25 questions on the topics that decide a grade 4: percentages, ratio, area and perimeter, linear equations, averages and probability.
🎯 Grade 4 pass booster
A grade 4 is the standard pass that colleges, apprenticeships and employers ask for, and most students who miss it do not miss it on the hard questions — they miss it by dropping marks across the accessible ones. This sheet gathers twenty-five of those: percentage of an amount and percentage change, sharing in a ratio, area and perimeter of rectangles, triangles and compound shapes, solving a linear equation, reading the mean, median, mode and range from a list or a table, and simple probability. Nothing here is a stretch topic, and all of it is worth marks on every paper, on both tiers. Work through it, mark it honestly, and treat anything you got wrong as a priority over anything further up the specification — a secured mark is worth more than an attempted one.
- 1.The price of a share falls by 10% on Monday and then rises by 10% on Tuesday. Work out the overall percentage change from Monday's starting price.
- 2.A metal alloy is made from copper and tin in the ratio 7:3. Work out the mass of tin in 250 g of the alloy.
- 3.Four angles meet at a point. Three of them measure 82°, 105° and 96°. Work out the size of the fourth angle.
- 4.In a test, Amelia answered 18 of the 24 questions correctly. Work out the percentage of the questions she answered correctly.
- 5.A regular nonagon has 9 sides. Work out the sum of its interior angles.
- 6.A charity fun run raises money through entry fees and donations. Entry fees raise £1,260, which is 60% of the total amount raised. Work out how much money was raised through donations.
- 7.Solve 5x − 9 = 16
- 8.The mean of the numbers x, 20 and 30 is equal to the mean of the numbers 15 and 25. Work out the value of x.
- 9.A parallelogram has an area of 136 cm² and a base of 17 cm. Work out the perpendicular height of the parallelogram.
- 10.A fish tank is a cuboid measuring 40 cm long, 25 cm wide and 30 cm high. Water is poured in to a depth of 20 cm. Work out the volume of water in the tank, in litres.
- 11.A cylinder has a volume of 942 cm³ and a height of 12 cm. Using π = 3.14, work out the radius of the cylinder.
- 12.A fair coin is flipped and a fair five-sided spinner, labelled 1 to 5, is spun. All the outcomes are listed as pairs, such as (H, 3). Work out the probability that the outcome is a tail and an odd number.
- 13.A garden water trough is a prism whose cross-section is a right-angled triangle with base 40 cm and height 30 cm. The trough is 120 cm long. Work out the volume of water needed to fill the trough completely.
- 14.Two angles are complementary. One of them is 35°. Work out the size of the other angle.
- 15.A straight line crosses a pair of parallel lines. One of the co-interior (allied) angles is 118°. Work out the size of the other co-interior angle.
- 16.Work out the difference between 45% of 70 and 35% of 80.
- 17.A fair four-sided dice, numbered 1 to 4, is rolled twice. Every outcome is listed as an ordered pair, first roll then second roll, such as (2, 3). Work out how many different outcomes are in the list.
- 18.The mean of three numbers is 50. A fourth number, 100, is added to the set. Work out the mean of the four numbers.
- 19.A ramp's sloped surface crosses two horizontal parallel rails. At the top rail, the angle between the ramp and the rail on the right of the ramp is (3x + 10)°. At the bottom rail, the angle between the ramp and the rail on the left of the ramp is (5x − 30)°. Work out the value of x.
- 20.A cylinder has a radius of 3 cm and a height of 10 cm. Using π = 3.14, work out the volume of the cylinder. Give your answer to the nearest whole number of cubic centimetres.
- 21.A jumper costs £45 at Shop A, where it is reduced by 20%. The same jumper costs £34 at Shop B, where a further 10% reduction is then applied. Work out the difference between the two reduced prices.
- 22.A parallelogram has a base of 4 cm and a perpendicular height of 17 cm. Work out the area of the parallelogram.
- 23.Order these three values from smallest to largest: 3/8, 0.43, 41%.
- 24.A regular polygon has 12 sides. Work out the size of one exterior angle of the polygon.
- 25.Four books have 120, 200, 160 and 80 pages. Work out the mean number of pages.
Answer key
- (b) −1% — Method: write each change as a multiplier and multiply them. A 10% fall is × 0.9 and a 10% rise is × 1.1. Working: 0.9 × 1.1 = 0.99, so the final price is 99% of the original, which is 1% less. Answer: an overall change of −1%. The distractors: 0% comes from assuming a 10% fall and a 10% rise cancel — they do not, because the rise is 10% of a smaller amount; +1% has the size right but the sign wrong, from reading the multiplier 0.99 as 1% above 1 instead of 1% below it; −2% comes from finding the 1% fall and then counting it once for each of the two changes.
