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 interior angles of a pentagon are 100°, 110°, 120°, x° and x°. Work out the size of each of the two angles marked x°.
- 2.At a point on a straight line, two angles are formed. One of them is 63°. Work out the size of the other angle.
- 3.A straight line crosses two parallel lines. One of the angles formed is (2x + 10)°, and the angle alternate to it is 74°. Work out the value of x.
- 4.Solve 3x − 1 = 8
- 5.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.
- 6.2.4 kg of cheese costs £36. Work out the cost of 1.5 kg of the same cheese.
- 7.A regular nonagon has 9 sides. Work out the sum of its interior angles.
- 8.Amelia's mean mark over four tests is 29. Her first three marks are 31, 26 and 26. Work out her fourth mark.
- 9.A bag contains red counters and blue counters in the ratio 5:3. There are 56 counters in the bag altogether. Work out how many counters are blue.
- 10.A parallelogram has a base of 9 cm and a perpendicular height of 6 cm. Work out the area of the parallelogram.
- 11.To solve 5x − 4 = 2x + 8, Chloe's working is shown. Line 1: 5x − 4 = 2x + 8. Line 2: 3x − 4 = 8. Line 3: 3x = 4. Line 4: x = 4/3. Chloe has made a mistake in her working. Which line contains the mistake?
- 12.A solid cube has a volume of 8 cm³. Work out the length of one edge of the cube.
- 13.Two ordinary fair dice are rolled and the two scores are added together. Work out the probability that the total is a prime number.
- 14.In a recipe the mass of chocolate to the mass of milk is in the ratio 1:4. Amelia uses 200 g of milk. Work out the mass of chocolate she needs.
- 15.The same jumper is sold at two shops. Shop A charges £40 and Shop B charges £50. Write down the price at Shop A as a percentage of the price at Shop B.
- 16.Solve 7x − 5 = 3x + 11
- 17.Three numbers are in the ratio 1:2:3. The three numbers add up to 72. Work out the largest of the three numbers.
- 18.Freya buys three items whose prices are in the ratio 2:3:5. Altogether she pays £400. Work out the price of the most expensive item.
- 19.A café sold 10 sandwiches on each of four days: 10, 10, 10, 10. Work out the range of the numbers sold.
- 20.A red fair dice and a blue fair dice are rolled together. Every outcome is listed as an ordered pair, red score then blue score. Work out the probability that the red score is less than the blue score.
- 21.Noah wrote down the numbers 9, 3, 7, 1, 5 and said, “The median is 7, because 7 is in the middle of my list.” Is Noah right? Give a reason for your answer.
- 22.A rectangular photograph has an area of 96 cm². One of its sides is 8 cm long. Work out the length of the other side.
- 23.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.
- 24.A trapezium has parallel sides of 5.6 cm and 8.4 cm, and a perpendicular height of 4 cm. Work out the area of the trapezium.
- 25.Three business partners share a profit of £48,000 in the ratio 3:5:4. Work out how much the partner with 5 parts receives.
Answer key
- (a) 105° — The interior angles of a pentagon add up to (5 − 2) × 180° = 540°. Subtracting the three known angles, 540 − 100 − 110 − 120 = 210°, and this 210° is shared equally between the two angles marked x°, so each one is 210 ÷ 2 = 105°. 210° stops one step early, giving the total of the two unknown angles instead of one of them. 108° is the interior angle of a regular pentagon, which does not apply here since this pentagon's angles are not all equal. 55° comes from halving one of the given angles, 110°, instead of halving the remaining total.
- (b) 117 — Method: two angles meeting at a point on a straight line add up to 180°. Working: 180 − 63 = 117. Answer: 117°. A candidate who thinks the two angles on a straight line must be equal gives 63. A candidate who uses 90° instead of 180°, working out 90 − 63, gets 27. A candidate who uses 360° instead of 180°, working out 360 − 63, gets 297.
- (c) 32 — Method: alternate angles between parallel lines are equal, so 2x + 10 = 74. Working: subtracting 10 from both sides gives 2x = 64; dividing by 2 gives x = 32. Answer: x = 32. A candidate who forgets to subtract 10 first and divides 74 by 2 directly gets 37. A candidate who treats the angles as co-interior instead of alternate, so that the two expressions add to 180° rather than being equal, gets 48 after solving. A candidate who makes a sign error and treats the equation as 2x equalling 10 minus 74 instead of 74 minus 10 gets −32.
- (d) x = 3 — Method: undo the operations on the left in reverse order — deal with the subtraction of 1 first, then with the multiplication by 3. Working: adding 1 to both sides gives 3x = 9, and dividing both sides by 3 gives x = 3. Answer: x = 3. The distractors: x = 9 comes from stopping at 3x = 9 and writing 9 as the value of x; x = 6 comes from subtracting 3 from 9 instead of dividing by 3; x = 27 comes from multiplying 9 by 3 instead of dividing by 3.
- (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.
