Sample paper · GCSE Foundation · grades 1–5
GCSE Foundation 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 Foundation qualification — number 25%, algebra 20%, ratio, proportion and rates of change 25%, geometry and measures 15%, probability 7.5%, statistics 7.5% — with no calculator-only items. AO1 / AO2 / AO3 at this tier: 50% / 25% / 25%.
Non-calculatorGCSE Foundation
GCSE Foundation sample Paper 1 (non-calculator)
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- 1.Work out 2 × 3 × 5 + 1 and decide whether the result is a prime number.
- 2.Sophie's mother is 34 years old. In x years' time she will be 50 years old. Work out the value of x.
- 3.A post 2 metres tall casts a shadow 3 metres long. At the same time a nearby tree casts a shadow 12 metres long. Work out the height of the tree.
- 4.The diagram shows a solid built from identical cubes. How many squares are in its plan view, the view looking straight down from above?
- 5.Two games at a school fete are played independently. The probability of winning game X is 3/8. The probability of winning game Y is 0.4. Which statement correctly compares the two games?
- 6.The mean of the four numbers 10, 15, 20 and x is 18. Work out the value of x.
- 7.Work out −(−3)⁴ + (−3)³
- 8.Expand and simplify 4(2x − 1) − 3(x + 2) − 5x
- 9.An oven is preheating. Its temperature, T °C, is plotted against time, t minutes, on a straight-line graph. The line passes through the points (2, 60) and (6, 140). Work out the gradient of the line.
- 10.To construct the perpendicular from a point P to a line l, where P is a point above l, an arc centred at P is drawn to cross l at two points, X and Y. What is the correct next step?
- 11.A biased spinner is spun 200 times. It lands on red 70 times, on blue 50 times, and on green 80 times. Using these results, work out the expected number of times the spinner does NOT land on red, in 500 spins of the same spinner.
- 12.Work out 20 − 8 ÷ 2 + 1
- 13.The nth term of a sequence is n² + 3. Work out the first term of the sequence that is greater than 50.
- 14.A train journey takes 2 hours 15 minutes. A bus journey to the same place takes 3 hours 45 minutes. Write the time for the train journey as a fraction of the time for the bus journey, giving your answer in its simplest form.
- 15.Write down the exact value of sin 60°.
- 16.At a swimming pool the ratio of adults to children is 3 : 5. Of the children, 2/5 are boys. What fraction of the people at the pool are boys?
- 17.A straight line has gradient −3 and passes through the point (0, 1). Work out the value of y when x = 2.
- 18.A recipe for 6 muffins needs 150 g of flour. Aisha wants to make 15 muffins using the same recipe. Work out how much flour she needs.
- 19.A number line runs from 0 to 1 and is divided into 4 equal parts. Which mark on the line represents 1/4?
- 20.A necklace is made using gold beads and silver beads in the ratio 7:3. There are 40 more gold beads than silver beads. Work out the total number of beads in the necklace.
Answer key
- (d) 31, which is prime — Method: work out the value, remembering that multiplication comes before addition, then test it for primality by dividing by each prime up to its square root. Working: 2 × 3 × 5 = 30, so the value is 30 + 1 = 31. Since 6² = 36 is larger than 31, only 2, 3 and 5 need testing: 31 is odd, 31 ÷ 3 leaves a remainder of 1, and 31 does not end in 0 or 5. It therefore has exactly two factors, 1 and itself. Answer: 31, which is prime. The distractors: 30, which is not prime comes from working out 2 × 3 × 5 and forgetting to add the 1; the claim that 31 = 1 × 31 makes it non-prime comes from treating any factor pair as proof, forgetting that a prime is allowed the pair 1 and itself; the claim that 31 is a multiple of 3 comes from assuming that a number containing the digit 3 divides by 3, when in fact 31 ÷ 3 leaves a remainder.
- (a) 16 — Method: the future age is the present age plus the number of years, so form that equation and solve it. Working: 34 + x = 50, and subtracting 34 from both sides gives x = 50 − 34 = 16. Answer: 16. The distractors: 84 comes from adding the two ages instead of subtracting, 34 + 50; 50 comes from giving her future age rather than the number of years that pass; 24 comes from a column subtraction in which the smaller digit is taken from the larger in each column, 4 − 0 and 5 − 3, instead of exchanging.
- (c) 8 m — Method: in the same sunlight every object has its height and its shadow in the same ratio, so write 2:3 = h:12, find the multiplier that takes 3 to 12 and apply it to the height. Working: 12 ÷ 3 = 4, so the tree's shadow is 4 times the post's shadow; the height must be scaled by the same 4, giving 4 × 2 = 8 m. Answer: 8 m. The distractors: 18 m comes from setting up the proportion upside down, 12 ÷ 2 × 3, which scales by shadow over height instead of height over shadow; 24 m comes from multiplying the 12 m shadow by the post's height of 2 m and never dividing by the post's shadow of 3 m; 4 m is the scale factor 12 ÷ 3, given as a length instead of being used to scale the 2 m post.
