Sample paper · GCSE Foundation · grades 1–5
GCSE Foundation sample Paper 2 (calculator)
The real Paper 2 is 1.5 hour 30 minutes and 80 marks, 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 a calculator allowed. AO1 / AO2 / AO3 at this tier: 50% / 25% / 25%.
CalculatorGCSE Foundation
GCSE Foundation sample Paper 2 (calculator)
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- 1.A square petri dish has an area of 121 mm². A scientist wants to know its side length. Work out the side length of the dish.
- 2.The formula for the volume of a cuboid is V = lwh, where l is the length, w is the width and h is the height. Make h the subject of the formula.
- 3.A ratio is 3 : 5. Write the first part of the ratio as a fraction of the whole.
- 4.A patio plan states: 'Side AB of the patio is parallel to side DC, and AB is longer than DC. Sides BC and AD are not parallel to each other and are not equal in length.' The four sides measure AB = 5.1 m, BC = 3.2 m, DC = 2.8 m and AD = 4.5 m. The tiler needs to order edging strip only for the sides that are not parallel to any other side. What total length of edging strip, in metres, should the tiler order?
- 5.A library recorded loans for 250 books over a week, using a frequency tree. The first branch splits the books into 160 fiction and 90 non-fiction. Of the fiction books, 112 were returned on time and the rest were returned late. All the non-fiction books were returned on time. Work out the probability that a book, chosen at random from the 250, was returned late. Give your answer as a fraction in its simplest form.
- 6.A vertical line chart shows the number of pets owned by each pupil in a class: 0 pets — 5 pupils, 1 pet — 4 pupils, 2 pets — 3 pupils, 3 pets — 3 pupils. Work out the total number of pupils in the class.
- 7.The attendance at a football match is 23,000, correct to the nearest thousand. Which of these attendances is not possible for the match?
- 8.Isla says that the lines y = 2x + 1 and y = −2x + 3 are parallel. Write down the statement that gives the correct answer for the correct reason.y = 2x + 1y = -2x + 3
- 9.y is inversely proportional to x. When x = 5, y = 8. Work out the value of y when x = 10.
- 10.Two straight lines cross at a point. Which statement about a pair of vertically opposite angles is always true?
- 11.Priya spins a fair spinner with P(red) = 0.3, and separately flips a fair coin with P(heads) = 0.5. Using a tree diagram, work out the probability that she gets red AND heads.
- 12.Round 6.283 to 1 significant figure.
- 13.Write down the expression that means the same as w ÷ 6.
- 14.Work out 50% of 60.
- 15.Triangles PQS and RQS share the common side QS. PQ equals RQ, and PS equals RS. Which condition proves that triangle PQS is congruent to triangle RQS?
- 16.A number line runs from 0 to 1 and is divided into 4 equal parts. Which mark on the line represents 1/4?
- 17.Write down the expression that means the same as 3 × y × y.
- 18.A delivery driver travels 45 km in the first 30 minutes of a journey, and then a further 75 km in the next 1 hour. Work out her average speed for the whole journey, in km/h.
- 19.Work out −(−3)⁴ + (−3)³
- 20.A shop sells fruit juice concentrate that must be diluted with water in the ratio 2 : 5 (concentrate : water) to make squash. Priya has 8 litres of concentrate. Work out how much water she needs to keep the same ratio.
Answer key
- (d) 11 mm — Method: for a square, the side length is the square root of the area. Working: 11 × 11 = 121, so the side length is 11 mm. 60.5 mm comes from working out 121 ÷ 2 = 60.5, halving the area instead of finding its square root. 242 mm comes from working out 121 × 2 = 242, doubling the area instead of finding its square root. 22 mm comes from working out 11 × 2 = 22, doubling the correct side length. Answer: 11 mm.
- (c) h = V / (lw) — Method: undo the multiplication by lw by dividing both sides by lw. Working: V = lwh, so dividing both sides by lw gives h = V / (lw). The value h = V − lw comes from subtracting lw instead of dividing by it. The value h = Vlw comes from multiplying by lw instead of dividing. The value h = lw / V comes from inverting the fraction, dividing lw by V instead of V by lw.
