Sample paper · GCSE Higher · grades 4–9
GCSE Higher sample Paper 2 (calculator)
The real Paper 2 is 1.5 hour 30 minutes and 80 marks, calculator, and all three papers carry equal weight. This sample is 20 original questions in the same content proportions as the Higher qualification — number 15%, algebra 30%, ratio, proportion and rates of change 20%, geometry and measures 20%, probability 7.5%, statistics 7.5% — with a calculator allowed. AO1 / AO2 / AO3 at this tier: 40% / 30% / 30%.
CalculatorGCSE Higher
GCSE Higher sample Paper 2 (calculator)
MathsUKwww.geekhero.co.uk
- 1.4ˣ = 64. Work out the value of x.
- 2.A point has coordinates (x, y). In which two quadrants is the product x × y positive?
- 3.Write the ratio 2/3 : 1/2 as a ratio of whole numbers, in its simplest form.
- 4.A, B, C and D are points on a circle, and ABCD is a cyclic quadrilateral with the vertices in that order around the circle. Angle ABC = 108° and angle ADC = 72°. Which circle theorem is the reason that angle ABC + angle ADC = 180°?
- 5.In a game, Zara says the probability of scoring is 3/8, the probability of missing is 5/12 and the probability of a rebound is 1/6, and that these are the only three outcomes. Is Zara correct that her three probabilities are valid?
- 6.In a spelling test the 20 pupils in Group A had a mean mark of 80, and the 30 pupils in Group B had a mean mark of 70. Work out the mean mark of all 50 pupils.
- 7.A measuring jug shows a volume of 340 ml, correct to the nearest 20 ml. Work out the smallest possible volume in the jug.
- 8.A rule multiplies the input by a fixed number and then adds a fixed number. An input of 1 gives an output of 5, and an input of 3 gives an output of 11. Work out the rule, writing the input as x and the output as y.
- 9.y is directly proportional to √x. When x = 25, y = 20. Construct the equation connecting x and y, then work out the value of x when y = 32.
- 10.A sector of a circle has radius 6 cm and angle 150°. Using π = 3.14, work out the perimeter of the sector.
- 11.A charity raffle has three types of ticket: winning, near-miss and losing, and every ticket is exactly one of these. The probability that a ticket is winning is 1/8 and the probability that it is a near-miss is 1/4. Work out the probability that a ticket is losing.
- 12.To estimate the cost of buying 38.7 m of rope at £21.40 per metre, both numbers are first rounded to 1 significant figure. Work out the estimate.
- 13.Solve x² − 5x + 6 = 0 by factorising.
- 14.A garden centre has 90 rose bushes in stock. It receives a delivery that increases its stock of rose bushes by 20%. In the same week it sells 150 lavender bushes. Work out the new number of rose bushes, then write it as a fraction of the number of lavender bushes sold, giving your answer in its simplest form.
- 15.Triangle T has a vertex at (2, 1). It is enlarged by a scale factor of 3, centre (0, 0). Work out the coordinates of the image of this vertex.
- 16.Solve the inequality 6.3x + 1.8 ≤ 40. Hence write down the greatest number with exactly one decimal place that satisfies the inequality.
- 17.A taxi journey of m miles costs C pounds, where C = 2.5m + 3. A driver says the ratio C : m is the same for every journey. Is the driver correct? Give a reason for your answer.
- 18.A rectangular name card has a perimeter of 30 cm. The card is 9 cm long. Work out the width of the card.
- 19.The simultaneous equations kx + 2y = 4 and 3x + y = 5 have no solution. Work out the value of k.
- 20.Write down the coefficient of xy in the expression 4x²y − 7xy + 2.
Answer key
- (b) 3 — Method: solving an index equation like this means finding how many factors of the base multiply together to give the number on the right. Working: 4¹ = 4, 4² = 16 and 4³ = 64, so three factors of 4 are needed. Answer: 3. The distractors: 4 comes from listing 4, 16 and 64 and counting the base itself as a step, which gives one more than the index; 6 comes from solving the equation with 2 as the base instead of 4, since 2⁶ = 64; 16 comes from dividing 64 by 4, treating the index as an instruction to divide.
