⏱️ Ten-minute starters — Foundation

Answer key

1. (d) 7/5 — The number of unbroken crayons is 60 − 25 = 35. The comparison is with the broken crayons, so the broken crayons are the denominator: 35/25. Both parts divide by 5: 35 ÷ 5 = 7 and 25 ÷ 5 = 5. The fraction is 7/5, which is greater than 1 because there are more unbroken crayons than broken ones.
2. (b) y = 2x − 1 — Method: in a rule that multiplies and then adds, the multiplier is the step in the outputs for each step of 1 in the input, and the number added on is the output when the input is 0. Working: the inputs 0, 1, 2 rise in ones while the outputs −1, 1, 3 rise by 2 each time, so the input is multiplied by 2; an input of 0 gives 2 × 0 = 0 and the output must be −1, so 1 is subtracted. Answer: y = 2x − 1, checked against the last pair by 2 × 2 − 1 = 3. The distractors: y = 2x + 1 comes from finding the multiplier 2 correctly and then reading the output at an input of 0 as +1 instead of −1; y = x − 1 comes from taking the multiplier as 1 because the inputs go up in ones, instead of using the step in the outputs; y = 3x − 1 comes from reading the largest output, 3, as the multiplier.
3. (b) (60 + 4.5π) cm² — Method: find the area of the rectangle and the area of the semicircle separately, then add them. Working: the rectangle has area 10 × 6 = 60 cm². The semicircle has radius 3 cm, so its area is half of π × 3² = half of 9π = 4.5π cm². Total area = (60 + 4.5π) cm². Answer: (60 + 4.5π) cm². (60 + 18π) cm² comes from using the diameter (6 cm) as the radius in the semicircle area formula: half of π × 6² = 18π. (60 + 9π) cm² comes from forgetting to halve the full circle's area: π × 3² = 9π. (60 + 3π) cm² comes from finding the semicircle's arc length instead of its area: half of 2 × π × 3 = 3π.
4. (a) (40, 0) — Method: a graph crosses the x-axis where y = 0, which in this context is the moment the tank holds no water, so substituting y = 0 and solving gives the time. Working: 0 = −5x + 200 gives 5x = 200, so x = 200 ÷ 5 = 40 and the crossing point is (40, 0); the tank is empty after 40 minutes. Answer: (40, 0). The distractors: (0, 200) is the y-axis crossing, the 200 litres in the tank at the start, found by substituting x = 0 instead of y = 0; (200, 0) comes from reading the constant 200 as the x-coordinate without dividing by the 5 litres lost each minute; (0, 40) has the right number in the wrong place, with the time written as a y-coordinate.
5. (a) 1/6 — Method: write the results of the two dice as ordered pairs, count the pairs whose scores add to the total asked for, divide by the number of ordered pairs there are, then cancel the fraction down. Working: there are 6 × 6 = 36 equally likely ordered pairs. The pairs whose scores add to 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1), which is 6 pairs, so the probability is 6/36. Dividing the top and the bottom by 6 gives 1/6. Answer: the probability is 1/6. The distractors: 7/36 comes from taking the number of favourable pairs to be 7 because 7 is the total asked for, confusing the size of a total with the number of ways of making it; 1/7 comes from using the 21 different combinations of two scores as the equally likely results, finding the 3 combinations 1 and 6, 2 and 5, 3 and 4, and cancelling 3/21; 6/11 comes from counting the 6 favourable pairs correctly but dividing by the 11 possible totals from 2 to 12 rather than by the 36 pairs.
6. (d) 20 — Gradient = change in T ÷ change in t = (140 − 60) ÷ (6 − 2) = 80 ÷ 4 = 20. A student who subtracts in the wrong order gets −20. A student who divides 80 by 2 instead of 4 gets 40. A student who wrongly treats the line as passing through the origin and uses the point (2, 60) on its own gets 60 ÷ 2 = 30.
7. (a) 4 — The length scale factor, cubed, gives the volume scale factor: L³ = 64, so L = the cube root of 64 = 4, since 4 × 4 × 4 = 64. 64 comes from using the volume scale factor itself as the length scale factor, without taking a root at all. 8 comes from taking the square root of 64 instead of the cube root — that would be the correct root for an area scale factor, not a volume one. 32 comes from halving the volume scale factor (64 ÷ 2), instead of cube-rooting it.
8. (c) 67 — Method: the mean is the total of the values divided by how many values there are, so add first and divide second. Working: the total is 50 + 83 + 68 = 201 points and three matches were played, so the mean is 201 ÷ 3 = 67 points. Answer: 67. The distractors: 68 comes from writing down the median, the middle value of 50, 68, 83, instead of the mean; 33 comes from working out the range, 83 − 50, which measures spread and not centre; 100.5 comes from dividing the total by 2 instead of by the 3 matches played.
9. (d) 5/4 — Work out the number of free seats: 54 − 24 = 30. Form the fraction 30/24; both numbers share a factor of 6, so 30 ÷ 6 = 5 and 24 ÷ 6 = 4, giving 5/4. 4/5 comes from writing the fraction the wrong way round, as reserved over free (24/30). 5/9 comes from comparing the free seats with the total number of seats (30/54), instead of with the reserved seats. 5/6 comes from miscalculating 54 − 24 as 20 instead of 30, then forming 20/24.
10. (c) Swapped the x and y components — The student's vector has the same two numbers, 5 and −2, but in swapped positions, so the error is swapping the x and y components rather than an error with signs or size. 'Reversed both signs' is wrong because the numbers 5 and −2 have not changed sign, only position. 'Reversed only the y sign' is wrong for the same reason — no sign has actually changed. 'Doubled the x component' is wrong because neither number has changed in size.
11. (a) y = 0 — Method: in the pair (x, y) the first coordinate measures how far left or right of the origin a point is and the second how far above or below the x-axis it is, so a point on an axis has one of those measurements equal to zero. Working: the x-axis is the horizontal line through the origin, so a point sitting on it is neither above nor below that line and its second coordinate is zero, while its first coordinate may be positive, negative or zero. Answer: y = 0. The distractors: x = 0 is the condition for lying on the y-axis, the other axis; x > 0 comes from assuming a point on the x-axis must be to the right of the origin, which is true only of part of that axis; x = y holds only at the origin, which is one point of the x-axis rather than a property shared by all of them.
12. (b) $\binom{8}{3}$ — The column vector is (end − start) in each coordinate: (9 − 1, 4 − 1) = (8, 3), written as $\binom{8}{3}$. $\binom{3}{8}$ swaps the horizontal and vertical components. $\binom{10}{5}$ comes from adding the coordinates instead of subtracting them. $\binom{−8}{−3}$ comes from working out start minus end instead of end minus start.