- (d) 3/8 — The ratio 3 : 5 has 3 + 5 = 8 parts in total. The first part as a fraction of the whole is 3 out of 8, or 3/8. Writing 3/5 gives the first part compared to the second part, not to the whole. Writing 5/8 gives the second part as a fraction of the whole, not the first. Writing 8/3 has the fraction upside down — the whole must be on the bottom.
- (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.
- (b) 24/125 — On the fiction branch, 160 − 112 = 48 books were returned late. None of the non-fiction books were late, so the total number of late books is 48, out of 250 books altogether: 48/250 = 24/125. Writing 3/10 is wrong because it divides the 48 late fiction books by the fiction total (160) instead of the whole library total (250). Writing 56/125 is wrong because 112/250 simplifies to 56/125, and 112 is the number of fiction books returned ON TIME, not late. Writing 9/25 is wrong because 90/250 simplifies to 9/25, and 90 is simply the number of non-fiction books, which has nothing to do with late returns. The probability is 24/125.
- (b) 15 pupils — The frequencies are 5, 4, 3 and 3 pupils, and 5 + 4 + 3 + 3 = 15, so there are 15 pupils in the class. Choosing 6 pupils comes from adding the numbers of pets, 0 + 1 + 2 + 3 = 6, instead of the frequencies. Choosing 10 pupils comes from leaving out the '0 pets' row: 4 + 3 + 3 = 10. Choosing 18 pupils comes from counting the '3 pets' row twice: 5 + 4 + 3 + 3 + 3 = 18.
- (d) 23,600 — 23,000 correct to the nearest thousand means the true attendance, a, satisfies 22,500 ≤ a < 23,500. 23,600 is outside this range, so it is not possible. 22,500 is the lower bound of the range, so it is possible. 23,499 is just below the upper bound, so it is possible. 23,000 is the given value itself, so it is possible.
- (b) No, because their gradients are 2 and −2 — Method: two lines are parallel exactly when their gradients are equal as signed numbers, so m is read from each equation written in the form y = mx + c and the two are compared. Working: y = 2x + 1 has gradient 2 and y = −2x + 3 has gradient −2; those are not equal, so the lines are not parallel, and indeed one slopes upwards while the other slopes downwards. Answer: No, because their gradients are 2 and −2. The distractors: saying yes because both gradients have size 2 comes from comparing the sizes of the gradients and ignoring their signs; saying yes because the gradients add to 0 comes from using a sum of zero as the test for parallel lines instead of equality of gradients; saying no because the y-intercepts are 1 and 3 reaches the right verdict by the wrong route, since the intercepts decide where the lines sit rather than whether they are parallel.
- (a) 4 — Method: for inverse proportion, x × y always stays the same value. Working: when x = 5 and y = 8, the constant is 5 × 8 = 40. When x = 10, y = 40 ÷ 10 = 4. So y = 4. Distractor 16 comes from treating the relationship as direct proportion instead of inverse, working out 8 × 10 ÷ 5. Distractor 3 comes from assuming y decreases by the same amount that x increases, an additive rather than proportional idea. Distractor 0.8 comes from dividing the given y-value, 8, by the new x-value, 10, without first finding the constant.
- (d) They are always equal to each other — Method: label the four angles a, b, c, d in order around the crossing point, then use the fact that neighbouring angles lie on a straight line. Working: a and b lie on a straight line, so a + b = 180°; b and c also lie on a straight line, so b + c = 180°. Both a and c are therefore 180° minus b, which forces a = c. Answer: a pair of vertically opposite angles is always equal to each other. The distractors: the claim that each is 90° holds only when the two lines happen to be perpendicular, so it is not always true; the claim that they add to 180° confuses the opposite pair with the neighbouring pair that lies along a straight line, and again holds only in the perpendicular case; the claim that they add to 360° uses the total of all four angles at the point rather than of the opposite pair.
