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.A rectangular plywood panel measures 2.4 m by 0.75 m. Work out the area of the panel in square metres, giving your answer as a fraction in its simplest form.
(a)4/5
(b)63/20
(c)9/5
(d)9/50
2.The function f(x) = x² for all real values of x has no inverse function, but g(x) = x² for x ≥ 0 does have one. Which statement correctly explains this?
y = x²
(a)Restricting any domain always creates an inverse
(b)Squares can never be reversed, under any conditions
(c)g's outputs are positive; f's could be negative
(d)g is one-to-one: f(3) = f(−3), removed by x ≥ 0
3.The amount of fuel left in a car's tank, F litres, is plotted against the distance travelled, d miles, and the points lie on a straight line. The line passes through (0, 45) and (150, 15). Work out the gradient of the line and say what it tells you.
(a)−5, the car uses 5 litres of fuel for each mile
(b)−30, the car uses 30 litres of fuel for each mile
(c)−0.2, the car uses 0.2 litres of fuel for each mile
(d)0.2, the car gains 0.2 litres of fuel for each mile
4.A cuboid has three different edge lengths: a length, a width and a height, all different from each other. How many rectangles make up its net in total?
(a)6
(b)5
(c)12
(d)3
5.A group of 80 pupils was asked whether they walk to school. 45 of the pupils are girls and 35 are boys. 30 of the girls walk to school and 14 of the boys walk to school. One of the girls is picked at random. Work out the probability that she walks to school. Give your answer in its simplest form.
(a)3/8
(b)11/20
(c)1/3
(d)2/3
6.A call centre records the length, t seconds, of 100 calls: 0 ≤ t < 20, 15 calls; 20 ≤ t < 30, 24 calls; 30 ≤ t < 50, 40 calls; 50 ≤ t < 80, 21 calls. The manager's target is for a call to be finished in under 35 seconds. Estimate the number of calls that met the target.
(a)49
(b)79
(c)69
(d)39
7.Work out an estimate for 588 ÷ 31, by rounding each number to 1 significant figure.
(a)20
(b)19
(c)17
(d)200
8.A rectangular garden has length (x + 7) m and width (x − 7) m. Work out an expression for the area of the garden, giving your answer in its simplest form.
(a)2x
(b)x² + 49
(c)x² − 14x − 49
(d)x² − 49
9.The speed of a car is given by v = d ÷ t. A car travels 180 km in 3 hours. Work out the average speed of the car in km/h.
(a)90 km/h
(b)540 km/h
(c)18 km/h
(d)60 km/h
10.Which of these statements about solving a triangle is correct?
(a)Two sides + a non-included angle: cosine rule gives the angle directly.
(b)Two sides + angle between them: use the sine rule.
(c)All three sides: use the sine rule for any angle.
(d)Two angles and a side: sine rule finds other sides.
11.At a fair, a game costs £2 to play. The probability of winning the game is 0.15, and each win pays out £10. Amir plays the game 200 times. Work out how much money Amir should expect to lose in total.
(a)£300
(b)£100
(c)£700
(d)£400
12.Two-digit numbers are made using the digits 1, 2, 3 and 4, and no digit may be used twice in the same number. Work out how many different two-digit numbers can be made.
(a)12
(b)6
(c)16
(d)8
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.There are 84 passengers on a double-decker bus. 35 of the passengers are adults and the rest are children. Write the number of adults as a fraction of the number of children. Give your answer in its simplest form.
(a)2/7
(b)5/12
(c)5/7
(d)7/5
15.Line 1 passes through (0, 1) and (2, 5). Line 2 passes through (3, 2) and (5, 6). Work out the gradient of each line. Then write down whether the two lines are parallel.
(a)Parallel, since both gradients are 2
(b)Not parallel, since the gradients are 2 and 4
(c)Parallel, since both lines pass through positive coordinates
(d)Not parallel, since the two lines cross the y-axis at different points
16.Solve the inequality 2(3x − 1) ≥ 4x + 8.
(a)x ≥ 4.5
(b)x ≥ 5
(c)x ≥ 1
(d)x ≥ 20
17.A straight-line graph shows the number of pages printed, p, plotted against the time in minutes, t, since a printer started a print job. The line passes through the points (2, 30) and (5, 90). Work out the gradient of the line.
