest of the three, giving the order 58%, 0.55, 3/5.
- (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.
- (a) £177.60 — Total raised = 240 × £1.85 = £444.00. The hospital receives 40% of this: £444.00 × 0.4 = £177.60. A candidate who works out the remaining 60% instead of the 40% given away gets £266.40. A candidate who forgets to find the percentage and gives the full total gets £444.00. A candidate who halves 40% by mistake and uses 20% gets £88.80.
- (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) 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.
- (a) 11:20 — Method: find the flight time using time = distance ÷ speed, then add this to the departure time. Working: 2340 ÷ 780 = 3 hours; 08:20 + 3 hours = 11:20. Answer: 11:20. 08:40 comes from dividing speed by distance instead of distance by speed, giving a flight time of 1/3 hour (20 minutes) rather than 3 hours. 11:00 comes from adding the 3-hour flight time to the hour of the departure time only, 8 + 3 = 11, and losing the 20 minutes. 03:00 comes from finding the flight time correctly but giving it as a clock time on its own, forgetting to add it to the departure time.
- (b) £13.48 — Method: an amount of money is written to the nearest penny, which is 2 decimal places, so the calculator display has to be rounded to 2 decimal places. Working: 53.90 ÷ 4 = 13.475, and the digit in the third decimal place is 5, so the penny digit goes up from 7 to 8. Answer: £13.48. The distractors: £13.47 comes from chopping the third decimal place off instead of rounding with it; £13.50 comes from rounding to the nearest 10p rather than to the nearest penny; £13.40 comes from cutting the display short at 1 decimal place, which is both the wrong degree of accuracy and a truncation rather than a rounding.
- (b) 18.5 ≤ T < 18.7 — Method: the error interval reaches half the rounding unit either side of the recorded value. Working: half of 0.2 is 0.1, so the interval runs from 18.6 − 0.1 to 18.6 + 0.1. Answer: 18.5 ≤ T < 18.7. (18.4 ≤ T < 18.8 comes from using the full rounding unit, 0.2, either side instead of half of it. 18.5 ≤ T ≤ 18.7 comes from including the upper bound with ≤ instead of excluding it with <. 18.6 ≤ T < 18.8 comes from treating the recorded value as the start of the interval and adding the whole rounding unit, 0.2, above it.)
- (a) 3.8 × 10⁻³ — 0.0038 is less than 1, so the power of 10 is negative. Moving the decimal point 3 places gives A = 3.8, so 0.0038 = 3.8 × 10⁻³. A candidate who wrote 3.8 × 10³ used a positive power, which is only correct for numbers of 10 or more. A candidate who wrote 38 × 10⁻⁴ used a value of A outside the required range. A candidate who wrote 3.8 × 10⁻⁴ counted one place too many when moving the decimal point.
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