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 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.
- 2.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.
- 3.A charity raises money from a raffle and a cake sale in the ratio 5 : 3. Altogether the charity raises £320. Work out how much money the cake sale raised.
- 4.A circle has a radius of 3 cm. Which of these is the exact area of the circle?
- 5.A café orders 340 bread rolls at 24p each and 85 cakes at £1.35 each. Work out the total cost of the order.
- 6.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.
- 7.Which of these is written correctly in standard form?
- 8.Ten players enter a chess tournament. Every player plays every other player exactly once. Work out how many games are played in the tournament.
- 9.Work out 35% of 180.
- 10.Write these numbers in order, starting with the smallest: −1.4, 5/4, −6/5, 1.3, 0
- 11.p = 24 and q = 6, each correct to the nearest integer. Work out the greatest degree of accuracy to which p ÷ q can be guaranteed correct.
- 12.Round 0.006482 to 2 significant figures.
- 13.The density of a metal is calculated using density = mass ÷ volume. A sample has a mass of 156 g, correct to the nearest gram, and a volume of 12 cm³, correct to the nearest cm³. Work out the minimum possible density, in g/cm³.
- 14.Write these three numbers in order, starting with the smallest: 7/20, 0.3, 32%
- 15.A cube-shaped storage box has edges of length 7 cm. A shelf can hold a total volume of 2,000 cm³. Work out the greatest number of these boxes that will fit in that volume.
Answer key
- (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) 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.
- (d) £120 — Total parts = 5 + 3 = 8, so one part is worth 320 ÷ 8 = 40 pounds. The cake sale is 3 parts, so it raised 3 × 40 = 120 pounds. £200 comes from working out the raffle's share, 5 × 40, instead of the cake sale's share. £40 comes from finding the value of one part but forgetting to multiply by 3. £192 comes from dividing the total by 5 instead of 8 to find the value of one part, 320 ÷ 5 = 64, then multiplying by 3, 3 × 64 = 192.
- (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) £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.
- (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.
- (a) 7.5 × 10⁴ — Standard form is A × 10ⁿ, where A is between 1 and 10 (1 ≤ A < 10) and n is an integer — 7.5 × 10⁴ satisfies all of this. 12 × 10³ fails because 12 is not below 10. 0.5 × 10⁴ fails because 0.5 is not at least 1. 6.2 × 4⁵ fails because standard form always uses a power of 10, not a power of 4.
- (c) 45 — Method: count the ordered pairings with the product rule and then correct for the fact that a game between two players is the same game whichever player it is counted from. Working: each of the 10 players meets 9 opponents, so 10 × 9 = 90 pairings are counted; every game has been counted twice, once from each player's side, so the number of games is 90 ÷ 2 = 45. Answer: 45. The distractors: 90 comes from stopping at 10 × 9 and never halving, so that each game is counted once for each of its two players; 55 comes from adding 10 + 9 + 8 + ... + 1 instead of 9 + 8 + ... + 1, which counts one extra round of games; 20 comes from multiplying the 10 players by the 2 players in each game rather than pairing the players with one another.
- (b) 63 — 10% of 180 = 18, so 5% = 9. 35% = (3 × 18) + 9 = 54 + 9 = 63. A candidate who uses 25% instead of 35% gets 45. A candidate who doubles 35% to get 70% by mistake gets 126. A candidate who subtracts 35 from 180 instead of finding a percentage gets 145.
- (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.
- (a) the nearest whole number — The error intervals are 23.5 ≤ p < 24.5 and 5.5 ≤ q < 6.5. The minimum of p ÷ q is 23.5 ÷ 6.5 ≈ 3.615, and the maximum is 24.5 ÷ 5.5 ≈ 4.455. Both of these round to 4 at the nearest whole number, so the answer is guaranteed correct to the nearest whole number — but not to the nearest 0.1, since 3.615 rounds to 3.6 while 4.455 rounds to 4.5, which do not agree. Claiming the nearest 0.1 assumes every figure a calculator shows is trustworthy, without checking whether the bounds actually agree that far. Claiming only the nearest 10 badly understates how much can be guaranteed here, since both bounds already round to 4, not merely to 0. Saying no degree of accuracy can be guaranteed gives up before checking whether the bounds agree at any level at all.
- (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.
- (c) 12.44 — The error intervals are 155.5 ≤ mass < 156.5 and 11.5 ≤ volume < 12.5. To make a quotient as small as possible, use the SMALLEST possible numerator together with the LARGEST possible denominator: 155.5 ÷ 12.5 = 12.44 g/cm³. Using the lower bound for both mass and volume, 155.5 ÷ 11.5 ≈ 13.52, forgets that dividing by a smaller number makes the result bigger, not smaller — that pairing does not give a minimum at all. Dividing the two given rounded values directly, 156 ÷ 12 = 13, ignores that both measurements have their own error interval. Using the upper bound of mass with the upper bound of volume, 156.5 ÷ 12.5 = 12.52, takes both bounds the same way round; it is neither the minimum nor the maximum, since the maximum needs the largest mass with the smallest volume, 156.5 ÷ 11.5 ≈ 13.61.
- (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.
- (c) 5 — Method: work out the volume of one box, divide the total volume by it, then round down since a partial box cannot fit. Working: volume of one box = 7³ = 343 cm³. 2000 ÷ 343 = 5.83 (2 d.p.). Since only whole boxes fit, the greatest number is 5. Answer: 5. (6 comes from rounding 5.83 up to the nearest whole number instead of rounding down to the number of boxes that actually fit. 343 comes from giving the volume of one box instead of the number of boxes. 5.8 comes from leaving the division as a decimal instead of rounding down to a whole number of boxes.)
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