🎯 Grade 4 pass booster

Answer key

1. (d) An obtuse angle — Method: compare the angle with the two markers that separate the angle names, a right angle at 90° and a straight line at 180°. Working: 108° is greater than 90° and smaller than 180°, so it lies between the right angle and the straight line. Answer: an obtuse angle. The distractors: an acute angle is one below 90°, and is chosen by a candidate who checks only that 108° is less than 180°; a reflex angle is one above 180°, and is chosen by a candidate who checks only that 108° is more than 90° and then takes the largest category; a right angle is exactly 90°, and is chosen by reading 108° as near enough to 90° instead of comparing it properly.
2. (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.
3. (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.
4. (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.
5. (c) 3/8 — Method: write out every result of the three coins as a string of three letters, H for heads and T for tails, count the results that match the description and divide by how many results the list holds. Working: each coin lands two ways and no coin affects another, so the list holds 2 × 2 × 2 = 8 equally likely results. Exactly two heads means one coin lands on tails and the other two on heads, so the results are HHT, HTH and THH — 3 of the 8. Answer: the probability is 3/8. The distractors: 4/8 comes from reading 'exactly two heads' as 'at least two heads' and counting HHH as well; 2/8 comes from a list made without a system, in which HHT and THH are written down and HTH, the result with the tail between the two heads, is missed; 6/8 comes from counting 3 × 2 = 6 ways of picking which two of the three coins show heads, which counts every pair of coins twice, once in each order.
6. (b) 60 cm² — Method: the area of a triangle is half the base multiplied by the perpendicular height, and here the perpendicular height is the 12 cm, not the sloping side. Working: 10 × 12 = 120, then 120 ÷ 2 = 60. Answer: 60 cm². The distractors: 65 cm² comes from using the sloping side of 13 cm as the height, (10 × 13) ÷ 2; 120 cm² comes from using the right two lengths but forgetting to halve, 10 × 12; 78 cm² comes from taking 13 cm and 12 cm as the base and the height and ignoring BC altogether, (13 × 12) ÷ 2.
7. (c) £478.40 — Method: apply the percentage increase, then apply the percentage decrease to the new price. Working: after the increase, the laptop costs £520 × 1.15. Multiplying this result by 0.80 gives the final price, £478.40. Answer: £478.40. £494 comes from combining the two percentages into a single net change (15% − 20% = −5%) and applying it directly, £520 × 0.95 = £494, instead of applying the two changes one after the other. £416 comes from applying only the 20% decrease to the original price, £520 × 0.80 = £416, forgetting the increase entirely. £598 comes from applying only the 15% increase and stopping there, forgetting to apply the decrease at all.
8. (b) 6 m — Volume of a cuboid = length × width × height, so height = volume ÷ (length × width) = 360 ÷ (12 × 5) = 360 ÷ 60 = 6 m. A pupil who divides the volume by the length only gets 360 ÷ 12 = 30 m. A pupil who divides the volume by the width only gets 360 ÷ 5 = 72 m. A pupil who subtracts length × width from the volume instead of dividing gets 360 − 60 = 300 m. The correct height is 6 m.
9. (a) 20 cm — Method: match the measurement you are given to its own part of the ratio, use it to find the value of one part, then multiply by the parts belonging to the measurement asked for. Working: the length is the second measurement listed, so it matches 4 parts and one part = 16 ÷ 4 = 4 cm; the height is 5 parts, so 5 × 4 = 20. Answer: 20 cm. The distractors: 12 cm is the width, which is the 3-part measurement; 4 cm is the value of one part only; 80 cm comes from multiplying the 16 cm by 5 without first dividing by the 4 parts the length is worth.
10. (c) 7 — Method: the median is the middle value when the data are written in order of size, and with an odd number of values there is exactly one middle value. Working: the numbers are already in order, 2, 4, 7, 12, 26, and there are 5 of them, so the middle position is the third and the value sitting there is 7. Answer: 7, with two values below it and two above it. The distractors: 10.2 comes from working out the mean, 51 ÷ 5, instead of the median; 14 comes from taking the value halfway between the smallest and the largest, (2 + 26) ÷ 2; 24 comes from working out the range, 26 − 2, which measures spread rather than centre.
