🎯 Grade 4 pass booster

Answer key

1. (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.
2. (a) 1/10 — There are 4 × 5 = 20 equally likely outcomes in total. The pairs whose product is 12 are Spinner A showing 3 with Spinner B showing 4, and Spinner A showing 4 with Spinner B showing 3, which is 2 outcomes, giving a probability of 2/20 = 1/10. Choosing 1/20 comes from finding only one of the two pairs, (3, 4), and missing (4, 3) as a separate outcome. Choosing 1/8 comes from using 16 as the total number of outcomes, 4 × 4, forgetting that Spinner B has 5 sections rather than 4. Choosing 1/5 comes from listing the factor pairs of 12 as 2 × 6 and 3 × 4 and counting each one in both orders, (2, 6), (6, 2), (3, 4) and (4, 3), giving 4 outcomes out of 20 without checking that neither spinner has a 6 on it.
3. (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.
4. (a) 1/3 — Method: for two independent spinners, multiply the probability of each separate outcome, but first work out each spinner's own probability correctly, using how many of its equal sections actually carry that result. Working: Spinner A has 2 even numbers, 2 and 4, out of 4 sections, so P(even) = 2/4 = 1/2. Spinner B has 2 red sections out of 3, so P(red) = 2/3. Multiplying gives 1/2 × 2/3, which cancels down to 1/3. Answer: 1/3. Watch out: writing down 1/4 treats Spinner B's two colours as equally likely and uses 1/2 for red, when in fact 2 of its 3 sections are red — the sections are not split evenly between the two colours. Writing down 1/6 undercounts Spinner A's even numbers as just one out of four instead of two. And writing down 5/6 applies the 'at least one' formula, P(A) + P(B) − P(A)×P(B), which answers a different question about EITHER spinner landing the right way, not both together.
5. (a) 180 cm³ — Method: the volume of a right prism is the area of its cross-section multiplied by its length, and the area of a triangle is half the base multiplied by the perpendicular height. Working: the cross-section has area (6 × 5) ÷ 2 = 15 cm², and 15 × 12 = 180. Answer: 180 cm³. The distractors: 360 cm³ comes from taking the cross-section as 6 × 5 = 30 and never halving it, which measures the rectangle around the triangular face rather than the face itself; 66 cm³ comes from adding the base and the perpendicular height and halving, (6 + 5) ÷ 2 = 5.5, which is the trapezium rule used where the triangle rule is needed, and then multiplying by the 12 cm length; 15 cm³ comes from working out the triangular cross-section correctly and stopping there, so the 12 cm length is never used and an area is handed in as a volume.
6. (b) 1/4 — There are 4 × 4 = 16 equally likely ordered outcomes for the two dice. The pairs that total 5 are: first dice 1 with second dice 4; first dice 2 with second dice 3; first dice 3 with second dice 2; and first dice 4 with second dice 1 — which is 4 outcomes, so the probability is 4/16 = 1/4. A candidate who answers 1/8 has listed only 2 of the four pairs, forgetting that first dice 1 with second dice 4 and first dice 4 with second dice 1 are separate outcomes because the dice are different. A candidate who answers 3/16 has found 3 pairs instead of 4, missing one from the list. A candidate who answers 1/16 has counted only a single pair, such as first dice 2 with second dice 3, and treated the order of the dice as not mattering.
7. (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.)
8. (a) 225 — Method: find the interior angle of the regular octagon, then subtract it from 360° to find the reflex angle at the same vertex, since the interior angle and the reflex angle together make a full turn. Working: there are 8 − 2 = 6 triangles' worth of angle in the octagon, so the interior angle = 6 × 180 ÷ 8 = 135; reflex angle = 360 − 135 = 225. Answer: 225°. A candidate who stops after finding the interior angle gives 135. A candidate who works out the exterior angle instead, 360 ÷ 8 = 45, gives 45. A candidate who subtracts the exterior angle from 360° instead of the interior angle, working out 360 − 45, gets 315.
9. (d) 5/18 — Method: list every ordered pair of dice scores whose difference is 1, then divide by the 36 equally likely pairs. Working: the pairs with a difference of 1 are (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6) and (6, 5), which is 10 pairs out of 36, cancelling down to 5/18. Answer: 5/18. Watch out: writing down 5/36 lists only the 5 pairs going up, (1, 2), (2, 3), (3, 4), (4, 5) and (5, 6), and misses that each one has a matching pair the other way round, such as (2, 1) — the two dice are different objects, and order matters, so each of those 5 gaps counts twice. Writing down 1/6 treats the six possible differences, 0 to 5, as equally likely and picks 1 out of 6 of them, but a difference of 1 is reached by far more pairs of scores than a difference of 5 is, so the six differences are not equally likely. And writing down 1/4 comes from adding the two dice's outcome counts instead of multiplying them, 6 + 6 = 12, and grouping the six scores into just three non-overlapping pairs one apart, {1, 2}, {3, 4} and {5, 6}, giving 3 out of that wrong pool of 12.
