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
Answer key: GCSE Foundation sample Paper 1 (non-calculator)
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- (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.
- (a) 2x + 24 — Method: multiply each term inside the bracket by the 2 in front of it. Working: 2 × x = 2x and 2 × 12 = 24, and the products are added because the bracket contains an addition. Answer: 2x + 24. The distractors: 2x + 12 comes from multiplying only the x by 2 and copying the 12 across unchanged; 24x comes from multiplying 2 by 12 and then attaching the letter to that product, as though the two terms inside the bracket could be joined into one; 2x + 14 comes from adding 2 and 12 instead of multiplying them.
- (a) 4 — The length scale factor, cubed, gives the volume scale factor: L³ = 64, so L = the cube root of 64 = 4, since 4 × 4 × 4 = 64. 64 comes from using the volume scale factor itself as the length scale factor, without taking a root at all. 8 comes from taking the square root of 64 instead of the cube root — that would be the correct root for an area scale factor, not a volume one. 32 comes from halving the volume scale factor (64 ÷ 2), instead of cube-rooting it.
- (a) (5, 7) — Method: a midpoint is the mean of the two end points, so for each coordinate (start + end) ÷ 2 = midpoint; rearranging that gives end = 2 × midpoint less the start. Working: for x, (1 + x) ÷ 2 = 3, so 1 + x = 6 and x = 5. For y, (3 + y) ÷ 2 = 5, so 3 + y = 10 and y = 7. B is therefore (5, 7). Answer: (5, 7). The distractors: (2, 2) comes from subtracting A from the midpoint, (3 − 1, 5 − 3), which gives the step from A to the midpoint and stops there instead of taking that same step a second time; (4, 8) comes from adding A to the midpoint, (3 + 1, 5 + 3), without doubling the midpoint first; (6, 10) comes from doubling the midpoint, (2 × 3, 2 × 5), and then forgetting to take A off.
- (c) 500 — Method: the expected number of successes is the number of trials multiplied by the probability, so to find the number of trials, divide the expected number by the probability. Working: let n be the number of seeds planted. Then n multiplied by 0.6 must come to 300, so n = 300 ÷ 0.6 = 500. Answer: the grower should plant 500 seeds. The distractors: 180 comes from multiplying instead of dividing, 300 × 0.6 = 180, which answers how many of 300 seeds would germinate; 750 comes from dividing by the probability of not germinating, 300 ÷ 0.4 = 750; 120 comes from multiplying by that same 0.4, 300 × 0.4 = 120.
- (c) 5 — Method: the mode is the value that occurs most often, so count how many times each different value appears and compare the counts. Working: 3 appears twice, 5 appears three times, 7 appears once and 8 appears once, so the highest frequency is three and the value carrying it is 5. Answer: 5. The distractors: 3 comes from writing down the frequency of the most common answer instead of the answer itself; 7 comes from taking the middle number of the list as it was written, which applies the median without ordering the data and without answering the question asked; 8 comes from picking the largest value, which confuses the mode with the maximum.
- (a) Fewer than 48,500 people attended — Method: a figure given to the nearest thousand lies within half of 1,000, that is 500, of the figure printed, so the attendance is at least 47,500 and below 48,500. Working: 48,000 − 500 = 47,500 and 48,000 + 500 = 48,500, and an attendance of 48,500 would have been reported as 49,000, so every possible attendance is below 48,500. Answer: Fewer than 48,500 people attended. The distractors: more than 48,500 turns the upper limit of the range into a minimum; fewer than 47,500 uses the lower limit as though it were the upper one; more than 48,000 assumes the printed figure was rounded down, when it could just as well have been rounded up from a smaller attendance.
- (a) 3ab — 3 × a × b means 3, a and b are all multiplied together, and in algebraic notation this is written with no multiplication signs: 3ab. Writing 3 + a + b turns every multiplication into an addition, giving a completely different expression. Writing a³b misreads the 3 as a power on a rather than as a coefficient in front of both letters. Writing 3a + b multiplies the 3 by a correctly but then adds b instead of also multiplying it in. The expression that means 3 × a × b is 3ab.
- (d) The line is straight and passes through the origin. — Direct proportion means y = kx for a constant k. This is a straight line, and when x = 0, y = 0, so it passes through the origin. For a positive k it slopes upward from left to right.
- (a) segment — A chord cuts a circle into two regions, and each of these regions is called a segment. A sector is a different region, bounded by two radii and an arc, not by a chord. An arc is a curved part of the circumference, a length, not a region. A radius is a straight line from the centre to the edge, also a length, not a region.
- (b) 0.45 — Winning a toy, winning a sweet and winning neither are mutually exclusive and exhaustive, so their probabilities sum to 1. P(toy) + P(sweet) = 0.15 + 0.4 = 0.55. P(neither) = 1 − 0.55 = 0.45. Adding 0.15 and 0.4 and stopping there gives 0.55, which is the probability of winning a toy or a sweet, not of winning neither. Subtracting only 0.15 from 1 gives 0.85, and ignores the sweet probability entirely. Subtracting only 0.4 from 1 gives 0.6, and ignores the toy probability entirely.
