🎯 Grade 4 pass booster

Answer key

1. (c) 40° — Method: the three angles of a triangle always add up to 180°, so split 180° into the number of parts shown by the ratio. Working: the ratio 2:3:4 has 2 + 3 + 4 = 9 parts, so one part is 180 ÷ 9 = 20°, and the smallest angle is 2 × 20 = 40°. So the smallest angle is 40°. Distractor 60° is the middle-sized angle, 3 parts. Distractor 80° is the largest angle, 4 parts. Distractor 20° is the value of one part, found correctly but never multiplied by 2.
2. (c) 1.44 m³ — The cross-section is a triangle, so its area = base × height ÷ 2. Base × height = 1.2 × 0.8 = 0.96 m², and half of that is 0.96 ÷ 2 = 0.48 m². The volume of the prism = cross-sectional area × length = 0.48 × 3 = 1.44 m³. A pupil who forgets to halve when finding the triangle's area gets 1.2 × 0.8 × 3 = 2.88 m³. A pupil who ignores the height altogether, treating the cross-section as base × length, gets 1.2 × 3 = 3.6 m³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length of the bed stops at 0.48 m³. The correct volume of soil is 1.44 m³.
3. (a) The means are equal, and Class B's marks are the more consistent because its range is smaller. — Method: comparing two distributions needs two things — a measure of average and a measure of spread — and each must be put into the context of the question. Working: both classes have a mean mark of 70, so on average the two classes scored the same; the range measures spread, and Class A's range of 30 marks is three times Class B's range of 10 marks, so Class B's marks sit closely around the mean while Class A's are far more spread out. Answer: the means are equal, and Class B's marks are the more consistent because its range is smaller. The distractors: the reply crediting Class A with more consistency reverses the meaning of the range, treating a larger range as tighter data when a larger range means more spread; the reply that Class A's mean mark is higher compares the wrong pair of figures, reading the range of 30 as an average; the reply that Class B's mean mark is higher reads the spread correctly but its claim about the means is false, since both means are 70.
4. (a) x = 4 — Method: undo the addition of 4 first, then undo the multiplication by 6. Working: subtracting 4 from both sides gives 6x = 24, and dividing both sides by 6 gives x = 4. Checking: 6 × 4 + 4 = 28. Answer: x = 4. The distractors: x = 24 comes from stopping at 6x = 24 and writing 24 as the value of x; x = 18 comes from subtracting 6 from 24 instead of dividing by 6; x = 144 comes from multiplying 24 by 6 instead of dividing by 6.
5. (c) 36 — Add the parts of the ratio: 3 + 4 = 7. Divide the total by the number of parts: 84 ÷ 7 = 12, so one part is worth 12. The smaller share has 3 parts: 3 × 12 = 36. (48 is the larger share, using 4 parts instead of 3. 42 comes from splitting 84 in half, treating the ratio as if it were 1:1. 28 comes from dividing 84 by 3 — one of the ratio numbers — instead of dividing by the total number of parts, 7.)
6. (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.
7. (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
8. (c) mean = 8, median = 8, mode = 8 — Method: work out each measure separately — the mean is the total divided by how many values there are, the median is the middle value once the data are in order, and the mode is the value that occurs most often. Working: the total is 8 + 8 + 8 + 8 = 32 and there are 4 marks, so the mean is 32 ÷ 4 = 8; in order the marks read 8, 8, 8, 8, and the mean of the middle pair is (8 + 8) ÷ 2 = 8; the value 8 occurs 4 times and no other value occurs at all, so the mode is 8. Answer: mean = 8, median = 8, mode = 8 — when every value in a data set is the same, all three measures of central tendency take that value. The distractors: a mean of 32 comes from stopping at the total and never dividing by 4; a mode of 4 comes from writing down how many times 8 occurs instead of the value that occurs; a mean of 2 comes from dividing a single value, 8, by the 4 marks instead of dividing the total by 4.
9. (b) An angle between 90° and 180° — Method: angle names are fixed by two markers, a right angle at 90° and a straight line at 180°. An obtuse angle is one that has opened past the right angle but has not reached the straight line. Working: an obtuse angle is greater than 90° and smaller than 180°. Answer: an angle between 90° and 180°. The distractors: an angle between 0° and 90° is the definition of an acute angle, chosen by a candidate who remembers the wrong side of the right-angle marker; an angle between 180° and 360° is the definition of a reflex angle, chosen by a candidate who knows the angle is past 90° and then takes the largest category; an angle of exactly 90° is the right angle itself, which is the marker an obtuse angle has passed rather than the obtuse angle.
