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.
Answer key: Number worksheet — GCSE Foundation
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- (a) 0 — Method: BIDMAS works through the index first, then the multiplication, then the subtraction. Working: 5² = 25, then 4 × 25 = 100, and finally 100 − 100 = 0. Answer: 0. The distractors: 2400 comes from working from left to right and subtracting first, giving (100 − 4) × 25 = 96 × 25 = 2400; −300 comes from multiplying before applying the index, giving (4 × 5)² = 20² = 400 and then 100 − 400 = −300; 60 comes from reading 5² as 5 × 2 = 10, so that 4 × 10 = 40 and 100 − 40 = 60.
- (a) 23 — Division undoes multiplication, so the missing number is 391 ÷ 17 = 23. Writing down 17 repeats the number already given instead of solving for the missing one. Subtracting instead of dividing gives 391 − 17 = 374. Multiplying instead of dividing gives 391 × 17 = 6647.
- (a) 3 — Method: the fourth root of a number is the positive value that gives that number when it is multiplied by itself four times. Working: 2 × 2 × 2 × 2 = 16, which is too small, and 3 × 3 × 3 × 3 = 9 × 9 = 81. Answer: 3. The distractors: 9 comes from taking the square root of 81 instead of its fourth root; 4.5 comes from taking the square root and then halving it, as though a fourth root were half a square root; 20.25 comes from dividing 81 by 4, treating the root's index as a divisor.
- (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.)
- (c) 6.3 × 10⁷ — To add numbers in standard form, first write them with the same power of 10. 6 × 10⁷ = 60 × 10⁶, so the sum is 60 × 10⁶ + 3 × 10⁶ = 63 × 10⁶ = 6.3 × 10⁷. A candidate who added the A values without adjusting for the different powers worked out 6 + 3 = 9 and kept the larger power, writing 9 × 10⁷. A candidate who added the powers of 10 as if multiplying wrote 9 × 10¹³. A candidate who added the A values but used the smaller power wrote 9 × 10⁶.
- (a) 300 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 8,900 rounds to 9,000 and 29 rounds to 30; cancelling a zero from each gives 900 ÷ 3. Answer: 300. The distractors: 450 comes from rounding 29 down to 20 instead of to the nearest ten, giving 9,000 ÷ 20; 3,000 comes from rounding 29 to 3 rather than to 30, a place-value slip that divides by a number ten times too small; 307 is the exact quotient rounded to the nearest whole number, worked out in full when the question asks for an estimate.
- (a) 12 — List the multiples of each number: multiples of 4 are 4, 8, 12, 16, 20, 24; multiples of 6 are 6, 12, 18, 24. The lowest number that appears in both lists is 12. Picking 24, a common multiple but not the lowest one, gives an answer that is too big. Picking 6, the larger of the two original numbers rather than a common multiple, ignores that the lowest common multiple must appear in both lists. Working out the highest common factor instead of the lowest common multiple gives 2. So the lowest common multiple of 4 and 6 is 12.
- (a) 1/2 — The ratio 1:2:3 has 1 + 2 + 3 = 6 parts in total. Amir and Bo together receive 1 + 2 = 3 of those parts, so together they receive 3/6 of the £60, which simplifies to 1/2. Using only Amir's single part, 1/6, ignores Bo's share entirely. Adding Bo's and Chen's parts instead of Amir's and Bo's, 2 + 3 = 5, gives 5/6. Comparing Amir and Bo's 3 parts to Chen's 3 parts, rather than to the total of 6 parts, gives 3/3 = 1.
- (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.
- (c) £80 — Method: a unit fraction acts as an operator, so finding 1/9 of a price means dividing that price by 9. Working: £720 ÷ 9 = £80. Answer: £80. The distractors: £6480 comes from multiplying by the denominator instead of dividing by it, giving £720 × 9 = £6480; £640 comes from working out what is left of the £720 once the case is paid for, £720 − £80, instead of the cost of the case itself; £72 comes from dividing by 10 instead of 9, treating one ninth as one tenth.
- (c) £70 — Method: 20% of an amount is 20/100 of it; a reliable route is to find 10% by dividing by 10 and then double it. Working: 10% of £350 is £350 ÷ 10 = £35, and 20% is twice as much, £35 × 2 = £70. Answer: £70. The distractors: £35 comes from finding 10% and stopping there; £17.50 comes from reading 20% as one twentieth and working out £350 ÷ 20 = £17.50; £280 comes from taking 20% off the money raised rather than finding 20% of it, giving £350 ÷ 5 = £70 and then £350 − £70 = £280.
- (b) 54 — Method: round each number to the nearest whole number, then subtract the rounded values. Working: 79.3 rounds to 79 (nearest whole number) and 24.6 rounds to 25 (nearest whole number). 79 − 25 = 54. Answer: 54. 54.7 is the exact value of 79.3 − 24.6, found without rounding first, so it is not an estimate. 55 comes from rounding 24.6 down to 24 instead of up to the nearest whole number, 25, giving 79 − 24. 59 comes from rounding 24.6 to the nearest 10, 20, instead of to the nearest whole number, 25, giving 79 − 20.
- (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.
- (b) 3.65 ≤ n < 3.75 — Rounding to 1 decimal place means n can be up to half of one decimal place, 0.05, below or above 3.7 before it would round to a different value. This gives a lower bound of 3.7 − 0.05 = 3.65 and an upper bound of 3.7 + 0.05 = 3.75. A value exactly at 3.75 would round up to 3.8, not 3.7, so the upper bound is excluded while the lower bound, 3.65, does still round to 3.7. Writing 3.65 ≤ n ≤ 3.75 wrongly includes 3.75. Writing 3.6 ≤ n < 3.8 uses a whole decimal place, 0.1, either side instead of half of one, 0.05. Writing 3.65 < n < 3.75 wrongly excludes 3.65, which does round to 3.7.
- (a) 4,000 nanometres — Method: the number of nanometres is the diameter divided by the length of one nanometre, and dividing powers of ten means subtracting the indices. Working: −6 − (−9) = 3, so 10⁻⁶ ÷ 10⁻⁹ = 10³, and the diameter is 4 × 10³ nanometres. Answer: 4,000 nanometres. The distractors: 400 nanometres comes from taking the difference between the indices as 2 instead of 3; 4 nanometres comes from changing the name of the unit without converting, leaving the coefficient untouched; 0.004 nanometres comes from dividing by 10³ instead of multiplying by it, as though a nanometre were the larger of the two units.
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