10. (d) 15 — Method: each of Spinner A's outcomes can be paired with each of Spinner B's outcomes, so the two counts are combined by multiplying, not adding. Working: Spinner A has 3 outcomes and Spinner B has 5, so there are 3 × 5 = 15 equally likely outcomes in total. Answer: 15. Watch out: adding the two counts, 3 + 5 = 8, undercounts drastically — every one of Spinner A's 3 outcomes pairs with all 5 of Spinner B's outcomes, not just one each. Writing down 3 counts Spinner A alone and leaves Spinner B out completely. And writing down 25 comes from squaring Spinner B's own outcome count, 5 × 5, and forgetting Spinner A altogether.
11. (a) 20% — Method: a percentage is a number of parts per hundred, so divide the numerator by the denominator and multiply by 100. Working: 1 ÷ 5 = 0.2, and 0.2 × 100 = 20; the same result comes from scaling 1/5 to 20/100. Answer: 20%. The distractors: 5% comes from reading the denominator 5 straight off as the percentage; 25% comes from confusing 1/5 with 1/4, the fraction met most often in percentage work; 50% comes from a place value slip in the division, writing 1 ÷ 5 as 0.5 instead of 0.2.
12. (c) £90 — Method: split the total amount into the number of parts shown by the ratio, then find Chloe's share. Working: the ratio 2:3 has 2 + 3 = 5 parts, so one part is £150 ÷ 5 = £30, and Chloe's share is 3 × £30 = £90. So Chloe receives £90. Distractor £60 is Ben's share, not Chloe's. Distractor £75 comes from splitting the money into two equal halves, ignoring the ratio. Distractor £30 is the value of one part, found correctly but never multiplied by 3.
13. (a) 5/36 — Method: list every result of the two dice as an ordered pair, first score then second score, count the pairs that give the total asked for and divide by how many pairs the list holds. Working: each dice can show 6 scores, so there are 6 × 6 = 36 equally likely ordered pairs. The pairs whose scores add to 6 are (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1), which is 5 pairs out of the 36. Answer: the probability is 5/36. The distractors: 4/36 comes from listing (1, 5), (5, 1), (2, 4) and (4, 2) and leaving (3, 3) out, because a double does not look like a pair that can be turned round; 5/12 comes from finding the 5 pairs but taking the number of possible results to be 6 + 6 = 12, adding the two dice instead of multiplying them; 1/11 comes from treating the eleven possible totals 2, 3, 4 and so on up to 12 as equally likely, so that a total of 6 is one result out of eleven.
14. (a) 70° — Method: the three angles of a triangle add up to 180°, so take the known corner away from 180° to find what is left for the other two corners, then share that remainder equally because those two corners are equal. Working: 180 − 40 = 140, and the two equal corners share that 140° between them, so 140 ÷ 2 = 70. Answer: 70°. The distractors: 140° comes from stopping after 180 − 40 and offering the combined total of the two equal corners as though it were the size of one of them; 20° comes from halving the 40° corner that was given instead of halving the 140° that the other two corners have to share; 50° comes from 90 − 40, using the two acute angles of a right-angled triangle as the fixed total rather than the 180° angle sum of the whole triangle.
15. (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.
16. (d) 23.57 °C — Method: add all seven temperatures, divide by the number of readings and round only at the end. Working: 22 + 24 + 23 + 25 + 26 + 21 + 24 = 165, and 165 ÷ 7 = 23.5714…, which rounds to 23.57 to 2 decimal places. Answer: 23.57 °C. The distractors: 24 °C comes from writing down the mode, the only temperature recorded twice, instead of the mean; 27.5 °C comes from dividing the total by 6 instead of by the 7 days recorded; 5 °C comes from working out the range, 26 − 21, which is a measure of spread and not an average.
17. (a) 30 — Method: an outlier is a value that lies far away from the pattern set by the rest of the data, so compare each value with the group the others form. Working: five of the counts, 5, 7, 8, 9 and 11, lie within 6 of one another and the steps between them are 2, 1, 1 and 2; the remaining count of 30 is 19 above the nearest of them, so it is the value that does not belong to the pattern. Answer: 30. The distractors: 5 comes from picking the smallest value, on the idea that the odd one out must be at the bottom of the list; 11 comes from ordering the data and stopping one value short, taking the largest of the counts that sit close together; 8.5 comes from working out the median, (8 + 9) ÷ 2, and giving a measure of centre where a value standing apart was asked for.
18. (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).)
19. (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.
20. (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.
21. (a) 20% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is 84 − 70 = 14 marks, and 14 ÷ 70 = 0.2, so 0.2 × 100 = 20. Answer: an increase of 20%. The distractors: 14% comes from quoting the 14 mark increase as though marks and per cent were the same thing; 17% comes from dividing the 14 by the new mean 84 instead of by the original 70, which gives 17% to the nearest per cent; 120% is the new mean written as a percentage of the old one, which is the whole of the new mean rather than the increase.
22. (a) £1,000 — Method: find the increase, then add it to the original price; the multiplier 1.25 does both steps at once. Working: 25% is one quarter, so 25% of £800 = £800 ÷ 4 = £200, and £800 + £200 = £1,000. Answer: £1,000. The distractors: £200 is the increase on its own, not the new price; £825 comes from adding £25 to £800, treating the 25% as £25; £600 comes from taking the 25% off the price instead of adding it on.
23. (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.
24. (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.)
25. (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.