- (c) 0.15 — Method: for two independent events, multiply along the branches of the tree to find the probability of both outcomes happening together. Working: P(red and heads) = P(red) × P(heads) = 0.3 × 0.5 = 0.15. Answer: 0.15. Watch out: adding the two probabilities, 0.3 + 0.5 = 0.8, does not give the probability of both — probabilities along one path of a tree are multiplied, not added. Writing down 0.5 ignores the spinner altogether and gives only the coin's probability. And writing down 0.65 is the probability of red OR heads, which is 0.3 + 0.5 − 0.15 = 0.65, a different question from the one asked here.
- (c) 6 — Method: the first significant figure is the first non-zero digit; round using the digit after it to decide whether to round up or down. Working: the first significant figure of 6.283 is the 6; the next digit is 2, which rounds down, so 6.283 rounds to 6. 6.3 comes from rounding to 2 significant figures instead of 1. 10 comes from rounding up to the nearest 10 instead of finding 1 significant figure of the number itself. 0.6 comes from misplacing the decimal point after rounding. Answer: 6.
- (c) w/6 — Method: a ÷ b is written as a fraction a/b, with the number being divided (w) on top. Working: w ÷ 6 = w/6. Answer: w/6. 6/w comes from writing the numbers the wrong way round, putting the 6 on top instead of w. 6w comes from reading the ÷ sign as ×, multiplying instead of dividing. w − 6 comes from reading ÷ as −, subtracting instead of dividing.
- (d) 30 — Method: 50% is one half, so 50% of a quantity is the quantity divided by 2. Working: 60 ÷ 2 = 30. Answer: 30. The distractors: 120 comes from multiplying by 2 instead of dividing; 3000 comes from multiplying by 50 without dividing by 100; 6 comes from finding 10% instead of 50%.
- (d) SSS, using shared side QS — PQ equals RQ and PS equals RS are two given pairs of equal sides, and QS is common to both triangles, so QS equals itself and gives a third pair of equal sides. Three pairs of equal sides is exactly the SSS condition, so 'SSS, using shared side QS' is correct. 'SAS, using the angle at Q' is wrong because no angle is given anywhere in this question; angle PQS and angle RQS are not stated to be equal, and assuming they are would be assuming the very thing being proved. 'Only two pairs of sides — not enough' is wrong because it forgets that the shared side QS is itself a third pair of equal sides. 'Cannot prove — no angle given' is wrong because SSS is one of the four basic congruence conditions and specifically requires no angle at all.
- (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) 3y² — There is one 3 in the product and the letter y appears twice. A number that multiplies a letter is written in front of it as the coefficient, so the 3 goes at the front. The two ys multiply each other, so the letter carries an index of 2. The shortest correct way of writing the product is therefore 3y². Note that (3y)² would mean 3 × y × 3 × y, which squares the coefficient as well, while 3 + y² turns a multiplication into an addition and y³ treats the coefficient as a third factor of y.
- (a) 80 km/h — Average speed = total distance ÷ total time. Total distance = 45 + 75 = 120 km. Total time = 30 minutes + 1 hour = 1.5 hours. 120 ÷ 1.5 = 80 km/h. 60 km/h comes from treating the 30 minutes as a whole hour, giving a total time of 2 hours instead of 1.5 (120 ÷ 2). 82.5 km/h comes from averaging the two separate speeds (45 ÷ 0.5 = 90 km/h and 75 ÷ 1 = 75 km/h, then (90 + 75) ÷ 2) instead of using total distance over total time. 75 km/h comes from using only the second part of the journey (75 km in 1 hour) and ignoring the first part.
- (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.
- (d) 20 litres — The ratio of concentrate to water is 2 : 5, so water = concentrate × 5 ÷ 2. 8 × 5 ÷ 2 = 20, so Priya needs 20 litres of water. Giving 40 litres multiplies by 5 but forgets to divide by 2 (8 × 5 = 40). Giving 3.2 litres uses the ratio inverted, multiplying by 2 ÷ 5 instead of 5 ÷ 2 (8 × 2 ÷ 5 = 3.2). Giving 11 litres uses additive reasoning instead of multiplicative: it adds the difference between the ratio parts, 5 − 2 = 3, onto the amount of concentrate (8 + 3 = 11), but ratios scale by multiplying, not by adding a fixed amount.
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.