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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Number worksheet — GCSE Foundation
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- 1.Light travels at 3 × 10⁸ metres per second. Work out how far light travels in 2 × 10⁻⁶ seconds. Give your answer in standard form.
- 2.A car is hired from 10:00 on Monday until 16:00 on Thursday. Work out the total length of the hire, in hours.
- 3.A charity raises money from a raffle and a cake sale in the ratio 5 : 3. Altogether the charity raises £320. Work out how much money the cake sale raised.
- 4.Work out (3 × 10²) × (2 × 10⁵). Give your answer in standard form.
- 5.Write these decimals in order, starting with the smallest: 0.6, 0.55, 0.601, 0.5, 0.56
- 6.The thickness of a sheet of card is 0.02384 cm. Write this thickness correct to 2 significant figures.
- 7.Two fair six-sided dice are rolled and their scores are added together. By listing the possible totals systematically, work out how many different totals are possible.
- 8.Four locations each record a temperature one winter morning. Which of these temperatures is the coldest?
- 9.The length of a pencil is 8.4 cm, correct to 1 decimal place. Using L for the length of the pencil in centimetres, write down the error interval for L.
- 10.A café is planning its lunch menu. It will offer one soup from 3 choices and one sandwich from 4 choices, but the mushroom soup, one of the 3 soups, will only be served with the cheese sandwich, one of the 4 sandwiches, on a normal day. Work out how many different soup-and-sandwich combinations are available on a normal day.
- 11.Write 260,000 in standard form.
- 12.Leah measures the length of her classroom with a tape measure marked in centimetres. She writes the length down as 7.3157 m. Give a reason why this is not an appropriate degree of accuracy.
- 13.Write 45,000 in standard form.
- 14.A recipe needs 0.485 kg of flour per cake. A bakery estimates its flour order by rounding this amount to 1 significant figure, then multiplying by the 60 cakes it plans to bake. Work out the bakery's estimate for the total flour needed, in kg.
- 15.A country's population is 8,340,000, and it is estimated that 1,950,000 of them live in the capital city. By rounding each number to 1 significant figure, work out an estimate for the number of people who do not live in the capital city.
Answer key
- (a) 6 × 10² metres — Method: distance = speed × time, so multiply the coefficients and add the indices. Working: 3 × 2 = 6 for the coefficients, and 8 + (−6) = 2 for the indices; 6 lies between 1 and 10, so the coefficient needs no adjustment. Answer: 6 × 10² metres, which is 600 metres. The distractors: 5 × 10² metres comes from adding the coefficients, 3 + 2, instead of multiplying them; 6 × 10¹⁴ metres comes from subtracting the indices, 8 − (−6), which is the rule for dividing rather than for multiplying; 6 × 10⁻⁴⁸ metres comes from multiplying the indices, 8 × (−6), instead of adding them.
- (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.
- (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.
- (d) 6 × 10⁷ — 3 × 2 = 6, and 2 + 5 = 7, so (3 × 10²) × (2 × 10⁵) = 6 × 10⁷. Multiplying the exponents instead of adding them gives 2 × 5 = 10, so 6 × 10¹⁰. Adding the coefficients instead of multiplying them gives 3 + 2 = 5, so 5 × 10⁷. Subtracting the exponents instead of adding them gives 5 − 2 = 3, so 6 × 10³.
- (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) 0.024 cm — Method: significant figures are counted from the first non-zero digit; the zeros in front of it only fix the place value and are not significant. Working: in 0.02384 the first significant figure is 2 and the second is 3, so the rounding is decided by the next digit, 8. As 8 is 5 or more, the second significant figure goes up from 3 to 4, in the same place value. Answer: 0.024 cm. The distractors: 0.023 cm comes from chopping the 84 off instead of rounding it; 0.02 cm comes from counting the leading zeros as significant figures, so the 2 is taken as the second figure and the rounding stops there; 0.0238 cm is 0.02384 correct to 3 significant figures, one figure too many.
- (a) 11 — Method: list every possible total from the smallest to the largest, and count how many different values there are. Working: the smallest total is 1+1=2 and the largest is 6+6=12, and every whole number total from 2 to 12 is possible: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 — that is 11 different totals. Answer: 11. 36 comes from counting the number of possible dice outcomes (6×6) instead of the number of different totals. 10 comes from listing the totals but missing one from the ends of the list, for example starting at 3 instead of 2. 6 comes from counting only the number of different scores on one die, not the totals of both dice together.
