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 water tank holds 80 litres when full. It currently contains 60 litres. Work out what fraction of the tank is empty.
- 2.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.
- 3.A restaurant adds a service charge of 15% to a bill of £120. Work out the service charge.
- 4.A jumper is reduced by 15% in a sale to a price of £42.50. Work out the original price.
- 5.Work out 10% of 30% of £200.
- 6.Put these numbers in order, starting with the smallest: 3.2 × 10⁴, 2.9 × 10⁵, 4.1 × 10³
- 7.A café orders 340 bread rolls at 24p each and 85 cakes at £1.35 each. Work out the total cost of the order.
- 8.Work out (6 × 10⁷) + (3 × 10⁶). Give your answer in standard form.
- 9.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.
- 10.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.
- 11.3/5 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.
- 12.2/3 of an amount of money is £30. Work out the amount.
- 13.Work out (3 × 10²) × (2 × 10⁵). Give your answer in standard form.
- 14.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.
- 15.Write these decimals in order, starting with the smallest: 0.6, 0.55, 0.601, 0.5, 0.56
Answer key
- (a) 1/4 — The empty part of the tank is 80 − 60 = 20 litres. As a fraction of the full capacity, this is 20/80, which simplifies to 1/4. Finding the fraction of the tank that is FULL instead of empty, 60/80, simplifies to 3/4 — the wrong quantity for the question asked. Writing the empty amount over the amount remaining instead of over the full capacity, 20/60, simplifies to 1/3. Comparing the empty amount to 100 instead of to the tank's actual capacity of 80, 20/100, gives 1/5.
- (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) £18 — Method: the percentage acts as an operator on the bill, and 15% can be built from 10% and 5%, where 5% is half of 10%. Working: 10% of £120 is £120 ÷ 10 = £12, and 5% is half of that, £6, so the charge is £12 + £6 = £18. Answer: £18. The distractors: £8 comes from reading 15% as one fifteenth and working out £120 ÷ 15 = £8; £12 comes from finding 10% of the bill and stopping there; £138 comes from finding the charge correctly and then giving the new total, £120 + £18, rather than the charge the question asks for.
- (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.
- (d) £6 — First find 30% of £200, which is £60, then find 10% of that: £60 × 0.1 = £6. Adding the two percentages together instead of applying them one after the other, 10% + 30% = 40%, and finding 40% of £200 gives £80. Finding 30% of £200 = £60 correctly but stopping before applying the second percentage leaves £60 as the final answer. Finding only 10% of the original £200, ignoring the 30% entirely, gives £20.
- (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.
- (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.
- (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.
- (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) 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.
- (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.
- (b) £45 — Since 2/3 of the amount is £30, one third is £30 ÷ 2 = £15, and the whole amount is three thirds: £15 × 3 = £45. Applying the fraction forwards to £30 instead of reversing it, £30 × 2/3 = £20, treats the given amount as the whole rather than as two thirds of it. Finding one third correctly as £15 but forgetting to multiply by 3 to get the whole amount leaves £15 as the final answer. Reading £30 as one third of the amount rather than as two thirds, and so multiplying straight by 3, gives £30 × 3 = £90.
- (d) 6 × 10⁷ — 3 × 2 = 6, and 2 + 5 = 7, so (3 × 10²) × (2 × 10⁵) = 6 × 10⁷. Multiplying the exponents instead of adding them gives 2 × 5 = 10, so 6 × 10¹⁰. Adding the coefficients instead of multiplying them gives 3 + 2 = 5, so 5 × 10⁷. Subtracting the exponents instead of adding them gives 5 − 2 = 3, so 6 × 10³.
- (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.
- (a) 0.5, 0.55, 0.56, 0.6, 0.601 — Compare the decimals by giving them all the same number of decimal places first: 0.600, 0.550, 0.601, 0.500, 0.560. Ordering these from smallest to largest gives 0.500, 0.550, 0.560, 0.600, 0.601, which is 0.5, 0.55, 0.56, 0.6, 0.601. Comparing the digits as though they were whole numbers, reading 0.601 as "601" and 0.5 as "5", without padding to the same number of decimal places, gives the wrong order 0.5, 0.6, 0.55, 0.56, 0.601, because it ignores the place value of each digit. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 0.601, 0.6, 0.56, 0.55, 0.5. Misreading the close values 0.55 and 0.56 and swapping them gives 0.5, 0.56, 0.55, 0.6, 0.601. So the correct order, smallest to largest, is 0.5, 0.55, 0.56, 0.6, 0.601.
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