- (b) 75 g — Method: split the total mass into the number of parts shown by the ratio, then find the mass of tin. Working: the ratio 7:3 has 7 + 3 = 10 parts, so one part is 250 ÷ 10 = 25 g, and the mass of tin is 3 × 25 = 75 g. So the alloy contains 75 g of tin. Distractor 175 g is the mass of copper, not tin. Distractor 125 g comes from splitting the alloy into two equal halves, ignoring the ratio. Distractor 25 g is the value of one part, found correctly but never multiplied by 3.
- (d) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.
- (a) 75% — Method: to express one quantity as a percentage of another, divide the part by the whole and multiply by 100. Working: 18 ÷ 24 = 0.75, and 0.75 × 100 = 75. Answer: 75%. The distractors: 25% is the percentage she got wrong, 6 out of 24; 133% comes from dividing the whole by the part, 24 ÷ 18; 18% comes from writing the number of correct answers with a percent sign.
- (a) 1260° — The sum of the interior angles of a polygon with n sides is (n − 2) × 180°. For a nonagon, n = 9, so the sum is (9 − 2) × 180° = 7 × 180° = 1260°. 1620° uses 9 × 180° without subtracting 2 from n first. 140° is the size of a single interior angle of a regular nonagon (1260° ÷ 9), not the sum of all nine. 1440° uses (n − 1) × 180° = 8 × 180° instead of (n − 2) × 180°.
- (c) £840 — Method: find the total amount raised using the reverse percentage, then subtract the entry fees to find the donations. Working: £1,260 is 60% of the total, so the total is £1,260 ÷ 0.6, and subtracting the entry fees from this total leaves £840 raised through donations. Answer: £840. £2,100 comes from correctly finding the total amount raised but then forgetting to subtract the entry fees, giving the total instead of the donations alone. £504 comes from working out 40% of the entry fees themselves, £1,260 × 0.4 = £504, instead of first finding the total amount raised. £1,890 comes from treating £1,260 as 40% of the total instead of 60%, dividing by 0.4 to get a total of £3,150, and then subtracting the entry fees from that incorrect total.
- (b) 5 — Method: add 9 to both sides, then divide by 5. Working: 5x = 16 + 9 = 25, so x = 25 ÷ 5 = 5. Answer: 5. 3.2 comes from dividing 16 by 5 directly, without adding 9 first. 1.4 comes from a sign error, subtracting 9 from 16 instead of adding it, then dividing by 5. 25 comes from correctly working out 5x = 25 but stopping there, without dividing by 5 to find x.
- (c) 10 — Method: work out the mean that can be found straight away, then use total = mean × number of values on the group of three to find the missing number. Working: the mean of 15 and 25 is (15 + 25) ÷ 2 = 40 ÷ 2 = 20, so the group of three must also have a mean of 20; three numbers with a mean of 20 have a total of 20 × 3 = 60, and 20 + 30 = 50 of that total is already accounted for, so x = 60 − 50 = 10. Answer: 10, and checking, (10 + 20 + 30) ÷ 3 = 20. The distractors: 20 comes from working out the mean the two groups share and writing that down as x; −10 comes from dividing the group of three by 2 instead of by 3, which gives x + 50 = 40; 70 comes from reading the total 15 + 25 = 40 as the mean of the pair, which sets the target total at 120 and leaves x = 70.
- (b) 8 cm — Area = base × height, so height = area ÷ base = 136 ÷ 17 = 8 cm. (16 cm comes from using the triangle formula, thinking area = 1/2 × base × height and so height = 2 × area ÷ base; 2312 cm comes from multiplying the area by the base instead of dividing; 119 cm comes from subtracting the base from the area instead of dividing the area by the base.)
- (a) 20 litres — Volume of water = length × width × depth of water = 40 × 25 × 20 = 20 000 cm³. Since 1000 cm³ = 1 litre, divide by 1000: 20 000 ÷ 1000 = 20 litres. A pupil who uses the full height of the tank, 30 cm, instead of the water depth, 20 cm, gets 40 × 25 × 30 = 30 000 cm³ = 30 litres. A pupil who forgets to convert cm³ to litres at all gives 20 000 litres. A pupil who divides by 1000 twice by mistake gets 20 000 ÷ 1000 ÷ 1000 = 0.02 litres. The correct volume of water is 20 litres.
- (c) 5 cm — Volume of a cylinder = πr²h, so r² = V ÷ (πh) = 942 ÷ (3.14 × 12) = 942 ÷ 37.68 = 25, and r = √25 = 5 cm. A pupil who finds r² = 25 but forgets to take the square root gives 25 cm. A pupil who forgets to divide by π, using r² = 942 ÷ 12 = 78.5, gets r = √78.5 ≈ 8.9 cm. A pupil who forgets to divide by the height, using r² = 942 ÷ 3.14 = 300, gets r = √300 ≈ 17.3 cm. The correct radius is 5 cm.