- (d) 51 is not prime, because 51 = 3 × 17. — Check 51 for small prime factors: 51 ÷ 3 = 17, and both 3 and 17 are themselves prime, so 51 = 3 × 17 and 51 is not a prime number — it has factors other than 1 and itself. Checking only 2, 3 and 5 and concluding wrongly that none of them divide 51 misses that 3 does divide it exactly, so the claim that 51 is prime because it avoids 2, 3 and 5 is false. Assuming any odd number must be prime ignores that 51 = 3 × 17 is a counterexample — plenty of odd numbers are not prime. Misreading 51 as the even number 52 leads to the false claim that it is divisible by 2; 51 itself is odd, and 2 is not one of its factors. So 51 is not prime, because 51 = 3 × 17.
- (d) 9 — Method: expand the brackets fully first, then collect the x-terms and constants before dividing. Working: 4(x − 3) = 4x − 12, so 4x − 12 = 2x + 6; 4x − 2x = 6 + 12; 2x = 18; x = 9. Answer: x = 9. 4.5 comes from expanding only the x-term in the bracket and forgetting to multiply the 3, giving 4x − 3 = 2x + 6, then 2x = 9, x = 4.5. −3 comes from a sign error when expanding, giving 4x + 12 = 2x + 6, then 2x = −6, x = −3. 3 comes from a sign error collecting the x-terms, adding instead of subtracting: 4x + 2x = 6 + 12, so 6x = 18, x = 3.
- (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.
- (b) 50° — Method: the three angles of a triangle add up to 180°, and the two angles opposite the equal sides are equal, so subtract the given angle from 180° and halve the remainder. Working: 180° − 80° = 100°, and 100° ÷ 2 = 50°. Answer: 50°. The distractors: 100° comes from subtracting from 180° and forgetting to halve, so it is the two equal angles together; 40° comes from halving the 80° that is given rather than halving what is left of the 180°; 80° comes from assuming that the two base angles must match the angle between the equal sides.
- (a) 7/8, 17/20, 0.8, 78% — Method: convert every value to a decimal, then order the decimals from largest to smallest. Working: 7/8 = 0.875, 17/20 = 0.85, 0.8 = 0.8, 78% = 0.78. Ordering from largest to smallest gives 7/8, 17/20, 0.8, 78%. Answer: 7/8, 17/20, 0.8, 78%. '78%, 0.8, 17/20, 7/8' comes from ordering the converted decimals from smallest to largest instead of largest to smallest. '17/20, 7/8, 0.8, 78%' comes from converting 17/20 incorrectly as larger than 7/8, for example treating 17/20 as 0.87 instead of 0.85, swapping the top two. '7/8, 0.8, 17/20, 78%' comes from converting 17/20 incorrectly as equal to 0.8, swapping the middle two.
- (a) 17 — Substitute a = 3 into 5a + 2: 5 × 3 + 2 = 15 + 2 = 17. 10 comes from treating 5a as 5 + a instead of 5 × a, giving 5 + 3 + 2. 25 comes from adding the 2 to a before multiplying by 5, 5 × (3 + 2). 15 comes from working out 5 × 3 correctly but forgetting to add the 2.
- (c) 10 g — Method: mass = density × volume. Working: 2 × 5 = 10, in grams because the density is in grams per cubic centimetre. Answer: 10 g. The distractors: 2.5 g comes from dividing the volume by the density, 5 ÷ 2, instead of multiplying; 0.4 g comes from dividing the density by the volume, 2 ÷ 5; 7 g comes from adding the density and the volume.
- (a) 300 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 8,900 rounds to 9,000 and 29 rounds to 30; cancelling a zero from each gives 900 ÷ 3. Answer: 300. The distractors: 450 comes from rounding 29 down to 20 instead of to the nearest ten, giving 9,000 ÷ 20; 3,000 comes from rounding 29 to 3 rather than to 30, a place-value slip that divides by a number ten times too small; 307 is the exact quotient rounded to the nearest whole number, worked out in full when the question asks for an estimate.
- (c) m = 6c — Method: the whole is the sum of the parts in the ratio, and the cement is 1 part, so one part weighs c kg. Working: the mix has 5 + 1 = 6 parts, each of mass c kg, so the total mass is 6 × c, giving m = 6c. Answer: m = 6c. The distractors: m = 5c uses the 5 gravel parts as the multiplier and forgets that the cement is in the mix too, so it gives the mass of the gravel and not the total; m = c + 5 comes from reading the ratio as '5 more than' and adding, which treats a number of parts as a mass in kilograms; m = c/6 turns the relationship upside down, as though the total were shared into the cement rather than the cement multiplied up to the total.
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.