Sample paper · GCSE Higher · grades 4–9

GCSE Higher sample Paper 1 (non-calculator)

The real Paper 1 is 1.5 hour 30 minutes and 80 marks, non-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 no calculator-only items. AO1 / AO2 / AO3 at this tier: 40% / 30% / 30%.

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GCSE Higher sample Paper 1 (non-calculator)

Maths · GCSE Higher · Paper 1 · Non-calculator · 20 questions

MathsUKwww.geekhero.co.uk

1.Given that 2⁵ = 32, work out 2⁶.
(a)34
(b)37
(c)192
(d)64
2.A hiker's distance-time graph shows the following: he walks 5 km in the first 1 hour at a steady speed, rests for 1 hour, then walks a further 9 km in the next 2 hours, also at a steady speed. Work out his average speed for the whole journey, in km/h.
(a)3.5 km/h
(b)4.75 km/h
(c)4.67 km/h
(d)2.25 km/h
3.A company's turnover this year is £180,000. Last year's turnover was £120,000. Write down this year's turnover as a percentage of last year's turnover.
(a)50%
(b)180%
(c)150%
(d)66.7%
4.A rhombus has all four sides equal in length, but it is not a square — its angles are not all 90°. How many lines of symmetry does this rhombus have?
(a)2
(b)1
(c)4
(d)0
5.A fair six-sided dice is rolled 90 times. Work out how many more times you would expect it to land on a number less than 4 than on a 6.
(a)45
(b)36
(c)30
(d)15
6.A sports shop in Cardiff sold 40 pairs of football boots last month: 4 pairs of size 6, 5 pairs of size 7, 8 pairs of size 8, 13 pairs of size 9 and 10 pairs of size 10. The manager will order 40 pairs for next month and wants as many pairs as possible to be in a size customers will buy. Work out the mean size and the modal size, and write down which of the two he should use.
(a)The modal size, 9, bought by more customers than any other
(b)The range of the sizes, 4, which covers every size sold
(c)The mean size, 8, the average of the five sizes on sale
(d)The mean size, 8.5, which takes account of all 40 pairs
7.Write 90 as a product of its prime factors.
(a)2 × 3² × 5
(b)2 × 45
(c)2 × 3 × 15
(d)3 × 3 × 10
8.Write down the expression that means the same as 'subtract 5 from n, then divide the result by 2'.
(a)(n − 5)/2
(b)n/2 − 5
(c)2(n − 5)
(d)(5 − n)/2
9.A mobile phone tariff is shown on a straight-line graph with the monthly cost, C pounds, on the vertical axis and the amount of data used, g gigabytes, on the horizontal axis. The line passes through (0, 10) and (8, 26). Work out the gradient and say what it represents.
(a)3.25, the cost in pounds of each extra gigabyte
(b)2, the cost in pounds of each extra gigabyte
(c)10, the cost in pounds of each extra gigabyte
(d)2, the fixed monthly charge in pounds
10.A square-based pyramid is cut by a flat plane parallel to its base, partway up between the base and the apex. What shape is the cross-section?
(a)Triangle
(b)Trapezium
(c)Square
(d)Pentagon
11.In a class experiment a fair six-sided dice is rolled 60 times and lands on a six 14 times. The whole school then rolls the same fair dice 3000 times. Work out the best estimate of the number of sixes the school should expect.
(a)600
(b)500
(c)2500
(d)700
12.Work out 6² − 4².
(a)2
(b)52
(c)20
(d)4
13.The nth term of a sequence is n² + 4n. Work out the term number, n, for which the term equals 45.
(a)11
(b)5
(c)7
(d)9
14.A shop's daily profit, in pounds, is plotted against the price charged for an item, in pounds. A tangent to the profit graph at a price of £15 passes through the points (12, 540) and (18, 480). Work out the instantaneous rate of change of profit with respect to price at £15, and state whether profit is increasing or decreasing there.
(a)Profit is increasing by £6 per £1 rise in price.
(b)Profit is increasing by £10 per £1 rise in price.
(c)Profit is decreasing by £60 per £1 rise in price.
(d)Profit is decreasing by £10 per £1 rise in price.
15.Shape S is enlarged by a scale factor of −3 about a fixed centre. Which statement correctly describes the image compared to the original shape?
(a)3 times the size, opposite side, rotated 180°
(b)3 times the size, same side as the original
(c)3 times the size, same orientation, just further away
(d)1/3 the size, opposite side of the centre
16.The curve y = −x² + 6x − 5 has a maximum point. Use completing the square to find its coordinates.
y = −x² + 6x − 5
(a)x = 3, y = −4
(b)x = 3, y = 4
(c)x = 6, y = 31
(d)x = −3, y = 4
17.y is directly proportional to x. When x = 7, the value of y is 21. Work out the value of x when y = 12.
(a)4
(b)36
(c)84
(d)9
18.A sailmaker cuts two triangular sail panels from a pattern. Panel X has a 10 m side, with angles of 50° and 75° at its two ends. Panel Y also has a 10 m side, with angles of 50° and 75° at its two ends, in the same arrangement. The sailmaker wants to check the panels will match exactly, without cutting a third measurement. Which condition proves the two panels are congruent?
(a)SAS
(b)RHS
(c)ASA
(d)SSS
19.The graph of y = mx + 6 passes through the point (−2, 0). Work out the value of m.
(a)−3
(b)−2
(c)3
(d)6
20.A candle is 30 cm tall when lit and burns at a constant rate. After burning for 5 minutes, its height is 20 cm. The height h cm after t minutes is given by h = 30 − mt. Work out the value of m.
(a)2
(b)4
(c)10
(d)0.5

