🧮 Paper 1 non-calculator warm-up — Higher

Answer key

1. (b) 12 m — sin 30° = opposite ÷ hypotenuse, where the opposite side is the height (6 m) and the hypotenuse is the string. So string = height ÷ sin 30° = 6 ÷ (1/2) = 12 m. The distractor 3 m comes from multiplying by sin 30° instead of dividing (6 × 1/2 = 3). The distractor 6√3 m comes from using tan 30° = 1/√3 instead of sin 30° (6 ÷ (1/√3) = 6√3). The distractor 4√3 m comes from using cos 30° = √3/2 instead of sin 30° (6 ÷ (√3/2) = 12/√3 = 4√3).
2. (c) 25 cm — Method: for a cube, the edge length is the cube root of the volume. Working: 25 × 25 × 25 = 15,625, so the edge length is 25 cm. 5 cm comes from cube-rooting 125 instead of 15,625, misreading the number of digits. 50 cm comes from working out 25 × 2 = 50, doubling the correct edge length. 125 cm comes from taking the square root of the volume instead of the cube root, since 125 × 125 = 15,625 — that would be the side of a SQUARE of area 15,625, not the edge of a cube of that volume. Answer: 25 cm.
3. (d) a/2 — Method: write both terms over the same denominator, subtract the numerators, then cancel the fraction down. Working: 6 is a multiple of 3, so a/3 is rewritten as 2a/6; the calculation becomes 5a/6 − 2a/6 = 3a/6, and dividing numerator and denominator by 3 gives a/2. Answer: a/2. The distractors: 4a/3 comes from subtracting the denominators as well as the numerators, giving (5 − 1)a over (6 − 3); 2a/3 comes from subtracting 1 from 5 without first rewriting a/3 as 2a/6, giving 4a/6; 7a/6 comes from adding the two fractions instead of subtracting them, giving 5a/6 + 2a/6.
4. (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.
5. (d) 1 — sin 30° = 1/2, so (sin 30°)² = 1/4. cos 30° = √3/2, so (cos 30°)² = 3/4. Adding these gives 1/4 + 3/4 = 1. '1/4' only calculates (sin 30°)² and forgets to add the cos 30° term. '3/4' only calculates (cos 30°)² and forgets to add the sin 30° term. '−1/2' comes from subtracting the two squared values instead of adding them: 1/4 − 3/4 = −1/2.
6. (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.
7. (a) x = 4 ± √13 — Method: halve the coefficient of x to form the bracket, subtract the square of that number to keep the expression equal to the original, then rearrange and take the square root of both sides. Working: half of −8 is −4, so x² − 8x + 3 = (x − 4)² − 16 + 3 = (x − 4)² − 13; setting this equal to zero gives (x − 4)² = 13, so x − 4 = ±√13 and x = 4 ± √13. Answer: x = 4 ± √13. The distractors: x = 8 ± √13 comes from putting the whole coefficient 8 inside the bracket instead of half of it; x = 4 ± √19 comes from adding 16 and 3 rather than subtracting 3 from 16; x = −4 ± √13 comes from writing the bracket as (x + 4)², which reverses the sign of the number that comes out of it.
8. (c) It has two different real solutions — Method: the discriminant is the quantity under the square root sign in the quadratic formula, so its sign decides how many real values the formula produces. Working: when the discriminant is positive its square root is a real number that is not zero, and the formula adds that root and subtracts it in turn, so the two results are different; a positive discriminant that is not a perfect square still gives two solutions, but they are not whole numbers. Answer: it has two different real solutions. The distractors: one repeated real solution is what a discriminant of zero gives, because adding and subtracting zero changes nothing; no real solutions is what a negative discriminant gives, because a negative number has no real square root; two whole-number solutions happens only when the discriminant is a perfect square and the division works out exactly, which a positive discriminant does not guarantee.
9. (c) (12 + 3π) cm — The perimeter of a quarter-circle is made up of two straight radii plus a quarter of the circumference. The two radii give 2 × 6 = 12 cm, and a quarter of the circumference is (1/4) × 2 × π × 6 = 3π cm, so the total perimeter is (12 + 3π) cm. Giving only the curved part, 3π cm, forgets the two straight edges entirely. Using the full circumference, 2 × π × 6 = 12π, instead of a quarter of it gives (12 + 12π) cm. Including only one radius instead of two gives (6 + 3π) cm.
10. (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.
11. (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.
12. (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.
13. (c) 5/11 — Let x = 0.45 recurring, so x = 0.454545... . Since two digits repeat, multiply by 100: 100x = 45.454545... . Subtracting the original x removes the recurring part exactly, because it lines up digit for digit: 100x − x = 45.454545... − 0.454545... = 45, so 99x = 45, giving x = 45/99 = 5/11. Treating the decimal as if it terminated at two places gives 45/100 = 9/20, which is only 0.45 and drops the repeating part entirely. Subtracting 10x instead of x — using 100x − 10x = 90x = 45 — is the wrong power of ten for a two-digit repeating block, and gives x = 45/90 = 1/2. Making an arithmetic slip in the numerator, 45 − 1 = 44 instead of 45, gives 44/99 = 4/9.
14. (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.
15. (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
16. (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.
17. (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.
18. (a) 25 — Method: squaring a square root removes the root, so a square root raised to the power 4 can be squared in two stages. Working: (√5)⁴ = ((√5)²)² = 5² = 5 × 5 = 25. Answer: 25. The distractors: 5 comes from squaring once and stopping, treating the fourth power as a square; 625 comes from raising 5 to the power 4 and ignoring the root sign altogether; 20 comes from multiplying 5 by the index 4 instead of raising 5 to that power.
19. (d) 72a⁷ — Method: a power outside brackets applies to every factor inside them, and multiplying two powers of the same letter adds their indices. Working: (3a²)² = 3² × a⁴ = 9a⁴, and (2a)³ = 2³ × a³ = 8a³. Multiplying the two results gives 9 × 8 = 72 for the number and 4 + 3 = 7 for the index of a. Answer: 72a⁷. The distractors: 36a⁷ comes from squaring the 2 in (2a)³ instead of cubing it, giving 4a³ and then 9 × 4; 72a¹² comes from multiplying the indices 4 and 3 when the two terms are multiplied, instead of adding them; 17a⁷ comes from adding the coefficients 9 and 8 rather than multiplying them.
20. (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.