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.At a school fête, a stall invites visitors to spin a fair spinner with 8 equal sections numbered 1 to 8, and separately toss a fair coin. A visitor wins a small prize only if the spinner lands on a multiple of 3 and the coin lands on heads. Work out the probability that a visitor wins a prize, giving your answer as a fraction in its simplest form.
- 2.In a test, Amelia answered 18 of the 24 questions correctly. Work out the percentage of the questions she answered correctly.
- 3.Grace asked 12 children how many brothers and sisters they have. Her results, in order, were 0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 4, 6. Work out the median number of brothers and sisters.
- 4.A wooden cube has edges of length 4 cm. Work out the total surface area of the cube.
- 5.Solve 3x − 5 = 4.
- 6.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.
- 7.The sum of the interior angles of a polygon is 1980°. Work out the number of sides of the polygon.
- 8.A parallelogram has a base of 20 cm and a sloping side of 13 cm. The perpendicular drawn from the top of that sloping side down to the base meets the base 5 cm from the foot of the sloping side, so the perpendicular, the 5 cm and the 13 cm sloping side form a right-angled triangle. Work out the area of the parallelogram.
- 9.Solve 2(x + 3) = 10
- 10.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?
- 11.The price of a jacket increases by 50% and then decreases by 50%. Describe the overall change from the original price.
- 12.A raffle has two independent draws. In the first, a winning ticket is picked at random from 5 red tickets and 3 blue tickets. In the second, a winning number is picked using a fair spinner with 4 equal sections, numbered 1 to 4. Work out the probability that the first winning ticket is blue and the second winning number is greater than 2.
- 13.A rectangular tile measures 4.5 cm by 6 cm. Work out the area of the tile.
- 14.Kwame flips a fair coin three times and writes down what it lands on each time. Work out the probability that it lands on heads all three times.
- 15.A pack contains 6 cards, numbered 1 to 6. A card is drawn at random from the pack, and a fair coin is flipped. Work out the probability of drawing an even-numbered card and the coin landing on tails.
- 16.The mean of the four numbers 10, 15, 20 and x is 18. Work out the value of x.
- 17.Spinner E has 4 equal sections, numbered 1 to 4. Spinner F has 6 equal sections, numbered 1 to 6. Both spinners are spun once. Work out the probability of getting at least one 3.
- 18.The numbers 2, 4, 6, 8, 10 and 12 have a mean of 7 and a median of 7. The value 100 is now added to the list. Which measure is changed more by adding 100, and why?
- 19.Three angles lie on a straight line. They measure 38°, 95° and x°. Work out the value of x.
- 20.There are 400 students at a school. 25% of them have a brother, 40% have a sister and 15% have both a brother and a sister. Work out how many of the students have neither a brother nor a sister.
- 21.A trapezium has parallel sides of 5 cm and 11 cm, and a perpendicular height of 4 cm. Work out the area of the trapezium.
- 22.Priya says that x = 6 is the solution of the equation 4x + 8 = 24. Priya is wrong. Work out the correct value of x.
- 23.Spinner C has 3 equal sections, numbered 2, 3 and 5. Spinner D has 6 equal sections, numbered 1 to 6. Both spinners are spun once, and the two numbers are multiplied together. Work out the probability that the product is a multiple of 5.
- 24.The mean mass of four parcels is 17 kg. Three of the parcels have masses 12 kg, 16 kg and 18 kg. Work out the mass of the fourth parcel.
- 25.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.
Answer key
- (d) 1/8 — The multiples of 3 from 1 to 8 are 3 and 6, so the probability of that event is 2/8, which simplifies to 1/4. The probability of the coin landing on heads is 1/2. Since the spin and the toss are independent, multiply the two probabilities: 1/4 × 1/2 = 1/8. A candidate who answers 1/4 has considered only the spinner and forgotten to combine it with the coin toss. A candidate who answers 1/2 has considered only the coin and forgotten the spinner condition entirely. A candidate who answers 1/16 has counted only one number, 6, as a multiple of 3 instead of two, giving 1/8 × 1/2.
