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.
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Number worksheet — GCSE Higher
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- 1.A café offers sandwiches with one filling from 5 choices and one type of bread from 4 choices. Cheese and mustard, which is one of the 5 fillings, is not available on gluten-free bread, which is one of the 4 breads. Work out how many different sandwiches are possible.
- 2.Write these three numbers in order, starting with the smallest: 7/20, 0.3, 32%
- 3.Write these three numbers in order, starting with the smallest: 3/5, 0.55, 58%
- 4.The radius of a circular pond is given as 3.2 m, correct to 1 decimal place. Calculate the upper bound for the area of the pond, giving your answer correct to 3 significant figures.
- 5.A plank of wood is 5 1/4 m long. Pieces of length 3/4 m are cut from it. Work out how many complete pieces of 3/4 m can be cut from the plank.
- 6.A florist has 5 different flowers. Freya picks 3 of them to make a small bunch, and the order in which she picks them does not matter. Work out how many different bunches she could make.
- 7.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.
- 8.A circle has a radius of 3 cm. Which of these is the exact area of the circle?
- 9.A driving instructor is working out how many current-style UK number plates are possible. The format is 2 letters, then 2 digits, then a space, then 3 letters (for example AB12 CDE). The letters I, O and Q are never used in any letter position, leaving 23 allowed letters, but every letter and every digit may repeat anywhere on the plate. Work out how many different number plates are possible, giving your answer in standard form to 2 significant figures.
- 10.By listing systematically, work out how many two-digit multiples of 5 can be made using the digits 0, 3 and 5, if each digit can be used at most once and the number cannot start with 0.
- 11.A bag holds 5 different sweets. Noah takes 2 of the sweets out of the bag together, so the order in which he takes them does not matter. Work out how many different pairs of sweets he could take.
- 12.Work out (8 × 10⁻⁵) × (5 × 10³). Give your answer in standard form.
- 13.A pair of trainers is priced at £80 online. The shop takes 25% off the price and then adds £10 for next-day delivery. Work out the total cost.
- 14.A jumper is reduced by 15% in a sale to a price of £42.50. Work out the original price.
- 15.A café orders 340 bread rolls at 24p each and 85 cakes at £1.35 each. Work out the total cost of the order.
Answer key
- (c) 19 — Without any restriction there would be 5 × 4 = 20 different sandwiches. The restriction removes exactly one combination, cheese and mustard on gluten-free bread, so subtract 1: 20 − 1 = 19. 20 comes from ignoring the restriction completely. 15 comes from removing the gluten-free bread altogether, as if none of the fillings were available on it, 5 × 3 = 15. 16 comes from removing the cheese and mustard filling completely, as if it were not available on any bread, 4 × 4 = 16.
- (b) 0.3, 32%, 7/20 — Method: convert every number to a decimal so they can be compared on the same scale. Working: 7/20 = 0.35, 0.3 stays as 0.3, and 32% = 0.32. Comparing 0.3, 0.32 and 0.35 in size gives the order 0.3, then 0.32, then 0.35. Answer: 0.3, 32%, 7/20. 7/20, 32%, 0.3 lists the numbers from largest to smallest instead of smallest to largest. 0.3, 7/20, 32% swaps 32% and 7/20, treating the fraction 7/20 as smaller even though 7/20 = 0.35 is bigger than 32% = 0.32. 32%, 0.3, 7/20 comes from moving the digits one place too far when converting the percentage, giving 0.032 instead of 0.32, which makes 32% look far smaller than it really is.
- (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) 33.2 — The radius was rounded to 1 decimal place, so its error interval is 3.15 ≤ r < 3.25. The upper bound for the area uses the upper bound of the radius, squared: area = π × 3.25² ≈ 33.183, which rounds to 33.2 m² (3 s.f.). Using the given value of the radius directly instead of its upper bound, π × 3.2² ≈ 32.2, ignores that the radius itself has a range of possible values. Bounding the radius correctly but forgetting to square it, using area = π × 3.25 ≈ 10.2 instead of π × 3.25², drops the whole squaring step from the area formula. Using the LOWER bound of the radius instead of the upper one, π × 3.15² ≈ 31.2, finds the lower bound of the area, not the upper one.
- (a) 7 — Convert the mixed number to an improper fraction: 5 1/4 = 21/4. Dividing by 3/4 means multiplying by its reciprocal, 4/3: 21/4 × 4/3 gives 84/12, which simplifies to 7. So exactly 7 complete pieces of 3/4 m can be cut. Ignoring the 1/4 m and dividing only the whole number, 5 ÷ 3/4, gives 20/3, which is 6 complete pieces with some wood left over. Multiplying by 3/4 instead of its reciprocal, 21/4 × 3/4, gives 63/16, which is 3 complete pieces. Misreading 5 1/4 as the fraction 5/4, then dividing by 3/4, gives 5/3, which is only 1 complete piece. So 7 complete pieces can be cut from the plank.
