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%.
Non-calculatorGCSE Higher
GCSE Higher sample Paper 1 (non-calculator)
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- 1.Meera says that 0.7 ÷ 0.1 = 0.07. Work out the correct value of 0.7 ÷ 0.1.
- 2.Solve the simultaneous equations 2x + y = 7 and x − y = 2.
- 3.There are 2500 electric cars registered in a town. The number is predicted to increase by 6% each year. Work out the predicted number of electric cars after 3 years, to the nearest whole number.
- 4.A children's slide is built as a right-angled triangle: the vertical ladder side is 2.4 m, the horizontal base along the ground is 3.2 m, and the sloping slide is the third side. Work out the exact length of the sloping slide.
- 5.An ordinary pack of 52 playing cards contains 13 hearts. One card is taken at random from the pack. Work out the probability that the card is not a heart. Give your answer as a percentage.
- 6.Two classes at a school in Coventry sit the same maths test, out of 20 marks. Class A has a mean mark of 14 and a range of 6. Class B has a mean mark of 14 and a range of 14. Write a sentence comparing the two classes, using the mean and the range.
- 7.A gym increases its membership price by 1/5. The original price is £60. Work out the new price.
- 8.A circle has centre (0, 0) and equation x² + y² = 64. Write down the radius of the circle.
- 9.Write the ratio 3/4 : 1/2 as a ratio of whole numbers in its simplest form.
- 10.In a kite, one pair of opposite angles are equal in size. In kite WXYZ, angle X = 40° and angle Z = 100° are the two angles that are NOT equal to each other. The other two angles, W and Y, are equal to each other. Work out the size of angle W.
- 11.Spinner E has 4 equal sections, numbered 1 to 4. Spinner F has 6 equal sections, numbered 1 to 6. Both spinners are spun once. Work out the probability of getting at least one 3.
- 12.Put these numbers in order, starting with the smallest: 3.2 × 10⁴, 2.9 × 10⁵, 4.1 × 10³
- 13.f(x) = 2x + 1. Work out the value of x for which f⁻¹(x) = 5.y = 2x + 1
- 14.Two mathematically similar triangular flags have areas in the ratio 4 : 25. The height of the smaller flag is 6 cm. Work out the height of the larger flag.
- 15.A circular plate has a diameter of 20 cm. Work out the area of the plate. Use π = 3.14.
- 16.A taxi charges a £3 fixed fee plus £2 for each mile travelled. Write an expression, in pounds, for the total cost of a journey of n miles.
- 17.Two hikers' journeys are shown on separate distance-time graphs. Hiker A's graph has gradient 4.5 km/h. Hiker B's graph has gradient 3.2 km/h. Work out how much faster Hiker A walks than Hiker B, in km/h.
- 18.Triangle PQR has vertices P(0, 0), Q(9, 0) and R(0, 4). Work out the area of triangle PQR.
- 19.A student is asked whether 3(x − 4) = 3x − 4 is an identity. Which statement gives the correct verdict and reason?
- 20.The formula connecting distance (d), speed (s) and time (t) is d = st. Make s the subject of the formula.
Answer key
- (a) 7 — Dividing by 0.1 is the same as multiplying by 10, so 0.7 ÷ 0.1 = 7. Meera's answer of 0.07 comes from dividing 0.7 by 10 instead of by 0.1, the wrong way round. A candidate who confuses 0.1 with 0.01 multiplies by 100 instead of 10 and gets 70. A candidate who thinks dividing by a number less than 1 does not change the value gets 0.7.
- (c) x = 3, y = 1 — Method: the y terms are +y and −y, so adding the two equations removes y and leaves an equation in x alone. Working: adding 2x + y = 7 and x − y = 2 gives 3x = 9, so x = 3; substituting x = 3 into x − y = 2 gives 3 − y = 2, so y = 1. Answer: x = 3, y = 1, which also satisfies 2 × 3 + 1 = 7. The distractors: x = 1, y = 3 comes from finding the two values correctly and then writing them against the wrong letters; x = 3, y = 2 comes from substituting x = 3 into 2x + y = 7 as 2 + 3 + y = 7, adding the coefficient instead of multiplying by it; x = 3, y = −1 comes from substituting into x − y = 2 as though it read x + y = 2.
- (a) 2978 — A rise of 6% is a multiplier of 1.06, applied once for each year. After year 1: 2500 × 1.06 = 2650. After year 2: 2650 × 1.06 = 2809. After year 3: 2809 × 1.06 = 2977.54, which is 2978 to the nearest whole number. Multiplying by 1.18 in one go would be wrong, because the second and third years grow from larger numbers than the first.