(a)18
(b)20
(c)0.05
(d)−20
18.A regular polygon has 12 sides. Work out the size of one exterior angle of the polygon.
(a)30°
(b)150°
(c)15°
(d)36°
19.f(x) = (x + 1)/2. Find f⁻¹(x).
(a)2x − 1
(b)(x − 1)/2
(c)x/2 − 1
(d)2x + 1
20.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

Answer key

(c) 9/5 — Method: the area of a rectangle is its length multiplied by its width; the decimal product is then written over a power of ten and cancelled. Working: 24 × 75 = 1800, and 2.4 and 0.75 have three decimal places between them, so 2.4 × 0.75 = 1.8; the area of the panel is 1.8 square metres, which is eighteen tenths, so it can be written as 18/10, and dividing the numerator and the denominator by 2 gives 9 over 5. Answer: 9/5. The distractors: 4/5 comes from converting only the digits after the decimal point and losing the whole one, turning 1.8 into eight tenths; 63/20 comes from adding the two sides instead of multiplying them, giving 3.15; 9/50 comes from misplacing the decimal point in the product and writing 0.18, which cancels to 9 over 50.
(d) g is one-to-one: f(3) = f(−3), removed by x ≥ 0 — A function has an inverse only if it is one-to-one: every output must come from exactly one input. f(3) = 9 and f(−3) = 9, so two different inputs give the same output, and there is no way to send 9 back to a single input — f is not one-to-one over all real x. Restricting the domain to x ≥ 0 removes one of the two inputs behind every such pair, so g is one-to-one and does have an inverse. 'g's outputs are positive; f's could be negative' is wrong because f(x) = x² also only gives outputs of 0 or more — the outputs of f and g are identical sets; it is the INPUTS that differ, not the outputs. 'Restricting any domain always creates an inverse' is wrong because a restriction only helps if it actually removes the repeated outputs: restricting f(x) = x² to x ≥ −3 still leaves f(1) = f(−1) = 1, so that restricted function is still not one-to-one and still has no inverse. 'Squares can never be reversed, under any conditions' is wrong because √9 = 3 does reverse 3² = 9 once you know the input was non-negative — a square root just cannot tell you WHICH of two inputs you started from unless the domain has already ruled one of them out.
(c) −0.2, the car uses 0.2 litres of fuel for each mile — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, which on this graph is a number of litres for each mile, and a negative gradient means the vertical quantity is going down. Working: from (0, 45) to (150, 15) the fuel changes by 15 − 45 = −30 litres while the distance changes by 150 − 0 = 150 miles, so the gradient is −30 ÷ 150 = −0.2, which says the tank loses 0.2 litres for every mile driven. Answer: −0.2, the car uses 0.2 litres of fuel for each mile. The distractors: '0.2, the car gains 0.2 litres of fuel for each mile' comes from subtracting the fuel values the other way round, 45 − 15 = 30, which drops the minus sign and reverses what the graph says; '−5, the car uses 5 litres of fuel for each mile' comes from dividing the change in distance by the change in fuel, 150 ÷ (−30), turning the gradient upside down; '−30, the car uses 30 litres of fuel for each mile' is the change in fuel on its own, never divided by the 150 miles travelled.
(a) 6 — A cuboid has six faces in total: a top, a bottom, a front, a back, and two ends — each one is a rectangle in the net, giving six rectangles altogether. Choosing 3 counts only the three PAIRS of congruent rectangles (top/bottom, front/back, two ends) rather than all six individual faces. Choosing 5 forgets one face, as if the net were missing its lid. Choosing 12 is the number of edges of a cuboid, not the number of rectangles in its net.
(d) 2/3 — Method: the pupil picked is known to be a girl, so the sample space is the 45 girls and not all 80 pupils; divide the number of girls who walk by the number of girls. Working: 30 of the 45 girls walk to school, which gives 30/45. Dividing the numerator and the denominator by 15 gives 2/3. Answer: the probability is 2/3. The distractors: 3/8 is 30/80, dividing the girls who walk by every pupil in the group, which is the commonest slip on a conditional probability because it keeps the restricted numerator but the whole denominator; 1/3 is 15/45, counting the 15 girls who do not walk to school and so answering the opposite event inside the correct group; 11/20 is 44/80, adding the 30 girls and the 14 boys who walk and dividing by the whole group, which throws away the information that the pupil picked is a girl.
