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.
Calculator
Number worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- 1.A plane flies 2340 km at an average speed of 780 km/h. It departs at 08:20. Work out the arrival time, using the 24-hour clock.
- 2.3.4 × 10⁵ is written as an ordinary number. Work out this number.
- 3.A machine fills bags of sugar and shows the mass of each bag to the nearest 10 g. A checker rejects any bag whose actual mass is less than 996 g. One bag shows a mass of 1,000 g on the machine. Decide whether this bag could be rejected, and give a reason for your answer.
- 4.Put these numbers in order, starting with the smallest: 3.2 × 10⁴, 2.9 × 10⁵, 4.1 × 10³
- 5.A carton of orange juice holds 1.35 litres. Ruby pours the juice equally into 4 identical glasses. Work out how much juice is in each glass, giving your answer as a fraction of a litre in its simplest form.
- 6.Two fair six-sided dice are rolled together. One die is red and the other is blue. Work out how many different outcomes are possible.
- 7.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.
- 8.A train journey takes 45 minutes, correct to the nearest 5 minutes. Using t for the actual time of the journey in minutes, write down the error interval for t.
- 9.A crowd of 8,400 people is recorded correct to the nearest 100. Work out the smallest possible number of people in the crowd.
- 10.After a price increase of 10%, a laptop costs £330. Work out the original price.
- 11.A café sells ice cream in three flavours: vanilla, mango and pistachio. Isla buys a cone with two scoops, using two different flavours, one for the bottom scoop and one for the top scoop. Work out how many different cones she could buy.
- 12.The thickness of a sheet of card is 0.02384 cm. Write this thickness correct to 2 significant figures.
- 13.Josh buys 7 pencils at 85p each. He pays with a £10 note. Work out his change.
- 14.Four locations each record a temperature one winter morning. Which of these temperatures is the coldest?
- 15.A circle has a radius of 3 cm. Which of these is the exact area of the circle?
Answer key
- (a) 11:20 — Method: find the flight time using time = distance ÷ speed, then add this to the departure time. Working: 2340 ÷ 780 = 3 hours; 08:20 + 3 hours = 11:20. Answer: 11:20. 08:40 comes from dividing speed by distance instead of distance by speed, giving a flight time of 1/3 hour (20 minutes) rather than 3 hours. 11:00 comes from adding the 3-hour flight time to the hour of the departure time only, 8 + 3 = 11, and losing the 20 minutes. 03:00 comes from finding the flight time correctly but giving it as a clock time on its own, forgetting to add it to the departure time.
- (d) 340,000 — 3.4 × 10⁵ means moving the decimal point in 3.4 five places to the right, giving 340,000. Moving it only four places gives 34,000, one place short. Moving it six places gives 3,400,000, one place too many. Treating the exponent as negative instead of positive moves the point the wrong way, giving 0.000034.
- (b) Yes — the actual mass could be as low as 995 g — Method: a mass shown to the nearest 10 g lies within half of 10 g, that is 5 g, of the figure on the display, so compare the smallest mass the bag can have with the checker's limit of 996 g. Working: 1,000 − 5 = 995, so the actual mass of the bag can be as low as 995 g, and 995 g is below the 996 g limit, so a bag showing 1,000 g on the machine can still be rejected. Answer: Yes — the actual mass could be as low as 995 g. The distractors: 990 g comes from going a whole 10 g below the display instead of half of it; 999.5 g comes from treating the display as being to the nearest gram, when it is to the nearest 10 g; the claim that the mass is exactly 1,000 g treats a rounded display as an exact measurement.
- (c) 4.1 × 10³, 3.2 × 10⁴, 2.9 × 10⁵ — The exponent decides the size first: 10³ is smaller than 10⁴, which is smaller than 10⁵, so the order is 4.1 × 10³, then 3.2 × 10⁴, then 2.9 × 10⁵. Reversing the whole list gives largest to smallest instead of smallest to largest. Comparing 3.2 × 10⁴ and 4.1 × 10³ by their coefficients alone, 3.2 against 4.1, and swapping them ignores that 10⁴ is bigger than 10³ regardless of the coefficient. Comparing 2.9 × 10⁵ and 3.2 × 10⁴ by their coefficients alone and swapping them makes the same mistake at the top of the list.
- (d) 27/80 — Method: write the total as a fraction of a litre, then divide by the number of glasses. Working: 1.35 = 27/20, so each glass holds 27/20 ÷ 4 = 27/80 of a litre. Answer: 27/80. 27/20 comes from converting the total correctly to a fraction but forgetting to divide by the number of glasses. 27/5 comes from multiplying the total by 4 instead of dividing. 17/50 comes from rounding 1.35 ÷ 4 to 0.34 before converting to a fraction.