- (c) £22.50 — Method: find the cost of 1 kg, then multiply by the mass wanted. Working: £36 ÷ 2.4 = £15 per kilogram, and 1.5 × £15 = £22.50. Answer: £22.50. The distractors: £15 is the price of 1 kg, which is the first step and not what the question asks for; £54 comes from multiplying the £36 by 1.5 without first reducing it to a price per kilogram; £57.60 comes from using the scale factor upside down, multiplying £36 by 2.4 ÷ 1.5 = 1.6 instead of by 1.5 ÷ 2.4.
- (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) 33 — Method: multiply the mean by the number of tests to get the total marks, then subtract the marks that are already known. Working: four tests with a mean of 29 give a total of 29 × 4 = 116 marks; the first three marks total 31 + 26 + 26 = 83; so the fourth mark is 116 − 83 = 33. Answer: 33, and checking, (31 + 26 + 26 + 33) ÷ 4 = 116 ÷ 4 = 29. The distractors: 116 comes from stopping at the total for all four tests; 29 comes from assuming the missing mark must be the mean itself; 4 comes from multiplying the mean by 3, the number of marks given, leaving 87 − 83 = 4.
- (d) 21 — Add the parts: 5 + 3 = 8. Divide the total by the number of parts: 56 ÷ 8 = 7, so one part is worth 7 counters. Blue has 3 parts: 3 × 7 = 21. (35 is the number of red counters, using 5 parts instead of 3. 28 comes from splitting 56 counters in half instead of in the ratio 5:3. 7 is the value of one part — the number of blue counters is 3 lots of this, not just one.)
- (b) 54 cm² — Area of a parallelogram = base × perpendicular height = 9 × 6 = 54 cm². (27 cm² comes from halving the product as if it were a triangle, 1/2 × 9 × 6; 15 cm² comes from adding the base and height instead of multiplying, 9 + 6; 30 cm² comes from doubling the sum of the base and height, as if finding a perimeter, 2 × (9 + 6).)
- (d) Line 3 — Method: check each line of Chloe's working against the correct algebraic step. Working: Line 1 to Line 2 is correct, subtracting 2x from both sides gives 3x − 4 = 8. But Line 2 to Line 3 should add 4 to both sides, giving 3x = 12, not 3x = 4 — the constant −4 has been dropped rather than removed correctly. Line 4 follows correctly from Chloe's own, incorrect, Line 3. Answer: Line 3. Line 1 is simply the original equation, copied out correctly. Line 2 correctly subtracts 2x from both sides of Line 1. Line 4 divides Chloe's own Line 3 by 3 correctly — the arithmetic there is fine, the mistake happened one line earlier.
- (b) 2 cm — Method: the volume of a cube is edge × edge × edge, so finding the edge from the volume means undoing a cube, not a square or a halving. Working: edge × edge × edge = 8; testing whole numbers, 1 × 1 × 1 = 1 is too small and the next whole number gives 2 × 2 × 2 = 8, which matches. Answer: 2 cm. The distractors: 8 cm comes from writing the volume down as the edge length and changing only the unit; 4 cm comes from halving the volume, 8 ÷ 2, as though a cube were undone by dividing by 2; 2.83 cm is the square root of 8, from taking a square root where a cube root is needed.
- (d) 5/12 — The possible totals are 2 to 12. The prime totals are 2, 3, 5, 7 and 11, with 1, 2, 4, 6 and 2 ways to make them, giving 1 + 2 + 4 + 6 + 2 = 15 outcomes out of 36, so the probability is 15/36 = 5/12. Choosing 7/18 comes from forgetting that a total of 2 counts as prime, giving 14/36 = 7/18 instead. Choosing 19/36 comes from wrongly including a total of 9 as a prime number, adding its 4 ways to the correct 15 to get 19/36. Choosing 13/36 comes from forgetting the 2 ways of making a total of 11, giving 13/36 instead of 15/36.
- (d) 50 g — Method: the milk is 4 parts of the ratio, so use the milk to find the value of one part, then read off the chocolate, which is 1 part. Working: one part = 200 ÷ 4 = 50, and the chocolate is one part. Answer: 50 g. The distractors: 40 g comes from treating the 200 g as the total mass of the mixture and splitting it into 1 + 4 = 5 parts; 250 g is the total mass of the finished mixture, the 200 g of milk plus the chocolate, rather than the chocolate on its own; 800 g comes from multiplying 200 by 4 instead of dividing, which scales the milk up rather than down to the chocolate.
- (b) 80% — Percentage = (40 ÷ 50) × 100 = 80%.
- (d) 4 — Method: collect the x terms on one side and the number terms on the other. Working: subtract 3x from both sides: 4x − 5 = 11. Add 5 to both sides: 4x = 16. Divide by 4: x = 4. Answer: 4. 0.6 comes from adding the x terms instead of subtracting when collecting them, 7x + 3x = 10x, and also subtracting the constants the wrong way round, 11 − 5 = 6, giving 10x = 6. 1.5 comes from correctly collecting the x terms as 4x but subtracting the constants the wrong way round, 11 − 5 instead of 11 + 5. −4 comes from moving the x terms to the wrong side, giving 3x − 7x instead of 7x − 3x, along with a matching sign error on the constants.