- (a) 3 squares — Method: the plan view shows only the floor positions that have at least one cube standing on them; height does not add extra squares to the plan. Working: the base row occupies three floor positions in a line. The two extra cubes stand on top of two of those same three positions, so they do not create any new floor position. Answer: 3 squares. The distractors: 5 squares comes from adding the total number of cubes used (3 + 2 = 5) instead of counting distinct floor positions. 2 squares comes from counting only the raised two-cube section and ignoring the single cube at the other end of the row. 4 squares comes from counting one of the shared positions twice.
- (c) Game Y — 0.4 is greater than 3/8. — Converting 3/8 to a decimal gives 3 ÷ 8 = 0.375, and 0.4 is greater than 0.375, so game Y gives the better chance of winning: 'Game Y — 0.4 is greater than 3/8.' Comparing the numerator 3 to the digit 0 in front of the decimal point in 0.4 compares two unrelated numbers and gives 'Game X — 3 is greater than 0, the units digit of 0.4.' Treating a larger denominator as meaning more chances, rather than smaller equal shares, gives 'Game X — 3/8 has the larger denominator.' Rounding 0.375 to 0.4 to one decimal place makes the two values look equal, but they are not exactly equal, so 'Equal — 3/8 rounds to 0.4 to one decimal place' mistakes a rounded match for a true one.
- (a) 27 — Method: turn the mean into a total using total = mean × number of values, then subtract the numbers that are already known. Working: four numbers with a mean of 18 have a total of 18 × 4 = 72; the three known numbers give 10 + 15 + 20 = 45; so x = 72 − 45 = 27. Answer: 27, and checking, (10 + 15 + 20 + 27) ÷ 4 = 72 ÷ 4 = 18. The distractors: 72 comes from stopping at the total the four numbers must reach and never subtracting the known three; 18 comes from assuming the missing number must equal the mean; 45 comes from stopping at the total of the three known numbers.
- (a) −108 — Method: a power is worked out before any minus sign written in front of it, while a minus sign inside the brackets is part of the base. Working: (−3)⁴ = 81, because four negative factors multiply to a positive result, so −(−3)⁴ = −81. (−3)³ = −27, because three negative factors multiply to a negative result. Adding gives −81 + (−27) = −108. Answer: −108. The distractors: 54 comes from attaching the leading minus sign to the base, working out (−(−3))⁴ = 81 and then adding −27; −54 comes from taking (−3)³ as +27, forgetting that an odd power keeps the negative sign; 108 comes from believing that any power of a negative number is positive and that the leading minus belongs to the base, giving 81 + 27.
- (a) −10 — Method: multiply each bracket out, treating the second bracket as being multiplied by −3 because it is subtracted, then collect like terms. Working: 4(2x − 1) = 8x − 4 and −3(x + 2) = −3x − 6, so the expression becomes 8x − 4 − 3x − 6 − 5x; the x terms give 8x − 3x − 5x = 0, so no term in x survives, and the numbers give −4 − 6 = −10. Answer: −10. The distractors: 2 comes from expanding −3(x + 2) as −3x + 6, leaving the numbers −4 + 6; 5x − 10 comes from forgetting the final −5x, so the x terms give 8x − 3x = 5x; −7 comes from multiplying the 4 over only the first term of its bracket, giving 8x − 1 and so the numbers −1 − 6.
- (d) 20 — Gradient = change in T ÷ change in t = (140 − 60) ÷ (6 − 2) = 80 ÷ 4 = 20. A student who subtracts in the wrong order gets −20. A student who divides 80 by 2 instead of 4 gets 40. A student who wrongly treats the line as passing through the origin and uses the point (2, 60) on its own gets 60 ÷ 2 = 30.
- (a) Draw equal arcs from X and Y, meeting below line l. — After the first arc marks two points X and Y on line l, compasses are opened to a new radius and arcs of equal radius are drawn centred at X and at Y, so that they meet on the opposite side of l from P; joining P to that meeting point gives the perpendicular. (Joining X and Y with a straight line only retraces part of line l itself, since X and Y both already lie on it; drawing an arc centred at P through only one of X or Y repeats part of the first step instead of moving on; drawing a circle through X, Y and P does not locate the new point needed to complete the perpendicular.)
- (c) 325 — Method: first find the relative frequency of NOT landing on red from the 200 spins, then scale that up to 500 spins. Working: non-red results = 50 + 80 = 130, out of 200 spins, so P(not red) = 130 ÷ 200 = 0.65. Expected non-red results in 500 spins = 500 × 0.65 = 325. Answer: 325. Watch out: writing down 175 finds the expected number of RED results instead, 70 ÷ 200 × 500 = 175, answering the opposite of what was asked. Writing down 250 assumes landing red or not landing red must be a fair 50-50 split, but the spinner is biased and the actual results do not split evenly. And writing down 130 stops after finding how many of the 200 spins were non-red and forgets to scale that figure up to the 500 spins asked for.