- (d) 4 m — The sloping slide is the hypotenuse of a right-angled triangle with the other two sides 2.4 m and 3.2 m. By Pythagoras' Theorem, hypotenuse² = 2.4² + 3.2² = 5.76 + 10.24 = 16. Square root: √16 = 4 m. Adding the two sides instead of squaring them gives 2.4 + 3.2 = 5.6 m. Truncating each squared side to a whole number before adding (5.76 to 5 and 10.24 to 10) gives √15 ≈ 3.87 m. Forgetting to take the square root and leaving the sum of the squares as the answer gives 5.76 + 10.24 = 16, written as 16 m.
- (d) 75% — Method: count how many cards are not hearts, write that count over the total number of cards, cancel the fraction down and then turn it into a percentage. Working: 52 − 13 = 39 cards are not hearts, so the probability is 39/52; dividing the numerator and the denominator by 13 gives 3/4, and 3/4 = 0.75, so 0.75 × 100 = 75. Answer: 75%, three quarters of the way along the 0 to 1 scale. The distractors: 25% comes from giving the probability that the card is a heart, 13 out of 52, which cancels to 1/4; 50% comes from reading 'not a heart' as 'not a red card' and halving the pack; 39% comes from writing the count of 39 cards straight down as the percentage without comparing it with the 52 cards in the pack.
- (a) Equal means; Class A is more consistent, smaller range. — Method: when two data sets share a measure of location, compare a measure of spread to say more about consistency. Working: both classes have the same mean mark, 14, so on average they performed equally well. Class A has the smaller range, 6, so its marks are more tightly grouped around 14 than Class B's marks, which vary by as much as 14. So Class A's marks were more consistent, even though neither class did better on average. Saying Class B did better because it has the bigger range confuses a wide spread with a high score — a big range describes variability, not performance. Saying Class A did better because it has the smaller range makes the same mistake in the other direction: the two classes are tied on the mean, so neither one 'did better'. Saying the classes cannot be compared because their means are equal misses the whole point of also comparing the range. Always compare both an average AND a spread before describing two data sets — either one alone tells only half the story.
- (b) £72 — One fifth of £60 = £12. New price = £60 + £12 = £72. A candidate who gives the increase instead of the new price gets £12. A candidate who subtracts the increase instead of adding it gets £60 − £12 = £48. A candidate who uses 1/4 instead of 1/5 gets £60 + £15 = £75.
- (b) 8 — Method: a circle centred on the origin has equation x² + y² = r², where r is the radius, so the number on the right-hand side is the square of the radius and not the radius itself. Working: comparing x² + y² = 64 with x² + y² = r² gives r² = 64, so r = √64 = 8. Answer: the radius is 8. The distractors: 64 is r² read straight off the equation as though the right-hand side were the radius, which is the commonest error on this form; 32 comes from halving 64, treating the right-hand side as a diameter that has to be halved; 16 is the diameter, 2 × 8, quoted in place of the radius.
- (b) 3:2 — Write both fractions over a common denominator of 4: 3/4 stays as 3/4, and 1/2 = 2/4. Comparing the numerators gives the ratio 3 : 2. Getting 2 : 3 swaps the two parts round. Getting 3 : 1 comes from using the numerator of the first fraction and the original numerator of the second fraction (1) without converting to a common denominator. Getting 2 : 1 comes from using only the denominators, 4 and 2, and simplifying those instead of the numerators.
- (a) 110° — The angles in a quadrilateral add up to 360°. So 40° + 100° + W + W = 360°, giving 2W = 360° − 140° = 220°, so W = 110°. A pupil who works out 2W = 220° but forgets to divide by 2, since there are two equal angles W, gives 220°. A pupil who mistakenly uses the angle sum of a triangle, 180°, instead of 360°, gets 180° − 140° = 40°. A pupil who simply adds the two given angles together instead of subtracting from 360° gets 40° + 100° = 140°. The correct answer is 110°.
- (a) 3/8 — There are 4 × 6 = 24 equally likely outcomes. The outcomes with no 3 at all have Spinner E showing 1, 2 or 4 and Spinner F showing 1, 2, 4, 5 or 6, giving 3 × 5 = 15 outcomes. So the outcomes with at least one 3 are 24 − 15 = 9, and the probability is 9/24 = 3/8. Choosing 5/12 comes from adding the two individual probabilities of a 3, 1/4 + 1/6, which counts the outcome where both spinners show 3 twice over. Choosing 1/4 comes from only counting the case where Spinner E shows 3, and forgetting the outcomes where Spinner F shows 3 instead. Choosing 5/8 comes from working out the probability of getting no 3 at all, 15/24 = 5/8, and giving that as the final answer instead of subtracting it from 1.
- (c) 4.1 × 10³, 3.2 × 10⁴, 2.9 × 10⁵ — The exponent decides the size first: 10³ is smaller than 10⁴, which is smaller than 10⁵, so the order is 4.1 × 10³, then 3.2 × 10⁴, then 2.9 × 10⁵. Reversing the whole list gives largest to smallest instead of smallest to largest. Comparing 3.2 × 10⁴ and 4.1 × 10³ by their coefficients alone, 3.2 against 4.1, and swapping them ignores that 10⁴ is bigger than 10³ regardless of the coefficient. Comparing 2.9 × 10⁵ and 3.2 × 10⁴ by their coefficients alone and swapping them makes the same mistake at the top of the list.