10. (d) £50 — Area = 1/2 × (3.5 + 6.5) × 4 = 1/2 × 10 × 4 = 20 m². Cost = 20 × £2.50 = £50. (£100 comes from forgetting to halve the trapezium area, giving 40 m² instead of 20 m²; £65 comes from using only the longer parallel side, 6.5 × 4 = 26 m², instead of the trapezium formula; £35 comes from adding all three given lengths, 3.5 + 6.5 + 4, and treating that total as the area in square metres.)
11. (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.
12. (b) £20 — Method: add the parts of the ratio, divide the amount by the number of parts to find the value of one part, then take the parts belonging to the person asked about. Working: 1 + 2 = 3 parts, £60 ÷ 3 = £20 for one part, and Isla has 1 part. Answer: £20. The distractors: £40 is the 2-part share, which belongs to her brother; £30 comes from reading 1:2 as one part out of two and halving the £60; £120 comes from multiplying the £60 by 2 instead of dividing it into 3 parts.
13. (b) 5 litres — Method: adding water changes the total volume but adds no salt, so work out the volume of salt, then the total volume that makes that salt 20% of the mixture, then the extra water. Working: 30% of 10 litres is 0.3 × 10 = 3 litres of salt. For the same 3 litres to be 20% of the new mixture, the new total volume is 3 ÷ 0.2 = 15 litres. The water added is the extra volume, 15 − 10 = 5 litres. Answer: 5 litres. The distractors: 3 litres is the volume of salt in the solution, which is the first step and not what the question asks for; 15 litres is the total volume of the new mixture, which counts the 10 litres already in the container as water that was poured in; 2 litres comes from taking 20% of the original 10 litres, applying the new percentage to the old volume instead of to the new one.
14. (d) £5.40 — Method: work out the reduced price at each shop separately, then subtract the smaller from the larger. Working: Shop A's reduced price is £45 × 0.8 = £36, and Shop B's reduced price is £34 × 0.9 = £30.60, so the difference is £36 − £30.60 = £5.40. Answer: £5.40. £11.00 comes from comparing the two ORIGINAL prices, £45 − £34, without applying either shop's reduction at all. £1.60 comes from finding Shop A's reduced price correctly, £36, but then subtracting Shop B's original (unreduced) price of £34 instead of its reduced price. £66.60 comes from adding the two reduced prices together, £36 + £30.60, instead of subtracting them.
15. (d) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.
16. (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.
17. (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.
18. (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.
19. (b) 50 cm² — Method: the area of a rectangle is length × width. Working: 2 × 25 = 50. Answer: 50 cm². The distractors: 54 cm comes from working out the perimeter, 2 × (2 + 25), which is a length and not an area; 27 cm comes from adding the two sides, 2 + 25, instead of multiplying them; 25 cm² comes from halving the product, 50 ÷ 2, which is the rule for the area of a triangle rather than of a rectangle.
20. (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.
21. (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.
22. (b) 12 — The difference between the parts of the ratio is 5 − 2 = 3 parts, and this is worth 18. Divide to find one part: 18 ÷ 3 = 6. Cats have 2 parts: 2 × 6 = 12. (30 is the number of dogs, using 5 parts instead of 2. 6 is the value of one part — the number of cats is 2 lots of this, not just one. 9 comes from dividing 18 by 2 and stopping there, instead of dividing by the difference in parts, 3, and then multiplying by 2.)
23. (d) 3/20 — The probability of taking a red ball from Bag A is 3/5, and the probability of taking a red ball from Bag B is 1/4. Since the two picks are independent, the probabilities are multiplied: 3/5 × 1/4 = 3/20. A candidate who answers 4/9 has added the numerators and added the denominators, (3+1)/(5+4), instead of multiplying the two fractions. A candidate who answers 17/20 has added the two fractions, 3/5 + 1/4, instead of multiplying them. A candidate who answers 3/25 has misread Bag B as also containing 5 balls, using 1/5 instead of 1/4.
24. (b) 8 cm — Area = base × height, so height = area ÷ base = 136 ÷ 17 = 8 cm. (16 cm comes from using the triangle formula, thinking area = 1/2 × base × height and so height = 2 × area ÷ base; 2312 cm comes from multiplying the area by the base instead of dividing; 119 cm comes from subtracting the base from the area instead of dividing the area by the base.)
25. (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.