- (d) 17 — 8 ÷ 2 = 4, then 20 − 4 = 16, then 16 + 1 = 17. Stopping after the subtraction and forgetting to add the final 1 leaves 16. Adding the 4 and the 1 together before subtracting gives 4 + 1 = 5, then 20 − 5 = 15 — the subtraction should use the 4 from the division, not a combined total. Working strictly left to right without giving division priority gives 20 − 8 = 12, then 12 ÷ 2 = 6, then 6 + 1 = 7.
- (a) 52 — Test successive terms: n=6 gives 6²+3=39, which is not greater than 50. n=7 gives 7²+3=52, which is greater than 50, so the first term greater than 50 is 52. A candidate who stops at n=6, before checking whether 39 actually exceeds 50, would give 39. A candidate who solves n²>50 instead of n²+3>50, ignoring the +3 in the search, would find n=8 is the first value with n²>50 (since 7²=49) and compute 8²+3=67. A candidate who computes n² by doubling n instead of squaring it would compute 2×7+3=17.
- (d) 3/5 — Convert both times to minutes: 2 hours 15 minutes = 135 minutes; 3 hours 45 minutes = 225 minutes. Put the train time over the bus time: 135/225. Divide both numbers by their highest common factor, 45: 135÷45 = 3, 225÷45 = 5, giving 3/5. (5/3 comes from writing the times the wrong way round. 2/5 comes from finding the difference, 225 − 135 = 90 minutes, and writing it as a fraction of the bus time, 90/225. 3/8 comes from comparing the train time to the total time for both journeys, 135/360.)
- (c) √3/2 — Method: sin 60° comes from half of an equilateral triangle. Cut an equilateral triangle of side 2 down its middle: each half is right-angled, with hypotenuse 2 and base 1, and Pythagoras gives its height as √3 because 2² − 1² = 3. Working: sine is the opposite side over the hypotenuse, the side opposite the 60° angle is the height √3, and the hypotenuse is 2, so sin 60° = √3 ÷ 2. Answer: sin 60° = √3/2. Swapping sine for cosine at this angle gives cos 60° = 1/2; taking the value from the other special triangle, the right-angled isosceles one, gives sin 45° = √2/2; and reading the tangent row instead of the sine row gives tan 60° = √3.
- (a) 1/4 — First find the fraction of the people who are children: the pool is 3 + 5 = 8 equal shares and the children take 5, so the children are 5/8 of the people. The boys are 2/5 of that, and 'of' means multiply: 2/5 × 5/8 = 10/40 = 1/4. Answering 2/5 gives the boys as a fraction of the children only, 3/20 multiplies by the adults' fraction 3/8 instead of the children's, and 3/5 is the fraction of the children who are girls.
- (b) −5 — Method: a point whose x-coordinate is 0 lies on the y-axis, so its y-coordinate is the constant c; once m and c are both known the equation can be written down and x substituted into it. Working: the line passes through (0, 1), so c = 1 and the equation is y = −3x + 1; substituting x = 2 gives y = −3 × 2 + 1 = −6 + 1 = −5. Answer: −5. The distractors: 7 comes from ignoring the minus sign on the gradient and working out 3 × 2 + 1; 5 comes from working out 3 × 2 = 6 and then using the minus sign to take the constant off the product, 6 − 1 = 5; −7 comes from subtracting the constant instead of adding it, −6 − 1 = −7.
- (a) 375 g — Find the amount of flour needed per muffin: 150 ÷ 6 = 25 g. Multiply by the new number of muffins: 25 × 15 = 375 g. (60 g comes from using the scale factor the wrong way round, 6/15 × 150. 150 g comes from not scaling the recipe at all. 300 g comes from rounding the scale factor, 15 ÷ 6, down to 2 before multiplying.)
- (d) the first mark after 0 — Method: when a unit length is split into equal parts, each gap is one part of the whole, so 1/4 is one gap along from 0. Working: four equal parts means each gap measures 1/4, so the marks after 0 stand for 1/4, 2/4, 3/4 and 4/4. One gap along from 0 is therefore the mark for 1/4. Answer: the first mark after 0. The distractors: the second mark after 0 comes from counting 0 itself as the first mark; the third mark after 0 comes from counting back from the 1 end instead of forward from 0; the fourth mark after 0 comes from reading the 4 in the denominator as the number of the mark rather than the number of parts.
- (a) 100 — The difference between the parts of the ratio is 7 − 3 = 4 parts, and this is worth 40 beads. Divide to find one part: 40 ÷ 4 = 10. The total number of parts is 7 + 3 = 10, so the total number of beads is 10 × 10 = 100. (40 is just the given difference between gold and silver, not the total. 70 is the number of gold beads only, using 7 parts. 30 is the number of silver beads only, using 3 parts.)
How the 20 questions are shared out
- Number — 5 questions (25% of the qualification)
- Algebra — 4 questions (20% of the qualification)
- Ratio, proportion and rates of change — 5 questions (25% of the qualification)
- Geometry and measures — 3 questions (15% 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.