(a) 49 — Method: add the frequencies of the classes that lie wholly below 35 seconds, then use linear interpolation for the class that 35 cuts through, assuming the calls in that class are spread evenly across it. Working: below 30 seconds there are 15 + 24 = 39 calls; the value 35 lies in the class 30 ≤ t < 50, which is 20 seconds wide and holds 40 calls, and 35 is 35 − 30 = 5 seconds into it, so the estimated share is (5 ÷ 20) × 40 = 10 calls; the estimate is 39 + 10 = 49. Answer: about 49 calls met the target. The distractors: 79 comes from adding the whole of the class 30 ≤ t < 50, 39 + 40, and so counting calls of up to 50 seconds as being under 35; 39 comes from stopping at the class boundary 30 and ignoring the part class altogether; 69 comes from measuring the part of the class from 35 up to 50 instead of from 30 up to 35, giving (15 ÷ 20) × 40 = 30 and then 39 + 30.
(a) 20 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 588 rounds to 600 (1 s.f.) and 31 rounds to 30 (1 s.f.). 600 ÷ 30 = 20. Answer: 20. 17 comes from cutting 588 down to 500, keeping the leading digit as it stands instead of rounding it up to 1 significant figure, 600, then dividing by the correctly rounded 30. 200 comes from misreading the rounded divisor 30 as 3, giving 600 ÷ 3 instead of 600 ÷ 30. 19 is the exact value of 588 ÷ 31 rounded to the nearest whole number, found without rounding the numbers first.
(d) x² − 49 — Area = length × width = (x + 7)(x − 7). Expanding: x × x = x², x × (−7) = −7x, 7 × x = 7x, 7 × (−7) = −49. The two middle terms −7x and 7x cancel, leaving x² − 49. A candidate who misremembers the difference-of-two-squares result as a sum gets x² + 49. A candidate who makes a sign error and treats both middle terms as −7x instead of cancelling gets x² − 14x − 49. A candidate who confuses area with perimeter and simply adds the length and width gets 2x.
(d) 60 km/h — Method: use the formula v = d ÷ t with the distance and time given. Working: 180 ÷ 3 = 60 km/h. So the average speed is 60 km/h. Distractor 540 km/h comes from multiplying the distance and time instead of dividing. Distractor 90 km/h comes from dividing by 2 instead of 3. Distractor 18 km/h comes from dividing by 10 instead of 3, a decimal-point slip.
(d) Two angles and a side: sine rule finds other sides. — Method: match the data you are given to the rule that needs it. Working: the sine rule a/sin A = b/sin B = c/sin C needs a complete angle-side pair to set up its ratio, so the statement that two angles and a side (AAS or ASA) let the sine rule find the other sides is the correct one — the third angle comes from the angle sum, and each unknown side is then opposite a known angle. The statement that two sides and the angle between them (SAS) call for the sine rule is wrong: no angle-side pair is complete, so the cosine rule is what works there. The statement that the sine rule finds any angle from three sides (SSS) is wrong for the same reason in reverse — no angle is known at all, so the cosine rule must find the first one. The statement that the cosine rule finds a missing angle directly from two sides and a non-included angle (SSA) is wrong: the cosine rule reports the angle enclosed by the two sides it uses, so with SSA it is the sine rule that reaches the missing angle, and the ambiguous case is then settled from the wording of the question.
(b) £100 — Method: find the expected number of wins, turn that into the expected pay out, then compare it with what the games cost. Working: the expected number of wins is 200 × 0.15 = 30. Each win pays £10, so the expected pay out is 30 × 10 = 300 pounds. Playing 200 times at £2 a go costs 200 × 2 = 400 pounds. The expected loss is 400 − 300 = 100 pounds. Answer: Amir should expect to be about £100 down. The distractors: £300 is the expected winnings on their own, with the cost of playing never taken off; £400 is the total cost of playing, with the winnings never taken off; £700 comes from adding the two totals, 400 + 300 = 700, instead of subtracting one from the other.