- (b) the first and the third — Method: the sign of a product depends only on the signs of the two numbers multiplied: like signs give a positive product and unlike signs a negative one, so look for the quadrants in which both coordinates carry the same sign. Working: in the first quadrant x and y are both positive, and positive times positive is positive; in the third quadrant both are negative, and negative times negative is positive as well; in the second quadrant x is negative while y is positive, and in the fourth x is positive while y is negative, so each of those gives a negative product. Answer: the first and the third. The distractors: 'the second and the fourth' comes from applying the rule for a NEGATIVE product, unlike signs, to a positive one; 'the first and the second' comes from testing only the y-coordinate and keeping the quadrants where the height is positive; 'the first and the fourth' comes from testing only the x-coordinate in the same way.
- (a) 4:3 — Multiply both parts by the lowest common denominator, 6: (2/3) × 6 = 4 and (1/2) × 6 = 3, giving the ratio 4 : 3, which is already in simplest form.
- (d) Opposite angles in a cyclic quadrilateral sum to 180° — ABCD is a cyclic quadrilateral, and angle ABC and angle ADC are a pair of opposite angles in it — they don't sit next to each other around the quadrilateral. The theorem that fixes the sum of a cyclic quadrilateral's opposite angles at 180° is exactly the one needed here. The other three theorems don't apply to this figure at all: the angle-in-a-semicircle theorem gives a 90° angle only when one side is a diameter, and no diameter is mentioned; the angle-at-the-centre theorem needs a centre point and an inscribed angle on the same arc, and no centre is given here; the tangent–radius theorem gives a 90° angle only where a tangent line touches the circle, and there is no tangent in this figure. Only the cyclic-quadrilateral theorem fits what is actually drawn.
- (d) No, because 3/8 + 5/12 + 1/6 = 23/24 — Using a common denominator of 24: 3/8 = 9/24, 5/12 = 10/24 and 1/6 = 4/24. Adding these numerators gives 9 + 10 + 4 = 23, so the three probabilities sum to 23/24, which is less than 1 — Zara is not correct. Adding the original numerators (3 + 5 + 1 = 9) over a denominator of 12 instead of converting each fraction properly gives 9/12 = 3/4, still less than 1 but the wrong fraction. Converting 1/6 to 5/24 instead of 4/24 (using the wrong scaling) makes the total 9/24 + 10/24 + 5/24 = 24/24 = 1, wrongly suggesting the probabilities are valid. Judging validity from the fact that each individual fraction lies between 0 and 1 ignores that an exhaustive set must sum to exactly 1, not merely contain valid individual values.
- (b) 74 marks — Method: the two groups are different sizes, so their means cannot simply be averaged — rebuild each group's total mark, add the totals and divide by all 50 pupils. Working: Group A scored 20 × 80 = 1600 marks and Group B scored 30 × 70 = 2100 marks, giving 1600 + 2100 = 3700 marks altogether, so the overall mean is 3700 ÷ 50 = 74 marks. Answer: 74 marks, which sits nearer to 70 than to 80 because the larger group scored 70. The distractors: 75 marks comes from averaging the two group means, (80 + 70) ÷ 2, as though the groups were the same size; 76 marks comes from attaching each mean to the other group's size, (20 × 70 + 30 × 80) ÷ 50; 150 marks comes from adding the two means together and never dividing at all.
- (a) 330 ml — Correct to the nearest 20 ml means the true volume could be up to 10 ml (half of 20) either side of 340 ml. The smallest possible volume is 340 − 10 = 330 ml. A candidate who subtracted the full 20 ml instead of half of it worked out 340 − 20 = 320 ml. A candidate who added instead of subtracted, finding the largest possible volume instead of the smallest, worked out 340 + 10 = 350 ml. A candidate who halved the interval again by mistake, using 5 ml instead of 10 ml, worked out 340 − 5 = 335 ml.
