🧮 Paper 1 non-calculator warm-up — Higher

Answer key

1. (a) x = 4 or x = −3 — Method: find two numbers that multiply to give −12 and add to give −1 — these are −4 and 3. So x² − x − 12 = (x − 4)(x + 3) = 0, giving x = 4 or x = −3. Distractor origins: x = −4 or x = 3 has the signs the wrong way round; x = 4 or x = 3 makes both roots positive, ignoring the sign of −12; x = 12 or x = −1 comes from reading off the coefficient and the constant directly instead of factorising.
2. (d) 1/16 — Method: a negative index means the reciprocal of the power, so 2⁻⁴ is 1 divided by 2⁴. Working: 2⁴ = 2 × 2 × 2 × 2 = 16, so the value is 1/16. Answer: 1/16. The distractors: −16 comes from reading the negative index as a minus sign on the result; 16 comes from ignoring the minus sign and working out 2⁴; 1/8 comes from multiplying the base by the index, 2 × 4, and writing 1 over that product.
3. (a) 3x/2 — Factorise top and bottom first: 6x³ − 15x² = 3x²(2x − 5), and 4x² − 10x = 2x(2x − 5). The bracket (2x − 5) is common to both, so it cancels, leaving 3x²/(2x); dividing the power of x, 3x² ÷ x = 3x, gives 3x/2. Writing 3x²/2 cancels the (2x − 5) correctly and removes the denominator's x, but never reduces the power of x left in the numerator — 3x² ÷ x should give 3x, not stay as 3x². Writing 3/2 cancels the (2x − 5) correctly but then drops the x from the numerator altogether, treating 3x² over x as if it cancelled completely to 3 instead of reducing to 3x. Writing −3x/2 comes from factorising the denominator with the wrong sign, as 2x(5 − 2x) instead of 2x(2x − 5); cancelling (5 − 2x) against the numerator's (2x − 5) then needs an extra minus sign, which flips the answer to −3x/2.
4. (a) 0.1666... — Method: a fraction bar means divide, so the decimal is found by dividing the numerator by the denominator; when a remainder comes back unchanged the division never ends, the digit it produces repeats for ever, and the exact value has to be written with that recurring digit rather than a rounded one. Working: 1 ÷ 6 is set out as 1.000 ÷ 6; six does not go into 1, and six goes into 10 tenths once with 4 left over, so the first decimal digit is 1; the 4 left over makes 40 hundredths, and six goes into 40 six times with 4 left over again; that same remainder of 4 returns at every step, so the digit 6 repeats without end. Answer: 0.1666... The distractors: 0.16 comes from carrying the division out to two decimal places and stopping there, as though the decimal terminated; 0.17 comes from rounding the division to two decimal places, which gives a value close to one sixth but not equal to it; 0.6 comes from writing the digit of the denominator straight after the decimal point, as though 1/6 meant six tenths.
5. (d) 6√3 cm — Method: AB lies alongside the 30° angle at A and AC is the hypotenuse, so the ratio needed is cosine: cos 30° = AB ÷ AC. Working: the exact value of cos 30° is √3/2, so AB = 12 × √3 ÷ 2, and half of 12 is 6. Answer: AB = 6√3 cm, which is about 10.4 cm. Using sine by mistake gives 12 × 1/2 = 6 cm, which is the length of BC rather than AB. Using tan 30° = 1/√3 gives 12 ÷ √3, which is 4√3 cm. Remembering cos 30° as √3 rather than as √3 halved gives 12√3 cm, longer than the hypotenuse and so impossible.
6. (a) 13/40 — Method: write the decimal over 1000 using its three decimal places, then simplify. Working: 0.325 = 325/1000 = 13/40 (dividing both numerator and denominator by 25). Answer: 13/40. 13/4 comes from writing the decimal over 100 instead of 1000, as if there were only two decimal places. 8/25 comes from rounding 0.325 down to 0.32 before converting. 3/8 comes from recalling the learned conversion 3/8 = 0.375 and matching it to 0.325 because both are three-place decimals beginning with 3, instead of converting the decimal given.
7. (a) 5√2 — Simplify each surd first: √8 = √4 × √2 = 2√2, and √18 = √9 × √2 = 3√2. Both terms are now multiples of √2, so they are like terms: 2√2 + 3√2 = 5√2. Adding the numbers under the two roots first, 8 + 18 = 26, and writing √26 treats unlike surds as if they combine under one root — they only combine once they share the same radicand, which is not how addition of surds works. Writing 9√2 for √18 instead of 3√2 (forgetting to root the 9) and then adding gives 2√2 + 9√2 = 11√2. Writing 4√2 for √8 instead of 2√2 (forgetting to root the 4) and adding gives 4√2 + 3√2 = 7√2.
8. (b) 24√2 — The side length of the tile is √72. Since 72 = 36 × 2, √72 = √36 × √2 = 6√2 cm. A square has four equal sides, so the perimeter is 4 × 6√2 = 24√2 cm. Simplifying √72 by writing the perfect-square factor itself as the coefficient instead of its root, 36√2 instead of 6√2, and then multiplying by 4 lands on 144√2. Working out the correct side length, 6√2 cm, but then giving that as the final answer without multiplying by 4 for the perimeter gives 6√2. Doubling the side length instead of quadrupling it, as if the perimeter were 2 × 6√2 rather than 4 × 6√2, gives 12√2.
