⏱️ Ten-minute starters — Foundation

Answer key

1. (d) 1/15 — The probability that the first bead is red is 3/10. Since the first bead is not put back, there are now only 2 red beads left out of 9 beads in total, so the probability that the second bead is also red is 2/9. Multiplying these, 3/10 × 2/9 = 6/90 = 1/15. A candidate who answers 9/100 has treated the beads as replaced, using 3/10 twice. A candidate who answers 3/50 has correctly reduced the red count to 2 for the second pick but forgotten that the total also falls to 9, using 2/10 instead. A candidate who answers 5/19 has added the numerators and added the denominators, (3+2)/(10+9), instead of multiplying.
2. (c) x = 4 — Method: undo the addition of 1 first, then undo the multiplication by 2. Working: subtracting 1 from both sides gives 2x = 8, and dividing both sides by 2 gives x = 4. Answer: x = 4. The distractors: x = 8 comes from stopping at 2x = 8 and writing 8 as the value of x; x = 5 comes from adding 1 to both sides instead of subtracting it, giving 2x = 10; x = 16 comes from multiplying 8 by 2 instead of dividing by 2.
3. (c) 0.0479 — Method: round each option to 2 significant figures and check which one gives 0.048. Working: for 0.0479, the first two significant figures are 4 and 7; the next digit is 9, so 7 rounds up to 8, giving 0.048. For 0.0485, the first two significant figures are 4 and 8; the next digit is 5, so 8 rounds up to 9, giving 0.049, not 0.048. 0.052 already has exactly 2 significant figures, 5 and 2, so it stays as 0.052 and does not round to 0.048 at all. 0.04 has only 1 significant figure, so it is already less precise than the 2 significant figures asked for. Answer: 0.0479.
4. (b) x = 7, y = 3 — Method: one equation contains +y and the other −y, so adding them removes y; the value found is then substituted back to get the other letter. Working: adding x + y = 10 and x − y = 4 gives 2x = 14, so x = 7; substituting into x + y = 10 gives 7 + y = 10, so y = 3. Answer: x = 7, y = 3, and 7 − 3 = 4 as required. The distractors: x = 7, y = 4 comes from finding x correctly and then taking the 4 in x − y = 4 to be the value of y; x = 5, y = 5 comes from splitting the total of 10 equally and never using the difference; x = 14, y = −4 comes from adding the equations to 2x = 14 and forgetting to halve, so that x is taken as 14 and y as 10 − 14.
5. (d) 15/64 — Method: because the first marble is put back, the second draw is made from the same 8 marbles, so the two draws are independent and their probabilities are multiplied. Working: the bag holds 8 marbles both times, so yellow first has probability 5/8 and blue second has probability 3/8. Multiplying fractions gives 5 × 3 = 15 on the top and 8 × 8 = 64 on the bottom. Answer: the probability is 15/64. The distractors: 9/64 comes from working out the probability that BOTH marbles are blue, 3/8 × 3/8, instead of one of each colour in the stated order; 15/16 comes from multiplying the numerators but adding the denominators; 5/8 comes from writing down the first draw only and never combining it with the second.
6. (d) 110 — Method: form the equation 15 + 0.08m = 23.80, where m is the number of minutes, then solve for m. Working: subtract the fixed fee: 0.08m = 23.80 − 15 = 8.80. Divide by the cost per minute: m = 8.80 ÷ 0.08 = 110. Answer: 110 minutes. 1.1 comes from using 8 instead of 0.08 as the cost per minute, forgetting to convert pence to pounds. 297.5 comes from dividing the whole bill by the cost per minute without subtracting the fixed fee first, 23.80 ÷ 0.08. 485 comes from adding the fixed fee to the bill instead of subtracting it, before dividing by the cost per minute, (23.80 + 15) ÷ 0.08.
