🧮 Paper 1 non-calculator warm-up — Higher

Answer key

1. (d) 7/20 — Method: write the decimal over the power of ten that matches the number of digits after the point, then divide the numerator and the denominator by their highest common factor. Working: 0.35 has two digits after the point, so it is 35 hundredths and can be written as 35/100; the highest common factor of 35 and 100 is 5, and 35 ÷ 5 = 7 with 100 ÷ 5 = 20. Answer: 7/20. The distractors: 3/10 comes from reading only the first digit after the point and converting 0.3; 7/25 comes from dividing the numerator by 5 but the denominator by 4, using a different factor on the top and on the bottom; 35/10 comes from counting one decimal place instead of two and writing the digits over 10.
2. (b) 0.83333... — Divide 5 by 6 using long division. 5.000... ÷ 6: 50 ÷ 6 = 8 remainder 2, giving the first decimal digit 8. Bring down a 0 to make 20, and 20 ÷ 6 = 3 remainder 2 — the remainder 2 has reappeared, so from here the digit 3 repeats forever. This gives 5/6 = 0.83333... . Stopping after two decimal places and writing 0.83 treats the division as if it terminated, when the remainder never reaches zero. Shifting the decimal point one place too far to the left gives 0.083333..., the same digits divided by an extra power of ten. A slip in the long division itself, misreading a remainder, can produce the wrong repeating digit, 0.85555... .
3. (d) sin 30°, tan 30°, cos 30° — sin 30° = 1/2 = 0.5, tan 30° = √3/3 ≈ 0.577 and cos 30° = √3/2 ≈ 0.866, so the correct order from smallest to largest is sin 30°, tan 30°, cos 30°. 'sin 30°, cos 30°, tan 30°' swaps the last two, wrongly putting cos 30° before tan 30°. 'cos 30°, tan 30°, sin 30°' is the correct list written backwards, from largest to smallest. 'tan 30°, sin 30°, cos 30°' wrongly swaps sin 30° and tan 30° at the start.
4. (a) 1/3 — To subtract these fractions, first write 3/4 with a denominator of 12: 3/4 = 9/12. Then 9/12 − 5/12 = 4/12, which simplifies to 1/3. Subtracting the numerators and the denominators separately, (3 − 5)/(4 − 12), gives −2/−8, which simplifies to 1/4. Changing 3/4 to twelfths by only changing the denominator, without scaling the numerator to match, gives 3/12 − 5/12 = −2/12, which simplifies to −1/6. Adding the fractions instead of subtracting them, 9/12 + 5/12, gives 14/12, which simplifies to 7/6.
5. (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.
6. (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.
7. (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.
8. (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.
9. (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.
10. (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.
11. (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.
12. (d) 2⁷ — When multiplying powers of the same base, the indices add: 3 + 4 = 7, so 2³ × 2⁴ = 2⁷. Multiplying the indices instead of adding them gives 3 × 4 = 12, so 2¹². Subtracting the indices instead of adding them gives 4 − 3 = 1, so 2¹. Multiplying the bases together as well as adding the indices gives 2 × 2 = 4, so 4⁷.
13. (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.
14. (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.
15. (d) (3x − 2)(x + 4) — For 3x² + 10x − 8, find two numbers multiplying to 3 × (−8) = −24 and adding to 10: these are 12 and −2. Rewrite: 3x² + 12x − 2x − 8 = 3x(x + 4) − 2(x + 4) = (3x − 2)(x + 4). Choosing (3x + 2)(x − 4) expands to 3x² − 10x − 8 — the correct factor pair but the wrong signs, giving the middle term the wrong sign. Choosing (x − 2)(3x + 4) expands to 3x² − 2x − 8 — the 3 is attached to the wrong bracket, changing which terms combine for the x-coefficient. Choosing (3x − 4)(x + 2) expands to 3x² + 2x − 8 — this uses 4 and 2 instead of the correct pair 12 and 2, so the middle term does not come to 10x.
16. (d) x = 2 or x = −5 — Method: find two numbers that multiply to give −10 and add to give 3 — these are 5 and −2. So x² + 3x − 10 = (x + 5)(x − 2) = 0, giving x = −5 or x = 2. Distractor origins: x = −2 or x = 5 swaps the signs of the two roots; x = 2 or x = 5 makes both roots positive, ignoring the sign of −10; x = −5 or x = −2 makes both roots negative.
17. (d) 6 m — The horizontal distance is adjacent to the 60° angle and the zip-wire is the hypotenuse, so horizontal distance = 12 × cos 60° = 12 × 1/2 = 6 m. 6√3 m comes from using sin 60° = √3/2 instead of cos 60°, which would give the vertical drop, not the horizontal distance. 4√3 m comes from treating 12 as the side adjacent to a tangent ratio and dividing by tan 60° = √3. 24 m comes from dividing 12 by cos 60° instead of multiplying by it.
18. (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.
19. (b) (x − 3)/(x + 2) — Factorise both: x² − 9 = (x − 3)(x + 3) (difference of two squares), and x² + 5x + 6 = (x + 2)(x + 3) (two numbers multiplying to 6 and adding to 5, namely 2 and 3). The factor (x + 3) is common to both, so it cancels, leaving (x − 3)/(x + 2). Choosing −9/(5x + 6) comes from cancelling the x² terms directly without factorising first — x² is not a common factor of the whole numerator or denominator. Choosing (x + 3)/(x + 2) cancels the (x − 3) factor instead of the shared (x + 3) factor, and (x − 3) does not appear in the denominator to cancel with. Choosing x − 3 cancels the whole denominator (x + 2) as though it were equal to 1.
20. (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.