- (d) 10 — Method: picking 3 flowers from 5 leaves 2 flowers behind, so counting the different pairs that could be left out counts the bunches, and those pairs can be listed systematically. Working: number the flowers 1 to 5; the first flower can be left out alongside any of the 4 flowers after it, the second alongside any of the 3 after it, the third alongside any of the 2 after it and the fourth alongside the last one, so the number of pairs left out is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 60 comes from working out 5 × 4 × 3 and treating the three picks as an ordered selection when the order does not matter; 30 comes from dividing that product by 2 instead of by the 6 orders in which three chosen flowers could have been picked; 15 comes from multiplying the 5 flowers by the 3 flowers picked instead of counting the selections.
- (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) 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.
- (b) 6.4 × 10⁸ — There are 5 letter positions, each with 23 choices, and 2 digit positions, each with 10 choices, and every position is independent because repeats are allowed. By the product rule, the total is 23⁵ × 10² = 6,436,343 × 100 = 643,634,300, which is 6.4 × 10⁸ to 2 significant figures. Using all 26 letters instead of the 23 that are actually allowed, ignoring the excluded letters entirely, gives 26⁵ × 10² = 1,188,137,600, which is 1.2 × 10⁹ to 2 significant figures. Adding the seven counts of choices instead of multiplying them, 23 + 23 + 10 + 10 + 23 + 23 + 23, gives 135, which is 1.4 × 10² to 2 significant figures — a total far too small for seven independent positions. Swapping which count of choices belongs to letters and which belongs to digits, working out 23² × 10⁵ instead of 23⁵ × 10², gives 52,900,000, which is 5.3 × 10⁷ to 2 significant figures.
- (b) 3 — Method: list all valid two-digit numbers that can be made without starting with 0, then keep only the ones that are multiples of 5. Working: the two-digit numbers possible are 30, 35, 50 and 53. A number is a multiple of 5 only if it ends in 0 or 5: 30 ends in 0, 35 ends in 5, 50 ends in 0, but 53 ends in 3. So there are 3 multiples of 5. Answer: 3. 4 comes from including 53 as a multiple of 5 without checking that its last digit is not 0 or 5. 2 comes from leaving out 50, wrongly assuming 0 cannot be used as the second digit either. 6 comes from listing every two-digit arrangement of the three digits, including ones that start with 0, without applying either restriction.
- (d) 10 — Method: list the pairs systematically, taking each sweet in turn and pairing it only with the sweets that come after it, so that no pair is written down twice. Working: numbering the sweets 1 to 5, the first sweet pairs with 4 others, the second pairs with 3 sweets that come after it, the third with 2 and the fourth with 1, so the total is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 20 comes from working out 5 × 4 and never halving, which counts each pair twice, once in each order; 25 comes from working out 5 × 5, which allows the same sweet to be chosen twice; 9 comes from adding the 5 choices and the 4 remaining choices instead of combining them as a selection of two.
- (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.
- (d) £70 — Method: the discount is a percentage of the original price only, so work it out, subtract it, and add the fixed delivery charge afterwards. Working: 25% of £80 is £80 ÷ 4 = £20, so the discounted price is £80 − £20 = £60, and the total is £60 + £10 = £70. Answer: £70. The distractors: £60 comes from working out the discounted price and stopping there, leaving the delivery charge out of the total; £65 comes from taking £25 off the price instead of 25% of it, giving £80 − £25 = £55 and then £55 + £10 = £65; £67.50 comes from adding the delivery charge before the discount and reducing the whole amount, giving 75% of £90 = £67.50.
- (d) £50.00 — The sale price is 85% of the original, so the original price = £42.50 ÷ 0.85 = £50.00. 15% of £42.50 is £6.375. A candidate who finds 15% of £42.50 and subtracts it from the sale price gets £42.50 − £6.375 = £36.125, which is £36.13 to the nearest penny. A candidate who adds 15% of £42.50 instead of reversing the decrease gets £42.50 + £6.375 = £48.875, which is £48.88 to the nearest penny. A candidate who divides by 0.15 instead of 0.85 gets £283.33.
- (c) £196.35 — Method: convert both prices to pounds, multiply each by its quantity, then add the two totals. Working: 340 rolls at £0.24 each = £81.60; 85 cakes at £1.35 each = £114.75; £81.60 + £114.75 = £196.35. Answer: £196.35. £81.60 comes from working out the cost of the rolls only and forgetting to add the cost of the cakes. £114.75 comes from working out the cost of the cakes only and forgetting to add the cost of the rolls. £122.91 comes from converting 24p to £0.024 instead of £0.24, a place value error of a factor of 10 in the price of the rolls, before adding the correctly worked out cost of the cakes.
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