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.Simplify x⁵ × x³ ÷ x², giving your answer as a single power of x.
- 2.A length, L cm, has the error interval 24.5 ≤ L < 25.5. Write down the degree of accuracy to which the length was measured.
- 3.Light travels at 2.998 × 10⁸ metres per second. A distant object in space is 3.1 × 10¹⁵ metres from Earth. Work out an estimate for the number of seconds light takes to travel from the object to Earth, by rounding each number to 1 significant figure.
- 4.Grace drinks 1/3 of a bottle of water in the morning and another 1/3 of the same bottle in the afternoon. Work out what fraction of the bottle she has drunk altogether.
- 5.A cake recipe needs 3/4 of a kilogram of sugar. Aisha wants to make half the recipe. Work out how much sugar she needs, giving your answer as a fraction of a kilogram in its simplest form.
- 6.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.
- 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.
- 8.2/3 of an amount of money is £30. Work out the amount.
- 9.A concrete mix is made from cement and sand in the ratio 3 : 5 by mass. A builder mixes 96 kg of concrete. Work out the mass of cement in the mix.
- 10.Work out −(−3)⁴ + (−3)³
- 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.Simplify 3² ÷ 3⁵, giving your answer as a single power of 3.
- 13.A photograph uses 4 × 10⁶ bytes of storage. A memory card holds 3.2 × 10¹⁰ bytes. Work out how many of these photographs the card can hold. Give your answer in standard form.
- 14.The length of a nail is 12 cm, correct to the nearest centimetre. Using L for the length of the nail in centimetres, write down the error interval for L.
- 15.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.
Answer key
- (a) x⁶ — Method: work through the powers in order — multiplying powers of the same base means adding indices, and dividing powers of the same base means subtracting indices. Working: first, x⁵ × x³ = x⁸ (adding 5 and 3); then x⁸ ÷ x² = x⁶ (subtracting 2 from 8). x⁴ comes from swapping the two rules — subtracting for the multiplication, 5 − 3 = 2, and then adding for the division, 2 + 2 = 4. x¹⁰ comes from adding all three indices, 5 + 3 + 2 = 10, treating the division the same as a multiplication. 6x comes from correctly reaching a total index of 6 but then writing it as a coefficient of x instead of as its power. Answer: x⁶.
- (a) to the nearest centimetre — Method: the error interval of a rounded measurement runs from half a unit below the stated value to half a unit above it, so the width of the interval is one whole unit of the accuracy used. Working: the interval runs from 24.5 to 25.5, a width of 25.5 − 24.5 = 1, so the unit of accuracy is 1 cm; the stated value is the midpoint, 25 cm, and 25 correct to the nearest centimetre is exactly what gives 24.5 ≤ L < 25.5. Answer: to the nearest centimetre. To the nearest 0.5 cm comes from reading the half-unit, 0.5, as the accuracy itself instead of doubling it back to the full unit. To 1 decimal place comes from seeing the bounds written with one decimal place and taking that as the accuracy, but the bounds of a value given to 1 decimal place would be only 0.05 either side. To the nearest 10 cm comes from confusing the size of the value, about 25, with the unit it was rounded to; rounding to the nearest 10 cm would give an interval 5 cm either side of the stated value.
- (d) 1 × 10⁷ seconds — Method: the time for a journey is the distance divided by the speed, so round each number to 1 significant figure and then divide; dividing numbers in standard form means dividing the coefficients and subtracting the indices. Working: 3.1 × 10¹⁵ rounds to 3 × 10¹⁵ and 2.998 × 10⁸ rounds to 3 × 10⁸; 3 ÷ 3 = 1 for the coefficients, and 15 − 8 = 7 for the indices. Answer: about 1 × 10⁷ seconds. The distractors: 3 × 10⁷ seconds comes from subtracting the indices correctly but leaving the coefficient as 3 instead of dividing 3 by 3; 1 × 10⁻⁷ seconds comes from dividing the speed by the distance instead of the distance by the speed; 9 × 10²³ seconds comes from multiplying the two quantities instead of dividing them, since 3 × 3 = 9 and 15 + 8 = 23.
- (c) 2/3 — Method: fractions with the same denominator are added by adding the numerators and leaving the denominator alone, because the parts are already the same size. Working: 1/3 + 1/3 has numerators 1 + 1 = 2 and the denominator stays as 3, giving 2/3. Answer: 2/3. The distractors: 2/6 comes from adding the denominators as well as the numerators, 1 + 1 over 3 + 3; 2/9 comes from adding the numerators but multiplying the denominators, 1 + 1 over 3 × 3; 1/9 comes from multiplying throughout instead of adding, 1 × 1 over 3 × 3.
