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.Work out 2⁵.
- 2.Write 5,000,000 in standard form.
- 3.In a class, 1/3 of the pupils are girls. There are 12 girls in the class. Work out how many pupils are in the class.
- 4.Work out 2 × (−3 + 7)
- 5.A charity bake sale sells 187 cakes at £2.95 each. By rounding each number to 1 significant figure, work out an estimate for the total amount raised.
- 6.Write the fraction 3/4 as a decimal.
- 7.Simplify 5⁻² × 5⁴, giving your answer as a single power of 5.
- 8.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.
- 9.Work out 3 + 4 × 2²
- 10.Round 0.006482 to 2 significant figures.
- 11.A cyclist rides at a steady speed of 8 metres per second. Work out this speed in kilometres per hour.
- 12.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.
- 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.A box holds 140 pens. 25% of the pens are red. Work out how many of the pens are red.
- 15.3/5 of a 5/6 litre bottle of juice is poured out. Work out the exact volume poured out, in litres.
Answer key
- (c) 32 — Method: a power tells you how many times to multiply the base by itself. Working: 2⁵ = 2 × 2 × 2 × 2 × 2 = 32. Answer: 32. (10 comes from multiplying the base by the power, 2 × 5, instead of using repeated multiplication. 25 comes from swapping the base and the power and working out 5² instead. 16 comes from using one fewer 2, effectively working out 2⁴.)
- (c) 5 × 10⁶ — Method: standard form is written as A × 10ⁿ, where A is at least 1 and less than 10 and n counts the places the decimal point moves. Working: the digits of 5,000,000 give a coefficient of A = 5, and the decimal point travels from the end of 5,000,000 until it sits just after the 5, a move of 6 places, so n = 6. Answer: 5 × 10⁶. The distractors: 50 × 10⁵ comes from stopping before the coefficient has been brought into range, and 50 is not less than 10, so it is not standard form; 5 × 10⁷ comes from counting the seven digits of 5,000,000 instead of the six places the decimal point moves; 5 × 10⁻⁶ comes from making the index negative because the decimal point was carried to the left, when a negative index belongs to a number smaller than 1.
- (d) 36 — Method: the fraction is acting as an operator on the whole class, so one third of the class equals 12; the operation has to be reversed, and the inverse of dividing by 3 is multiplying by 3. Working: 1/3 × (number of pupils) = 12, so the number of pupils = 12 × 3 = 36. Answer: 36 pupils. The distractors: 4 comes from applying the operator instead of reversing it, working out 12 ÷ 3 = 4; 18 comes from reading the 12 girls as two thirds of the class, giving 12 ÷ 2 × 3 = 18; 24 comes from working out the number of boys, the other two thirds, as 2 × 12 = 24 and giving that instead of the size of the class.
- (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.
- (a) £600 — Method: round the number of cakes and the price of each cake to 1 significant figure, then multiply the rounded values. Working: 187 rounds to 200, and £2.95 rounds to £3 (the digit after the first, 9, rounds the 2 up to 3), so the estimate is 200 × £3 = £600. £400 comes from rounding £2.95 down to £2 instead of up to £3, giving 200 × £2 = £400. £561 comes from rounding only the price and using the exact number of cakes, 187 × £3 = £561. £570 comes from rounding 187 to the nearest 10 as 190 instead of to 1 significant figure as 200, giving 190 × £3 = £570. Answer: £600.
- (c) 0.75 — Method: a fraction means the numerator divided by the denominator, and carrying out that division gives the decimal. Working: 3 ÷ 4 is worked out by writing 3 as 3.00; 4 divides into 30 tenths 7 times with 2 tenths left over, and 4 divides into the remaining 20 hundredths exactly 5 times, giving 7 tenths and 5 hundredths. Answer: 0.75. The distractors: 0.34 comes from writing the numerator and the denominator as the two digits after the point instead of dividing; 0.7 comes from adding the numerator and the denominator to get 7 and reading that as seven tenths; 0.25 comes from converting the quarter that is left over rather than the three quarters the question asks about.
- (b) 5² — Method: multiplying two powers of the same base adds their indices, and a negative index is added as a negative number. Working: −2 + 4 = 2, so 5⁻² × 5⁴ = 5². Answer: 5². The distractors: 5⁶ comes from adding the sizes of the indices, 2 + 4, and ignoring the minus sign; 5⁻⁸ comes from multiplying the indices, −2 × 4, instead of adding them; 5⁻⁶ comes from subtracting the indices, −2 − 4, as though the powers were being divided.
- (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.
- (d) 19 — 2² = 4, then 4 × 4 = 16, then 3 + 16 = 19. Adding before multiplying gives 3 + 4 = 7, then 7 × 4 = 28 — multiplication comes before addition. Squaring the product instead of just the 2 gives 4 × 2 = 8, then 8² = 64, then 3 + 64 = 67. Working strictly left to right throughout gives 3 + 4 = 7, then 7 × 2 = 14, then 14² = 196.
- (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.
- (a) 28.8 km/h — A compound unit is converted one part at a time. There are 3600 seconds in an hour, so in one hour the cyclist travels 8 × 3600 = 28 800 metres. There are 1000 metres in a kilometre, so 28 800 m = 28 800 ÷ 1000 = 28.8 km/h. 28 800 km/h leaves the distance in metres, 0.48 km/h converts the seconds to minutes rather than to hours, and 2.22 km/h divides by 3.6 instead of multiplying.
- (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) 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) 35 — Method: 25% is 25/100, which cancels to 1/4, so finding 25% of an amount means dividing it by 4. Working: 25% = 25/100 = 1/4, and 140 ÷ 4 = 35. Answer: 35 pens. The distractors: 70 comes from halving instead of quartering, confusing 25% with 50%; 105 comes from working out the pens that are not red, which is 75% of 140, instead of the pens that are; 25 comes from ignoring the percent sign and reading the 25% as a count of 25 pens.
- (a) 1/2 — To find a fraction of an amount, multiply the fractions together: 3/5 × 5/6 = 15/30, which simplifies to 1/2 litre. Adding the fractions instead of multiplying them, using a common denominator of 30, gives 18/30 + 25/30 = 43/30, a value greater than the whole bottle. Dividing by 5/6 instead of multiplying by it, using its reciprocal 6/5, gives 3/5 × 6/5 = 18/25. Multiplying 5/6 by itself instead of by 3/5 gives 25/36.
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