🧮 Paper 1 non-calculator warm-up — Higher

Answer key

1. (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.
2. (d) 1/4 — Method: deal with the fractional index first, then the negative sign. Working: $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$. A negative index means take the reciprocal of that result, so $8^{-2/3} = \frac{1}{8^{2/3}} = \frac{1}{4}$. Answer: 1/4. A candidate who evaluates $8^{2/3}$ correctly but forgets the negative sign entirely gets 4 — they have dropped the instruction to take a reciprocal. A candidate who takes the reciprocal step but applies it as a sign change to the finished number instead of inverting it gets −4. A candidate who multiplies 8 by −2/3, treating the index as an ordinary factor rather than a power, gets −16/3.
3. (c) 2/5 — To find a fraction of a fraction, multiply them together: 3/5 × 2/3 = 6/15, which simplifies to 2/5. Multiplying only the numerators, 3 × 2 = 6, but adding the denominators, 5 + 3 = 8, instead of multiplying them gives 6/8, which simplifies to 3/4. Using only the fraction who study French, 3/5, and ignoring that a further fraction of them also study Spanish gives 3/5. Dividing by 2/3 instead of multiplying by it, using its reciprocal 3/2, gives 3/5 × 3/2 = 9/10.
4. (a) x = 6 or x = −4 — We need two numbers that multiply to −24 and add to −2: these are −6 and 4, since −6 × 4 = −24 and −6 + 4 = −2. So x² − 2x − 24 = (x − 6)(x + 4) = 0, giving x = 6 or x = −4. A candidate who swaps the signs, using 6 and −4 the wrong way round in the brackets, gets x = −6 or x = 4. A candidate who picks the wrong factor pair, 8 and −3 (which multiply to −24 but add to +5, not −2), gets (x + 8)(x − 3) = 0 and answers x = −8 or x = 3. A candidate who makes both factors negative gets x = −6 or x = −4, which would require the constant term to be +24, not −24.
5. (c) 1 — sin 30° = 1/2 and cos 60° = 1/2, so sin 30° + cos 60° = 1/2 + 1/2 = 1. 0 comes from subtracting the two values instead of adding them (1/2 − 1/2). 1/2 comes from writing down only one of the two exact values and forgetting to add the other. √3 comes from swapping the two angles and working out sin 60° + cos 30° = √3/2 + √3/2 = √3.
6. (d) 12 m — Method: give the width a letter, write the length in terms of it and use length × width = area to form a quadratic. Working: with a width of x metres the length is x + 4, so x(x + 4) = 96, which rearranges to x² + 4x − 96 = 0; factorising gives (x + 12)(x − 8) = 0, and a width must be positive, so x = 8 and the length is 8 + 4 = 12. Answer: 12 m, and 12 × 8 = 96 as the area requires. The distractors: 8 m is the width rather than the length asked for; 16 m comes from taking the other root as 12 and adding 4 to it, ignoring that the root −12 cannot be a width; 24 m comes from dividing the area by the 4 in the question instead of forming an equation.
7. (a) 0.1 m — Method: the sloping surface is the hypotenuse and the vertical rise is the side opposite the 30° angle, so rise = 4.8 × sin 30°; then compare that rise with the limit. Working: the exact value of sin 30° is one half, so the rise = 4.8 × 1/2 = 2.4 m. The limit is 2.5 m, and 2.5 − 2.4 = 0.1. Answer: the ramp is 0.1 m below the limit. Working out the rise and stopping there gives 2.4 m, which answers a question that was not asked. Dividing by sin 30° instead of multiplying gives 4.8 ÷ 0.5 = 9.6 and then 9.6 − 2.5 = 7.1 m. Treating sine as proportional to the angle, so that sin 30° is a third of sin 90°, gives 4.8 ÷ 3 = 1.6 and then 2.5 − 1.6 = 0.9 m.
8. (c) 9π cm² — The area of a circle is π × r². With a radius of 3 cm this is π × 3² = 9π cm², and this is exact because π has not been replaced by any approximation. Writing 28.3 cm² replaces π with a rounded decimal value, 3.14, and then rounds the result again, so it is only an approximation. Writing 28.26 cm² uses π ≈ 3.14 without a final rounding step, but this is still only an approximation of 9π, not the exact value. Writing 27 cm² comes from replacing π with the rough approximation 3, which is even further from the true value.
9. (c) 5/8 — Method: write the decimal over 1000 using its three decimal places, then simplify. Working: 0.625 = 625/1000 = 5/8 (dividing both numerator and denominator by 125). Answer: 5/8. 25/4 comes from writing the decimal over 100 instead of 1000, as if there were only two decimal places. 31/50 comes from rounding 0.625 to 0.62 before converting. 8/5 comes from simplifying correctly to 5/8 and then writing the fraction upside down.
10. (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.
11. (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.
12. (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.
13. (c) 5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising.
14. (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.
15. (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.
16. (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.
17. (a) y¹² — Method: when a power is raised to another power, multiply the two indices. Working: (y³)⁴ means y³ × y³ × y³ × y³, which is four lots of three y's multiplied together, so the index is 3 × 4 = 12 and (y³)⁴ = y¹². y⁷ comes from adding the indices, 3 + 4 = 7, which is the rule for multiplying two separate powers, not for raising a power to a power. y⁸¹ comes from working out 3⁴ = 81 and using that as the index, raising the inner index to the outer power instead of multiplying the two indices. 12y comes from multiplying the indices to make 12 but then treating y as a coefficient instead of a power. Answer: y¹².
18. (c) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
19. (c) 2 — First write 1 1/2 as an improper fraction, 3/2. To divide by 3/4, multiply by its reciprocal, 4/3: 3/2 × 4/3 = 12/6 = 2. Dropping the whole number and dividing only the fractional part, 1/2 ÷ 3/4 = 1/2 × 4/3, gives 2/3. Multiplying by 3/4 directly instead of using its reciprocal, 3/2 × 3/4, gives 9/8. Using the reciprocal of the first fraction instead of the second, 2/3 × 3/4, gives 1/2.
20. (b) 9 — Method: a negative index means take the reciprocal of the base first and then apply the positive power. Working: the reciprocal of 1/3 is 3, so (1/3)⁻² = 3² = 3 × 3 = 9. Answer: 9. The distractors: 1/9 comes from ignoring the minus sign and squaring 1/3 as it stands; −9 comes from reading the negative index as a minus sign on the result; 6 comes from multiplying the denominator by the index, 3 × 2, instead of squaring the reciprocal.