- (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) 1.5 — Method: with an even number of values the median is the mean of the two middle values, which for 12 values are the 6th and the 7th once the data are in order. Working: the results are already in order, and 12 ÷ 2 = 6, so the middle pair are the 6th value, 1, and the 7th value, 2; the median is (1 + 2) ÷ 2 = 1.5. Answer: 1.5 brothers and sisters. The distractors: 1 comes from reading the 6th value and stopping there instead of averaging the middle pair; 2 comes from working out the mean, 24 ÷ 12, instead of the median; 6 comes from working out the range, 6 − 0, which measures spread rather than centre.
- (a) 96 cm² — Method: a cube has six identical square faces, so the total surface area is six times the area of one face. Working: one face has area 4 × 4 = 16 cm², and 6 × 16 = 96. Answer: 96 cm². The distractors: 16 cm² is the area of a single face, from stopping before multiplying by the six faces; 64 cm³ comes from working out the volume, 4 × 4 × 4, which is a different measure and carries a different unit; 24 cm² comes from multiplying the six faces by the edge length, 6 × 4, instead of by the area of a face.
- (d) 3 — Method: add the constant term to both sides first, then divide by the coefficient of x. Working: 3x = 4 + 5 = 9; x = 9 ÷ 3 = 3. Answer: x = 3. −1/3 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 4 − 5 = −1, then x = −1/3. 6 comes from subtracting the coefficient 3 instead of dividing by it: 9 − 3 = 6. 9 comes from correctly finding 3x = 9 but forgetting to divide by 3.
- (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.
- (b) 13 — Using the sum of interior angles formula, (n − 2) × 180° = 1980°, so n − 2 = 1980 ÷ 180 = 11, and n = 11 + 2 = 13. 11 stops after the division, forgetting to add 2 back to find n. 15 adds 2 twice by mistake, giving 11 + 2 + 2. 22 divides 1980 by 90 instead of 180.
- (d) 240 cm² — Method: the area of a parallelogram is base × perpendicular height, and the perpendicular height is not the sloping side, so it must be found first from the right-angled triangle. Working: the sloping side is the hypotenuse, so the height squared is 13² − 5² = 169 − 25 = 144, giving a height of √144 = 12 cm; then 20 × 12 = 240. Answer: 240 cm². The distractors: 260 cm² comes from using the 13 cm sloping side as the height, 20 × 13, without going through the right-angled triangle at all; 120 cm² comes from finding the height of 12 cm correctly and then halving the product, (20 × 12) ÷ 2, which is the rule for a triangle and not for a parallelogram; 100 cm² comes from using the 5 cm along the base as the height, 20 × 5.
- (d) x = 2 — Method: expand the bracket by multiplying both terms inside it by 2, then undo the addition and the multiplication in turn. Working: expanding gives 2x + 6 = 10; subtracting 6 from both sides gives 2x = 4; dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 5 comes from dividing both sides by 2 first, reaching x + 3 = 5 and writing 5 without taking the 3 away; x = 8 comes from adding 6 to both sides instead of subtracting it, giving 2x = 16; x = 3.5 comes from expanding 2(x + 3) as 2x + 3, multiplying only the x by the 2, which leads to 2x = 7.
- (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.
- (c) a decrease of 25% — Method: use multipliers. An increase of 50% is × 1.5 and a decrease of 50% is × 0.5. Working: 1.5 × 0.5 = 0.75, so the final price is 75% of the original. Answer: a decrease of 25%. The distractors: no change comes from assuming +50% and −50% cancel; a decrease of 50% comes from applying only the second change; an increase of 25% has the direction wrong.
- (d) 3/16 — Method: work out each draw's own probability first, then multiply them together since the two draws are independent. Working: there are 8 tickets in all, 3 of them blue, so P(blue) = 3/8. P(spinner number greater than 2) = 2/4 = 1/2, since 3 and 4 qualify. Multiplying gives 3/8 × 1/2, which comes to 3/16. Answer: 3/16. Watch out: writing down 3/8 stops after the first draw and never brings in the spinner at all. Writing down 1/2 does the opposite, using only the spinner and ignoring the ticket draw. And writing down 5/16 uses 5/8, the probability of a RED ticket, instead of 3/8 for blue — reading the wrong colour off the raffle.
