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.
Number worksheet — GCSE Foundation
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- 1.Work out 250 ÷ 1000.
- 2.A digital timer truncates every time to 1 decimal place. It shows a swimmer's time for one length as 12.3 seconds. Using t for the swimmer's actual time in seconds, write down the error interval for t.
- 3.Which of these numbers lies between −4 and −1 on a number line?
- 4.In a fruit basket the ratio of bananas to oranges is 2 : 5. What fraction of the fruit are oranges?
- 5.In the number 3.472, work out the value of the digit 7.
- 6.Work out ∛64.
- 7.Work out (8 × 10⁻⁵) × (5 × 10³). Give your answer in standard form.
- 8.A speed camera measures a car's speed as 34 mph, correct to the nearest mph. Work out the smallest possible actual speed of the car, in mph.
- 9.Work out the highest common factor of 12 and 18.
- 10.Work out the value of √(16 + 9)
- 11.A school orders 187 packed lunches for a trip. Each packed lunch costs £4.85. The school has £900 to spend. By rounding each number to 1 significant figure, work out an estimate for the total cost and decide whether £900 is enough.
- 12.Work out √49
- 13.The number of pages in a book is 240, correct to the nearest 10 pages. Write down the error interval for the actual number of pages, n.
- 14.A train journey takes 45 minutes, correct to the nearest 5 minutes. Using t for the actual time of the journey in minutes, write down the error interval for t.
- 15.At midday the temperature in the Lake District was 3 °C. By midnight it had fallen to −6 °C. Work out the fall in temperature.
Answer key
- (d) 0.25 — Dividing by 1000 moves every digit three place-value columns, so 250 ÷ 1000 = 0.25. A candidate who divides by 100 instead of 1000 gets 2.5. A candidate who divides by 10,000 instead of 1000 gets 0.025. A candidate who divides by 10 instead of 1000 gets 25.
- (c) 12.3 ≤ t < 12.4 — Method: truncating cuts the later digits off instead of rounding them, so nothing is ever pushed upwards. The displayed value is therefore the smallest the time can be, and the time can run up to, but not reach, the next value the display can show. Working: the display reads 12.3, so the actual time is at least 12.3 seconds; as soon as the time reaches 12.3 + 0.1 = 12.4 seconds the display would read 12.4, so 12.4 is not included. Answer: 12.3 ≤ t < 12.4. The distractors: 12.25 ≤ t < 12.35 is the interval for a time rounded to 1 decimal place, and this display does not round; 12.3 < t ≤ 12.4 excludes the one value the display certainly allows and includes the one it rules out; 12.3 ≤ t ≤ 12.4 treats 12.4 seconds as possible, but at 12.4 seconds the display would no longer read 12.3.
- (d) −2 — Method: place the two end values on a number line and list the integers that sit strictly between them. Working: reading from left to right the integers run −4, −3, −2, −1, so the values strictly between the ends are −3 and −2. Only one of those is offered. Answer: −2. The distractors: −5 comes from ordering negatives by the size of their digits, which wrongly places −5 to the right of −4; 0 comes from carrying on past −1 instead of stopping at it; 2 comes from ignoring the minus signs and choosing a number between 1 and 4.
- (d) 5/7 — Add the parts of the ratio to find the total: 2 + 5 = 7. Oranges are 5 of those 7 parts, so the fraction of the fruit that are oranges is 5/7. 2/7 comes from finding the fraction of bananas instead of oranges. 5/2 comes from writing the ratio itself as a fraction, without adding the parts to find the total. 7/5 comes from putting the total number of parts over the number of oranges instead of the number of oranges over the total.
- (c) 0.07 — Each digit after the decimal point has a place value: the first digit is tenths, the second is hundredths, the third is thousandths. In 3.472, the 4 is in the tenths place and the 7 is in the hundredths place, so it is worth 0.07. Reading it as 7 ignores place value altogether, treating it as if it were a whole number. Reading it as 0.7 puts it one place too big, in the tenths place. Reading it as 0.007 puts it one place too small, in the thousandths place. The digit 7 in 3.472 is worth 0.07.
- (b) 4 — Method: the cube root of a number is the value that multiplies by itself three times to give that number. Working: 4 × 4 × 4 = 64, so ∛64 = 4. Answer: 4. (8 comes from finding the square root of 64 instead of the cube root. 192 comes from multiplying 64 by 3 instead of finding the number that cubes to 64. 21.3 comes from dividing 64 by 3 instead of finding its cube root.)
