⏱️ Ten-minute starters — Foundation

Answer key

1. (c) (x + 7)/4 — Method: dividing an expression by a number is written as a fraction, with the whole expression on top. Working: (x + 7) ÷ 4 = (x + 7)/4. Answer: (x + 7)/4. 4/(x + 7) comes from writing the numbers the wrong way round, putting 4 on top. 4(x + 7) comes from reading ÷ as ×, multiplying instead of dividing. (x + 7) − 4 comes from reading ÷ as −, subtracting instead of dividing.
2. (a) They cannot both be describing the same path — Jon's measurement means the true length, l, satisfies 11.5 m ≤ l < 12.5 m. Mia's measurement means the true length satisfies 12.55 m ≤ l < 12.65 m. These two ranges do not overlap, so the two measurements cannot both be describing the same path. 'They must both be describing the same path' ignores that the two ranges do not overlap at all. 'Jon's measurement must be wrong' wrongly assumes Jon is the one at fault, when the mismatch does not show which measurement, if either, is wrong. 'Mia's measurement must be wrong' makes the same unjustified assumption in the other direction.
3. (a) Square — Method: two conditions are being asked for at once, so test each shape against both — the two diagonals must always be the same length as each other, and they must always meet at 90°. Working: in a rectangle the diagonals are equal but they meet at 90° only in the special case where the rectangle is also a rhombus; in a rhombus the diagonals do meet at 90° but they are of different lengths unless the rhombus is also a rectangle; the shape that satisfies both conditions for every example of it is the one that is both, and its diagonals are equal and perpendicular. Answer: the square. The distractors: the rectangle is where a candidate stops who tests only the equal-length condition and never checks the angle at the crossing; the rhombus is where a candidate stops who tests only the right-angle condition and never checks the two lengths; the parallelogram is chosen by a candidate who remembers that the diagonals of a parallelogram bisect each other and treats bisecting each other as being equal to each other, which is a different property.
4. (c) Dodecagon — Deca- means 10, so Decagon names a 10-sided polygon, two sides short of this one, so it is wrong. Hendeca- means 11, one side short of 12, so Hendecagon is also wrong. Icosa- means 20, double the number of sides given here, so Icosagon is wrong too. Dodeca- means 12, so Dodecagon correctly names this polygon.
5. (d) The cyclist's speed, in kilometres per hour — Method: on any straight-line graph the gradient is the change in the quantity on the vertical axis for each 1 unit of the quantity on the horizontal axis, so its meaning is read off the two axis labels. Working: the vertical axis is distance in kilometres and the horizontal axis is time in hours, so the gradient counts kilometres for each hour, and distance for each hour is speed. Answer: the cyclist's speed, in kilometres per hour. The distractors: 'the total distance the cyclist travels, in kilometres' reads the gradient as a value taken off the vertical axis, but a gradient is a rate and no end point of the journey has been given; 'the time the cyclist takes, in hours' names the horizontal axis, which is the quantity the gradient divides by rather than the gradient itself; 'the distance the cyclist travels in 18 hours' treats the 18 as a value of t, when 18 is the steepness of the line and not a point on it.
6. (b) x ≥ 5 — Expand the bracket: 2(3x − 1) = 6x − 2, so the inequality is 6x − 2 ≥ 4x + 8. Subtract 4x from both sides and add 2 to both sides: 2x ≥ 10. Divide both sides by 2: x ≥ 5. A candidate who only multiplies the 3x by 2 and forgets to multiply the −1 gets 6x − 1 ≥ 4x + 8, leading to x ≥ 4.5. A candidate who adds 4x instead of subtracting it gets 10x ≥ 10, leading to x ≥ 1. A candidate who multiplies by 2 instead of dividing gets x ≥ 20.
7. (a) 4,000 nanometres — Method: the number of nanometres is the diameter divided by the length of one nanometre, and dividing powers of ten means subtracting the indices. Working: −6 − (−9) = 3, so 10⁻⁶ ÷ 10⁻⁹ = 10³, and the diameter is 4 × 10³ nanometres. Answer: 4,000 nanometres. The distractors: 400 nanometres comes from taking the difference between the indices as 2 instead of 3; 4 nanometres comes from changing the name of the unit without converting, leaving the coefficient untouched; 0.004 nanometres comes from dividing by 10³ instead of multiplying by it, as though a nanometre were the larger of the two units.
8. (a) 3/4 — 1 litre = 1000 ml, so 750 ml is 750/1000 of a litre. Dividing both the numerator and denominator by 250 simplifies this to 3/4. Writing the fraction upside down, as the litre out of the 750 ml, gives 4/3. Finding the fraction of the litre that is NOT filled, 250/1000, gives 1/4. Dividing the numerator by 250 but the denominator by only 100, an inconsistent simplification, gives 3/10.
9. (b) Nonagon — Method: match the number of sides given to the correct polygon name. Working: a polygon with 9 sides is called a nonagon. An octagon has 8 sides, a decagon has 10 sides, and a heptagon has 7 sides, so each names a different number of sides from the 9 given. Answer: nonagon.
10. (a) £4.20 — 350 g is 3.5 lots of 100 g, since 350 ÷ 100 = 3.5, so the cost is 3.5 × £1.20 = £4.20. Multiplying the mass in grams directly by the price, without dividing by 100 first, gives £420.00. Working out 100 ÷ 350 instead of 350 ÷ 100 inverts the ratio and gives about £0.34. Rounding 350 g down to 300 g gives 3 × £1.20 = £3.60.
11. (d) 20 litres — The ratio of concentrate to water is 2 : 5, so water = concentrate × 5 ÷ 2. 8 × 5 ÷ 2 = 20, so Priya needs 20 litres of water. Giving 40 litres multiplies by 5 but forgets to divide by 2 (8 × 5 = 40). Giving 3.2 litres uses the ratio inverted, multiplying by 2 ÷ 5 instead of 5 ÷ 2 (8 × 2 ÷ 5 = 3.2). Giving 11 litres uses additive reasoning instead of multiplicative: it adds the difference between the ratio parts, 5 − 2 = 3, onto the amount of concentrate (8 + 3 = 11), but ratios scale by multiplying, not by adding a fixed amount.
12. (b) No, the true mass could be as high as 852.5 kg — Method: find the upper bound of the true mass and compare it with the weight limit. Working: the display is correct to the nearest 5 kg, so half of 5 kg is 2.5 kg, and the true mass, m kg, satisfies 847.5 ≤ m < 852.5. Part of that interval lies above 850 kg, so the parcels are not definitely within the limit. Answer: the true mass could be as high as 852.5 kg, which is above the limit. ("Yes, the display reads 850 kg, which is not above the limit" compares the limit with the displayed value instead of with the largest value the true mass could take. "Yes, the true mass is at least 847.5 kg and at most 850 kg" uses the correct half unit below but caps the interval at the limit instead of at 852.5 kg. "No, 850 kg on the display rounds up to 855 kg" wrongly treats the displayed value as if it rounds again.)