- (c) 27 cm² — Area of a rectangle = length × width = 4.5 × 6 = 27 cm². A pupil who adds the two sides instead of multiplying gets 4.5 + 6 = 10.5 cm². A pupil who works out the perimeter instead of the area gets 2 × (4.5 + 6) = 21 cm². A pupil who rounds 4.5 up to 5 before multiplying gets 5 × 6 = 30 cm². The correct area is 27 cm².
- (d) 1/8 — Method: a run of flips of a fair coin gives equally likely sequences of heads and tails, so count the sequences that match and divide by how many sequences there are. Working: each flip lands two ways and no flip affects another, so three flips give 2 × 2 × 2 = 8 equally likely sequences: HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. Only HHH has a head at every flip, so 1 sequence of the 8 matches. Answer: the probability is 1/8. The distractors: 1/4 comes from treating 'three heads', 'two heads', 'one head' and 'no heads' as four equally likely results, which they are not, since one sequence gives three heads and three sequences give two; 1/6 comes from taking the number of sequences to be 2 + 2 + 2 = 6, adding the two ways each flip can land instead of multiplying them; 1/2 comes from reading the first flip only and giving the probability of a head on one flip, without combining it with the other two.
- (b) 1/4 — There are 6 × 2 = 12 equally likely outcomes. The even-numbered cards are 2, 4 and 6, so there are 3 × 1 = 3 outcomes with an even card and tails, giving a probability of 3/12 = 1/4. Choosing 1/2 comes from working out only the probability of drawing an even card, 3/6, and forgetting to combine it with the coin landing on tails. Choosing 1/12 comes from treating only one specific outcome, such as card 6 with tails, as the only one that counts, instead of all three even cards paired with tails. Choosing 1/8 comes from doubling the coin stage when counting the total, using 6 × 2 × 2 = 24 outcomes instead of 6 × 2 = 12, and giving 3/24 = 1/8.
- (a) 27 — Method: turn the mean into a total using total = mean × number of values, then subtract the numbers that are already known. Working: four numbers with a mean of 18 have a total of 18 × 4 = 72; the three known numbers give 10 + 15 + 20 = 45; so x = 72 − 45 = 27. Answer: 27, and checking, (10 + 15 + 20 + 27) ÷ 4 = 72 ÷ 4 = 18. The distractors: 72 comes from stopping at the total the four numbers must reach and never subtracting the known three; 18 comes from assuming the missing number must equal the mean; 45 comes from stopping at the total of the three known numbers.
- (a) 3/8 — There are 4 × 6 = 24 equally likely outcomes. The outcomes with no 3 at all have Spinner E showing 1, 2 or 4 and Spinner F showing 1, 2, 4, 5 or 6, giving 3 × 5 = 15 outcomes. So the outcomes with at least one 3 are 24 − 15 = 9, and the probability is 9/24 = 3/8. Choosing 5/12 comes from adding the two individual probabilities of a 3, 1/4 + 1/6, which counts the outcome where both spinners show 3 twice over. Choosing 1/4 comes from only counting the case where Spinner E shows 3, and forgetting the outcomes where Spinner F shows 3 instead. Choosing 5/8 comes from working out the probability of getting no 3 at all, 15/24 = 5/8, and giving that as the final answer instead of subtracting it from 1.
