⏱️ Ten-minute starters — Foundation

Answer key

1. (c) 47 — Method: fill the first pair of branches of the frequency tree, then the failures on each branch, then add only the failing end branches. Working: 70% of 200 is 140, so 140 cars were more than 3 years old and 200 − 140 = 60 cars were 3 years old or less. One quarter of the older cars failed: 140 ÷ 4 = 35. The newer branch gives 12 failures. Adding the two failing branches gives 35 + 12 = 47. Answer: 47 of the cars failed the test. The distractors: 35 is the older branch on its own, with the 12 newer failures never added; 62 comes from taking 1 in 4 of all the cars, 200 ÷ 4 = 50, and then adding the 12; 153 is 200 − 47 and counts the cars that passed.
2. (a) 5 — Method: apply the machine's operation to the input, and treat zero as an input like any other. Working: the machine gives 0 + 5, and counting on 5 from zero leaves the 5 unchanged. Answer: 5. The distractors: 0 comes from assuming that an input of 0 must give an output of 0, which is true for a machine that multiplies but not for one that adds; −5 comes from subtracting 5 instead of adding it; 6 comes from treating the input as 1 rather than 0 and working out 1 + 5.
3. (a) Game B, by £8 — Game A's expected total is 40 × 0.2 × £5 = £40. Game B's expected total is 40 × 0.1 × £12 = £48. Game B is higher, by £48 − £40 = £8. Writing 'Game A, by £8' is wrong because it has the right difference but the wrong game — Game A's total (£40) is actually LOWER than Game B's, not higher. Writing 'Game B, by £48' is wrong because £48 is Game B's whole expected total, not the DIFFERENCE between the two games. Writing 'Game A, by £40' is wrong in the same way, using Game A's whole total as if it were the margin, and naming the wrong game as the winner. Game B gives the higher expected total, by £8.
4. (d) 600 — Method: the number of seats is the number of rows multiplied by the number of seats in each row, so round each number to 1 significant figure and then multiply the rounded values, which is quick because a product of two multiples of ten is found by multiplying the non-zero digits and attaching the zeros. Working: 21 rounds to 20 and 29 rounds to 30; 2 × 3 = 6, and 20 and 30 carry one zero each, so two zeros follow the 6. Answer: about 600 seats. The distractors: 50 comes from adding the two rounded numbers instead of multiplying them, 20 + 30; 60 comes from multiplying 20 by the 3 of 30 and forgetting the zero in 30; 6,000 comes from attaching three zeros to 2 × 3 when 20 and 30 provide only two between them.
5. (b) 5 m — A scale of 1 : 500 means 1 cm on the plan represents 500 cm in real life. Converting to metres: 500 ÷ 100 = 5 m. 500 m comes from forgetting to convert the 500 cm into metres at all. 50 m comes from dividing by 10 instead of 100 when converting centimetres to metres: 500 ÷ 10 = 50. 0.5 m comes from dividing by 1000 instead of 100, as if converting to kilometres instead of metres: 500 ÷ 1000 = 0.5.
6. (a) 2/3 — Total parts = 6 + 4 + 5 = 15. Sopranos and altos together are not tenors: 6 + 4 = 10 parts, so the fraction is 10/15, which simplifies to 2/3. 1/3 comes from finding the fraction of tenors instead of the fraction that is not tenors. 2/5 comes from counting only the sopranos as not tenors and leaving the altos out. 4/9 comes from leaving sopranos out of the total, 4 + 5 = 9, and then using only the altos as the fraction that is not tenors.
7. (d) 1200 cm — Method: multiply the drawing length by the scale factor. Working: 4 × 300 = 1200 cm. Wrong options: 0.013 cm comes from dividing instead of multiplying (4 ÷ 300); 304 cm comes from adding the scale factor to the drawing length instead of multiplying; 600 cm comes from using half the scale factor (150) instead of 300.
8. (b) 17 — Roots and powers are worked out first: √25 = 5 and 4² = 16. Division comes next: 12 ÷ 3 = 4. Then addition and subtraction, left to right: 5 + 16 − 4 = 17. A candidate who treated 4² as 4 × 2 = 8, multiplying the base by the exponent instead of squaring it, worked out 5 + 8 − 4 = 9. A candidate who did not evaluate the root and used 25 itself worked out 25 + 16 − 4 = 37. A candidate who ignored the priority of division and worked through 5 + 16 − 12 ÷ 3 strictly left to right got 5 + 16 = 21, then 21 − 12 = 9, then 9 ÷ 3 = 3.
9. (b) Provider Y — £21.00 against Provider X's £23.00 — Provider X's gradient is (35 − 15) ÷ 100 = 0.2, so cost = 15 + 0.2 × 40 = 15 + 8 = £23.00. Provider Y's gradient is (45 − 5) ÷ 100 = 0.4, so cost = 5 + 0.4 × 40 = 5 + 16 = £21.00. £21.00 is less than £23.00, so Provider Y is cheaper: 'Provider Y — £21.00 against Provider X's £23.00'. Getting both costs right but naming Provider X as cheaper compares the two numbers the wrong way round — £23.00 is more than £21.00, not less. Comparing only the fixed fees, £15.00 and £5.00, ignores the cost of the 40 gigabytes actually used. Reading off the costs at 100 gigabytes, £35.00 and £45.00, directly from the graph answers a different usage from the 40 gigabytes the question asks about.
10. (c) (0, 0) — A curve crosses the y-axis where x = 0. Substituting x = 0 into y = x³ − 4x gives y = 0³ − 4(0) = 0 − 0 = 0, so the curve crosses the y-axis at (0, 0). A candidate who reads off the coefficient of x as the y-intercept, instead of working out the constant term, might write (0, −4). A candidate who swaps the coordinates might write (4, 0). A candidate who takes the coefficient of x but drops its sign might write (0, 4).
11. (c) 13 cm — Method: the two given sides meet at the right angle, so they are the shorter pair and the hypotenuse comes from Pythagoras' theorem, a² + b² = c². Working: c² = 5² + 12² = 25 + 144 = 169, so c = √169 = 13. Answer: 13 cm. The distractors: 17 cm comes from adding the two sides, 5 + 12, rather than adding their squares; 60 cm comes from multiplying them, 5 × 12, which is twice the area of the triangle and not a length; 7 cm comes from subtracting, 12 − 5, as though the hypotenuse were the difference of the two shorter sides.
12. (b) −5 — Method: a point whose x-coordinate is 0 lies on the y-axis, so its y-coordinate is the constant c; once m and c are both known the equation can be written down and x substituted into it. Working: the line passes through (0, 1), so c = 1 and the equation is y = −3x + 1; substituting x = 2 gives y = −3 × 2 + 1 = −6 + 1 = −5. Answer: −5. The distractors: 7 comes from ignoring the minus sign on the gradient and working out 3 × 2 + 1; 5 comes from working out 3 × 2 = 6 and then using the minus sign to take the constant off the product, 6 − 1 = 5; −7 comes from subtracting the constant instead of adding it, −6 − 1 = −7.