7. (b) 20 km — The scale 2 cm : 5 km means each 1 cm on the map represents 5 ÷ 2 = 2.5 km in real life. The footpath is 8 cm on the map, so its real length is 8 × 2.5 = 20 km. 40 km comes from multiplying 8 by 5 directly, ignoring that the scale's '2 cm' has to be divided out first: 8 × 5 = 40. 3.2 km comes from dividing 8 by 2.5 instead of multiplying: 8 ÷ 2.5 = 3.2. 5 km comes from multiplying 2.5 by the scale's '2' instead of by the footpath's 8 cm: 2.5 × 2 = 5.
8. (d) 19 — 2² = 4, then 4 × 4 = 16, then 3 + 16 = 19. Adding before multiplying gives 3 + 4 = 7, then 7 × 4 = 28 — multiplication comes before addition. Squaring the product instead of just the 2 gives 4 × 2 = 8, then 8² = 64, then 3 + 64 = 67. Working strictly left to right throughout gives 3 + 4 = 7, then 7 × 2 = 14, then 14² = 196.
9. (d) All angles equal and all sides equal — Congruent triangles are identical in both shape and size, so every corresponding angle is equal and every corresponding side is equal, so 'all angles equal and all sides equal' is correct. 'All angles equal, sides may differ' describes 'similar' triangles, which have equal angles but sides in the same ratio rather than necessarily equal, not congruent ones. 'All sides equal, angles may differ' is not geometrically possible for triangles, since equal sides throughout force the angles to match too. 'Same area, angles and sides may differ' is wrong because equal area alone does not guarantee congruence; two triangles can share an area with completely different shapes.
10. (c) 4n − 2 — Method: find the common difference, then find the constant that fits the first term. Working: 6 − 2 = 4, 10 − 6 = 4, 14 − 10 = 4, so the terms increase by 4 each time and the nth term has the form 4n + c. Substituting n = 1: 4(1) + c = 2, so c = −2. Answer: the correct nth term is 4n − 2. The value 4n is Priya's value, which comes from using only the common difference and leaving out the constant. The value 4n + 2 comes from a sign error when finding the constant. The value 2n + 2 comes from using the first term, 2, as the coefficient of n instead of the common difference, and then attaching +2 rather than working the constant out.
11. (a) 1,400 and 1,240, so combine the samples for one estimate — Method: scale each sample up to the whole stock, then use the fact that a larger sample gives a more reliable estimate than a smaller one. Working: the first sample gives 35 ÷ 50 = 0.7 and 0.7 × 2,000 = 1,400 paperbacks; the second gives 31 ÷ 50 = 0.62 and 0.62 × 2,000 = 1,240 paperbacks. Two random samples of the same size are expected to differ a little, so neither estimate is wrong. Putting the two together gives 35 + 31 = 66 paperbacks in 100 books, and 66 ÷ 100 = 0.66 with 0.66 × 2,000 = 1,320, an estimate resting on twice as many books as either volunteer checked. Answer: 1,400 and 1,240, so combine the samples for one estimate. The distractors: keeping 1,400 because it is larger picks an estimate by its size, when both samples held 50 books and neither has a stronger claim; saying a volunteer must have miscounted assumes two random samples ought to agree exactly, which is precisely what random sampling does not promise; 1,750 and 1,550 come from 35 × 50 = 1,750 and 31 × 50 = 1,550, multiplying each count by the size of the sample instead of scaling by 2,000 ÷ 50.
12. (a) 70 km/h — Method: average speed for a whole journey is the total distance divided by the total time, not the mean of the separate speeds. Working: 3 × 80 = 240 km and 1 × 40 = 40 km, giving 280 km in 3 + 1 = 4 hours, so 280 ÷ 4 = 70. Answer: 70 km/h. The distractors: 60 km/h comes from taking the mean of 80 and 40, which would only be right if equal times were spent at each speed; 280 km/h is the total distance written with a speed unit, from forgetting to divide by the total time; 80 km/h comes from quoting the speed of the longer leg as the average for the whole journey.