🧮 Paper 1 non-calculator warm-up — Higher

Answer key

1. (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.
2. (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
3. (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.
4. (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.
5. (a) 2√3/3 — tan 30° = √3/3, so tan 30° + tan 30° = 2 × √3/3 = 2√3/3. √3 comes from wrongly treating tan 30° + tan 30° as tan(30° + 30°) = tan 60° = √3 — adding angles is not the same as adding ratios. √3/3 comes from forgetting to double the value and just writing down tan 30° on its own. 2√3 comes from doubling the numerator of √3/3 but forgetting to keep the denominator of 3.
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) x¹¹ — Method: when multiplying powers of the same base, add the indices. Working: 7 + 4 = 11, so x⁷ × x⁴ = x¹¹. x²⁸ comes from multiplying the indices, 7 × 4 = 28, instead of adding them. x³ comes from working out 7 − 4 = 3, which is the rule for dividing powers, not multiplying them. 11x comes from adding the indices to make 11 but then treating x as a coefficient instead of a power. Answer: x¹¹.
8. (d) 2/25 — Method: write the decimal over 100 using its two decimal places, then simplify. Working: 0.08 = 8/100 = 2/25 (dividing both numerator and denominator by 4). Answer: 2/25. The student's fraction, 8/10, comes from ignoring the zero in the tenths column and reading 0.08 as though it were 0.8; it simplifies to 4/5. 1/125 comes from writing the decimal over 1000 instead of 100, as if there were three decimal places. 25/2 comes from flipping the correct fraction upside down.
9. (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.
10. (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.
11. (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.
12. (a) 1/2 — To find a fraction of an amount, multiply the fractions together: 3/5 × 5/6 = 15/30, which simplifies to 1/2 litre. Adding the fractions instead of multiplying them, using a common denominator of 30, gives 18/30 + 25/30 = 43/30, a value greater than the whole bottle. Dividing by 5/6 instead of multiplying by it, using its reciprocal 6/5, gives 3/5 × 6/5 = 18/25. Multiplying 5/6 by itself instead of by 3/5 gives 25/36.
13. (c) 7/30 — Let x = 0.2333... . Because only the 3 recurs, use two multiples of x that line up the recurring part exactly: 10x = 2.333... and 100x = 23.333... . Subtracting removes the recurring tail completely: 100x − 10x = 23.333... − 2.333... = 21, so 90x = 21, giving x = 21/90 = 7/30. Treating the decimal as if it terminated after two places, writing 0.23 as 23/100, ignores that the 3 carries on forever. Misreading which digits recur — treating 0.2333... as if the block '23' repeated, giving 0.232323... — leads to x = 23/99, which is a different, larger recurring decimal from the one given. A numerator slip in the subtraction, computing 22 instead of 21, gives x = 22/90 = 11/45.
14. (b) (3x + 1)(2x − 3) — To factorise 6x² − 7x − 3, look for two numbers that multiply to 6 × (−3) = −18 and add to −7: these are −9 and 2. Splitting the middle term gives 6x² − 9x + 2x − 3, which groups to 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3). Writing (3x − 1)(2x + 3) comes from using the pair 9 and −2 instead — the right product but the wrong signs, giving +7x instead of −7x. Writing (6x − 1)(x + 3) comes from picking the factor pair 18 and −1: it multiplies to −18 correctly, but it adds to 17, not −7, so checking only the product splits the middle term as 6x² + 18x − x − 3 and pairs the wrong factors of 6 and 3 together. Writing (x − 3)(6x + 1) comes from that same unchecked pair used the other way round, splitting the middle term as 6x² + x − 18x − 3.
15. (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.
16. (d) 3√5 — Split 45 into a perfect square times a factor: 45 = 9 × 5. Take the square root of each part separately: √45 = √9 × √5 = 3√5, since √9 = 3. Writing the perfect-square factor itself (9) as the coefficient instead of its root would give 9√5 — that trap comes from forgetting the last step, rooting 9. Multiplying 3 and 5 together instead of keeping them as coefficient and radicand gives 15, which throws away the surd entirely. Doubling the correct coefficient by mistake gives 6√5.
17. (d) 6√3 cm — Method: AB lies alongside the 30° angle at A and AC is the hypotenuse, so the ratio needed is cosine: cos 30° = AB ÷ AC. Working: the exact value of cos 30° is √3/2, so AB = 12 × √3 ÷ 2, and half of 12 is 6. Answer: AB = 6√3 cm, which is about 10.4 cm. Using sine by mistake gives 12 × 1/2 = 6 cm, which is the length of BC rather than AB. Using tan 30° = 1/√3 gives 12 ÷ √3, which is 4√3 cm. Remembering cos 30° as √3 rather than as √3 halved gives 12√3 cm, longer than the hypotenuse and so impossible.
18. (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.
19. (d) 6 — By Pythagoras' theorem, the square of the hypotenuse equals the sum of the squares of the other two sides: (√12)² + (√24)² = 12 + 24 = 36. The hypotenuse is √36 = 6 cm. Adding the two side lengths directly instead of squaring them first, treating the theorem as if it were a straight sum of the sides, gives √12 + √24 = 2√3 + 2√6. Multiplying the two squared values, 12 × 24 = 288, instead of adding them, then taking the root, gives √288 = 12√2. Adding the squares correctly to get 36 but forgetting to take the square root at the end leaves 36 as the answer instead of the hypotenuse itself.
20. (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.