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) −4 — To subtract a negative number, add its positive equivalent: −7 − (−3) becomes −7 + 3. Work out −7 + 3 to get −4. Treating "− (−3)" as simply "−3" without flipping the sign gives the wrong working −7 − 3, which is −10. Ignoring the negative sign on −7 and just subtracting the values, 7 − 3, gives 4, which loses the sign of the starting number. Flipping the sign of both numbers, 7 + 3, gives 10, which changes more than the double negative allows. So −7 − (−3) = −4.
- (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) £200 — Method: add the parts of the ratio, divide the total paid by the number of parts to find the value of one part, then multiply by the parts in the most expensive item. Working: 2 + 3 + 5 = 10 parts, £400 ÷ 10 = £40 for one part, and the most expensive item is 5 parts, so 5 × £40 = £200. Answer: £200. The distractors: £40 is the value of one part; £80 is the 2-part item, the cheapest of the three; £120 is the 3-part item.
- (b) No — the angle given is not the included angle — Method: check whether the given angle sits between the two given sides, since SAS requires the included angle. Working: sides AB and BC meet at vertex B, so the angle between them is angle B — but the angle given is angle A, which is not between the two given sides, and the same mismatch happens in triangle DEF. Options: 'two sides and one angle match' restates SAS's ingredients without checking their positions, which is exactly Meera's mistake; 'SSS needs three equal sides' is a true fact about a different condition, but it is not the reason Meera is wrong here; 'SAS allows any equal angle' states a rule that is not how SAS works, since the angle must be the included one. Answer: no, the angle given is not the included angle.
- (a) 1/16 — The two spins are independent, so multiply the probability of blue on each spin: 1/4 × 1/4 = 1/16. Choosing 1/2 comes from adding the two probabilities instead of multiplying, 1/4 + 1/4 = 1/2. Choosing 1/8 comes from wrongly treating the second spin as having a 1/2 chance of blue instead of the given 1/4, giving 1/4 × 1/2 = 1/8. Choosing 1/4 comes from giving the probability for a single spin and forgetting that the spinner is spun twice.
- (a) The median, as the one very large wage does not move it — Method: an average describes a population well when it sits close to most of the values, so compare what each average does when one value lies far from the rest. Working: in order the wages are 420, 440, 460, 480 and 1,500, so the median is the third of the five, £460. The mean uses every wage: 420 + 440 + 460 + 480 + 1,500 = 3,300 and 3,300 ÷ 5 = 660, so the mean is £660. Four of the five people earn less than £660, and the nearest of those four wages is £180 below it, so £660 describes nobody at the garage; £460 sits inside the group of four similar wages. Answer: the median, as the one very large wage does not move it, while that same wage drags the mean £200 above the median. The distractors: saying the median is always larger than the mean is an invented rule, and here the median £460 is smaller than the mean £660; saying the mean is the only average that uses all five wages is true as far as it goes, but using a value and being dragged by it are the same thing when that value is £1,500; saying £660 lies between the smallest and largest wage is true of every mean ever calculated, so it proves nothing about whether this one is typical.
- (b) 245 ≤ m < 255 — Method: a mass given to the nearest 10 g lies within half of 10 g, that is 5 g, of the figure written down. The lower limit is included because it rounds up to that figure, and the upper limit is excluded because it rounds up to the next multiple of 10. Working: 250 − 5 = 245 and 250 + 5 = 255, and a mass of 245 g rounds to 250 g while a mass of 255 g rounds to 260 g. Answer: 245 ≤ m < 255. The distractors: 240 ≤ m < 260 goes a whole 10 g either side instead of half of it; 245 < m ≤ 255 has the two limits the wrong way round; 240 ≤ m < 250 treats the stated 250 g as the largest mass possible, as though rounding were always upwards.
- (a) Tuesday, by 2 km/h — Monday's speed is 15 ÷ 2.5 = 6 km/h and Tuesday's speed is 12 ÷ 1.5 = 8 km/h, so Tuesday was faster, by 8 − 6 = 2 km/h. A candidate who works out the correct speeds but mislabels which day is faster gets Monday, by 2 km/h. A candidate who divides 15 ÷ 2.5 incorrectly as 5 instead of 6 gets a difference of 8 − 5 = 3 km/h, still crediting Tuesday. A candidate who forgets to find Monday's speed and gives Tuesday's speed itself as the difference states Tuesday, by 8 km/h.
- (d) 600 g — Find the flour used for one person: 240 ÷ 6 = 40 g per person. Then scale up to 15 people: 40 × 15 = 600 g. Dividing 240 by 15 and then multiplying by 6 gives 96 g — that uses the ratio the wrong way round, scaling down instead of up. Stopping at 40 g only gives the amount for one person, not for 15. Adding 15 − 6 = 9 to 240 gives 249 g, which treats the number of people and the grams of flour as if they were the same kind of quantity — proportion means scaling, not adding. The recipe needs 600 g of flour.
- (d) 6 cm — Volume = πr²h, so height = volume ÷ (πr²) = 471 ÷ (3.14 × 25) = 471 ÷ 78.5 = 6 cm. (30 cm comes from dividing by πr instead of πr², missing one factor of the radius; 150 cm comes from dividing by π only, without using r² at all; 24 cm comes from treating the given 5 cm as a diameter and using a radius of 2.5 cm instead.)