- (c) y = 3x + 2 — Method: divide the change in the outputs by the change in the inputs to find the multiplier, then put one pair of values into the rule to find the number added on. Working: the output rises by 11 − 5 = 6 while the input rises by 3 − 1 = 2, so the multiplier is 6 ÷ 2 = 3; with an input of 1, 3 × 1 = 3 and the output is 5, so 2 is added. Answer: y = 3x + 2, checked against the second pair by 3 × 3 + 2 = 11. The distractors: y = 3x − 2 comes from finding the multiplier 3 and then subtracting the 2 instead of adding it; y = 2x + 3 comes from swapping the multiplier and the number added on; y = x + 4 comes from assuming the input is multiplied by 1 and using 5 − 1 = 4 as the number added on.
- (d) 64 — Since y is directly proportional to √x, y = k√x. Using x = 25, y = 20: √25 = 5, so 20 = k × 5, giving k = 20 ÷ 5 = 4. The equation is y = 4√x. When y = 32: √x = 32 ÷ 4 = 8, and x = 8² = 64. Stopping at √x = 8 without squaring leaves the square root of x, not x itself. Treating the relationship as if y were proportional to x itself gives k = 20 ÷ 25 = 0.8 and then x = 32 ÷ 0.8 = 40, which is a different relationship entirely. Multiplying instead of dividing when isolating √x gives √x = 32 × 4 = 128, far too large to be a square root here. When y = 32, x = 64.
- (c) 27.7 cm — Arc length = (150 ÷ 360) × 2 × 3.14 × 6 = (5 ÷ 12) × 37.68 = 15.7 cm. The perimeter of a sector also includes the two straight radii, so perimeter = 15.7 + 6 + 6 = 27.7 cm. (15.7 cm comes from stopping after the arc length and forgetting the two straight edges; 21.7 cm comes from adding only one radius instead of two; 31.4 cm comes from doubling the arc length instead of adding the two straight edges.)
- (a) 5/8 — Winning, near-miss and losing are exhaustive, so the three probabilities sum to 1. Writing 1/4 as 2/8 so every fraction has the same denominator, 1 − 1/8 − 2/8 = 8/8 − 1/8 − 2/8 = 5/8. Subtracting only the winning probability and forgetting the near-miss probability gives 1 − 1/8 = 7/8. Subtracting only the near-miss probability and forgetting the winning probability gives 1 − 1/4 = 3/4. Adding the two given probabilities and stopping there gives 1/8 + 2/8 = 3/8, the probability that a ticket is winning or a near-miss, not the probability that it is losing.
- (d) £800 — Rounding 38.7 to 1 significant figure gives 40, and rounding 21.40 to 1 significant figure gives 20. Multiplying the rounded values gives an estimate of 40 × 20 = £800. Rounding 21.40 to the nearest whole number instead of to 1 significant figure gives 21, and 40 × 21 = £840, one place value too fine for the price. Adding the rounded values instead of multiplying them gives 40 + 20 = £60. Rounding both numbers to 2 significant figures instead of 1, giving 39 and 21, produces 39 × 21 = £819.
- (c) x = 2 or x = 3 — Method: factorise into two brackets whose numbers multiply to the constant term and add to the coefficient of x, then set each bracket equal to zero. Working: two numbers that multiply to 6 and add to −5 are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3) = 0; then x − 2 = 0 gives x = 2 and x − 3 = 0 gives x = 3. Answer: x = 2 or x = 3. The distractors: x = −2 or x = −3 comes from reading the numbers inside the brackets as the solutions instead of changing their signs; x = 1 or x = 6 comes from taking the first factor pair of 6 without checking that the pair adds to −5; x = 5 or x = 6 comes from reading the solutions straight off the 5 and the 6 in the equation.
- (a) 18/25 — First find the new number of rose bushes: 90 × 1.2 = 108 (a 20% increase multiplies by 1.2). Then write 108 over 150 and divide top and bottom by 6 to get 18/25. Choosing 3/5 comes from using the original 90 rose bushes without applying the 20% increase (90/150 = 3/5). Choosing 25/18 comes from writing the number of lavender bushes over the new number of rose bushes, the wrong way round. Choosing 3/25 comes from multiplying 90 by 0.2 instead of 1.2, finding only the increase (18) rather than the new total, then writing 18/150 = 3/25.
