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.A small pack contains 5 cards, numbered 1 to 5. One card is drawn at random from the pack, and a fair coin is flipped. Every outcome is listed as a pair, such as (3, H). Work out how many equally likely outcomes there are in total.
- 2.Solve 2(x + 3) = 10
- 3.A café offers 3 types of sandwich, cheese, ham and egg, and 4 types of drink, tea, coffee, juice and water. A customer chooses one sandwich and one drink at random. Work out the probability that the customer chooses egg and water.
- 4.A company's turnover this year is £180,000. Last year's turnover was £120,000. Write down this year's turnover as a percentage of last year's turnover.
- 5.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.
- 6.2.4 kg of cheese costs £36. Work out the cost of 1.5 kg of the same cheese.
- 7.In triangle ABC, angle A is 2x°, angle B is 3x° and angle C is 4x°. Work out the size of angle B.
- 8.Write 3/4 as a percentage.
- 9.A parallelogram has an area of 84 cm² and a base of 12 cm. Work out the perpendicular height of the parallelogram.
- 10.The price of a jacket increases by 50% and then decreases by 50%. Describe the overall change from the original price.
- 11.Seven pupils were asked how many books they had read last month. Their answers were 3, 5, 5, 7, 8, 5, 3. Write down the mode.
- 12.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.
- 13.The number of members of a running club increases from 45 to 54. Work out the percentage increase.
- 14.Two ordinary fair dice are rolled and the two scores are added together. Work out the probability that the total is a prime number.
- 15.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.
- 16.Angle ABC is 130°. The line BF divides angle ABC into two equal parts. Work out the size of angle FBC.
- 17.Write 0.25 as a percentage.
- 18.A solid cube has a volume of 8 cm³. Work out the length of one edge of the cube.
- 19.A regular hexagon is divided into six identical triangles by joining its centre to each of the six vertices. Work out the size of the angle of one of these triangles at the centre of the hexagon.
- 20.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.
- 21.A spinner has three equal sections, coloured red, blue and green. It is spun twice, and each outcome is listed as an ordered pair of colours, such as (red, blue). Work out how many outcomes are in the full list.
- 22.Three numbers are in the ratio 1:2:3. The three numbers add up to 72. Work out the largest of the three numbers.
- 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.A regular polygon has an interior angle of 156°. Work out the number of sides of the polygon.
- 25.Solve 6x − 5 = 3x + 13.
Answer key
- (c) 10 — There are 5 possible cards and 2 possible coin results, so listing every pair gives 5 × 2 = 10 equally likely outcomes. Choosing 5 comes from listing only the card outcomes and forgetting the coin flip adds a second stage to each one. Choosing 7 comes from adding the two stages instead of combining them, 5 + 2 = 7, rather than pairing every card with every coin result. Choosing 20 comes from counting each coin result twice for every card, 5 × 2 × 2 = 20, effectively pairing every card with the coin twice over.
- (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) 1/12 — Method: list the full possibility space of sandwich-and-drink pairs, then divide the one matching pair by the size of the whole space. Working: there are 3 × 4 = 12 equally likely sandwich-and-drink pairs, and exactly one of them is egg and water. Answer: 1/12. Watch out: writing down 1/7 comes from adding the two counts, 3 + 4 = 7, instead of multiplying them to build the possibility space. Writing down 1/3 uses only the chance of choosing egg out of 3 sandwiches and ignores the drink altogether. And writing down 1/4 uses only the chance of choosing water out of 4 drinks and ignores the sandwich altogether.
- (c) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (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.
- (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) 60° — Method: the angles of a triangle add up to 180°, so add the three expressions, solve for x and then substitute back into the expression for angle B. Working: 2x + 3x + 4x = 9x, so 9x = 180 and x = 20. Angle B is 3x, so angle B = 3 × 20 = 60. Answer: 60°. The distractors: 20° is the value of x, from stopping as soon as the equation is solved instead of substituting back; 120° comes from using 360° as the angle sum, which gives x = 40 and 3x = 120; 80° is 4x, the angle at C, from substituting into the wrong expression.
- (d) 75% — Method: change a fraction to a percentage by dividing the numerator by the denominator and multiplying by 100. Working: 3 ÷ 4 = 0.75, and 0.75 × 100 = 75. Answer: 75%. The distractors: 34% comes from reading the digits 3 and 4 straight off as a percentage; 0.75% comes from dividing but forgetting to multiply by 100; 133% comes from inverting the fraction and working out 4 ÷ 3 instead.