- (a) −4.5 °C — Order the temperatures by their actual value on a number line, remembering that a more negative number is further below zero and therefore colder: −4.5 °C is the coldest, since it is further below zero than −4.05 °C, −3.8 °C or 2 °C. Comparing the digits 405 and 45 as though the decimal points lined up, without padding −4.5 to match the number of decimal places in −4.05 first, makes −4.05 °C look like it has the bigger size, so it gets picked as the coldest by mistake — in fact −4.05 °C is closer to zero than −4.5 °C, not further from it. Picking −3.8 °C comes from choosing the negative reading with the smallest absolute value, forgetting that for negative numbers, a smaller absolute value means a warmer, less negative temperature, not a colder one. Picking 2 °C comes from ignoring the negative signs on the other three readings altogether and comparing raw digit sizes, when in fact any negative temperature is colder than any positive temperature. So the coldest temperature is −4.5 °C.
- (a) 8.35 ≤ L < 8.45 — Method: a length rounded to 1 decimal place lies within half of 0.1 cm, that is 0.05 cm, of the value written down. The lower limit is included because it rounds up to that value, and the upper limit is excluded because it rounds up to the next value instead. Working: 8.4 − 0.05 = 8.35 and 8.4 + 0.05 = 8.45, so a length of 8.35 cm still rounds to 8.4 cm while a length of 8.45 cm rounds to 8.5 cm. Answer: 8.35 ≤ L < 8.45. The distractors: 8.3 ≤ L < 8.5 goes a whole 0.1 cm either side instead of half of it; 8.35 < L ≤ 8.45 has the two limits the wrong way round, excluding the length that does round to 8.4 cm and including the one that does not; 8.4 ≤ L < 8.5 is the interval for a length truncated to 1 decimal place, not one rounded to it.
- (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.
- (c) 2.6 × 10⁵ — In standard form, A must satisfy 1 ≤ A < 10. Moving the decimal point in 260,000 to just after the 2 gives A = 2.6, and the decimal point moved 5 places, so 260,000 = 2.6 × 10⁵. A candidate who wrote 26 × 10⁴ used a value of A outside the required range, even though it has the same overall value. A candidate who wrote 2.6 × 10⁶ counted one place too many when moving the decimal point. A candidate who wrote 0.26 × 10⁶ used a value of A below 1, again outside the required range.
- (a) The tape can only give the length to the nearest centimetre — Method: a measurement should never be written to a finer degree of accuracy than the instrument used can read. Working: the tape is marked in centimetres, so the smallest division Leah can read is 1 cm, which is 0.01 m and two decimal places in metres; writing 7.3157 m claims the length to the nearest tenth of a millimetre, four decimal places, which the markings cannot support. A record of 7.32 m, to the nearest centimetre, is what this tape justifies. Answer: The tape can only give the length to the nearest centimetre. The distractors: the nearest millimetre contradicts the markings described in the question, which are centimetres, and would still claim more accuracy than the tape offers; the rule that a length in metres must be written to 2 decimal places borrows the habit of writing money to the penny, when the accuracy of a length depends on the instrument; rounding to the nearest metre would throw away accuracy the tape genuinely provides.
- (a) 4.5 × 10⁴ — 45,000 = 4.5 × 10,000 = 4.5 × 10⁴, with the coefficient between 1 and 10 as standard form requires. Writing 45 × 10³ keeps the coefficient too large — 45 is not between 1 and 10. Writing 4.5 × 10³ undercounts the places moved, giving only 4,500. Writing 4.5 × 10⁵ overcounts the places moved, giving 450,000.
- (d) 30 kg — Round 0.485 kg to 1 significant figure: 0.5 kg. Multiply by the 60 cakes: 0.5 × 60 = 30 kg. A candidate who rounded to 2 significant figures instead of 1 used 0.49 kg, giving 0.49 × 60 = 29.4 kg. A candidate who used the unrounded amount instead of the estimate worked out 0.485 × 60 = 29.1 kg. A candidate who rounded 0.485 down to 0.4 kg instead of up to 0.5 kg worked out 0.4 × 60 = 24 kg.
- (d) 6,000,000 — Method: round each number to 1 significant figure, then subtract. Working: 8,340,000 rounds to 8,000,000 (1 s.f.); 1,950,000 rounds to 2,000,000 (1 s.f.); 8,000,000 − 2,000,000 = 6,000,000. Answer: 6,000,000. 6,390,000 is the exact difference, found without rounding the numbers first. 8,000,000 comes from rounding the population correctly but forgetting to subtract the capital's population at all. 6,300,000 comes from rounding 8,340,000 to the nearest hundred thousand, 8,300,000, instead of to 1 significant figure, then subtracting the correctly rounded 2,000,000.
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