- (d) 12 — There are 120 − 70 = 50 female members. 38 of them play netball, so the rest play football: 50 − 38 = 12. Writing 45 is wrong because that is the number of MALE members who play football, not female. Writing 38 again is wrong because that is the female netball total, not the female football total — the question needs the total minus 38, not 38 itself. Writing 25 is wrong because that comes from the male branch (70 − 45 = 25 male netball players), not the female branch. 12 female members play football.
- (a) 20 — Method: round each number to 1 significant figure, then divide. Working: 397 rounds to 400 and 21 rounds to 20, so the estimate is 400 ÷ 20 = 20. Answer: 20. The distractors: 19 comes from not estimating at all, since 397 ÷ 21 = 18.9 to 1 decimal place, which rounds to 19, while the question asks for an estimate from rounded numbers; 200 comes from dividing 400 by 2 instead of by 20, losing the place value of the rounded divisor; 2 comes from dividing the leading digits only and losing the place value of both numbers.
- (a) 2/5 — Method: compare the equation with y = mx + c, where m is the gradient. Working: in y = (2x/5) − 3, the coefficient of x is 2/5. Answer: the gradient is 2/5. −3 comes from confusing the gradient with the y-intercept. 5/2 comes from inverting the fraction that multiplies x. −2/5 comes from wrongly carrying the negative sign from the −3 term onto the coefficient of x.
- (d) 8/5 — Two masses can only be compared once they are in the same unit. Since 1 kg is 1000 g, the recipe needs 1200 g. The recipe's mass is being written as a fraction of Dan's mass, so 1200 goes on the top and 750 on the bottom, giving 1200/750. The highest common factor of the two is 150: 1200 ÷ 150 = 8 and 750 ÷ 150 = 5. The fraction is 8/5, which is greater than 1 because the recipe needs more flour than Dan has.
- (c) 62° — Co-interior (allied) angles on parallel lines sum to 180°, so the co-interior angle is 180° − 118° = 62°. 118° comes from treating the angles as corresponding angles, which are equal, instead of co-interior angles, which sum to 180°. 59° comes from halving the given angle. 236° comes from doubling the given angle.
- (c) 3.50 kg — Method: a mass given to the nearest kilogram lies within half a kilogram of the stated value, and a mass exactly halfway is rounded up. Working: rounding each mass to the nearest kilogram, 2.50 kg rounds up to 3 kg, 2.90 kg rounds to 3 kg and 3.49 kg rounds to 3 kg, so each of those could be the parcel; 3.50 kg is exactly halfway between 3 kg and 4 kg and so rounds up to 4 kg, which is not what the parcel was recorded as. Answer: 3.50 kg. The distractors: 2.50 kg is chosen by a candidate who rounds a value exactly halfway downwards, when the convention is to round it up; 2.90 kg is chosen by a candidate who thinks any mass below 3 kg must round down to 2 kg; 3.49 kg is chosen by a candidate who rounds twice, taking 3.49 to 3.5 first and then on to 4.
- (c) x = 3 and x = −5 — To factorise x² + 2x − 15, find two numbers that multiply to −15 and add to 2: these are 5 and −3, since 5 × (−3) = −15 and 5 + (−3) = 2. So x² + 2x − 15 = (x + 5)(x − 3). Setting each factor to zero gives x = −5 and x = 3. Choosing x = −3 and x = 5 comes from swapping the signs of the correct roots. Choosing x = 5 and x = 3 uses the right pair of numbers, 5 and 3, but forgets that one of them must be negative for the product to equal −15. Choosing x = −15 and x = 1 mistakes the constant term, −15, for one of the roots, and pairs it oddly with x = 1.
- (c) 5 : 9 — Write the ratio mass : cost = 3 : 5.4. Multiply both parts by 10 to clear the decimal: 30 : 54. Both numbers share a factor of 6, so 30 ÷ 6 = 5 and 54 ÷ 6 = 9, giving 5 : 9. 3 : 5 comes from ignoring the decimal point and treating £5.40 as £5. 9 : 5 comes from writing the ratio the wrong way round, cost to mass instead of mass to cost. 1 : 18 comes from multiplying only the cost by 10 instead of both parts, giving 3 : 54, and then cancelling that correctly to 1 : 18 — the cancelling is fine, but the ratio being cancelled is not the right one.
- (a) 1/3 — To subtract these fractions, first write 3/4 with a denominator of 12: 3/4 = 9/12. Then 9/12 − 5/12 = 4/12, which simplifies to 1/3. Subtracting the numerators and the denominators separately, (3 − 5)/(4 − 12), gives −2/−8, which simplifies to 1/4. Changing 3/4 to twelfths by only changing the denominator, without scaling the numerator to match, gives 3/12 − 5/12 = −2/12, which simplifies to −1/6. Adding the fractions instead of subtracting them, 9/12 + 5/12, gives 14/12, which simplifies to 7/6.
- (b) 35 N/m² — Pressure = force ÷ area. 84 ÷ 2.4 = 35 N/m². 201.6 N/m² comes from multiplying the force by the area instead of dividing (84 × 2.4). 81.6 N/m² comes from subtracting the area from the force (84 − 2.4) instead of dividing. 0.03 N/m² comes from dividing the area by the force instead of the force by the area (2.4 ÷ 84).
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.