(a) 12 — Method: build the number one place at a time, listing systematically: fix the tens digit, then run through every units digit that is still available. Working: any of the 4 digits can go in the tens place, and once it has been used only 3 digits are left for the units place, so there are 4 × 3 = 12 numbers; listing the numbers that begin with 1 gives 12, 13 and 14, and each of the other three starting digits gives 3 numbers in the same way. Answer: 12. The distractors: 16 comes from working out 4 × 4, which allows a digit to be used twice; 8 comes from multiplying the 4 digits by the 2 places in the number instead of multiplying the choices available at each place; 6 comes from treating a number and its reverse as the same, counting only the unordered pairs of digits.
(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.
(c) 5/7 — First find the number of children: 84 − 35 = 49. The question compares the adults with the children, not with everyone on the bus, so the denominator is 49 and the numerator is 35, giving 35/49. Both parts divide by 7: 35 ÷ 7 = 5 and 49 ÷ 7 = 7. In its simplest form the fraction is 5/7.
(a) Parallel, since both gradients are 2 — Gradient of line 1 = (5 − 1) ÷ (2 − 0) = 4 ÷ 2 = 2. Gradient of line 2 = (6 − 2) ÷ (5 − 3) = 4 ÷ 2 = 2. The two gradients are equal, so the lines are parallel. "Not parallel, since the gradients are 2 and 4" uses 4 for line 2, which comes from dividing the change in y, 4, by 1 rather than by the change in x, 5 − 3 = 2. "Not parallel, since the two lines cross the y-axis at different points" confuses parallel lines with the same line repeated — parallel lines are distinct lines, and where a line crosses the y-axis has no effect on its gradient. "Parallel, since both lines pass through positive coordinates" reaches the right conclusion for the wrong reason: lines are parallel because their gradients are equal, not because the coordinates given happen to be positive.
(b) x ≥ 5 — Expand the bracket: 2(3x − 1) = 6x − 2, so the inequality is 6x − 2 ≥ 4x + 8. Subtract 4x from both sides and add 2 to both sides: 2x ≥ 10. Divide both sides by 2: x ≥ 5. A candidate who only multiplies the 3x by 2 and forgets to multiply the −1 gets 6x − 1 ≥ 4x + 8, leading to x ≥ 4.5. A candidate who adds 4x instead of subtracting it gets 10x ≥ 10, leading to x ≥ 1. A candidate who multiplies by 2 instead of dividing gets x ≥ 20.
(b) 20 — The gradient is the change in p divided by the change in t: (90 − 30) ÷ (5 − 2) = 60 ÷ 3 = 20. Choosing 0.05 comes from dividing the change in t by the change in p, the wrong way round (3 ÷ 60). Choosing −20 comes from subtracting the coordinates in the wrong order for one part of the calculation, for example (30 − 90) ÷ (5 − 2), giving a negative value. Choosing 18 comes from dividing the second p-coordinate by the second t-coordinate directly (90 ÷ 5) instead of using the change between the two points.
(a) 30° — Method: the exterior angles of any convex polygon add up to 360°, and in a regular polygon they are all equal, so divide 360° by the number of sides. Working: 360 ÷ 12 = 30. Answer: 30°. The distractors: 150° is the interior angle, 180 − 30, which answers for the wrong angle at the vertex; 15° comes from dividing 180 by 12, using the angles on a straight line instead of the full turn; 36° comes from dividing 360 by 12 − 2 = 10, carrying the subtraction of 2 out of the interior angle sum formula into a calculation that does not need it.
(a) 2x − 1 — Swap x and y: x = (y + 1)/2. Multiply both sides by 2: 2x = y + 1. Subtract 1 from both sides: y = 2x − 1, so f⁻¹(x) = 2x − 1. Writing 2x + 1 comes from not flipping the sign on the 1 when it is moved across the equals sign. Writing (x − 1)/2 comes from reversing the sign of the 1 but leaving the ÷2 from the original rule in place, instead of turning it into ×2. Writing x/2 − 1 comes from dividing only the x by 2 and treating the 1 as already outside the fraction.
(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.

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.