🧮 Paper 1 non-calculator warm-up — Higher

Answer key

1. (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.
2. (c) 7 + 4√3 — Expand the brackets fully: (2 + √3)² = 2² + 2 × 2 × √3 + (√3)² = 4 + 4√3 + 3. Adding the two whole-number terms, 4 + 3 = 7, gives 7 + 4√3. Using (a + b)² = a² + b² and skipping the middle cross term entirely gives just 4 + 3 = 7, with no surd term at all. Treating (√3)² as if it stayed √3 rather than becoming 3, then merging it with the existing surd term, gives 4 + 5√3. Squaring only the surd term correctly but carrying the whole-number term as 2 instead of squaring it to 4 gives 2 + 3 + 4√3 = 5 + 4√3.
3. (b) 3/50 — Method: write the decimal over the power of ten that matches the number of digits after the point, counting every digit including a zero, then divide the numerator and the denominator by their highest common factor. Working: 0.06 has two digits after the point, so it is 6 hundredths and can be written as 6/100; the highest common factor of 6 and 100 is 2, and 6 ÷ 2 = 3 with 100 ÷ 2 = 50. Answer: 3/50. The distractors: 3/5 comes from ignoring the zero straight after the point and converting 0.6 instead, giving 6/10, which cancels to 3/5; 3/500 comes from counting three decimal places instead of two and writing 6/1000, which cancels to 3/500; 1/6 comes from putting 1 over the digits after the point, as though 0.06 meant one sixth.
4. (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.
5. (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.
6. (b) 9/25 — Method: write the decimal over the matching power of ten, then divide the numerator and the denominator by their highest common factor. Working: 0.36 has two digits after the point, so 0.36 = 36/100; the highest common factor of 36 and 100 is 4, and 36 ÷ 4 = 9 with 100 ÷ 4 = 25; since 9 and 25 share no factor greater than 1, the fraction is fully cancelled. Answer: 9/25. The distractors: 3/10 comes from reading only the first digit after the point and converting 0.3; 9/50 comes from dividing the numerator by 4 but the denominator by only 2; 36/10 comes from counting one decimal place instead of two and writing the digits over 10.
7. (c) tan 45° — Method: replace each ratio by its exact value, then compare. Working: a right-angled triangle with a 45° angle is isosceles, so its opposite and adjacent sides are equal and the tangent of 45° is exactly 1. The others are cos 30° = √3/2, about 0.87; sin 45° = √2/2, about 0.71; and cos 60° = 1/2. Answer: tan 45°, the only one of the four that reaches 1. Reading √3/2 as though it were √3, about 1.73, makes cos 30° look the largest, but the division by 2 is part of the value. Ranking by the size of the angle also fails here, because the cosine of an angle falls as the angle grows.
8. (c) 25 — Method: read off a, b and c with their signs and substitute them into b² − 4ac. Working: for 2x² + 3x − 2 = 0, a = 2, b = 3 and c = −2, so b² − 4ac = 3² − 4 × 2 × (−2) = 9 − (−16) = 9 + 16 = 25. Answer: 25. The distractors: −7 comes from taking c as +2, which gives 9 − 16; 22 comes from working b² as 2 × 3 = 6 and then 6 + 16; 13 comes from leaving the 4 out of 4ac and working 9 − 2 × (−2).
9. (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.
10. (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.
11. (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.
12. (d) x = 4 or x = −4 — Method: divide both sides by 2 to get x² = 16, then take the square root of both sides: x = 4 or x = −4. Distractor origins: x = 4 forgets the negative root; x = 8 comes from halving 16 instead of taking its square root; x = 16 stops at x² = 16 without ever taking the square root.
13. (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.
14. (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.
15. (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.
16. (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.
17. (a) 2√3 — Multiply the top and bottom of the fraction by √3, since √3 × √3 = 3: 6/√3 = (6 × √3)/(√3 × √3) = 6√3/3. Dividing 6 by 3 gives 2, so the fraction simplifies to 2√3. Multiplying only the numerator by √3 and then cancelling the surd in the denominator against it as if they were the same term, without properly squaring the denominator, leads to 6. Dividing 6 by 3 as 3 instead of 2 gives 3√3 — a slip in the final division. Simplifying 6√3/3 by cancelling the whole numerator's 3 with the denominator's 3, including the surd, gives 2, which loses the surd altogether.
18. (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.
19. (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.
20. (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.