- (d) 3/8 — Method: making half the recipe means dividing the quantity of sugar by 2. Working: 3/4 ÷ 2 = 3/8. Answer: 3/8. 3/2 comes from multiplying by 2 instead of dividing, as if doubling the recipe. 5/4 comes from adding 1/2 to 3/4 instead of halving it, confusing "half of" with "plus a half". 3/4 comes from leaving the amount unchanged, forgetting to halve it for the smaller recipe.
- (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) 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.
- (b) £45 — Since 2/3 of the amount is £30, one third is £30 ÷ 2 = £15, and the whole amount is three thirds: £15 × 3 = £45. Applying the fraction forwards to £30 instead of reversing it, £30 × 2/3 = £20, treats the given amount as the whole rather than as two thirds of it. Finding one third correctly as £15 but forgetting to multiply by 3 to get the whole amount leaves £15 as the final answer. Reading £30 as one third of the amount rather than as two thirds, and so multiplying straight by 3, gives £30 × 3 = £90.
- (a) 36 kg — The mix is 3 + 5 = 8 equal shares, so the cement is 3/8 of the mass. One share is 96 ÷ 8 = 12 kg, and the cement is 3 shares: 3 × 12 = 36 kg. Working out 3/5 of 96 gives 57.6 kg, which uses the sand as the denominator instead of the whole mix; 60 kg is the mass of the sand; 12 kg is one share only.
- (a) −108 — Method: a power is worked out before any minus sign written in front of it, while a minus sign inside the brackets is part of the base. Working: (−3)⁴ = 81, because four negative factors multiply to a positive result, so −(−3)⁴ = −81. (−3)³ = −27, because three negative factors multiply to a negative result. Adding gives −81 + (−27) = −108. Answer: −108. The distractors: 54 comes from attaching the leading minus sign to the base, working out (−(−3))⁴ = 81 and then adding −27; −54 comes from taking (−3)³ as +27, forgetting that an odd power keeps the negative sign; 108 comes from believing that any power of a negative number is positive and that the leading minus belongs to the base, giving 81 + 27.
- (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) 3⁻³ — Method: when dividing powers of the same base, subtract the index of the number you are dividing by from the index of the number being divided, keeping them in the order the question writes them. Working: 2 − 5 = −3, so 3² ÷ 3⁵ = 3⁻³. It is worth checking this against the numbers: 3² = 9 and 3⁵ = 243, and 9 ÷ 243 = 1/27, which is 3⁻³. 3³ comes from subtracting the other way round, 5 − 2 = 3, which reverses the sign of the index and gives 27 instead of 1/27. 3⁷ comes from working out 2 + 5 = 7, which is the rule for multiplying powers, not dividing them. 3¹⁰ comes from multiplying the indices, 2 × 5 = 10, instead of subtracting them. Answer: 3⁻³.
- (a) 8 × 10³ — Method: divide the capacity of the card by the size of one photograph, dividing the coefficients and subtracting the indices, then bring the coefficient back into the range 1 to 10. Working: 3.2 ÷ 4 = 0.8 and 10 − 6 = 4, which gives 0.8 × 10⁴; a coefficient of 0.8 is smaller than 1, so the decimal point moves one place to the right and the index falls by 1. Answer: 8 × 10³. The distractors: 8 × 10⁴ comes from correcting 0.8 to 8 without reducing the index, which makes the answer ten times too large; 1.28 × 10¹⁷ comes from multiplying the two numbers instead of dividing them, since 3.2 × 4 = 12.8 and 10 + 6 = 16; 8 × 10¹⁵ comes from dividing the coefficients but adding the indices instead of subtracting them.
- (a) 11.5 ≤ L < 12.5 — Rounding to the nearest centimetre means L can be up to half a centimetre below or above 12 before it would round to a different whole number. Half of 1 cm is 0.5 cm, so the lower bound is 12 − 0.5 = 11.5 and the upper bound is 12 + 0.5 = 12.5. A value exactly at the upper bound, 12.5, would round up to 13, not 12, so 12.5 itself is excluded, giving 11.5 ≤ L < 12.5. Writing 11.5 ≤ L ≤ 12.5 wrongly includes 12.5 on both ends. Writing 11 ≤ L < 13 uses a whole centimetre either side instead of half a centimetre. Writing 11.5 < L < 12.5 wrongly excludes the lower bound, which is a value that does round to 12.
- (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.
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