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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- (c) 3 — Method: the index counts how many times the coefficient has been multiplied by 10, which is the number of places the decimal point moves from the end of the number to just after the first significant digit. Working: 2,000 = 2 × 1,000, and 1,000 = 10 × 10 × 10, which is three tens. Answer: 3. The distractors: 4 comes from counting the four digits of 2,000 rather than the three places the decimal point moves; 2 comes from copying the coefficient 2 into the index; −3 comes from making the index negative, which would describe a number smaller than 1 rather than two thousand.
- (d) an equation — Method: recall what each piece of algebraic vocabulary means and match it to the description given. Working: a statement with an equals sign, one unknown letter, and only particular value(s) making it true is, by definition, an equation. Answer: an equation. A formula also has an equals sign, but it links two or more different letters so that one quantity can be calculated from the others — here there is only one letter. An identity is a statement with an equals sign that is true for every value of the letter, not just particular ones. An expression has no equals sign at all, so it cannot be 'true' or 'false'.
- (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) 7.7 m — AB is parallel to DC, so those two sides are each parallel to another side. BC and AD are stated to be not parallel to each other, so neither one is parallel to any other side — these are the two sides that need edging. Adding these: 3.2 + 4.5 = 7.7 m, so 7.7 m is correct. 7.6 m comes from an arithmetic slip when adding 3.2 and 4.5. 15.4 m comes from doubling the correct total, mistakenly assuming edging strip is needed along both faces of each side. 4.5 m comes from using only the longer of the two non-parallel sides and forgetting to add the shorter one.
- (c) 4/5 — Add the counts of everyone who owns a cat, a dog, or both: 12 + 15 + 5 = 32 out of 40 people, which simplifies to 4/5. Leaving out the 5 people who own both, and adding only the two only-groups, gives 27/40. Using the 8 people who own neither, instead of everyone who owns at least one pet, gives 8/40 = 1/5. Counting the 5 people who own both twice, once alongside each only-group as well as on their own, gives 12 + 15 + 5 + 5 = 37 out of 40, or 37/40.
- (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.
- (c) 3 : 5 — The club is 8 equal shares and the girls take 3 of them, so the boys take 8 − 3 = 5 shares. A ratio compares one part with the other part, so girls : boys = 3 : 5. Writing 3 : 8 compares the girls with the whole club rather than with the boys, 5 : 3 puts the two parts in the wrong order, and 8 : 3 compares the whole club with the girls.
- (c) 5 — Method: substitute the value into the formula, work out the index first and the addition afterwards. Working: x² = (−2) × (−2) = 4, so y = 4 + 1 = 5. Answer: 5. The distractors: −3 comes from squaring only the 2 and keeping the minus sign outside the index, giving −4 + 1; 1 comes from adding the 1 to x before squaring, giving (−2 + 1)²; 4 comes from squaring correctly but stopping there and never adding the 1.
- (b) £50 — Method: add the parts of the ratio, divide the amount by the number of parts to find the value of one part, then multiply by the parts in the largest share. Working: 3 + 4 + 5 = 12 parts, £120 ÷ 12 = £10 for one part, and the largest share is 5 parts, so 5 × £10 = £50. Answer: £50. The distractors: £10 is the value of one part only; £30 is the 3-part share, which is the smallest one; £40 is the 4-part share, the middle one.
- (b) 2.8 m — Use Pythagoras' Theorem: the plank is the hypotenuse (3.5 m) of a right-angled triangle formed with the wall and the ground (2.1 m). height² = 3.5² − 2.1² = 12.25 − 4.41 = 7.84. height = √7.84 = 2.8 m. A student who subtracts the two given lengths directly instead of using Pythagoras gets 3.5 − 2.1 = 1.4 m. A student who doubles the distance from the wall by mistake gets 2.1 × 2 = 4.2 m.
- (a) 0.36 — Relative frequency is the number of times the event happened divided by the total number of trials: 18 ÷ 50 = 0.36. Dividing by 100 instead of the actual 50 spins gives 18 ÷ 100 = 0.18. Finding the relative frequency of NOT landing on green, using 50 − 18 = 32 spins, gives 32 ÷ 50 = 0.64. Misplacing the decimal point in the division, so that 18 ÷ 50 is carried out as 18 ÷ 500, gives 0.036 — a tenth of the correct value.
- (c) 9π cm² — The area of a circle is π × r². With a radius of 3 cm this is π × 3² = 9π cm², and this is exact because π has not been replaced by any approximation. Writing 28.3 cm² replaces π with a rounded decimal value, 3.14, and then rounds the result again, so it is only an approximation. Writing 28.26 cm² uses π ≈ 3.14 without a final rounding step, but this is still only an approximation of 9π, not the exact value. Writing 27 cm² comes from replacing π with the rough approximation 3, which is even further from the true value.