- (a) 36 — Method: add the parts of the ratio, divide the total by the number of parts to find the value of one part, then multiply by the number of parts in the share asked for. Working: 1 + 2 + 3 = 6 parts, 72 ÷ 6 = 12 for one part, and the largest number is 3 parts, so 3 × 12 = 36. Answer: 36. The distractors: 12 is the value of one part, which is the smallest of the three numbers rather than the largest; 24 is 2 parts, the middle number; 216 comes from multiplying 72 by 3 instead of dividing 72 by the 6 parts first.
- (d) £200 — Method: add the parts of the ratio, divide the total paid by the number of parts to find the value of one part, then multiply by the parts in the most expensive item. Working: 2 + 3 + 5 = 10 parts, £400 ÷ 10 = £40 for one part, and the most expensive item is 5 parts, so 5 × £40 = £200. Answer: £200. The distractors: £40 is the value of one part; £80 is the 2-part item, the cheapest of the three; £120 is the 3-part item.
- (d) 0 — Method: the range is the largest value minus the smallest value, whatever those two values turn out to be. Working: every value is 10, so the largest value is 10 and the smallest value is 10 as well, and the range is 10 − 10 = 0. Answer: 0 — a range of nothing says the data do not vary at all. The distractors: 10 comes from writing down the repeated value itself instead of the difference between the extremes; 20 comes from adding the largest and the smallest, 10 + 10, instead of subtracting; 40 comes from adding all four values, which gives the total sold and not a measure of spread.
- (b) 5/12 — There are 36 equally likely ordered pairs. Listing the pairs where the red score is smaller than the blue score gives 15 outcomes, so the probability is 15/36 = 5/12. Choosing 1/2 comes from assuming the 36 outcomes split evenly between 'red smaller' and 'red bigger', which ignores the 6 outcomes where the two dice show the same score. Choosing 7/12 comes from including the outcomes where the two scores are equal, 15 + 6 = 21, giving 21/36 = 7/12, instead of using 'less than' strictly. Choosing 5/36 comes from counting only the outcomes where the two scores differ by exactly 1, such as (1, 2) and (2, 3), which is 5 outcomes, and missing the pairs that differ by more than 1.
- (b) No — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. — Method: the median is the middle value of the data in order of size, so the data must be sorted before any position is read off. Working: Noah's list 9, 3, 7, 1, 5 is not in order; sorted it becomes 1, 3, 5, 7, 9, and with 5 values the middle position is the third, which now holds 5 rather than 7. Noah has read the third value of the unsorted list. Answer: no — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. The distractors: the reply giving 3 as the median sorts the data correctly but then reads the value in the second place instead of the third; the reply that 7 is the third number he wrote accepts a position in the unsorted list, which is exactly the mistake the question is about; the reply using the mean claims a value of 7 for it, but the mean is 25 ÷ 5 = 5, so that reasoning is false as well.
- (b) 12 cm — Method: the area of a rectangle is one side multiplied by the other, so when the area and one side are known the other side is found by reversing that multiplication — divide the area by the side that is known. Working: 96 ÷ 8 = 12. Answer: 12 cm. The distractors: 88 cm comes from 96 − 8, subtracting the known side as though the area had been made by adding the two sides together; 768 cm comes from 96 × 8, running the area rule forwards on the two numbers given instead of reversing it; 40 cm comes from reading the 96 as a perimeter — halving it to 48 and taking the 8 cm side away — which reverses the perimeter rule rather than the area rule.
- (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) 28 cm² — Area of a trapezium = (sum of parallel sides) ÷ 2 × height. Sum of parallel sides = 5.6 + 8.4 = 14 cm. Half of that is 14 ÷ 2 = 7 cm. Area = 7 × 4 = 28 cm². A pupil who forgets to halve gets 14 × 4 = 56 cm². A pupil who uses only the longer parallel side, as if this were a rectangle, gets 8.4 × 4 = 33.6 cm². A pupil who subtracts the parallel sides instead of adding them gets (8.4 − 5.6) ÷ 2 × 4 = 1.4 × 4 = 5.6 cm². The correct area is 28 cm².
- (a) £20,000 — Method: add the parts of the ratio, divide the profit by the number of parts to find the value of one part, then multiply by the parts belonging to the partner asked about. Working: 3 + 5 + 4 = 12 parts, £48,000 ÷ 12 = £4,000 for one part, and that partner has 5 parts, so 5 × £4,000 = £20,000. Answer: £20,000. The distractors: £4,000 is the value of one part; £12,000 is the 3-part share; £16,000 is the 4-part share, which is what a candidate reaches by assuming the biggest share must go with the number written last in the ratio.
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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