11. (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.
12. (c) 80 — Method: the range is a measure of spread and is found by subtracting the smallest value from the largest. Working: the largest number sold is 100 and the smallest is 20, so the range is 100 − 20 = 80. Answer: 80. The distractors: 100 comes from writing down the largest value and never subtracting the smallest; 60 comes from working out the median, the middle value of 20, 40, 60, 70, 100, instead of the range; 58 comes from working out the mean, 290 ÷ 5, which measures centre rather than spread.
13. (b) 74 marks — Method: the two groups are different sizes, so their means cannot simply be averaged — rebuild each group's total mark, add the totals and divide by all 50 pupils. Working: Group A scored 20 × 80 = 1600 marks and Group B scored 30 × 70 = 2100 marks, giving 1600 + 2100 = 3700 marks altogether, so the overall mean is 3700 ÷ 50 = 74 marks. Answer: 74 marks, which sits nearer to 70 than to 80 because the larger group scored 70. The distractors: 75 marks comes from averaging the two group means, (80 + 70) ÷ 2, as though the groups were the same size; 76 marks comes from attaching each mean to the other group's size, (20 × 70 + 30 × 80) ÷ 50; 150 marks comes from adding the two means together and never dividing at all.
14. (d) 25 — Method: add the parts of the ratio, divide the total membership by the number of parts to find the value of one part, then multiply by the parts belonging to the group asked for. Working: 3 + 5 = 8 parts, 40 ÷ 8 = 5 members in one part, and the seniors are 5 parts, so 5 × 5 = 25. Answer: 25 seniors. The distractors: 15 is the number of juniors, which is the 3-part group; 5 is the size of one part only; 24 comes from dividing the 40 by 5, the seniors' number in the ratio, to get 8 and then multiplying that by 3.
15. (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.
16. (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
17. (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.
18. (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.
19. (d) 60 — Method: 75% is three quarters, so divide by 4 to find one quarter and then multiply by 3. Working: 80 ÷ 4 = 20 for one quarter, and 3 × 20 = 60. Answer: 60 milk chocolates. The distractors: 20 is one quarter of 80, which is 25% and not 75%; 55 comes from subtracting 25 from 80 rather than taking 25% of 80 away from it; 75 comes from quoting the percentage itself as a number of chocolates.
20. (d) 1/4 — Method: a product is odd only when BOTH factors are odd, so list the ordered pairs where both scores are odd and divide by 36. Working: the odd scores on a dice are 1, 3 and 5, so there are 3 × 3 = 9 ordered pairs where both scores are odd, out of the 36 equally likely pairs, cancelling down to 1/4. Answer: 1/4. Watch out: writing down 3/4 finds the probability that AT LEAST ONE score is odd, 1 minus the probability both are even, which is a different, easier condition to meet than both being odd. Considering only the first dice's score and ignoring the second gives 1/2, since 3 of the first dice's 6 scores are odd — but the product also depends on what the second dice shows. And writing down 1/12 comes from counting only the pairs where the SAME odd number appears twice, (1, 1), (3, 3) and (5, 5), missing pairs like (1, 3) and (5, 1) where the two odd scores differ.
21. (c) 12 m — Method: the perimeter of a rectangle is twice the sum of the length and the width, so half the perimeter is one length plus one width; split that half using the ratio. Working: half of 64 is 64 ÷ 2 = 32 m, the ratio 5:3 has 5 + 3 = 8 parts, so one part is 32 ÷ 8 = 4 m, and the width is 3 × 4 = 12 m. So the width is 12 m. Distractor 20 m is the length, 5 parts, not the width. Distractor 24 m comes from splitting the whole perimeter, 64 m, into 8 parts and multiplying by 3, forgetting to halve the perimeter first. Distractor 8 m comes from the same slip stopped one step earlier: splitting the whole perimeter into 8 parts, 64 ÷ 8 = 8, and giving that instead of the width.
22. (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.
23. (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.
24. (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.
25. (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.