🧮 Paper 1 non-calculator warm-up — Higher

Answer key

1. (b) 9/25 — Method: write the decimal over the matching power of ten, then divide the numerator and the denominator by their highest common factor. Working: 0.36 has two digits after the point, so 0.36 = 36/100; the highest common factor of 36 and 100 is 4, and 36 ÷ 4 = 9 with 100 ÷ 4 = 25; since 9 and 25 share no factor greater than 1, the fraction is fully cancelled. Answer: 9/25. The distractors: 3/10 comes from reading only the first digit after the point and converting 0.3; 9/50 comes from dividing the numerator by 4 but the denominator by only 2; 36/10 comes from counting one decimal place instead of two and writing the digits over 10.
2. (d) 636 g — Let x = 0.636363... . Since two digits repeat, multiply by 100: 100x = 63.636363... . Subtracting removes the recurring part exactly: 100x − x = 63.636363... − 0.636363... = 63, so 99x = 63, giving x = 63/99 = 7/11 kg. Converting to grams: 7/11 × 1000 = 7000/11 = 636.3636... g, which rounds to 636 g. Treating the decimal as if it terminated, writing 0.63 as 63/100 kg, gives 630 g when multiplied by 1000 — this drops the recurring part entirely. Subtracting 10x instead of x, using 100x − 10x = 90x = 63, is the wrong power of ten for a two-digit block, giving x = 63/90 = 7/10 kg, which is 700 g. A numerator slip in the subtraction, 63 − 1 = 62 instead of 63, gives x = 62/99 kg, which is 62000/99 = 626.26... g, rounding to 626 g.
3. (d) 2x/(x − 2) — Factorise x² − 4 as (x − 2)(x + 2) first — it's a difference of two squares. The (x + 2) in the numerator then cancels with the (x + 2) in the factorised denominator, and 6x² ÷ 3x simplifies to 2x, leaving 2x/(x − 2). Writing 2/(x − 2) comes from over-cancelling 6x² ÷ 3x as 2 instead of 2x, dropping the x that should remain. Writing 2x/(x + 2) comes from factorising x² − 4 as (x + 2)² instead of (x − 2)(x + 2), a difference of two squares always has one plus and one minus bracket. Writing 2x²/(x − 2) comes from simplifying 6x² ÷ 3x as 2x² instead of 2x, dividing the coefficients but not reducing the power of x.
4. (d) 8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8.
5. (c) 5/11 — Let x = 0.45 recurring, so x = 0.454545... . Since two digits repeat, multiply by 100: 100x = 45.454545... . Subtracting the original x removes the recurring part exactly, because it lines up digit for digit: 100x − x = 45.454545... − 0.454545... = 45, so 99x = 45, giving x = 45/99 = 5/11. Treating the decimal as if it terminated at two places gives 45/100 = 9/20, which is only 0.45 and drops the repeating part entirely. Subtracting 10x instead of x — using 100x − 10x = 90x = 45 — is the wrong power of ten for a two-digit repeating block, and gives x = 45/90 = 1/2. Making an arithmetic slip in the numerator, 45 − 1 = 44 instead of 45, gives 44/99 = 4/9.
6. (b) 3/50 — Method: write the decimal over the power of ten that matches the number of digits after the point, counting every digit including a zero, then divide the numerator and the denominator by their highest common factor. Working: 0.06 has two digits after the point, so it is 6 hundredths and can be written as 6/100; the highest common factor of 6 and 100 is 2, and 6 ÷ 2 = 3 with 100 ÷ 2 = 50. Answer: 3/50. The distractors: 3/5 comes from ignoring the zero straight after the point and converting 0.6 instead, giving 6/10, which cancels to 3/5; 3/500 comes from counting three decimal places instead of two and writing 6/1000, which cancels to 3/500; 1/6 comes from putting 1 over the digits after the point, as though 0.06 meant one sixth.
7. (d) 1 — sin 30° = 1/2, so (sin 30°)² = 1/4. cos 30° = √3/2, so (cos 30°)² = 3/4. Adding these gives 1/4 + 3/4 = 1. '1/4' only calculates (sin 30°)² and forgets to add the cos 30° term. '3/4' only calculates (cos 30°)² and forgets to add the sin 30° term. '−1/2' comes from subtracting the two squared values instead of adding them: 1/4 − 3/4 = −1/2.
8. (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
9. (a) 4 + √6 — Multiply top and bottom by the conjugate, 4 + √6. The denominator becomes (4 − √6)(4 + √6) = 4² − (√6)² = 16 − 6 = 10. The numerator becomes 10 × (4 + √6) = 40 + 10√6. So the fraction is (40 + 10√6)/10 = 4 + √6, since both terms in the numerator divide by 10. Distributing the conjugate to only the whole-number term of the numerator, and forgetting the surd term entirely, leaves just 4. Rationalising by multiplying the numerator by the conjugate but leaving the ORIGINAL denominator's sign unchanged instead of squaring it lands on 4 − √6, with the surd's sign never actually flipping to positive. Dividing only the whole-number part of the numerator by 10 and forgetting to divide the surd term too leaves 4 + 10√6.
10. (a) 1/2 — cos 0° = 1 and sin 30° = 1/2, so cos 0° − sin 30° = 1 − 1/2 = 1/2. 1 comes from writing down cos 0° alone and forgetting to subtract sin 30°. 3/2 comes from adding the two values instead of subtracting. −1/2 comes from working out sin 30° − cos 0°, the two terms the wrong way round.
