⏱️ Ten-minute starters — Foundation

Answer key

1. (d) 21 − 3n — Method: find the common difference, then find the constant that fits the first term. Working: 15 − 18 = −3, 12 − 15 = −3, 9 − 12 = −3, so the terms decrease by 3 each time and the nth term has the form c − 3n. Substituting n = 1: c − 3(1) = 18, so c = 21. Answer: the nth term is 21 − 3n. The value 18 − 3n comes from using the first term as c directly, without adding back the difference that was removed. The value 3n − 21 comes from a sign error that flips the whole expression. The value −3n comes from using only the common difference and leaving out the constant.
2. (a) 62.5 — Method: a mean cannot be averaged with a new value — rebuild the total, add the new value to it, then divide by the new count. Working: three numbers with a mean of 50 have a total of 50 × 3 = 150; adding 100 makes the total 150 + 100 = 250; there are now 4 numbers, so the new mean is 250 ÷ 4 = 62.5. Answer: 62.5. The distractors: 75 comes from averaging the old mean with the new value, (50 + 100) ÷ 2, which ignores that three numbers pull against one; 50 comes from assuming an extra value leaves the mean unchanged; 37.5 comes from dividing the old total of 150 by the new count of 4, adding the new value to the count but not to the total.
3. (a) Draw equal arcs from X and Y, meeting below line l. — After the first arc marks two points X and Y on line l, compasses are opened to a new radius and arcs of equal radius are drawn centred at X and at Y, so that they meet on the opposite side of l from P; joining P to that meeting point gives the perpendicular. (Joining X and Y with a straight line only retraces part of line l itself, since X and Y both already lie on it; drawing an arc centred at P through only one of X or Y repeats part of the first step instead of moving on; drawing a circle through X, Y and P does not locate the new point needed to complete the perpendicular.)
4. (c) y = 3x + 2 — Method: divide the change in the outputs by the change in the inputs to find the multiplier, then put one pair of values into the rule to find the number added on. Working: the output rises by 11 − 5 = 6 while the input rises by 3 − 1 = 2, so the multiplier is 6 ÷ 2 = 3; with an input of 1, 3 × 1 = 3 and the output is 5, so 2 is added. Answer: y = 3x + 2, checked against the second pair by 3 × 3 + 2 = 11. The distractors: y = 3x − 2 comes from finding the multiplier 3 and then subtracting the 2 instead of adding it; y = 2x + 3 comes from swapping the multiplier and the number added on; y = x + 4 comes from assuming the input is multiplied by 1 and using 5 − 1 = 4 as the number added on.
5. (c) 2.6 × 10⁵ — In standard form, A must satisfy 1 ≤ A < 10. Moving the decimal point in 260,000 to just after the 2 gives A = 2.6, and the decimal point moved 5 places, so 260,000 = 2.6 × 10⁵. A candidate who wrote 26 × 10⁴ used a value of A outside the required range, even though it has the same overall value. A candidate who wrote 2.6 × 10⁶ counted one place too many when moving the decimal point. A candidate who wrote 0.26 × 10⁶ used a value of A below 1, again outside the required range.
6. (d) 7.7 m — The horizontal distance is adjacent to the 50° angle and the zip wire is the hypotenuse, so horizontal distance = 12 × cos 50° = 12 × 0.6428... = 7.71...≈ 7.7 m. "9.2 m" uses the sine ratio instead of cosine, 12 × sin 50° = 9.19...≈ 9.2 m, which actually finds the vertical drop of the zip wire, not the horizontal distance. "15.7 m" comes from dividing by the sine ratio instead of multiplying by the cosine ratio, 12 ÷ sin 50° = 15.66...≈ 15.7 m, both the wrong operation and the wrong ratio. "12.0 m" simply uses the length of the zip wire itself as the horizontal distance, ignoring the angle of 50° altogether.
7. (c) 12 — Method: a square root asks which positive number multiplied by itself gives the number under the root sign. Working: 11 × 11 = 121, which is too small, and 12 × 12 = 144. Answer: 12. The distractors: 14 comes from recalling the wrong square fact, since 14 × 14 = 196; 36 comes from dividing 144 by 4 instead of taking a square root; 72 comes from halving 144, treating a square root as a halving.
8. (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.
9. (c) 2 — First write 1 1/2 as an improper fraction, 3/2. To divide by 3/4, multiply by its reciprocal, 4/3: 3/2 × 4/3 = 12/6 = 2. Dropping the whole number and dividing only the fractional part, 1/2 ÷ 3/4 = 1/2 × 4/3, gives 2/3. Multiplying by 3/4 directly instead of using its reciprocal, 3/2 × 3/4, gives 9/8. Using the reciprocal of the first fraction instead of the second, 2/3 × 3/4, gives 1/2.
10. (d) 10.3 m — By Pythagoras' theorem, the hypotenuse = √(5² + 9²) = √(25 + 81) = √106 = 10.29...≈ 10.3 m. "106 m" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "14 m" comes from adding the two shorter sides, 5 + 9, instead of using Pythagoras' theorem at all. "10.2 m" comes from rounding 10.29...m down to 10.2 instead of correctly rounding it up to 10.3.
11. (a) x < 7 — Subtract 5 from both sides: x < 12 − 5, so x < 7. A candidate who adds 5 instead of subtracting gets x < 17. A candidate who subtracts the wrong way round gets x < −7. A candidate who correctly finds 7 but wrongly flips the inequality (as if dividing by a negative had happened) writes x > 7.
12. (b) 5.5 kg — Method: with an even number of values the median is the mean of the two middle values, taken once the data are in order of size. Working: the eight masses are already in order and 8 ÷ 2 = 4, so the middle pair are the 4th and 5th values, 5 kg and 6 kg; the median is (5 + 6) ÷ 2 = 5.5 kg. Answer: 5.5 kg. The distractors: 5 kg comes from reading the 4th value and stopping there instead of averaging the middle pair; 8 kg comes from working out the range, 11 − 3, which measures spread rather than centre; 4 kg comes from writing down the modal mass, the only value that occurs twice, instead of the median.