Printable · GCSE Higher · ages 14-16
Number worksheet — GCSE Higher
Fifteen questions across the number statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Number worksheet — GCSE Higher
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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 country has an area of 25,000,000 hectares. Write this area in standard form.
- 3.Maya buys 4 plants at £3.20 each and a bag of compost for £6.75. She pays with a £20 note. Work out her change.
- 4.A quarter-circle has a radius of 6 cm. Work out the exact perimeter of the quarter-circle, giving your answer in terms of π.
- 5.A plumber charges a call-out fee of £45 and then £28 for each hour she works. She is at a house from 09:15 until 12:45. Work out the total charge.
- 6.Round 0.006482 to 2 significant figures.
- 7.After a price increase of 10%, a laptop costs £330. Work out the original price.
- 8.Write these numbers in order, starting with the smallest: −1.4, 5/4, −6/5, 1.3, 0
- 9.A board game has 5 different character pieces and 4 different colour tokens. The dragon piece can only be used with the gold token. Work out how many different combinations of one character piece and one colour token are possible.
- 10.A car is hired from 10:00 on Monday until 16:00 on Thursday. Work out the total length of the hire, in hours.
- 11.A van has a mass of 2,000 kg, correct to 1 significant figure. Using m for the mass of the van in kilograms, write down the error interval for m.
- 12.Without using a calculator, estimate the value of √70 × ∛65, giving your answer to 1 significant figure.
- 13.Write 12 as a product of its prime factors.
- 14.Work out 3 1/5 + 1 2/3. Give your answer as a mixed number in its simplest form.
- 15.√700 lies between which two consecutive integers?
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.
- (a) 2.5 × 10⁷ — Method: place the decimal point so that the coefficient is at least 1 and less than 10, then count the places it has moved. Working: the digits give a coefficient of 2.5, and the decimal point travels from the end of 25,000,000 until it sits between the 2 and the 5, a move of 7 places. Answer: 2.5 × 10⁷. The distractors: 25 × 10⁶ is the same area but not in standard form, because the coefficient must be less than 10; 2.5 × 10⁸ comes from counting the eight digits of 25,000,000 instead of the seven places the decimal point moves; 2.5 × 10⁻⁷ comes from making the index negative because the decimal point was carried to the left.
- (c) £0.45 — Method: find the total cost, then subtract from £20. Working: 4 × £3.20 = £12.80. £12.80 + £6.75 = £19.55. Change = £20.00 − £19.55 = £0.45. Answer: £0.45. (£7.20 comes from forgetting to include the compost and subtracting only the plants' cost from £20. £1.45 comes from dropping the carry when adding the pence: 80p + 75p = £1.55, but only the 55p is written down, giving £18.55 instead of £19.55. £10.05 comes from buying only one plant instead of four, using £3.20 + £6.75.)
- (c) (12 + 3π) cm — The perimeter of a quarter-circle is made up of two straight radii plus a quarter of the circumference. The two radii give 2 × 6 = 12 cm, and a quarter of the circumference is (1/4) × 2 × π × 6 = 3π cm, so the total perimeter is (12 + 3π) cm. Giving only the curved part, 3π cm, forgets the two straight edges entirely. Using the full circumference, 2 × π × 6 = 12π, instead of a quarter of it gives (12 + 12π) cm. Including only one radius instead of two gives (6 + 3π) cm.
- (d) £143 — From 09:15 to 12:45 is 3 hours 30 minutes, which is 3.5 hours. The charge for the time is 3.5 × £28 = £98, and the call-out fee is added on once: £98 + £45 = £143. £98 leaves out the call-out fee altogether, £129 charges for 3 hours instead of 3.5 hours, and £255.50 adds the fee to the hourly rate before multiplying, which charges the £45 for every hour.
- (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.
- (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.
- (d) −1.4, −6/5, 0, 5/4, 1.3 — Method: convert the fractions 5/4 and −6/5 to decimals so every number is written the same way, then compare all five decimals. Working: 5/4 = 1.25 and −6/5 = −1.2. Comparing −1.4, −1.2, 0, 1.25 and 1.3 in size gives the order −1.4, −1.2, 0, 1.25, 1.3. Answer: −1.4, −6/5, 0, 5/4, 1.3. −6/5, −1.4, 0, 5/4, 1.3 swaps the two negative numbers, treating −6/5 as more negative than −1.4 even though −1.2 is closer to zero than −1.4. 1.3, 5/4, 0, −6/5, −1.4 lists the numbers from largest to smallest instead of smallest to largest. −1.4, −6/5, 0, 1.3, 5/4 swaps 5/4 and 1.3, comparing the numerator 5 directly with 1.3 instead of converting 5/4 to the decimal 1.25 first.