- (d) 3/10 — Method: list the full possibility space as pairs of coin and spinner results, then count how many pairs satisfy both conditions and divide by the size of the whole space. Working: the coin gives 2 outcomes and the spinner gives 5, so the full space has 2 × 5 = 10 equally likely pairs. The pairs with a tail and an odd number are (T,1), (T,3) and (T,5), which is 3 out of 10. Answer: 3/10. Watch out: writing down 1/2 uses only the coin's own chance of a tail and ignores that the spinner also has to land on an odd number. Writing down 3/5 uses only the spinner's chance of landing on an odd number and ignores the coin altogether. And writing down 1/10 counts just one matching outcome, such as (T,1), instead of all three pairs that satisfy both conditions.
- (b) 72 000 cm³ — Cross-sectional area = 1/2 × 40 × 30 = 600 cm². Volume = cross-sectional area × length = 600 × 120 = 72 000 cm³. (144 000 cm³ comes from forgetting the 1/2 in the triangle's area, using 40 × 30 as the cross-section; 36 000 cm³ comes from halving the correct volume again, as if the 1/2 applied a second time; 720 cm³ comes from adding the cross-sectional area and the length, 600 + 120, instead of multiplying them.)
- (b) 55° — Method: complementary angles are a pair that add up to 90°, so subtract the known angle from 90°. Working: 90 − 35 = 55. Answer: 55°. The distractors: 145° comes from subtracting from 180°, which is the rule for supplementary angles and not for complementary ones; 325° comes from subtracting from 360°, the total of the angles at a point; 90° is the total that the pair must make, written down in place of the missing angle.
- (b) 62 — Method: co-interior (allied) angles between parallel lines add up to 180°. Working: 180 − 118 = 62. Answer: 62°. A candidate who treats co-interior angles as equal, like corresponding angles, gives 118. A candidate who uses 360° instead of 180°, working out 360 − 118, gets 242. A candidate who subtracts as if the angles were complementary, working out 118 − 90, gets 28.
- (b) 3.5 — Method: work out each percentage of its number separately, then subtract the smaller result from the larger one. Working: 45% of 70 = 31.5, and 35% of 80 = 28, so the difference is 31.5 − 28 = 3.5. Answer: 3.5. 11.5 comes from pairing the percentages with the wrong numbers, working out 35% of 70 = 24.5 and 45% of 80 = 36, and finding their difference. 59.5 comes from adding the two correct results, 31.5 + 28, instead of subtracting them. 10 comes from simply subtracting the two percentages themselves, 45 − 35, without applying them to the numbers at all.
- (a) 16 — Method: keep the two rolls in order, first roll then second roll, and count every ordered pair the dice can land on, so a (2, 3) result is counted separately from a (3, 2) result. Working: the first roll can land on any of 4 numbers, and for each of those the second roll can also land on any of 4 numbers, giving 4 × 4 = 16 ordered pairs. Answer: 16. Watch out: writing down 8 comes from adding the two rolls' counts, 4 + 4, instead of multiplying them. Writing down 10 comes from listing unordered pairs, so (2, 3) and (3, 2) are counted as the same entry — that gives the 4 doubles plus 6 mixed pairs, 10 in total, instead of 16 ordered pairs. And writing down 4 comes from counting only one roll's outcomes and never pairing the two rolls together at all.
- (a) 62.5 — Method: a mean cannot be averaged with a new value — rebuild the total, add the new value to it, then divide by the new count. Working: three numbers with a mean of 50 have a total of 50 × 3 = 150; adding 100 makes the total 150 + 100 = 250; there are now 4 numbers, so the new mean is 250 ÷ 4 = 62.5. Answer: 62.5. The distractors: 75 comes from averaging the old mean with the new value, (50 + 100) ÷ 2, which ignores that three numbers pull against one; 50 comes from assuming an extra value leaves the mean unchanged; 37.5 comes from dividing the old total of 150 by the new count of 4, adding the new value to the count but not to the total.
- (d) 20 — Alternate angles between parallel lines are equal, so 3x + 10 = 5x − 30. Rearranging, 10 + 30 = 5x − 3x, so 40 = 2x, and x = 20. −10 comes from a sign error when rearranging, moving a term to the wrong side and getting −20 = 2x instead. 25 comes from wrongly treating the two angles as co-interior and adding them to 180°: (3x + 10) + (5x − 30) = 180 gives 8x − 20 = 180, so x = 25. 47.5 makes the same co-interior mistake but sets the sum equal to 360° instead of 180°, giving 8x − 20 = 360 and x = 47.5.