- (c) 4 × 10⁻¹ — Multiply the A values: 8 × 5 = 40. Add the powers of 10: −5 + 3 = −2, giving 40 × 10⁻². Since A must satisfy 1 ≤ A < 10, rewrite 40 as 4 × 10¹, so 40 × 10⁻² = 4 × 10¹ × 10⁻² = 4 × 10⁻¹. A candidate who stopped at 40 × 10⁻² did the index arithmetic correctly but left the answer outside standard form, since 40 is not between 1 and 10. A candidate who adjusted the A value to 4 correctly but then took the power of 10 by subtracting the two given powers, −5 − 3 = −8, wrote 4 × 10⁻⁸. A candidate who adjusted the A value to 4 but multiplied the two given powers, −5 × 3 = −15, wrote 4 × 10⁻¹⁵. Both of these forgot that multiplying in standard form means adding the powers.
- (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) 6 — List the factors of each number: the factors of 12 are 1, 2, 3, 4, 6 and 12; the factors of 18 are 1, 2, 3, 6, 9 and 18. The common factors are 1, 2, 3 and 6, and the highest of these is 6. Picking 2, a common factor but not the largest, gives an answer that is too small. Picking 3, also a common factor but still not the largest, gives another answer that is too small. Working out the lowest common multiple instead of the highest common factor gives 36. So the highest common factor of 12 and 18 is 6.
- (a) 5 — 16 + 9 = 25, then √25 = 5. Splitting the root over the addition instead gives √16 = 4 and √9 = 3, then 4 + 3 = 7 — but a root does not split over a sum like this. Multiplying those two roots instead of adding them gives 4 × 3 = 12. Taking the negative square root instead of the positive one gives −5.
- (b) £1,000, so £900 is not enough — Method: round each number to 1 significant figure, multiply to estimate the total cost, then compare the estimate with the money available. Working: 187 rounds to 200 and £4.85 rounds to £5, so the estimate is 200 × 5 = 1,000, and £1,000 is more than the £900 the school has. Answer: £1,000, so £900 is not enough. The distractors: £800 comes from cutting £4.85 down to £4 instead of rounding it up to £5, giving 200 × 4 = 800, and that estimate wrongly suggests the money stretches; £935 comes from rounding the price only and keeping 187 lunches, giving 187 × 5 = 935; £950 comes from rounding 187 to the nearest 10 rather than to 1 significant figure, giving 190 × 5 = 950.
- (c) 7 — Method: a square root asks which positive number multiplied by itself gives the number under the root sign, so work up through the square numbers until one of them is 49. Working: 5 × 5 = 25, 6 × 6 = 36 and 7 × 7 = 49. Answer: 7. The distractors: 9 comes from recalling the wrong square fact and pairing 49 with 9, when 9 × 9 = 81; 24.5 comes from treating a square root as a halving and working out 49 ÷ 2; 2401 comes from squaring 49 instead of square-rooting it, applying the inverse operation the wrong way round.
- (b) 235 ≤ n < 245 — Method: the error interval stretches half the rounding unit either side of the rounded value, with the upper bound excluded because it would round up to the next value. Working: half of 10 is 5, so the interval runs from 240 − 5 to 240 + 5. Answer: 235 ≤ n < 245. (230 ≤ n < 250 comes from using the whole rounding unit, 10, either side instead of half of it. 235 ≤ n ≤ 245 comes from including the upper bound with ≤ instead of excluding it with <. 239.5 ≤ n < 240.5 comes from rounding to the nearest whole number instead of the nearest 10, so half of 1 is used in place of half of 10.)
- (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.
- (a) 9 °C — Method: the fall is the difference between the two readings, so subtract the lower reading from the higher one; subtracting a negative number is the same as adding its positive. Working: 3 − (−6) = 3 + 6 = 9. Counting it out, the temperature drops 3 degrees to reach zero and a further 6 degrees below zero. Answer: 9 °C. The distractors: −9 °C comes from subtracting the readings the wrong way round, as −6 − 3, and reporting a fall as a negative amount; 3 °C comes from ignoring the minus sign and working out 6 − 3; 6 °C comes from counting only the part of the fall that happens below zero and forgetting the 3 degrees above it.
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