- (d) 17 — Without the restriction there would be 5 × 4 = 20 combinations. The dragon piece can only be paired with the gold token, so of the 4 tokens, 3 are not allowed with the dragon piece, giving 20 − 3 = 17 valid combinations. 20 comes from ignoring the restriction completely. 19 comes from subtracting only 1 of the 3 invalid dragon combinations instead of all 3, 20 − 1 = 19. 16 comes from multiplying only the 4 non-dragon pieces by the 4 tokens, 4 × 4 = 16, and forgetting to add back the one valid combination of the dragon piece with the gold token.
- (b) 78 hours — From 10:00 on Monday to 10:00 on Thursday is exactly 3 complete days, which is 3 × 24 = 72 hours. From 10:00 to 16:00 on the Thursday is a further 6 hours, giving a total of 72 + 6 = 78 hours. Counting Monday to Thursday as 4 full calendar days instead of 3 complete 24-hour periods gives 4 × 24 = 96 hours. Undercounting the number of complete days as 2 instead of 3 gives 2 × 24 + 6 = 54 hours. Subtracting the extra 6 hours instead of adding them to the 3 complete days gives 72 − 6 = 66 hours.
- (a) 1,500 ≤ m < 2,500 — Method: a four-digit figure written to 1 significant figure has been rounded to the nearest 1,000, so the mass lies within half of 1,000, that is 500, of the figure given. Working: 2,000 − 500 = 1,500 and 2,000 + 500 = 2,500. The lower limit is included, because 1,500 kg rounds up to 2,000 kg to 1 significant figure, while 2,500 kg rounds up to 3,000 kg, so the upper limit is not. Answer: 1,500 ≤ m < 2,500. The distractors: 1,950 ≤ m < 2,050 comes from rounding to the nearest 100 instead of to 1 significant figure; 1,000 ≤ m < 3,000 goes a whole 1,000 either side instead of half of it; 1,500 < m ≤ 2,500 has the two limits the wrong way round.
- (b) 30 — 70 is close to the perfect square 64, so √70 ≈ 8. 65 is close to the perfect cube 64, so ∛65 ≈ 4. Multiplying these estimates: 8 × 4 = 32, which rounds to 30 to 1 significant figure. Estimating ∛65 as 5 instead of 4, perhaps by confusing it with the nearby cube 125 = 5³ rather than the much closer 64 = 4³, and then multiplying by 8, gives 8 × 5 = 40. Adding the two estimates instead of multiplying them, 8 + 4 = 12, rounds to 10 to 1 significant figure. Rounding both estimates up to the next whole number using the wrong nearby power for each, taking √70 as 9 and ∛65 as 5, gives 9 × 5 = 45, which rounds to 50 to 1 significant figure.
- (a) 2² × 3 — Method: divide repeatedly by the smallest prime that goes in, until 1 is reached, then write the primes used as a product with indices. Working: 12 ÷ 2 = 6, 6 ÷ 2 = 3 and 3 ÷ 3 = 1, so the primes used are 2, 2 and 3, which is written as 2² × 3. Answer: 2² × 3. The distractors: 2 × 6 comes from stopping at the first factor pair without splitting the 6, which is not prime; 2 × 3 comes from listing each prime once and losing the repeat, and it multiplies to 6 rather than 12; 2 × 3² puts the index on the wrong prime and multiplies to 18.
- (d) 4 13/15 — Convert to fifteenths: 1/5 is equivalent to 3/15 (multiply by 3/3), and 2/3 is equivalent to 10/15 (multiply by 5/5), so 3 1/5 is equivalent to 3 3/15 and 1 2/3 is equivalent to 1 10/15. Add the whole numbers (3 + 1 = 4) and the fractions (3/15 + 10/15 = 13/15), giving 4 13/15. A candidate who adds the numerators and denominators straight across, treating 1/5 + 2/3 as (1+2)/(5+3), gets a fraction part of 3/8, giving 4 3/8. A candidate who adds the fraction parts correctly but forgets to add the second whole number gets 3 13/15. A candidate who adds the whole numbers but copies the first fraction across without ever adding 2/3 to it gets 4 1/5.
- (d) 26 and 27 — Find the two consecutive perfect squares either side of 700: 26² = 676 and 27² = 729. Since 676 < 700 < 729, √700 lies between 26 and 27. Answering 7 and 8 comes from stripping the two zeros off 700 and using the 7 itself as the size of the root, instead of comparing 700 with the perfect squares around it — dividing the number under the root by 100 divides the root by 10, so the digits do not simply carry across. Answering 25 and 26 comes from checking 25² = 625, seeing that it is less than 700, and stopping there without also checking the square directly above it. Answering 35 and 36 comes from halving 700 to 350 and then treating that halved value as if it were ten times the true root, drifting into the thirties instead of the twenties.
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