Answer key

(d) 64 — Method: each time the power increases by 1, the value doubles, since one more 2 is multiplied in. Working: 2⁶ = 2⁵ × 2 = 32 × 2 = 64. Answer: 64. (37 comes from adding the power to the value, 32 + 5, instead of multiplying by the base. 192 comes from multiplying 32 by the new power, 6, instead of by the base, 2. 34 comes from adding the base to the value, 32 + 2, instead of multiplying.)
(a) 3.5 km/h — Average speed is total distance ÷ total time. Total distance = 5 + 9 = 14 km. Total time, including the rest, = 1 + 1 + 2 = 4 hours. So average speed = 14 ÷ 4 = 3.5 km/h. Leaving out the 1 hour rest and dividing by only the 3 hours of walking gives 14 ÷ 3 = 4.67 km/h. Averaging the two separate speeds, 5 km/h and 4.5 km/h, instead of using total distance over total time, gives (5 + 4.5) ÷ 2 = 4.75 km/h. Dividing only the second leg's distance by the total time, 9 ÷ 4 = 2.25 km/h, ignores the first leg's distance.
(c) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
(a) 2 — Method: recall that the diagonals of a rhombus are always lines of symmetry, whatever its angles are. Working: a rhombus (all sides equal) always has its two diagonals as lines of symmetry, giving 2 lines of symmetry, whether or not the angles are 90°. Options: 0 wrongly assumes a non-square rhombus has no symmetry at all; 4 comes from the number of lines of symmetry a square has, mistaking this rhombus for a square; 1 comes from treating the rhombus like a kite, which has only one diagonal as a line of symmetry. Answer: 2.
(c) 30 — The numbers less than 4 are 1, 2 and 3, so the probability of that event is 3/6, and the expected count in 90 rolls is 90 × 3/6 = 45. The probability of rolling a 6 is 1/6, and the expected count is 90 × 1/6 = 15. The difference between the two expected counts is 45 − 15 = 30. A candidate who answers 45 has given the expected count for 'less than 4' only, forgetting to subtract the other expected count. A candidate who answers 15 has given the expected count for '6' only. A candidate who answers 36 has used a dice with 5 possible numbers instead of 6, giving 90 × 3/5 = 54 and 90 × 1/5 = 18, a difference of 36.
(a) The modal size, 9, bought by more customers than any other — Method: work out both averages from the frequencies, then choose the one the shop can act on. Working: for the mean, multiply each size by the number of pairs sold at it and add: 6 × 4 + 7 × 5 + 8 × 8 + 9 × 13 + 10 × 10 = 340, and 340 ÷ 40 = 8.5, so the mean size is 8.5. The largest frequency is 13, which belongs to size 9, so the modal size is 9. The mean 8.5 is a size no customer in the record asked for, so 40 pairs of it would sit unsold, while 13 of the 40 customers wanted size 9, more than wanted any other size. Answer: the modal size, 9, bought by more customers than any other. The distractors: the mean size 8.5 does take account of all 40 pairs, but a mean of sizes is a summary figure and not a size the month's customers were buying; the mean size 8 comes from averaging the five sizes on sale, 6 + 7 + 8 + 9 + 10 = 40 and 40 ÷ 5 = 8, which ignores how many pairs were sold at each size and so treats the 4 pairs of size 6 as equal in weight to the 13 pairs of size 9; the range 4 comes from 10 − 6 and measures spread, so it says how wide a set of sizes the shop must stock, not which size to stock most of.
(a) 2 × 3² × 5 — Method: divide repeatedly by the smallest prime number until only prime factors remain. Working: 90 ÷ 2 = 45, 45 ÷ 3 = 15, 15 ÷ 3 = 5, and 5 is prime, so 90 = 2 × 3 × 3 × 5, written as 2 × 3² × 5. 2 × 3 × 15 stops before the 15 is broken down into 3 × 5, so it is not fully factorised. 3 × 3 × 10 stops before the 10 is broken down into 2 × 5. 2 × 45 stops after only one division. Answer: 2 × 3² × 5.
(a) (n − 5)/2 — Method: do the subtraction first, keep it together as a single bracket, then divide that whole bracket by 2. Working: 'subtract 5 from n' is (n − 5); 'divide the result by 2' means the whole bracket goes over 2, giving (n − 5)/2. Answer: (n − 5)/2. n/2 − 5 comes from dividing n by 2 first and only then subtracting 5, the wrong order. 2(n − 5) comes from multiplying by 2 instead of dividing. (5 − n)/2 comes from subtracting n from 5 instead of subtracting 5 from n, the wrong way round.