11. (a) 3 m — Height = sloping length × sin 45° = 3√2 × √2/2 = (3 × 2)/2 = 3 m, since √2 × √2 = 2. 3√2 m comes from forgetting to multiply by sin 45° at all. 3√2/2 m comes from using sin 30° = 1/2 instead of sin 45° = √2/2. 6 m comes from using √2 instead of √2/2 for sin 45°, dropping the denominator of the exact value: 3√2 × √2 = 6.
12. (d) x³ + 6x² + 11x + 6 — (x + 1)(x + 2) = x² + 3x + 2. Multiplying by (x + 3): (x² + 3x + 2)(x + 3) = x³ + 3x² + 2x + 3x² + 9x + 6, which simplifies to x³ + 6x² + 11x + 6. Choosing x³ + 6x² + 6x + 6 has the right x² and constant terms but adds 1 + 2 + 3 = 6 for the x-coefficient instead of the correct sum of pairwise products 1×2 + 1×3 + 2×3 = 11. Choosing x³ + 5x² + 11x + 6 sums only two of the three constants (2 + 3 = 5) for the x² coefficient, leaving out the 1. Choosing x³ + 6x² + 11x + 5 adds the last two constants (2 + 3 = 5) instead of multiplying all three (1 × 2 × 3 = 6) for the constant term.
13. (c) 0.001 — Method: a negative index means 'one over' the positive power, so $10^{-3}$ means one over $10^{3}$. Working: ten cubed is 1000, and one over 1000 is 0.001. 1000 comes from ignoring the negative sign and working out ten cubed instead of its reciprocal. −1000 comes from ignoring what the negative index does to the size, while still writing a negative sign on the large value. −0.001 comes from correctly finding the size, 0.001, but wrongly keeping a negative sign because the index was negative. Answer: 0.001.
14. (b) 16π — Use the circumference formula C = 2πr to find the radius: 8π = 2πr, so r = 8π ÷ 2π = 4 cm. Then use the area formula A = πr²: A = π × 4² = 16π cm². Using C = πr instead of C = 2πr gives r = 8, and squaring that gives 64π. Finding r = 4 correctly but then substituting it back into the circumference formula instead of the area formula gives 2π × 4 = 8π. Finding r = 4 correctly but forgetting to square it in the area formula, using A = πr instead of A = πr², gives 4π.
15. (a) 2/11 — Let x = 0.18 recurring, so x = 0.181818... . Since two digits repeat, multiply by 100: 100x = 18.181818... . Subtracting the original x removes the recurring part, because the digits line up exactly: 100x − x = 18.181818... − 0.181818... = 18, so 99x = 18, giving x = 18/99 = 2/11. Treating the decimal as if it terminated at two places gives 18/100 = 9/50, which is only 0.18 and drops the repeating part entirely. Subtracting 10x instead of x — using 100x − 10x = 90x = 18 — is the wrong power of ten for a two-digit repeating block, and gives x = 18/90 = 1/5. Making an arithmetic slip in the numerator, 18 − 1 = 17 instead of 18, gives 17/99.
16. (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
17. (c) 3⁻³ — Method: when dividing powers of the same base, subtract the index of the number you are dividing by from the index of the number being divided, keeping them in the order the question writes them. Working: 2 − 5 = −3, so 3² ÷ 3⁵ = 3⁻³. It is worth checking this against the numbers: 3² = 9 and 3⁵ = 243, and 9 ÷ 243 = 1/27, which is 3⁻³. 3³ comes from subtracting the other way round, 5 − 2 = 3, which reverses the sign of the index and gives 27 instead of 1/27. 3⁷ comes from working out 2 + 5 = 7, which is the rule for multiplying powers, not dividing them. 3¹⁰ comes from multiplying the indices, 2 × 5 = 10, instead of subtracting them. Answer: 3⁻³.
18. (d) x = 0 or x = 6 — Method: factorise by taking out the common factor x: x(x − 6) = 0, so x = 0 or x − 6 = 0, giving x = 0 or x = 6. Distractor origins: x = 6 comes from dividing both sides by x, which loses the solution x = 0; x = 0 stops after finding only one factor; x = 3 comes from halving the coefficient of x instead of factorising.
19. (c) 7/30 — Let x = 0.2333... . Because only the 3 recurs, use two multiples of x that line up the recurring part exactly: 10x = 2.333... and 100x = 23.333... . Subtracting removes the recurring tail completely: 100x − 10x = 23.333... − 2.333... = 21, so 90x = 21, giving x = 21/90 = 7/30. Treating the decimal as if it terminated after two places, writing 0.23 as 23/100, ignores that the 3 carries on forever. Misreading which digits recur — treating 0.2333... as if the block '23' repeated, giving 0.232323... — leads to x = 23/99, which is a different, larger recurring decimal from the one given. A numerator slip in the subtraction, computing 22 instead of 21, gives x = 22/90 = 11/45.
20. (b) (x − 5)(x + 2) — We need two numbers that multiply to −10 and add to −3: these are −5 and 2, since −5 × 2 = −10 and −5 + 2 = −3. So x² − 3x − 10 = (x − 5)(x + 2). A candidate who swaps the signs, using +5 and −2, gets (x + 5)(x − 2), which expands to x² + 3x − 10 — the wrong middle term. A candidate who picks the factor pair 1 and 10 instead of 2 and 5 gets (x − 10)(x + 1), which expands to x² − 9x − 10. A candidate who makes both factors negative gets (x − 5)(x − 2), which expands to x² − 7x + 10 — the wrong sign on the constant term.