🧮 Paper 1 non-calculator warm-up — Higher

Answer key

1. (c) x = 0 or x = −7 — Factorising: x² + 7x = x(x + 7) = 0, so x = 0 or x + 7 = 0, giving x = 0 or x = −7. A candidate who divides both sides of the original equation by x, which loses the solution x = 0, gets only x = −7. A candidate who makes a sign error solving x + 7 = 0 gets x = 0 or x = 7. A candidate who misreads the coefficient and doubles it gets x = 0 or x = −14.
2. (a) k = 3 — Method: a solution of an equation makes both sides balance, so substitute it in and solve the equation in k that is left. Working: putting x = 3 gives 3² − (k + 1) × 3 + k = 0, that is 9 − 3k − 3 + k = 0, so 6 − 2k = 0 and k = 3; the equation is then x² − 4x + 3 = 0, whose solutions are 3 and 1. Answer: k = 3. The distractors: k = −3 comes from solving 6 − 2k = 0 as though it gave 2k = −6; k = 4 comes from expanding −3(k + 1) as −3k − 1, multiplying only the k by 3; k = 1.5 comes from working 3² as 3 × 2 = 6, which leaves 3 − 2k = 0.
3. (d) x⁽¹⁄²⁾ — Method: dividing two powers of the same letter subtracts the index of the divisor from the index of the term being divided, and fractional indices are subtracted like any other fractions. Working: 3/4 − 1/4 = 2/4, which simplifies to 1/2, so the result is x⁽¹⁄²⁾. Answer: x⁽¹⁄²⁾. The distractors: x comes from adding the indices, 3/4 + 1/4 = 1, as though the powers were being multiplied; x³ comes from dividing the indices, so that 3/4 divided by 1/4 gives 3; x⁽³⁄¹⁶⁾ comes from multiplying the indices, 3/4 × 1/4.
4. (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.
5. (c) 0.7, 0.72, 3/4 — Method: numbers written in different forms cannot be compared as they stand, so every fraction is turned into a decimal by dividing the numerator by the denominator, and the decimals are then compared place by place from the left. Working: 3/4 means 3 ÷ 4 = 0.75, so the three values to compare are 0.7, 0.75 and 0.72; written to two decimal places they are 0.70, 0.75 and 0.72, and the hundredths digits 0, 5 and 2 put 0.70 first, 0.72 next and 0.75 last; written again in the forms the question used, the order from smallest is 0.7, then 0.72, then 3/4. Answer: 0.7, 0.72, 3/4. The distractors: 3/4, 0.7, 0.72 comes from turning 3/4 into 0.34 by writing the numerator and the denominator as the two digits after the point, which makes the fraction the smallest of the three; 0.72, 3/4, 0.7 comes from the belief that the more digits a decimal has the smaller it must be, which puts both 0.72 and 0.75 below 0.7 and 0.72 below 0.75; 3/4, 0.72, 0.7 comes from comparing the three values correctly but listing them largest first, against an instruction to start with the smallest.
6. (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.
7. (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.
8. (b) x = 5 — Area = length × width, so x(x + 3) = 40, which rearranges to x² + 3x − 40 = 0. This factorises as (x + 8)(x − 5) = 0: the two numbers in the brackets must multiply to −40 and add to +3, and the pair 8 and −5 does both. This gives x = −8 or x = 5. Since x is a length, it cannot be negative, so x = 5. A candidate who gives both solutions without rejecting the negative one, which cannot be a length, answers x = 5 or x = −8. A candidate who picks the wrong factor pair of 40, such as 10 and −4 instead of 8 and −5, gets (x + 10)(x − 4) = 0 and answers x = 4. A candidate who rejects the wrong root, keeping the negative solution instead of the positive one, answers x = −8.
9. (c) x = −2 or x = 4 — The solutions of x² − 2x − 8 = 0 are the x-values where y = 0. From the table, y = 0 when x = −2 and when x = 4, so those are the two solutions. Distractor origins: x = −1 or x = 3 picks the pair of x-values that give equal (but non-zero) y-values instead of y = 0; x = 0 or x = −8 mixes up an x-value with its corresponding y-value; x = −2 only reads off one of the two roots and misses the other.
10. (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.
11. (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.
12. (a) 6 — Use √a × √b = √(ab): √3 × √12 = √(3 × 12) = √36 = 6. Adding the numbers under the roots instead of multiplying them, 3 + 12 = 15, gives √15 — that comes from applying the rule for adding surds to a multiplication question. Multiplying the two numbers under the roots but then forgetting to take the square root at the end leaves 36. Simplifying only √12 to 2√3 and then dropping the other √3 factor entirely gives 2√3.
13. (c) 5√3 m — The cable, the pole and the ground form a right-angled triangle: the ground distance (5 m) is adjacent to the 60° angle, and the height of the pole is opposite it, so height = 5 × tan 60° = 5 × √3 = 5√3 m. 5√3/2 m comes from using sin 60° = √3/2 instead of tan 60°. 5/√3 m comes from using tan 30° = 1/√3, the reciprocal-angle value, instead of tan 60°. 10√3 m comes from doubling the correct height by mistake.
14. (b) (60 + 4.5π) cm² — Method: find the area of the rectangle and the area of the semicircle separately, then add them. Working: the rectangle has area 10 × 6 = 60 cm². The semicircle has radius 3 cm, so its area is half of π × 3² = half of 9π = 4.5π cm². Total area = (60 + 4.5π) cm². Answer: (60 + 4.5π) cm². (60 + 18π) cm² comes from using the diameter (6 cm) as the radius in the semicircle area formula: half of π × 6² = 18π. (60 + 9π) cm² comes from forgetting to halve the full circle's area: π × 3² = 9π. (60 + 3π) cm² comes from finding the semicircle's arc length instead of its area: half of 2 × π × 3 = 3π.
15. (b) 8π cm² — A diameter of 8 cm gives a radius of 4 cm. The area of a full circle would be π × r² = π × 4² = 16π cm², and a semicircle is exactly half of this, giving 16π ÷ 2 = 8π cm². Forgetting to halve the area for the semicircle gives 16π cm², the area of the whole circle. Halving the diameter twice, using a radius of 2 instead of 4, gives π × 2² = 4π cm². Using the diameter itself as the radius, so π × 8² = 64π, and then halving that for the semicircle gives 32π cm².
16. (a) x⁶ — Method: work through the powers in order — multiplying powers of the same base means adding indices, and dividing powers of the same base means subtracting indices. Working: first, x⁵ × x³ = x⁸ (adding 5 and 3); then x⁸ ÷ x² = x⁶ (subtracting 2 from 8). x⁴ comes from swapping the two rules — subtracting for the multiplication, 5 − 3 = 2, and then adding for the division, 2 + 2 = 4. x¹⁰ comes from adding all three indices, 5 + 3 + 2 = 10, treating the division the same as a multiplication. 6x comes from correctly reaching a total index of 6 but then writing it as a coefficient of x instead of as its power. Answer: x⁶.
17. (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.
18. (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.
19. (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.
20. (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.