🧮 Paper 1 non-calculator warm-up — Higher

Answer key

1. (b) 2⁵ — Method: a power outside brackets multiplies the index of every factor inside them, and dividing powers of the same base subtracts their indices. Working: (5² × 2³)³ = 5⁶ × 2⁹ and (5³ × 2²)² = 5⁶ × 2⁴. Dividing gives 5⁶⁻⁶ × 2⁹⁻⁴, and since 5⁰ = 1 the whole expression reduces to 2⁵. Answer: 2⁵. The distractors: 2² comes from adding the outside index to each inside index instead of multiplying, which gives 5⁵ × 2⁶ over 5⁵ × 2⁴; 2¹³ comes from adding the indices of 2 when dividing, 9 + 4, instead of subtracting them; 2 comes from applying the outside power to the first factor inside each bracket only, leaving 5⁶ × 2³ over 5⁶ × 2².
2. (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.
3. (a) −10 — Method: multiply each bracket out, treating the second bracket as being multiplied by −3 because it is subtracted, then collect like terms. Working: 4(2x − 1) = 8x − 4 and −3(x + 2) = −3x − 6, so the expression becomes 8x − 4 − 3x − 6 − 5x; the x terms give 8x − 3x − 5x = 0, so no term in x survives, and the numbers give −4 − 6 = −10. Answer: −10. The distractors: 2 comes from expanding −3(x + 2) as −3x + 6, leaving the numbers −4 + 6; 5x − 10 comes from forgetting the final −5x, so the x terms give 8x − 3x = 5x; −7 comes from multiplying the 4 over only the first term of its bracket, giving 8x − 1 and so the numbers −1 − 6.
4. (a) 17 — Method: work out each power separately, then combine them as the question asks. Working: $2^3 = 8$ and $3^2 = 9$, and 8 + 9 = 17. 72 comes from working out 8 × 9 = 72, multiplying the two powers instead of adding them. 12 comes from misreading the powers as repeated multiplication of the base by the index, 2 × 3 + 3 × 2 = 6 + 6 = 12. −1 comes from working out 8 − 9 = −1, subtracting the powers instead of adding them. Answer: 17.
5. (b) 5² — Method: multiplying two powers of the same base adds their indices, and a negative index is added as a negative number. Working: −2 + 4 = 2, so 5⁻² × 5⁴ = 5². Answer: 5². The distractors: 5⁶ comes from adding the sizes of the indices, 2 + 4, and ignoring the minus sign; 5⁻⁸ comes from multiplying the indices, −2 × 4, instead of adding them; 5⁻⁶ comes from subtracting the indices, −2 − 4, as though the powers were being divided.
6. (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.
7. (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.
8. (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.
9. (d) 6 + 2√3 — Multiply √3 by each term in the bracket separately. First term: √3 × 2 = 2√3. Second term: √3 × √12 = √(3 × 12) = √36 = 6. Adding the two results in the order they were found, and writing the whole-number term first, gives 6 + 2√3. Adding the numbers under the root for the second term instead of multiplying them (3 + 12 = 15) gives √15 in place of 6, leading to √15 + 2√3. Multiplying √3 by the 2 but never distributing to the √12 term at all leaves just 2√3. Treating √3 × 2 as if the 3 were multiplied by the 2 inside the root, √3 × 2 → √6, while still getting the second term correct, gives 6 + √6.
10. (b) 6√3 m — The horizontal distance covered by one support is adjacent to the 30° angle, so it equals 6 × cos 30° = 6 × √3/2 = 3√3 m. The total base width is made up of both supports, so it is 2 × 3√3 = 6√3 m. '3√3 m' gives only one support's horizontal distance and forgets to double it for the total width. '6 m' comes from using sin 30° instead of cos 30° for the horizontal distance (6 × sin 30° = 3, doubled to 6). '12 m' comes from doubling the full sloping length of 6 m without using any trigonometry at all.