- (d) 7 cm — Area of a parallelogram = base × height, so height = area ÷ base = 84 ÷ 12 = 7 cm. A pupil who multiplies instead of dividing gets 84 × 12 = 1008 cm. A pupil who divides the base by the area instead of the area by the base gets 12 ÷ 84 ≈ 0.14 cm. A pupil who mistakenly halves the area first, as if this were a triangle, gets (84 ÷ 2) ÷ 12 = 3.5 cm. The correct height is 7 cm.
- (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.
- (c) 5 — Method: the mode is the value that occurs most often, so count how many times each different value appears and compare the counts. Working: 3 appears twice, 5 appears three times, 7 appears once and 8 appears once, so the highest frequency is three and the value carrying it is 5. Answer: 5. The distractors: 3 comes from writing down the frequency of the most common answer instead of the answer itself; 7 comes from taking the middle number of the list as it was written, which applies the median without ordering the data and without answering the question asked; 8 comes from picking the largest value, which confuses the mode with the maximum.
- (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.
- (c) 20% — Method: percentage increase = (increase ÷ original) × 100. Working: the increase is 54 − 45 = 9, and 9 ÷ 45 = 0.2, so the percentage increase is 0.2 × 100 = 20. Answer: 20%. The distractors: 9% comes from writing the actual increase as a percentage; 16.7% comes from dividing by the new value 54 instead of the original 45; 120% is the multiplier 1.2 written as a change rather than the change itself.
- (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) 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) 65° — Method: a line that divides an angle into two equal parts gives each part half of the original angle, so halve 130°. Working: 130 ÷ 2 = 65. Answer: 65°. The distractors: 130° is the whole of angle ABC, written down without halving it; 50° comes from working out 180 − 130, using the angles on a straight line instead of dividing the angle in two; 32.5° comes from dividing by 4 instead of by 2, as though the line split the angle into four equal parts.
- (d) 25% — Method: to convert a decimal to a percentage, multiply by 100. Working: 0.25 × 100 = 25. Answer: 25%. The distractors: 0.25% comes from writing a percent sign after the decimal without multiplying; 2.5% comes from multiplying by 10 instead of 100; 250% comes from multiplying by 1000.
- (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) 60° — Method: the six angles at the centre together make one complete turn of 360°, and because the hexagon is regular they are all equal, so divide 360° by 6. Working: 360 ÷ 6 = 60. Answer: 60°. The distractors: 120° is the interior angle of a regular hexagon, 720 ÷ 6, which is the angle at a vertex and not the angle at the centre; 45° comes from dividing 360 by 8, treating the hexagon as though it had eight sides; 30° comes from halving the angle at the centre, as though each of the six triangles were split again by a line of symmetry.
- (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.
- (d) 9 — Method: each of the first spin's 3 outcomes can be paired with each of the second spin's 3 outcomes, so the two counts are multiplied together. Working: 3 × 3 = 9 ordered pairs. Answer: 9. Watch out: writing down 6 comes either from adding the two spins' counts, 3 + 3, instead of multiplying them, or from listing colour pairs without order, treating (red, blue) as the same entry as (blue, red) — both shortcuts miss outcomes that the ordered list contains. Writing down 3 lists only a single spin's outcomes and never pairs the two spins up at all. And writing down 27 treats the spinner as though it had been spun three times, 3 × 3 × 3, one spin too many.
- (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.
- (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) 15 — The exterior angle is 180° − 156° = 24°, and the number of sides of a regular polygon is 360° divided by the exterior angle, so 360 ÷ 24 = 15. 24° is the exterior angle itself, stopping one step before the final division. 17 comes from finding 15 correctly and then adding 2, muddling the exterior angle rule with the (n − 2) that appears in the interior angle sum formula. 7.5 comes from dividing 180 by the exterior angle instead of 360, using the angles on a straight line rather than the total of the exterior angles of a polygon.
- (d) 6 — Method: collect the x-terms on one side and the constants on the other, then divide by the remaining coefficient of x. Working: 6x − 3x = 13 + 5, so 3x = 18, x = 18 ÷ 3 = 6. Answer: x = 6. 2.67 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 13 − 5 = 8, x = 8 ÷ 3 ≈ 2.67. 2 comes from a sign error when moving the x-term, adding instead of subtracting: 9x = 18, x = 2. 18 comes from correctly finding 3x = 18 but forgetting to divide by 3.
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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