⏱️ Ten-minute starters — Foundation

Answer key

1. (d) 8 — A cuboid is a 3D shape with 6 rectangular faces, 12 edges and 8 vertices (corners). Counting the corners of a cuboid gives 8. A candidate who mixes up vertices with faces answers 6. A candidate who mixes up vertices with edges answers 12. A candidate who counts only the 4 corners of the top face, forgetting the 4 corners of the bottom face, answers 4. The correct number of vertices is 8.
2. (d) 180 — Method: bearings are measured clockwise from north and written using three figures. Working: south is a half turn (180°) clockwise from north. A student who answers 090 has confused south with east. A student who answers 270 has confused south with west. A student who answers 018 has written the correct digits in the wrong order. Answer: 180.
3. (b) 24 — Method: let one part of the ratio be worth x, write both groups in terms of x, and use the fact that the two groups end up equal. Working: the boys are 3x and the girls are 4x; after the 6 boys join, 3x + 6 = 4x, so x = 6; the girls are 4 parts, so 4 × 6 = 24. Answer: 24 girls. The distractors: 18 is the number of boys before the 6 join, which is 3 × 6; 30 comes from adding the 6 new members to the girls as well as to the boys; 42 is the total number of members in the choir before the 6 boys join, the 18 boys and the girls together.
4. (c) 38° — The interior angles of a hexagon sum to (6 − 2) × 180° = 720°. 130° + 142° + 125° + 150° + 135° = 682°. x = 720° − 682° = 38°. A student who uses 6 × 180° instead of (6 − 2) × 180°, forgetting to subtract the 2, gets 1080° − 682° = 398°. A student who mis-adds the five given angles as 680° instead of 682° gets x = 720° − 680° = 40°. A student who answers with one of the given angles instead of solving for x might write 142°.
5. (c) 2/3 — Method: fractions with the same denominator are added by adding the numerators and leaving the denominator alone, because the parts are already the same size. Working: 1/3 + 1/3 has numerators 1 + 1 = 2 and the denominator stays as 3, giving 2/3. Answer: 2/3. The distractors: 2/6 comes from adding the denominators as well as the numerators, 1 + 1 over 3 + 3; 2/9 comes from adding the numerators but multiplying the denominators, 1 + 1 over 3 × 3; 1/9 comes from multiplying throughout instead of adding, 1 × 1 over 3 × 3.
6. (c) 43 — Method: work out each power separately before adding. Working: 3³ = 27 and 2⁴ = 16, so 3³ + 2⁴ = 27 + 16 = 43. Answer: 43. (25 comes from using 3² instead of 3³, giving 9 + 16. 35 comes from working out 2⁴ as 2 × 4 = 8 instead of 2 × 2 × 2 × 2, giving 27 + 8. 432 comes from multiplying the two powers together instead of adding them.)
7. (a) 8π cm — Circumference = 2πr. Substitute r = 4: circumference = 2 × π × 4 = 8π cm. Using r in place of 2r (halving the formula) gives 4π cm. Using the area formula πr² in place of the circumference formula gives π × 4² = 16π cm. Multiplying 2 × 4 without including π at all gives 8 cm.
8. (a) 0.0069 — Leading zeros are not significant, so the significant figures in 0.006852 start at 6: 6, 8, 5, 2. Rounding to 2 significant figures means keeping 6 and 8, and looking at the next digit, 5, to decide whether to round up. Since 5 rounds up, the second significant figure increases from 8 to 9: 0.006852 rounds to 0.0069. A candidate who rounded to 1 significant figure instead of 2 wrote 0.007. A candidate who rounded to 3 significant figures instead of 2 wrote 0.00685. A candidate who did not round up despite the next digit being 5 wrote 0.0068.
9. (c) (−2, 10) — To translate M(3, 8) by $\binom{−5}{2}$, add −5 to the x-coordinate and 2 to the y-coordinate: (3 + (−5), 8 + 2) = (−2, 10). (5, 3) comes from swapping the two components of the vector before applying them. (−2, 8) applies only the x-component and forgets to add the y-component. (3, 10) applies only the y-component and forgets to add the x-component.
10. (d) 5 — Method: running a machine backwards means undoing its operation, and the operation that undoes multiplying by 3 is dividing by 3. Working: the input multiplied by 3 gives 15, so the input is 15 ÷ 3. Answer: 5, which checks because 3 × 5 = 15. The distractors: 45 comes from running the machine forwards on the output, 15 × 3; 12 comes from subtracting 3 from 15 instead of dividing; 18 comes from adding 3 to 15, which would undo a machine that subtracts rather than one that multiplies.
11. (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
12. (d) 1.2 m² — Method: an area in square metres needs lengths in metres, so convert first and then multiply. Working: 100 cm = 1 m, so 150 cm = 1.5 m and 80 cm = 0.8 m, and the area = 1.5 × 0.8 = 1.2 m². Answer: 1.2 m². The same result comes from working in centimetres: 150 × 80 = 12 000 cm², and a square metre is a square of side 100 cm, so 100 × 100 = 10 000 cm² make one square metre and 12 000 ÷ 10 000 = 1.2. Dividing the 12 000 cm² by 100 instead, as though a square metre held only 100 square centimetres, gives 120 m²; dividing by 1000 gives 12 m². Working out the perimeter rather than the area gives 1.5 + 0.8 + 1.5 + 0.8 = 4.6, which is a length and not an area.