- (b) c = k/6 — k = 6c means c has been multiplied by 6, so to make c the subject, divide both sides by 6: c = k/6. Writing c = 6k multiplies by 6 again instead of undoing the multiplication. Writing c = k − 6 mistakes multiplying for adding, and subtracts 6 rather than dividing. Writing c = 6 − k reverses the order as well as the operation. The correct rearrangement is c = k/6.
- (c) 9:15:21 — 6 : 10 : 14 simplifies to 3 : 5 : 7 (divide every part by 2). Multiplying every part of 3 : 5 : 7 by 3 gives 9 : 15 : 21, so 9 : 15 : 21 is equivalent to 6 : 10 : 14. Adding 2 to every part of 6 : 10 : 14 gives 8 : 12 : 16, which is not equivalent — ratios are equivalent when every part is multiplied by the same number, not when the same number is added to every part. Doubling only the first two parts, 6 × 2 = 12 and 10 × 2 = 20, but leaving the third part unchanged at 14, gives 12 : 20 : 14 — a scaling applied to two parts and not the third. Cancelling the first two parts correctly, 6 ÷ 2 = 3 and 10 ÷ 2 = 5, then treating the three numbers as a sequence and making the third part the sum of the first two, 3 + 5 = 8, gives 3 : 5 : 8 — the third part was never divided by 2 at all.
- (d) (6, 2) — Applying the first vector: (2, 1) + (5, −3) = (7, −2), which is the warehouse. Applying the second vector: (7, −2) + (−1, 4) = (6, 2), the delivery address. '(7, −2)' stops at the warehouse and forgets the second flight. '(8, −6)' comes from adding (1, −4) instead of (−1, 4) for the second vector, getting both signs wrong. '(11, −3)' comes from swapping the components of the second vector to (4, −1) before adding.
- (a) 3/13 — Vegetables are 9 of the 9 + 4 = 13 parts, so vegetables are 9/13 of the plot. Potatoes are a third of the vegetable section, so potatoes are 1/3 of 9/13, which is 9/39, simplifying to 3/13, of the whole plot. 9/13 comes from stopping after finding the fraction of the plot that is vegetables, without taking the further third for potatoes. 1/3 gives the fraction of the vegetable section that is potatoes, not the fraction of the whole plot. 4/39 comes from taking a third of the flowers' fraction, 4/13, instead of the vegetables' fraction.
- (a) 1 — Method: the value of y when x = 0 is where the line meets the y-axis, which is the constant c in y = mx + c, so the gradient is worked out from the two given points first and the constant follows by substituting one of them. Working: m = (16 − 7) ÷ (5 − 2) = 9 ÷ 3 = 3, so the line is y = 3x + c; substituting x = 2 and y = 7 gives 7 = 3 × 2 + c, so c = 7 − 6 = 1, and the value of y when x = 0 is that constant. Answer: 1. The distractors: 3 comes from stopping at the gradient and offering it as the intercept; 4 comes from stepping back from x = 2 to x = 0 by one unit of x instead of two, 7 − 3 = 4; −1 comes from working the constant out as mx − y, 3 × 2 − 7 = −1, instead of y − mx.
- (d) 35 — Method: in direct proportion the ratio y : x is the same for every pair, so find the constant and substitute the new value of x. Working: k = 20 ÷ 8 = 2.5, so y = 2.5x; when x = 14, y = 2.5 × 14 = 35. Answer: 35. The distractors: 26 comes from additive thinking — x rises by 6, so 6 is added to y — which would keep the difference constant rather than the ratio; 28 comes from rounding the constant 2.5 down to 2 and working out 2 × 14, which loses the half in the constant; 5.6 comes from using the constant upside down, 8 ÷ 20 = 0.4, and working out 0.4 × 14.
- (a) 5 — Method: each index is worked out first, and subtracting a negative number is the same as adding the positive. Working: (−2)³ = (−2) × (−2) × (−2) = −8 and (−3)² = (−3) × (−3) = 9, while − (−4) becomes + 4, so the calculation becomes −8 + 9 + 4 = 5. Answer: 5. The distractors: −13 comes from taking (−3)² as −9, giving −8 − 9 + 4 = −13; −3 comes from reading − (−4) as − 4, giving −8 + 9 − 4 = −3; 21 comes from treating every power of a negative number as positive, so that (−2)³ is taken as 8 and the calculation becomes 8 + 9 + 4 = 21.
- (d) 75 pages — Method: find the number of pages printed in one minute, then scale up to 10 minutes. Working: 45 ÷ 6 = 7.5 pages per minute, so 7.5 × 10 = 75 pages. Answer: 75 pages. 27 pages comes from using the ratio upside down, 45 × 6 ÷ 10, instead of finding the rate per minute first. 55 pages comes from simply adding the extra minutes, 10, onto the original number of pages, 45. 70 pages comes from rounding the rate down to 7 pages per minute before multiplying by 10, instead of using the exact rate of 7.5.
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.