- (d) (6, 3) — For an enlargement centred at the origin, each coordinate is multiplied by the scale factor: (2, 1) → (2 × 3, 1 × 3) = (6, 3). ((5, 4) comes from adding the scale factor to each coordinate instead of multiplying; (6, 1) comes from multiplying only the x-coordinate by 3 and leaving the y-coordinate unchanged; (2, 3) comes from multiplying only the y-coordinate by 3 and leaving the x-coordinate unchanged.)
- (d) 6.0 — Subtract 1.8 from both sides: 6.3x ≤ 38.2. Divide both sides by 6.3: x ≤ 6.0634… . The question asks for the greatest value to 1 decimal place that still satisfies the inequality. Testing 6.1: 6.3 × 6.1 + 1.8 = 40.23, which is more than 40, so 6.1 fails. Testing 6.0: 6.3 × 6.0 + 1.8 = 39.6, which is no more than 40, so 6.0 works and is the greatest such value. A candidate who simply rounds 6.0634… to 1 decimal place answers 6.1, without checking that it satisfies the inequality. A candidate who adds 1.8 instead of subtracting works out 41.8 ÷ 6.3 and answers 6.6. A candidate who forgets the 1.8 altogether divides 40 by 6.3 and answers 6.3.
- (b) The ratio C : m is not constant because the formula includes a fixed charge of £3 as well as the charge per mile. — For the ratio C : m to stay constant, C must be directly proportional to m, i.e. C = km with no constant term. Because of the +3 fixed charge, C is not directly proportional to m: for example m = 1 gives C = 5.5 (ratio 5.5 : 1), while m = 10 gives C = 28 (ratio 2.8 : 1) — the ratio has changed.
- (a) 6 cm — Method: a rectangle is made of two lengths and two widths, so the perimeter is 2 × (length + width). Half the perimeter is therefore one length plus one width, and the width is what is left of that half once the length is taken away. Working: 30 ÷ 2 = 15 cm is one length added to one width, and 15 − 9 = 6. Answer: 6 cm. The distractors: 21 cm comes from taking the length off the whole perimeter, 30 − 9, so three sides are still counted in the figure that is written down; 12 cm comes from removing both lengths correctly, 30 − 2 × 9 = 12, and then recording that remainder as the width instead of halving it, because it is in fact the two widths together; 7.5 cm comes from dividing the perimeter by four, 30 ÷ 4, which is the calculation for the side of a square and throws away the length that was given.
- (b) k = 6 — Method: two simultaneous linear equations have no solution when the lines they describe are parallel, so write each equation in the form y = mx + c and make the gradients equal. Working: kx + 2y = 4 rearranges to y = −(k/2)x + 2, so its gradient is −k/2, and 3x + y = 5 rearranges to y = −3x + 5, so its gradient is −3; setting −k/2 = −3 gives k = 6, and the first equation is then 6x + 2y = 4, which simplifies to 3x + y = 2 and can never agree with 3x + y = 5. Answer: k = 6. The distractors: k = −6 comes from reading the gradient of kx + 2y = 4 as +k/2 and solving k/2 = −3; k = 3 comes from making the x terms identical instead of making the gradients equal; k = 2/3 comes from writing the gradient of 3x + y = 5 upside down as −1/3 and solving −k/2 = −1/3.
- (a) −7 — The coefficient of xy is the number multiplying that exact term, including its sign: in 4x²y − 7xy + 2, the term in xy is −7xy, so the coefficient is −7. Answering 7 correctly picks out the number but drops the minus sign that belongs to it. Answering 4 picks the coefficient of the x²y term instead of the xy term — x²y and xy are different terms because their powers of x differ. Answering −5 combines −7 with the constant term +2, as though the coefficient of xy included the number with no letter attached to it. The coefficient of xy is −7.
How the 20 questions are shared out
- Number — 3 questions (15% of the qualification)
- Algebra — 6 questions (30% of the qualification)
- Ratio, proportion and rates of change — 4 questions (20% of the qualification)
- Geometry and measures — 4 questions (20% of the qualification)
- Probability — 2 questions (7.5% of the qualification)
- Statistics — 1 question (7.5% of the qualification)
Where an area has fewer printable questions than its share, the shortfall is filled from the other areas. These are original questions, not past papers.