(b) 2, the cost in pounds of each extra gigabyte — Method: the gradient is the change in cost divided by the change in data, so it is the cost of each extra gigabyte; the value where the line meets the vertical axis is the charge before any data is used, which is a different quantity. Working: from (0, 10) to (8, 26) the cost rises by 26 − 10 = 16 pounds while the data rises by 8 − 0 = 8 gigabytes, so the gradient is 16 ÷ 8 = 2, meaning each extra gigabyte costs £2. Answer: 2, the cost in pounds of each extra gigabyte. The distractors: '10, the cost in pounds of each extra gigabyte' reads the intercept as the gradient, but 10 is what the tariff costs when no data at all has been used; '3.25, the cost in pounds of each extra gigabyte' comes from 26 ÷ 8, treating the line as though it passed through the origin when it starts at 10; '2, the fixed monthly charge in pounds' has the gradient right but describes the intercept, and the fixed charge on this tariff is £10.
(c) Square — Because the cut is parallel to the square base, it produces a shape that is similar to the base — the same shape, scaled down — so it is also a square, with four equal sides, smaller than the base. A candidate who looks at the sloped triangular faces instead of at the horizontal cut answers triangle. A candidate who pictures a slanted cut, which would shorten one pair of sides, answers trapezium, but a cut parallel to the base scales all four sides by the same amount. A candidate who counts an extra side, perhaps including an edge of the base, answers pentagon. The cross-section is a square.
(b) 500 — Method: when a dice is known to be fair, the theoretical probability is the best thing to work from, and the more trials there are the closer the results tend to it. Working: for a fair dice the probability of a six is 1/6, so the expected number of sixes in 3000 rolls is 3000 × 1 ÷ 6 = 500. The class experiment gave a relative frequency of 14/60, but 60 trials is far too few to overturn a known theoretical value, and the school's 3000 rolls will tend towards 1/6 in any case. Answer: about 500 sixes. The distractors: 700 comes from using the class relative frequency instead of the theory, 3000 × 14 ÷ 60 = 700; 600 comes from splitting the difference between the two, since 1/6 is about 0.167 and 14/60 is about 0.233, whose mean is 0.2, and 3000 × 0.2 = 600; 2500 uses 5/6 instead of 1/6 and counts the rolls expected not to be a six.
(c) 20 — Method: work out each power separately before subtracting. Working: 6² = 36 and 4² = 16, so 6² − 4² = 36 − 16 = 20. Answer: 20. (4 comes from subtracting first, 6 − 4 = 2, and then squaring that result, instead of squaring each number first. 52 comes from adding the two squares, 36 + 16, instead of subtracting them. 2 comes from subtracting the two numbers, 6 − 4, and forgetting to square at all.)
(b) 5 — Set n² + 4n = 45, so n² + 4n − 45 = 0. This factorises as (n + 9)(n − 5) = 0, giving n = −9 or n = 5. Since a term number must be positive, n = 5. Taking the magnitude of the rejected negative solution, 9, instead of discarding it, gives 9. Dividing 45 by the coefficient of n and ignoring the n² term entirely, 45 ÷ 4 = 11.25, rounded to the nearest whole number, gives 11. Dropping the linear term 4n and solving n² = 45 instead, the nearest whole number to √45 = 6.708 is 7.
(d) Profit is decreasing by £10 per £1 rise in price. — The gradient of a tangent gives the instantaneous rate of change of profit with respect to price, found from the change in profit divided by the change in price between two points on the tangent. Here the tangent passes through (12, 540) and (18, 480), so the change in profit is 480 − 540 = −60 and the change in price is 18 − 12 = 6. The gradient is −60 ÷ 6 = −10. A negative gradient means profit is decreasing as price increases, so profit is decreasing at an instantaneous rate of £10 for every £1 rise in price. Subtracting the profits in the wrong order, 540 − 480 = 60, and dividing by the same change in price, 60 ÷ 6 = 10, gives a positive value and the wrong direction — profit is not increasing at £15. Stopping after finding only the change in profit, 480 − 540 = −60, without dividing by the change in price, is not a rate at all. Reading off the change in price, 6, and calling it the rate gives £6 per £1 rise in price, but 6 is only the width of the price interval — it is not a change in profit at all, and profit falls across that interval, so the direction is wrong too. The instantaneous rate of change of profit with respect to price at £15 is a decrease of £10 per £1 rise in price.
