⏱️ Ten-minute starters — Foundation

Answer key

1. (b) A right-angled triangle — The right angle at Y makes this a right-angled triangle, so that description is correct. Since XY is twice YZ, those two sides cannot be equal. The third side XZ is opposite the right angle, so it is the hypotenuse and is longer than either XY or YZ, so it cannot equal either of them. No two sides are equal, which rules out 'an isosceles triangle' and 'a right-angled isosceles triangle', both of which wrongly assume two equal sides. 'An equilateral triangle' would need all three sides equal, which contradicts XY being twice YZ, so it is wrong too.
2. (a) 5:3 — Divide both prices by their highest common factor, 3: 15 ÷ 3 = 5 and 9 ÷ 3 = 3, giving the ratio 5:3. Choosing 3:5 comes from writing the ratio the wrong way round, as child price to adult price. Choosing 2:3 comes from using the difference between the two prices (15 − 9 = 6) as the first part of the ratio instead of the adult price, then simplifying 6:9 by dividing by 3. Choosing 5:8 comes from comparing the adult price with the total cost of both tickets (£15 out of £24) instead of comparing it with the child price.
3. (a) £14.50 — Method: the first 200 minutes are already covered by the £12, so only 250 − 200 = 50 minutes are charged extra. Extra charge = 50 × £0.05 = £2.50. Total cost = £12 + £2.50 = £14.50. Distractor origins: £24.50 charges 5p for all 250 minutes instead of only the 50 minutes over 200 (250 × £0.05 = £12.50, plus £12 = £24.50); £12.50 makes that same slip of charging all 250 minutes but then forgets to add the £12 monthly fee; £13.50 works out 250 − 200 wrongly as 30 extra minutes instead of 50 (30 × £0.05 = £1.50, plus £12 = £13.50).
4. (d) an equation — Method: recall what each piece of algebraic vocabulary means and match it to the description given. Working: a statement with an equals sign, one unknown letter, and only particular value(s) making it true is, by definition, an equation. Answer: an equation. A formula also has an equals sign, but it links two or more different letters so that one quantity can be calculated from the others — here there is only one letter. An identity is a statement with an equals sign that is true for every value of the letter, not just particular ones. An expression has no equals sign at all, so it cannot be 'true' or 'false'.
5. (d) 12/25 — The group holds 40 of the 250 tickets, so for any one prize the probability the group wins it is 40/250 = 4/25. There are 3 prizes and the group has the same chance at each one, so the expected number won is 3 × 4/25 = 12/25. Writing 4/25 is wrong because it is the chance of winning just ONE prize, without multiplying by the 3 prizes available. Writing 4/75 is wrong because it divides by the 3 prizes instead of multiplying (4/25 ÷ 3 = 4/75), which would mean the group did worse the more prizes were on offer. Writing 64/15625 is wrong because it multiplies the single-prize probability by itself three times, (4/25)³, as though all three prizes had to be won together, instead of adding up the expected number across the three separate prizes. The expected number of prizes won by the group is 12/25.
6. (d) 6 cm — Volume = πr²h, so height = volume ÷ (πr²) = 471 ÷ (3.14 × 25) = 471 ÷ 78.5 = 6 cm. (30 cm comes from dividing by πr instead of πr², missing one factor of the radius; 150 cm comes from dividing by π only, without using r² at all; 24 cm comes from treating the given 5 cm as a diameter and using a radius of 2.5 cm instead.)
7. (b) 15 — Find the multiplier connecting y to x: 10 ÷ 4 = 2.5. Then apply it to the new value of x: 2.5 × 6 = 15. Working out 10 + (6 − 4) = 12 adds the change in x straight onto y instead of scaling proportionally. Working out 10 × 6 = 60 multiplies the given y-value by the new x-value directly, without finding the multiplier first. Writing 10 keeps y the same as before, not realising it must change with x. When x = 6, y = 15.
8. (d) 3/5 — Method: a decimal with one digit after the point is a number of tenths, so it is written over 10 and then cancelled. Working: 0.6 is six tenths, so 0.6 = 6/10; the highest common factor of 6 and 10 is 2, and 6 ÷ 2 = 3 with 10 ÷ 2 = 5. Answer: 3/5. The distractors: 2/3 comes from confusing 0.6 with the recurring decimal 0.666..., which is the one that equals 2/3; 1/6 comes from putting 1 over the single digit after the point; 3/50 comes from using hundredths for a one-place decimal, giving 6/100, which then cancels by 2 to 3/50.
9. (d) 3200 × 0.88³ — A fall of 12% leaves 88% of the value, because 100 − 12 = 88, and 88% written as a decimal multiplier is 0.88. Decay repeats that multiplier once for each year, so over 3 years it is applied three times: 0.88 × 0.88 × 0.88, which is written 0.88³. The calculation is therefore 3200 × 0.88³. Adding the percentages to make a single fall of 36% would be wrong, because each year's fall is taken from a smaller value than the year before.
10. (b) £8 — Rounding to 1 significant figure: £1.85 rounds to £2, and 3.6 kg rounds to 4 kg. The estimate is £2 × 4 = £8. A candidate who used the unrounded values instead of estimating worked out 1.85 × 3.6 = £6.66. A candidate who rounded only the mass and used the exact price worked out 1.85 × 4 = £7.40. A candidate who rounded the price to the nearest 10p instead of 1 significant figure worked out 1.9 × 4 = £7.60.
11. (c) 6 — A cuboid has six flat faces: a top, a bottom and four sides. Count each flat surface once: top, bottom, front, back, left, right — six faces in total, so the answer is 6. Choosing 8 counts the vertices (corners) instead of the faces. Choosing 12 counts the edges instead of the faces. Choosing 4 counts only the four side faces and forgets the top and the bottom.
12. (d) 400 m — The distance travelled equals the area under the speed-time graph. The first 10 seconds form a triangle with base 10 and height 20, giving an area of 0.5 × 10 × 20 = 100 m. The next 15 seconds form a rectangle with base 15 and height 20, giving an area of 15 × 20 = 300 m. The total distance is 100 + 300 = 400 m. The option 500 m treats the whole 25 seconds as travelled at the constant 20 m/s, ignoring that the speed was building up during the first 10 seconds: 25 × 20 = 500. The option 300 m only counts the constant-speed section and forgets the triangle section entirely. The option 200 m comes from working out the triangle's area without halving it (10 × 20 = 200) and forgetting the rectangle altogether.