- (c) 36 — Method: set the outcomes out in a grid with one die along the top and the other down the side, so that every cell of the grid is one outcome, and count the cells. Working: the red die can land in 6 ways, so the grid has 6 columns, and the blue die can also land in 6 ways, so the grid has 6 rows; the number of cells is 6 × 6 = 36. Answer: 36. The distractors: 12 comes from adding 6 and 6 instead of multiplying them; 6 comes from counting the outcomes of a single die and forgetting that the second die also has to land; 21 comes from treating the two dice as indistinguishable, so that a red 2 with a blue 3 and a red 3 with a blue 2 are counted as one outcome.
- (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) 42.5 ≤ t < 47.5 — Rounding to the nearest 5 minutes means the actual time can be up to half of 5 minutes, 2.5 minutes, below or above 45 before it would round to a different multiple of 5. The lower bound is 45 − 2.5 = 42.5 and the upper bound is 45 + 2.5 = 47.5. A time of exactly 47.5 minutes would round up to 50, not 45, so 47.5 is excluded while 42.5 does still round to 45. Writing 42.5 ≤ t ≤ 47.5 wrongly includes 47.5. Writing 40 ≤ t < 50 uses a whole rounding unit, 5, either side instead of half of it. Writing 44.5 ≤ t < 45.5 treats the rounding unit as 1 minute instead of 5 minutes.
- (a) 8,350 — Rounding to the nearest 100 means the true number can be up to half of 100, which is 50, below the recorded figure before it would round down to a lower hundred. The smallest possible number is therefore 8,400 − 50 = 8,350. Adding 50 instead of subtracting it gives 8,450, which is the upper end of the interval rather than the smallest value, and 8,450 is not itself possible because it would round up to 8,500. Subtracting a whole 100 instead of half of it gives 8,300, going too far below the recorded value. Subtracting 10 instead of half of the rounding unit gives 8,390, treating the rounding unit as 100 but the tolerance as only 10.
- (b) £300 — The increased price is 110% of the original, so the original price = £330 ÷ 1.1 = £300. A candidate who finds 10% of £330 and subtracts it, wrongly treating £330 as the original, gets £330 − £33 = £297. A candidate who adds 10% of £330 again instead of reversing the increase gets £330 + £33 = £363. A candidate who divides by 0.1 instead of 1.1 gets £3,300.
- (a) 6 — Method: the two scoops sit in different places on the cone, so a cone is an ordered choice; the possibilities can be listed systematically or counted by multiplying the choices available at each stage. Working: there are 3 flavours for the bottom scoop, and once that flavour is used only 2 flavours remain for the top scoop, so there are 3 × 2 = 6 cones; listing them confirms this, since vanilla on the bottom allows mango or pistachio on top, mango on the bottom allows vanilla or pistachio, and pistachio on the bottom allows vanilla or mango. Answer: 6. The distractors: 3 comes from treating the two scoops as interchangeable, so that vanilla under mango and mango under vanilla are counted as one cone; 9 comes from allowing the same flavour to be used for both scoops, giving 3 × 3; 5 comes from adding the 3 choices for the bottom scoop to the 2 choices left for the top scoop instead of multiplying them.
- (d) 0.024 cm — Method: significant figures are counted from the first non-zero digit; the zeros in front of it only fix the place value and are not significant. Working: in 0.02384 the first significant figure is 2 and the second is 3, so the rounding is decided by the next digit, 8. As 8 is 5 or more, the second significant figure goes up from 3 to 4, in the same place value. Answer: 0.024 cm. The distractors: 0.023 cm comes from chopping the 84 off instead of rounding it; 0.02 cm comes from counting the leading zeros as significant figures, so the 2 is taken as the second figure and the rounding stops there; 0.0238 cm is 0.02384 correct to 3 significant figures, one figure too many.
- (c) £4.05 — The total cost is 7 × 85p = £5.95. Josh's change is £10.00 − £5.95 = £4.05. Taking off the pounds and then only the 5p digit, £10.00 − £5.00 = £5.00 followed by £5.00 − £0.05, ignoring the 90p altogether, gives £4.95. Rounding the cost up to £6.00 and never accounting for the extra 5p at all gives £4.00. Multiplying 7 × 85 incorrectly as 605 (a slip in 7 × 5) gives a total cost of £6.05, and correctly subtracting that from £10.00 gives £3.95.
- (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.
- (c) 9π cm² — The area of a circle is π × r². With a radius of 3 cm this is π × 3² = 9π cm², and this is exact because π has not been replaced by any approximation. Writing 28.3 cm² replaces π with a rounded decimal value, 3.14, and then rounds the result again, so it is only an approximation. Writing 28.26 cm² uses π ≈ 3.14 without a final rounding step, but this is still only an approximation of 9π, not the exact value. Writing 27 cm² comes from replacing π with the rough approximation 3, which is even further from the true value.
Build your own mix at the worksheet builder.