- (d) 283 cm³ — Volume of a cylinder = πr²h. Using π = 3.14: V = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6, which rounds to 283 cm³. A pupil who uses the diameter, 6 cm, instead of the radius gets 3.14 × 6² × 10 = 1130.4 ≈ 1130 cm³. A pupil who forgets to square the radius gets 3.14 × 3 × 10 = 94.2 ≈ 94 cm³. A pupil who uses the circumference formula 2πr instead of πr² gets 2 × 3.14 × 3 × 10 = 188.4 ≈ 188 cm³. The correct volume is 283 cm³.
- (d) £5.40 — Method: work out the reduced price at each shop separately, then subtract the smaller from the larger. Working: Shop A's reduced price is £45 × 0.8 = £36, and Shop B's reduced price is £34 × 0.9 = £30.60, so the difference is £36 − £30.60 = £5.40. Answer: £5.40. £11.00 comes from comparing the two ORIGINAL prices, £45 − £34, without applying either shop's reduction at all. £1.60 comes from finding Shop A's reduced price correctly, £36, but then subtracting Shop B's original (unreduced) price of £34 instead of its reduced price. £66.60 comes from adding the two reduced prices together, £36 + £30.60, instead of subtracting them.
- (d) 68 cm² — Method: the area of a parallelogram is base × perpendicular height. Working: 4 × 17 = 68. Answer: 68 cm². The distractors: 34 cm² comes from halving the product, which is the rule for a triangle and not for a parallelogram; 42 cm comes from treating the two given lengths as the sides of the shape and working out a perimeter, 2 × (4 + 17), which is a length and not an area; 21 cm comes from adding the base and the height, 4 + 17, instead of multiplying them.
- (a) 3/8, 41%, 0.43 — Method: convert every value to a decimal so they can be compared on the same scale, then order them. Working: 3/8 = 0.375, 41% = 0.41, and 0.43 stays as 0.43, so from smallest to largest the decimals are 0.375, 0.41, 0.43, giving the order 3/8, 41%, 0.43. Answer: 3/8, 41%, 0.43. The order 3/8, 0.43, 41% comes from comparing 0.43 and 41% as raw digits (43 versus 41) without converting 41% into the decimal 0.41 first, wrongly placing 0.43 before 41%. The order 0.43, 41%, 3/8 comes from placing the values in completely reversed order, from largest to smallest instead of smallest to largest. The order 41%, 0.43, 3/8 comes from ordering the values by their TYPE (percentage, then decimal, then fraction) rather than by their actual size.
- (a) 30° — Method: the exterior angles of any convex polygon add up to 360°, and in a regular polygon they are all equal, so divide 360° by the number of sides. Working: 360 ÷ 12 = 30. Answer: 30°. The distractors: 150° is the interior angle, 180 − 30, which answers for the wrong angle at the vertex; 15° comes from dividing 180 by 12, using the angles on a straight line instead of the full turn; 36° comes from dividing 360 by 12 − 2 = 10, carrying the subtraction of 2 out of the interior angle sum formula into a calculation that does not need it.
- (c) 140 — Method: the mean shares the total equally between the items, so add the values and then divide by how many there are. Working: the total is 120 + 200 + 160 + 80 = 560 pages and there are 4 books, so the mean is 560 ÷ 4 = 140 pages. Answer: 140. The distractors: 560 comes from stopping at the total number of pages and never dividing by 4; 280 comes from dividing the total by 2 instead of by the 4 books; 120 comes from working out the range, 200 − 80, which measures spread instead of centre.
What is on this worksheet?
The sheet holds 25 questions drawn from the MathsUK bank — the content areas covered: Ratio, proportion and rates of change, Geometry and measures, Algebra, Statistics, Probability (statements R5, R9, G16, G3, A17, S4, P7). It is pitched at GCSE Foundation and takes about 40 minutes to work through in full. It is built for independent practice, with full answers at the end for self-marking.
How to use the sheet well
- Print it or open it on screen — both work. Printing is A4; the screen view fits phones and tablets.
- Do all 25 questions before checking — about 40 minutes is the guide, but there is no time pressure.
- Check the answers — press “Show answers” or print the answer page separately.
- Redo the questions you got wrong — twice as effective as doing 25 fresh ones.
- “New questions” — builds a fresh sheet on the same statements, so you can practise again without repeats.
Why this sheet helps
MathsUK worksheets use questions graded by difficulty and a fair spread of correct-answer positions (the answer is not always (a)) — so the student really has to think about each question rather than guess a pattern. Every question is tagged to a DfE content statement and checked before it enters the bank. The answers come with a step-by-step explanation, not just a value — so a wrong answer becomes a lesson.
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