- (d) 11 — f⁻¹(x) = 5 means x = f(5), since applying f to both sides undoes the inverse. f(5) = 2 × 5 + 1 = 11. Writing 2 comes from confusing f⁻¹(x) = 5 with f(x) = 5, and solving 2x + 1 = 5 instead: 2x = 4, x = 2. Writing 9 comes from finding f⁻¹(x) with a sign error, f⁻¹(x) = (x + 1)/2 instead of (x − 1)/2, then setting this equal to 5: x + 1 = 10, x = 9. Writing 6 comes from finding f⁻¹(x) without dividing by 2 at all, f⁻¹(x) = x − 1, then setting this equal to 5: x = 6.
- (a) 15 cm — Take the square root of each part of the area ratio to find the length ratio: the square root of 4 is 2 and the square root of 25 is 5, giving a length ratio of 2 : 5. Multiply the smaller flag's height by the scale factor 5 ÷ 2 = 2.5: 6 × 2.5 = 15, so the larger flag is 15 cm tall. Giving 37.5 cm uses the area ratio, 25 ÷ 4 = 6.25, directly as the scale factor without square-rooting it first (6 × 6.25 = 37.5). Giving 2.4 cm applies the length ratio the wrong way round, scaling the smaller flag down by 2 ÷ 5 instead of up by 5 ÷ 2 (6 × 0.4 = 2.4). Giving 27 cm adds the difference between the two area-ratio numbers, 25 − 4 = 21, onto the smaller height instead of using it as a scale factor (6 + 21 = 27).
- (a) 314 cm² — Method: the area of a circle is πr², and the radius is half the diameter, so halve the 20 cm before squaring. Working: r = 20 ÷ 2 = 10 cm, so the area is 3.14 × 10² = 3.14 × 100 = 314. Answer: 314 cm². The distractors: 1256 cm² comes from putting the diameter straight into πr² without halving it, 3.14 × 20²; 628 cm² comes from halving correctly but then using 2πr², a mixture of the circumference and area formulae; 62.8 cm² comes from working out πd = 3.14 × 20, which is the circumference of the plate rather than its area.
- (b) 2n + 3 — Method: multiply the cost per mile by the number of miles to get an expression, then add the fixed fee as a separate term. Working: n miles at £2 each is 2n; add the £3 fixed fee: 2n + 3. Answer: 2n + 3. 3n + 2 comes from swapping the fee and the rate round, treating £3 as the rate per mile and £2 as the fixed fee. 5n comes from adding the fee and the rate together first (3 + 2 = 5) and multiplying the result by n, instead of keeping the fixed fee as its own term. 2n − 3 comes from subtracting the fixed fee instead of adding it.
- (a) 1.3 — The gradient of a distance-time graph gives the speed. Difference in speed = 4.5 − 3.2 = 1.3 km/h. A student who subtracts the speeds the wrong way round, 3.2 − 4.5, gets −1.3. A student who adds the two speeds instead of comparing them gets 7.7. A student who multiplies the two gradients gets 14.4.
- (a) 18 — PQ lies along the x-axis with length 9, and PR lies along the y-axis with length 4, and these two sides meet at right angles at P, so they can be used as the base and height of the triangle. Area = 1/2 × base × height = 1/2 × 9 × 4 = 18. 36 comes from multiplying the base and height but forgetting to halve the result. 13 comes from adding the two lengths, 9 + 4, instead of multiplying them. 26 comes from the perimeter-style calculation 2 × (9 + 4) instead of the triangle area formula.
- (b) It is not even an ordinary equation with a solution: expanding the left-hand side gives 3x − 12, and 3x − 12 = 3x − 4 would require −12 = −4, which is never true. — Expanding the left-hand side, 3(x − 4) = 3x − 12. Setting this equal to the right-hand side, 3x − 12 = 3x − 4, gives −12 = −4 once the 3x terms are removed from both sides — a statement that is never true, so no value of x satisfies the equation at all, and it is certainly not an identity. The option about substituting a specific value misunderstands algebraic expansion, which holds for every x, not one chosen value. The option matching the first term wrongly assumes that is enough to prove equivalence. The option about multiplying the 4 by 3 on both sides is nonsensical, since there is only one bracket to expand, on the left-hand side.
- (b) s = d / t — Method: undo the multiplication by t by dividing both sides by t. Working: d = st, so dividing both sides by t gives s = d / t. The value s = dt comes from multiplying by t instead of dividing. The value s = t / d comes from inverting the fraction, dividing t by d instead of d by t. The value s = d − t comes from subtracting t instead of dividing by it.
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.