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 the reciprocal of (2 + 3)
- 2.Work out the value of 5³.
- 3.A block of butter has a mass of 250 g, correct to the nearest 10 g. Using m for the mass of the block in grams, write down the error interval for m.
- 4.Work out an estimate for 397 ÷ 21, by rounding each number to 1 significant figure.
- 5.Work out −6 × (−3).
- 6.Write 0.00072 in standard form.
- 7.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.
- 8.A charity raises 8 × 10⁶ pounds. This total is shared equally among 2 × 10² local projects. Work out how much each project receives, in standard form.
- 9.Work out the value of ∛8.
- 10.In a school long jump competition Priya's jump is recorded as 3.8 m and Nadia's jump is recorded as 3.9 m, each correct to the nearest 0.1 m. Priya says she may have jumped further than Nadia. Decide whether Priya is right, and give a reason for your answer.
- 11.Work out √49
- 12.Work out 3 + 4 × (−2).
- 13.3/5 of the students in a year group walk to school. 90 students walk to school. Work out the total number of students in the year group.
- 14.Round 4,685 to the nearest 100.
- 15.Round 0.006482 to 2 significant figures.
Answer key
- (a) 1/5 — Work out the bracket first: 2 + 3 = 5. The reciprocal of 5 is 1/5. A candidate who forgot to take the reciprocal and just gave the value of the bracket wrote 5. A candidate who took the reciprocal but made a sign error wrote −1/5. A candidate who found the reciprocal of each number separately and added them, treating reciprocal as if it distributes over addition, worked out 1/2 + 1/3 = 5/6.
- (c) 125 — Method: a number cubed means multiplying the number by itself three times. Working: 5 × 5 × 5 = 125. 15 comes from working out 5 × 3 = 15, multiplying by the index instead of cubing. 8 comes from working out 5 + 3 = 8 instead of cubing. 53 comes from writing the base and the index next to each other instead of carrying out the calculation. Answer: 125.
- (b) 245 ≤ m < 255 — Method: a mass given to the nearest 10 g lies within half of 10 g, that is 5 g, of the figure written down. The lower limit is included because it rounds up to that figure, and the upper limit is excluded because it rounds up to the next multiple of 10. Working: 250 − 5 = 245 and 250 + 5 = 255, and a mass of 245 g rounds to 250 g while a mass of 255 g rounds to 260 g. Answer: 245 ≤ m < 255. The distractors: 240 ≤ m < 260 goes a whole 10 g either side instead of half of it; 245 < m ≤ 255 has the two limits the wrong way round; 240 ≤ m < 250 treats the stated 250 g as the largest mass possible, as though rounding were always upwards.
- (a) 20 — Method: round each number to 1 significant figure, then divide. Working: 397 rounds to 400 and 21 rounds to 20, so the estimate is 400 ÷ 20 = 20. Answer: 20. The distractors: 19 comes from not estimating at all, since 397 ÷ 21 = 18.9 to 1 decimal place, which rounds to 19, while the question asks for an estimate from rounded numbers; 200 comes from dividing 400 by 2 instead of by 20, losing the place value of the rounded divisor; 2 comes from dividing the leading digits only and losing the place value of both numbers.
- (b) 18 — Method: multiply the two numbers ignoring their signs, then apply the rule that a negative number multiplied by a negative number gives a positive answer. Working: 6 × 3 = 18; since both −6 and −3 are negative, the product is positive. Answer: 18. −18 comes from keeping the answer negative, as if only one of the two negative signs affects the sign of the product. −9 comes from adding the two numbers instead of multiplying them, −6 + (−3) = −9. 9 comes from adding 6 and 3 as if both numbers were positive, ignoring the negative signs entirely.
- (a) 7.2 × 10⁻⁴ — 0.00072 = 7.2 × 10⁻⁴, moving the decimal point 4 places to the right to reach 7.2, so the exponent is negative. Writing 7.2 × 10⁴ uses a positive exponent, which would give a number far larger than 1, not a small decimal. Writing 7.2 × 10⁻⁵ moves the point one place too many. Writing 0.72 × 10⁻³ leaves the coefficient below 1, which standard form does not allow.
- (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.
- (c) 4 × 10⁴ — 8 ÷ 2 = 4, and 6 − 2 = 4, so each project receives 4 × 10⁴ pounds. Multiplying the exponents instead of subtracting them gives 6 × 2 = 12, so 4 × 10¹². Adding the exponents instead of subtracting them gives 6 + 2 = 8, so 4 × 10⁸. Subtracting the coefficients instead of dividing them gives 8 − 2 = 6, so 6 × 10⁴.
- (b) 2 — Method: the cube root of a number is the value that multiplies by itself three times to give that number. Working: 2 × 2 × 2 = 8, so ∛8 = 2. 4 comes from working out 8 ÷ 2 = 4, halving the number instead of finding its cube root. 24 comes from working out 8 × 3 = 24, multiplying by 3 instead of cube-rooting it. 64 is 8², the square of 8, not its cube root. Answer: 2.
- (b) No — their possible jump lengths do not overlap — Method: each recorded jump stands for the lengths within half of 0.1 m, that is 0.05 m, of the figure recorded, and Priya is right only if the two ranges overlap. Working: Priya's jump is at least 3.8 − 0.05 = 3.75 m and below 3.85 m, because a jump of 3.85 m would have been recorded as 3.9 m; Nadia's jump is at least 3.85 m and below 3.9 + 0.05 = 3.95 m. Every length Priya could have jumped is below 3.85 m and every length Nadia could have jumped is at least 3.85 m, so Nadia jumped further whatever the exact lengths were. Answer: No — their possible jump lengths do not overlap. The distractors: the reason that a recorded jump is exactly the length jumped reaches the same verdict by treating a rounded record as exact, which is the idea this question tests; both jumps being 3.85 m would put 3.85 m inside Priya's range, when a jump of that length is recorded as 3.9 m; Priya jumping up to 3.9 m goes a whole 0.1 m above her record instead of half of it.
- (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) −5 — Using the order of operations, work out the multiplication first: 4 × (−2) = −8. Then 3 + (−8) = −5. A candidate who adds before multiplying gets (3 + 4) × (−2) = −14. A candidate who drops the negative sign on the multiplication gets 3 + 4 × 2 = 11. A candidate who works out the multiplication correctly but gives that as the final answer, forgetting to combine it with the 3, gets −8.
- (c) 150 — Since 90 students represent 3 of the 5 equal parts, one part is 90 ÷ 3 = 30, and the whole year group is five parts: 30 × 5 = 150. Applying the fraction forwards to 90 instead of reversing it, 90 × 3/5 = 54, treats the given number as the whole rather than as three fifths of it. Finding one part correctly as 30 but forgetting to scale up to the whole year group leaves 30 as the final answer. Treating 90 as the whole year group and adding on 2/5 of 90 for the students who do not walk, 90 + (90 × 2/5) = 126, applies the missing fraction to the wrong base amount.
- (b) 4,700 — To round to the nearest 100, look at the digit in the tens column, which decides whether the hundreds column rounds up or stays the same. In 4,685 that digit is 8, and since 8 is 5 or more, the 6 in the hundreds column rounds up to 7, giving 4,700. Simply changing the last two digits to zero without checking the tens digit gives 4,600, which rounds down when it should round up. Rounding to the nearest 10 instead of the nearest 100 gives 4,690. Rounding to the nearest 1,000 instead gives 5,000, one place value too coarse.
- (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.
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