- (c) The mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3. — Method: work each measure out before the extra value is added and again afterwards, then compare the size of the two changes. Working: before, the six values total 42, so the mean is 42 ÷ 6 = 7, and the middle pair 6 and 8 give a median of (6 + 8) ÷ 2 = 7; after, the seven values total 142, so the mean is 142 ÷ 7 = 20.29 to 2 decimal places, while the median is now the 4th of the seven ordered values, which is 8; the mean has moved by about 13.3 and the median by 1. Answer: the mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3 — this is why the median is often preferred when a data set contains an outlier. The distractors: the reply that the mean rises by 100 adds the extra value to the mean instead of adding it to the total; the reply that the median moves to 12 takes the largest of the original values as the new middle instead of counting to the 4th of the seven values; the reply about even and odd counts quotes a rule that does not exist, since the median moved because a very large value was added, not because the count of values changed.
- (b) 47 — Method: angles on a straight line add up to 180°. Working: 180 − 38 − 95 = 47. Answer: x = 47. A candidate who uses 360° instead of 180° gets 227. A candidate who subtracts only one of the two given angles from 180° gets 142. A candidate who adds the two given angles instead of subtracting them from 180° gets 133.
- (c) 200 students — Method: find the percentage who have a brother or a sister, taking care that the students with both are not counted twice, then take that percentage from 100% and apply the result to the 400 students. Working: 25% + 40% = 65%, but the 15% with both has been counted in each of those figures, so 65% − 15% = 50% have a brother or a sister; that leaves 100% − 50% = 50%, and 50% of 400 = 200. Answer: 200 students. The distractors: 260 students is 65% of 400, the number with a brother or a sister when the 15% overlap is counted twice; 140 students comes from taking that same uncorrected 65% away from the 400; 300 students comes from subtracting only the 25% with a brother and ignoring the sisters altogether.
- (a) 32 cm² — Area of a trapezium = 1/2 × (sum of parallel sides) × height = 1/2 × (5 + 11) × 4 = 1/2 × 16 × 4 = 32 cm². (64 cm² comes from forgetting to halve the product; 44 cm² comes from using only the longer parallel side, 11 × 4, instead of both parallel sides; 20 cm² comes from adding all three given lengths together, 5 + 11 + 4, instead of using the area formula.)
- (b) 4 — Method: subtract the constant term from both sides first, then divide by the coefficient of x. Working: 4x = 24 − 8 = 16; x = 16 ÷ 4 = 4. Answer: x = 4. Priya's 6 comes from dividing 24 by 4 without first subtracting the 8, ignoring the constant term; substituting it back gives 4 × 6 + 8 = 32, not 24, which is why she is wrong. 16 comes from correctly finding 4x = 16 but forgetting to divide by 4. 12 comes from subtracting the coefficient 4 instead of dividing by it: 16 − 4 = 12. 8 comes from correctly finding 4x = 16 but then halving instead of dividing by 4.
- (a) 4/9 — There are 3 × 6 = 18 equally likely outcomes. The product is a multiple of 5 whenever Spinner C shows 5, whatever Spinner D shows, which is 6 outcomes, or whenever Spinner D shows 5 and Spinner C shows 2 or 3, which is 2 more outcomes. This gives 6 + 2 = 8 outcomes, so the probability is 8/18 = 4/9. Choosing 1/3 comes from only counting the 6 outcomes where Spinner C shows 5 and forgetting the 2 outcomes where Spinner D shows 5 instead. Choosing 1/9 comes from only counting the 2 outcomes where Spinner D shows 5 and forgetting the 6 outcomes where Spinner C shows 5. Choosing 1/2 comes from counting 9 outcomes instead of 8, by listing the case where Spinner C shows 5 and Spinner D shows 5 twice over.
- (c) 22 kg — Method: multiply the mean by the number of parcels to rebuild the total mass, then subtract the masses that are known. Working: four parcels with a mean mass of 17 kg have a total mass of 17 × 4 = 68 kg; the three known parcels total 12 + 16 + 18 = 46 kg; so the fourth parcel has mass 68 − 46 = 22 kg. Answer: 22 kg. The distractors: 68 kg comes from stopping at the total mass of all four parcels; 17 kg comes from assuming the missing parcel must have the mean mass; 5 kg comes from multiplying the mean by 3, the number of parcels whose mass is given, leaving 51 − 46 = 5.
- (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.
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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