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.In a car park the ratio of vans to cars is 2 : 7. Work out the number of vans as a fraction of the number of cars.
- 2.Work out 2 × (−3 + 7)
- 3.In standard form, 2,000 is written as 2 × 10ⁿ. Write down the value of n.
- 4.Work out 3 + 4 × (−2).
- 5.Work out an estimate for 8,900 ÷ 29, by rounding each number to 1 significant figure.
- 6.Four friends share a restaurant bill of £53.90 equally. A calculator gives each share as 13.475. Work out how much each friend should pay.
- 7.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.
- 8.Work out √144 − 2 × 3 + √25
- 9.Work out √225 ÷ 3.
- 10.A bag of flour is labelled 1.5 kg, correct to the nearest 0.1 kg. The true mass of the flour is m kg. Which inequality gives all the possible values of m?
- 11.A grain of fine sand has a mass of 0.001 grams. Write this mass in standard form.
- 12.In a choir, 2/9 of the members are boys and the rest are girls. Write down the ratio of girls to boys, in its simplest form.
- 13.Two bags of mixed nuts are combined. Bag A has peanuts and cashews in the ratio 2 : 3. Bag B has peanuts and cashews in the ratio 1 : 4. Both bags contain the same total number of nuts. Work out the fraction of the combined mixture that is peanuts.
- 14.Work out 3³ + 2⁴.
- 15.Work out (6 × 10⁷) + (3 × 10⁶). Give your answer in standard form.
Answer key
- (b) 2/7 — The comparison asked for is part with part, not part with whole, so the number of cars goes on the bottom. The ratio gives 2 shares of vans for every 7 shares of cars, so the vans are 2/7 of the cars. 2/9 uses the 9 shares in the whole car park as the denominator, 7/2 compares the cars with the vans, and 7/9 is the fraction of the vehicles that are cars.
- (d) 8 — Method: whatever is inside the brackets is worked out first, and the multiplication is carried out afterwards. Working: inside the brackets, −3 + 7 = 4, and then 2 × 4 = 8. Answer: 8. The distractors: 1 comes from ignoring the brackets and multiplying first, giving 2 × (−3) = −6 and then −6 + 7 = 1; −20 comes from reading −3 + 7 as −(3 + 7) = −10, so that 2 × (−10) = −20; 6 comes from working the brackets out correctly and then adding the 2 instead of multiplying by it, giving 2 + 4 = 6.
- (c) 3 — Method: the index counts how many times the coefficient has been multiplied by 10, which is the number of places the decimal point moves from the end of the number to just after the first significant digit. Working: 2,000 = 2 × 1,000, and 1,000 = 10 × 10 × 10, which is three tens. Answer: 3. The distractors: 4 comes from counting the four digits of 2,000 rather than the three places the decimal point moves; 2 comes from copying the coefficient 2 into the index; −3 comes from making the index negative, which would describe a number smaller than 1 rather than two thousand.
- (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.
- (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.
- (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.
- (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.
- (b) 11 — Method: roots and the multiplication are worked out before the addition and subtraction, and what is left is then worked through from left to right. Working: √144 = 12, √25 = 5 and 2 × 3 = 6, so the calculation becomes 12 − 6 + 5, which gives 6 + 5 = 11. Answer: 11. The distractors: 1 comes from carrying out the addition before the subtraction, giving 12 − (6 + 5) = 12 − 11 = 1; 35 comes from working from left to right with no priority, giving 12 − 2 = 10, then 10 × 3 = 30 and 30 + 5 = 35; 7 comes from combining the two roots as √(144 + 25) = √169 = 13 and then subtracting the product, giving 13 − 6 = 7.
- (b) 5 — Method: find the square root first, then divide. Working: √225 = 15, and 15 ÷ 3 = 5. Answer: 5. (75 comes from dividing 225 by 3 first and forgetting to take the square root at all. 8.7 comes from dividing 225 by 3 inside the root, √(225 ÷ 3) ≈ 8.7, instead of taking the root first. 45 comes from misreading the divisor as 5 instead of 3, working out 225 ÷ 5 = 45.)
- (b) 1.45 ≤ m < 1.55 — The flour's mass is labelled 1.5 kg, correct to the nearest 0.1 kg, so half of 0.1 kg is added to and subtracted from 1.5 kg to find the interval: 1.5 − 0.05 = 1.45 and 1.5 + 0.05 = 1.55, giving 1.45 ≤ m < 1.55. '1.4 ≤ m < 1.6' comes from taking the whole 0.1 kg as the margin either side, instead of half of it. '1.45 < m ≤ 1.55' comes from writing the inequality signs the wrong way round — the lower bound should be included and the upper bound excluded, not the other way round. '1.45 ≤ m ≤ 1.55' comes from including the upper bound, when the convention is that the upper bound is never actually reached.
- (d) 1 × 10⁻³ — Method: for a number smaller than 1 the index is negative, and it counts the places the decimal point moves to the right to leave a coefficient between 1 and 10. Working: the only significant digit is 1, so the coefficient is 1; the decimal point in 0.001 moves three places to the right to reach that 1, so the index is −3. Answer: 1 × 10⁻³. The distractors: 1 × 10³ comes from taking the index as positive, which describes one thousand grams rather than one thousandth of a gram; 0.1 × 10⁻² is the same mass written with a coefficient of 0.1, which is smaller than 1 and so is not standard form; 1 × 10⁻⁴ comes from counting the zero in front of the decimal point as well as the places after it.
- (b) 7:2 — If 2/9 of the choir are boys, the remaining 7/9 must be girls, since the two fractions together make the whole choir. The ratio of girls to boys compares these two parts to each other, giving 7:2. Writing the ratio the wrong way round, boys to girls instead of girls to boys, gives 2:7. Comparing the number of girls to the whole choir instead of to the number of boys gives 7:9. Simply rewriting the given fraction, 2/9, as a ratio without working out how many are girls gives 2:9.
- (b) 3/10 — Since each bag's ratio has 5 parts and both bags contain the same total number of nuts, imagine each bag has 5 nuts: Bag A has 2 peanuts and Bag B has 1 peanut, so together there are 2 + 1 = 3 peanuts out of a combined 5 + 5 = 10 nuts, giving 3/10. 1/5 comes from using only Bag A's peanuts, 2 out of 10, without adding Bag B's peanuts. 1/10 comes from using only Bag B's peanut, without adding Bag A's peanuts. 3/5 comes from writing the combined peanuts over the number of parts in one bag instead of the combined total number of nuts.
- (c) 43 — Method: work out each power separately before adding. Working: 3³ = 27 and 2⁴ = 16, so 3³ + 2⁴ = 27 + 16 = 43. Answer: 43. (25 comes from using 3² instead of 3³, giving 9 + 16. 35 comes from working out 2⁴ as 2 × 4 = 8 instead of 2 × 2 × 2 × 2, giving 27 + 8. 432 comes from multiplying the two powers together instead of adding them.)
- (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.
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