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.A shop buys boxes of pens for £4 each. It sells each box on to a school for £3 more than it paid, and charges an extra £2 delivery fee for the whole order. A school orders 5 boxes. Work out the total cost of the order.
- 2.A red blood cell has a diameter of about 7 × 10⁻⁶ metres. A virus has a diameter about 100 times smaller. Work out the diameter of the virus. Give your answer in standard form.
- 3.Write these three numbers in order, starting with the smallest: 3/5, 0.55, 58%
- 4.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.
- 5.A tin of soup has a mass of 450 g. A bag of potatoes has a mass of 2.6 kg. Work out the total mass, in kilograms.
- 6.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.
- 7.A cyclist rides at a steady speed of 8 metres per second. Work out this speed in kilometres per hour.
- 8.A circle has a radius of 3 cm. Which of these is the exact area of the circle?
- 9.Work out (8 × 10⁻⁵) × (5 × 10³). Give your answer in standard form.
- 10.Work out (6 × 10⁷) + (3 × 10⁶). Give your answer in standard form.
- 11.To estimate the cost of buying 38.7 m of rope at £21.40 per metre, both numbers are first rounded to 1 significant figure. Work out the estimate.
- 12.The rainfall in a town during April is recorded as 62.4 mm, correct to 1 decimal place. Write down the error interval for the actual rainfall, r mm.
- 13.A number, y, is equal to 8.2 when rounded to 1 decimal place. Write down the error interval for y.
- 14.3/5 of the students in a year group walk to school. 90 students walk to school. Work out the total number of students in the year group.
- 15.Write 45,000 in standard form.
Answer key
- (b) £37 — One box costs £4 + £3 = £7. Five boxes cost 5 × £7 = £35. Adding the single £2 delivery fee gives £35 + £2 = £37. A candidate who added the £2 delivery fee to each box instead of once for the whole order worked out 5 × (£7 + £2) = 5 × £9 = £45. A candidate who forgot the £3 markup and used the shop's buying price worked out 5 × £4 + £2 = £22. A candidate who added the £3 markup only once, after multiplying the buying price by 5, worked out 5 × £4 + £3 + £2 = £25.
- (c) 7 × 10⁻⁸ — '100 times smaller' means dividing by 100 = 10². Dividing 7 × 10⁻⁶ by 10² means subtracting 2 from the exponent: −6 − 2 = −8, giving 7 × 10⁻⁸. A candidate who multiplied by 100 instead of dividing added 2 to the exponent, getting 7 × 10⁻⁴. A candidate who divided by 10 instead of 100 subtracted only 1 from the exponent, getting 7 × 10⁻⁵. A candidate who did not apply the scale factor at all left the diameter as 7 × 10⁻⁶, the same as the red blood cell.
- (a) 0.55, 58%, 3/5 — Converting all three to decimals: 3/5 = 0.6, 0.55 stays as 0.55, and 58% = 0.58. In order from smallest to largest, this is 0.55, then 58%, then 3/5. Writing the numbers in the reverse order, largest to smallest, gives 3/5, 58%, 0.55. Misconverting 3/5 as 0.5 instead of 0.6 makes it appear smaller than both other values, giving the order 3/5, 0.55, 58%. Misconverting 58% as 0.058 instead of 0.58, by moving the decimal point two extra places, makes it appear smallest of the three, giving the order 58%, 0.55, 3/5.
- (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.
- (c) 3.05 kg — There are 1000 g in 1 kg, so 450 g = 450 ÷ 1000 = 0.45 kg. The total mass is 0.45 + 2.6 = 3.05 kg. 7.1 kg divides the grams by 100 instead of 1000, 2.645 kg divides them by 10 000, and 452.6 kg adds the two numbers without converting the grams at all.
- (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.
- (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.
- (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.
- (c) 4 × 10⁻¹ — Multiply the A values: 8 × 5 = 40. Add the powers of 10: −5 + 3 = −2, giving 40 × 10⁻². Since A must satisfy 1 ≤ A < 10, rewrite 40 as 4 × 10¹, so 40 × 10⁻² = 4 × 10¹ × 10⁻² = 4 × 10⁻¹. A candidate who stopped at 40 × 10⁻² did the index arithmetic correctly but left the answer outside standard form, since 40 is not between 1 and 10. A candidate who adjusted the A value to 4 correctly but then took the power of 10 by subtracting the two given powers, −5 − 3 = −8, wrote 4 × 10⁻⁸. A candidate who adjusted the A value to 4 but multiplied the two given powers, −5 × 3 = −15, wrote 4 × 10⁻¹⁵. Both of these forgot that multiplying in standard form means adding the powers.
- (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.
- (d) £800 — Rounding 38.7 to 1 significant figure gives 40, and rounding 21.40 to 1 significant figure gives 20. Multiplying the rounded values gives an estimate of 40 × 20 = £800. Rounding 21.40 to the nearest whole number instead of to 1 significant figure gives 21, and 40 × 21 = £840, one place value too fine for the price. Adding the rounded values instead of multiplying them gives 40 + 20 = £60. Rounding both numbers to 2 significant figures instead of 1, giving 39 and 21, produces 39 × 21 = £819.
- (c) 62.35 ≤ r < 62.45 — Method: with a value rounded to 1 decimal place, the error interval reaches half of 0.1 either side. Working: half of 0.1 is 0.05, so the interval runs from 62.4 − 0.05 to 62.4 + 0.05. Answer: 62.35 ≤ r < 62.45. (62 ≤ r < 63 comes from rounding to the nearest whole number instead of 1 decimal place. 62.35 ≤ r ≤ 62.45 comes from including the upper bound with ≤ instead of excluding it with <. 62.3 ≤ r < 62.5 comes from using 0.1 either side instead of half of it.)
- (a) 8.15 ≤ y < 8.25 — Rounding to 1 decimal place means the error interval spans half of 0.1, so 0.05, either side of 8.2: 8.2 − 0.05 = 8.15 and 8.2 + 0.05 = 8.25. The lower bound uses ≤ because 8.15 itself rounds to 8.2, but the upper bound uses < because 8.25 would round up to 8.3. So the error interval is 8.15 ≤ y < 8.25. A candidate who used the wrong rounding band gave 8.1 ≤ y < 8.2. A candidate who used a strict inequality at both ends wrote 8.15 < y < 8.25, wrongly excluding 8.15 itself. A candidate who added the full 0.1 instead of half of it wrote 8.2 ≤ y < 8.3.
- (c) 150 — Since 90 students represent 3 of the 5 equal parts, one part is 90 ÷ 3 = 30, and the whole year group is five parts: 30 × 5 = 150. Applying the fraction forwards to 90 instead of reversing it, 90 × 3/5 = 54, treats the given number as the whole rather than as three fifths of it. Finding one part correctly as 30 but forgetting to scale up to the whole year group leaves 30 as the final answer. Treating 90 as the whole year group and adding on 2/5 of 90 for the students who do not walk, 90 + (90 × 2/5) = 126, applies the missing fraction to the wrong base amount.
- (a) 4.5 × 10⁴ — 45,000 = 4.5 × 10,000 = 4.5 × 10⁴, with the coefficient between 1 and 10 as standard form requires. Writing 45 × 10³ keeps the coefficient too large — 45 is not between 1 and 10. Writing 4.5 × 10³ undercounts the places moved, giving only 4,500. Writing 4.5 × 10⁵ overcounts the places moved, giving 450,000.
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