9. (b) 1/9 — Method: let x stand for the recurring decimal, multiply by the power of ten that moves exactly one repeating block past the decimal point, subtract the original equation so that the recurring tail cancels, and solve the equation that is left. Working: let x = 0.111...; the repeating block is one digit long, so multiply by 10 to give 10x = 1.111...; subtracting the first equation from the second gives 10x − x = 1.111... − 0.111..., that is 9x = 1; dividing both sides by 9 gives x = 1/9. Answer: 1/9. The distractors: 1/10 comes from dividing by the multiplier 10 at the last step instead of by the 9 that is left in front of x; 1/11 comes from recalling the elevenths family instead of the ninths, although 1/11 = 0.0909... has a two-digit repeating block rather than a one-digit one; 11/100 comes from stopping the decimal after two digits and converting 0.11 into hundredths.
10. (a) x = 10 or x = −10 — Rearranging, x² = 100. Taking the square root of both sides gives x = ±10, i.e. x = 10 or x = −10. A candidate who forgets the negative root gives only x = 10. A candidate who halves 100 instead of taking its square root gets x = 50. A candidate who applies the ± sign to 100 itself instead of to its square root gets x = 100 or x = −100.
11. (d) 2/25 — Method: write the decimal over 100 using its two decimal places, then simplify. Working: 0.08 = 8/100 = 2/25 (dividing both numerator and denominator by 4). Answer: 2/25. The student's fraction, 8/10, comes from ignoring the zero in the tenths column and reading 0.08 as though it were 0.8; it simplifies to 4/5. 1/125 comes from writing the decimal over 1000 instead of 100, as if there were three decimal places. 25/2 comes from flipping the correct fraction upside down.
12. (b) 6x + 6 — Perimeter = 2[(2x + 5) + (x − 2)] = 2(3x + 3) = 6x + 6. A candidate who adds the length and width but forgets to double for the perimeter gets 3x + 3. A candidate who makes a sign error and adds 2 instead of subtracting it before doubling gets 2[(2x + 5) + (x + 2)] = 6x + 14. A candidate who multiplies the length and width instead of adding them, confusing the perimeter formula with the area formula, and then doubles that product, gets 2(2x + 5)(x − 2) = 4x² + 2x − 20.
13. (b) k = 9 — Method: complete the square, because a squared bracket is equal to zero for exactly one value of x. Working: x² − 6x = (x − 3)² − 9, so the equation becomes (x − 3)² − 9 + k = 0, that is (x − 3)² = 9 − k; there is exactly one solution when 9 − k = 0, so k = 9 and the equation reads x² − 6x + 9 = 0 with the repeated solution x = 3. Answer: k = 9. The distractors: k = 3 comes from halving the 6 and not squaring the result; k = 36 comes from squaring the whole of 6 instead of half of it; k = −9 comes from solving 9 − k = 0 with the sign of k the wrong way round.
14. (b) (2x + 1)/(x − 3) — Factorise both the numerator and the denominator before you cancel anything. The numerator 2x² + 7x + 3 factorises to (2x + 1)(x + 3), and the denominator x² − 9 is a difference of two squares, factorising to (x − 3)(x + 3). The (x + 3) factor is common to both, so it cancels, leaving (2x + 1)/(x − 3). Writing (2x + 1)/(x + 3) comes from factorising x² − 9 as (x + 3)² instead of (x − 3)(x + 3) — a difference of two squares always has one plus and one minus bracket. Writing (2x + 3)/(x − 3) comes from mis-factorising the numerator as (2x + 3)(x + 1) and then wrongly cancelling the (x + 1) against the denominator's (x + 3) as though they were the same bracket. Writing 2x + 1 comes from cancelling the (x + 3) factor correctly but then dropping the remaining (x − 3) on the denominator altogether.
15. (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.
16. (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.
17. (d) 26 — Method: expand both brackets, treating the second as multiplication by −2 so that both of its terms change sign, then collect like terms. Working: 2(a + 6) = 2a + 12 and −2(a − 7) = −2a + 14, so the expression becomes 2a + 12 − 2a + 14; the a terms give 2a − 2a = 0, so no term in a survives, and the numbers give 12 + 14 = 26. Answer: 26. The distractors: −2 comes from expanding the second bracket as −2a − 14, so that the numbers give 12 − 14; 4a − 2 comes from adding 2(a − 7) instead of subtracting it, giving 2a + 12 + 2a − 14; 13 comes from multiplying the 2 over only the first term of each bracket, giving 2a + 6 − 2a + 7.
18. (b) 7 cm — The side opposite the 30° angle is found using sin 30° = opposite/hypotenuse, so opposite = 14 × sin 30° = 14 × 1/2 = 7 cm. 7√3 cm comes from using cos 30° = √3/2 instead of sin 30° (mixing up the opposite and adjacent sides). 14/√3 cm comes from using tan 30° = 1/√3 instead of sin 30°. 28 cm comes from dividing 14 by sin 30° instead of multiplying by it.
19. (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.
20. (d) 27/80 — Method: write the total as a fraction of a litre, then divide by the number of glasses. Working: 1.35 = 27/20, so each glass holds 27/20 ÷ 4 = 27/80 of a litre. Answer: 27/80. 27/20 comes from converting the total correctly to a fraction but forgetting to divide by the number of glasses. 27/5 comes from multiplying the total by 4 instead of dividing. 17/50 comes from rounding 1.35 ÷ 4 to 0.34 before converting to a fraction.