(a) 3 times the size, opposite side, rotated 180° — Method: for any enlargement, the MAGNITUDE of the scale factor gives the size ratio between image and object, while the SIGN decides which side of the centre the image falls on; a negative scale factor puts the image on the opposite side, which is equivalent to a 180° rotation about the centre. Working: the scale factor is −3, so the size ratio is the magnitude, which is 3, and the negative sign puts the image on the opposite side of the centre, rotated 180° relative to the original. Answer: 3 times the size, opposite side, rotated 180°. The magnitude of the scale factor controls the SIZE only: do not let the sign leak into it and turn 3 into 1/3. The sign controls the SIDE and orientation, which a positive-only view of enlargement, just 'further away' with the same orientation, misses entirely.
(b) x = 3, y = 4 — −x² + 6x − 5 = −(x² − 6x) − 5 = −[(x − 3)² − 9] − 5 = −(x − 3)² + 9 − 5 = −(x − 3)² + 4. Because the coefficient of x² is negative, −(x − 3)² is at most zero, so this turning point is a maximum. Substituting x = 3 gives (x − 3)² = 0, so y = 4, confirming the maximum value 4 at x = 3: turning point (3, 4). Using 6 instead of half of 6 inside the bracket — forgetting to halve before completing the square — lands on turning point (6, 31), wrong, because only half the coefficient belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (−3, 4) — wrong, because (x − 3)² is zero at x = 3, not x = −3. Computing 5 − 9 = −4 instead of 9 − 5 = 4 flips the sign of the constant, giving (3, −4) — wrong, since the constant must be evaluated as 9 minus 5, not 5 minus 9. Whenever the leading coefficient is negative, the turning point is a maximum, not a minimum — check by substituting back into the original equation.
(a) 4 — Method: find the constant of proportionality from the pair given, write the equation, then substitute the new value of y and solve. Working: k = 21 ÷ 7 = 3, so y = 3x; putting y = 12 gives 12 = 3x, and x = 12 ÷ 3 = 4. Answer: 4. The distractors: 36 comes from multiplying by the constant instead of dividing by it, 12 × 3, which is the proportion set up upside down; 84 comes from multiplying 12 by the 7 from the first pair, using a value of x as though it were the constant; 9 comes from working out 12 − 3, treating the equation as y = x + 3 rather than y = 3x.
(c) ASA — Method: check which condition matches two angles and the side between them, since that is all the sailmaker has measured. Working: the 10 m side lies between the 50° and 75° angles in both panels, so this is two Angles and the included Side, ASA. Options: SAS would need two sides and the angle between them, but only one side has been measured here; SSS would need three sides, but only one is known; RHS needs a right angle and a hypotenuse, and neither panel has a stated right angle. Answer: ASA.
(c) 3 — Method: a point that lies on a graph makes its equation true, so substituting the coordinates into y = mx + 6 leaves an equation in m alone. Working: substituting x = −2 and y = 0 gives 0 = m × (−2) + 6, which rearranges to −2m = −6, so m = (−6) ÷ (−2) = 3. Answer: 3. The distractors: −3 comes from dividing −6 by 2 and keeping the minus sign, overlooking that the divisor is negative too, so the two signs cancel; −2 comes from writing down the x-coordinate of the given point in place of the gradient; 6 comes from reading the constant in y = mx + 6 as the gradient, confusing m with c.
(a) 2 — The candle's height falls from 30 cm to 20 cm, a drop of 10 cm, over 5 minutes, so m = 10 ÷ 5 = 2. 10 comes from using the drop in height but forgetting to divide by the time. 0.5 comes from dividing the time by the drop instead of the drop by the time (5 ÷ 10). 4 comes from dividing the final height by the time, 20 ÷ 5, instead of using the drop in height.

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.