11. (d) 16π cm² — Method: for a circle, area = π × radius². Working: area = π × 4² = π × 16 = 16π cm². Answer: 16π cm². The student squared the radius but left out the π, which is why 16 cm² is not the exact area. 4π cm² comes from multiplying by the radius once instead of squaring it: π × 4 = 4π. 8π cm² comes from using the circumference formula 2 × π × radius instead of the area formula: 2 × π × 4 = 8π. 64π cm² comes from using the diameter (8 cm) as the radius in the area formula: π × 8² = 64π.
12. (b) 3²⁰ is larger — Method: two powers with different bases and different indices can be compared once they are rewritten with a common index, which is possible whenever the indices share a factor. Working: 30 and 20 have a highest common factor of 10, so 2³⁰ = (2³)¹⁰ = 8¹⁰ and 3²⁰ = (3²)¹⁰ = 9¹⁰. Both are now tenth powers, and since 9 is larger than 8, 9¹⁰ is larger than 8¹⁰. Answer: 3²⁰ is larger. The distractors: 2³⁰ is larger comes from comparing only the indices and choosing the power with the bigger index; They are equal comes from multiplying base by index, 2 × 30 and 3 × 20, and finding 60 each time; They cannot be compared without a calculator comes from assuming that powers this large can only be ranked by evaluating them in full.
13. (d) 3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3.
14. (c) 1/2 — sin 45° = √2/2 and cos 45° = √2/2, so sin 45° × cos 45° = √2/2 × √2/2 = 2/4 = 1/2. √2/2 comes from writing down only one of the two factors and forgetting to multiply by the other. √2 comes from adding the two exact values instead of multiplying them: √2/2 + √2/2 = √2. 1 comes from wrongly treating sin 45° × cos 45° as sin(45° + 45°) = sin 90° = 1 — multiplying two ratios is not the same as adding their angles.
15. (d) x² − 6x + 9 — Method: squaring a bracket means multiplying that bracket by itself, so expand (x − 3)(x − 3) term by term and then collect like terms. Working: x × x = x², x × (−3) = −3x, (−3) × x = −3x and (−3) × (−3) = 9, giving x² − 3x − 3x + 9, and the two middle terms collect to −6x. Answer: x² − 6x + 9. The distractors: x² − 3x + 9 comes from writing down only one of the two middle products instead of both; x² + 6x + 9 comes from treating (−3) × x as +3x, so the middle terms are added rather than subtracted; x² − 9 comes from treating the square as the difference of two squares (x − 3)(x + 3).
16. (c) 3/11 — Method: let a letter stand for the recurring decimal, multiply by the power of ten that shifts exactly one repeating block past the point, subtract the original equation so that the recurring tail cancels, then solve and cancel. Working: let x = 0.272727...; the repeating block is two digits long, so multiply by 100 to give 100x = 27.272727...; subtracting gives 99x = 27, so x = 27/99; the highest common factor of 27 and 99 is 9, and 27 ÷ 9 = 3 with 99 ÷ 9 = 11. Answer: 3/11. The distractors: 27/100 comes from writing the repeating block over 100 instead of over 99, forgetting that subtracting x leaves 99x rather than 100x; 3/10 comes from rounding the decimal to one place and converting 0.3; 2/9 comes from treating only the 2 as recurring and converting 0.222... instead.
17. (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.
18. (d) k < 4 — Method: the number of real solutions of ax² + bx + c = 0 is decided by the discriminant b² − 4ac, and two different real solutions need it to be greater than zero. Working: here a = 1, b = −4 and c = k, so b² − 4ac = 16 − 4k; the condition is 16 − 4k > 0, which gives 16 > 4k and then k < 4. Answer: k < 4; for example k = 3 gives x² − 4x + 3 = 0, whose solutions are 1 and 3. The distractors: k ≤ 4 comes from using b² − 4ac ≥ 0, which also allows the single repeated solution at k = 4; k > 4 comes from dividing −4k > −16 by −4 without reversing the inequality sign; k < 16 comes from leaving the factor 4 out of 4ac and solving 16 − k > 0.
19. (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.
20. (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.