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%.

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GCSE Foundation sample Paper 2 (calculator)

Maths · GCSE Foundation · Paper 2 · Calculator · 20 questions

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1.A shop buys boxes of pens for £4 each. It sells each box on to a school for £3 more than it paid, and charges an extra £2 delivery fee for the whole order. A school orders 5 boxes. Work out the total cost of the order.
(a)£45
(b)£37
(c)£25
(d)£22
2.The graph of y = x² − 5x + 6 crosses the x-axis at two points. By factorising, work out the x-coordinates of these two points.
y = x² − 5x + 6
(a)x = −1 and x = −6
(b)x = 2 and x = −3
(c)x = 2 and x = 3
(d)x = −2 and x = −3
3.Two numbers are in the ratio 3 : 7. Their sum is 60. Work out the positive difference between the two numbers.
(a)80
(b)60
(c)4
(d)24
4.Triangle ABC is isosceles, with AB = AC. Angle B = 70°. Which reason correctly explains why angle C = 70°?
(a)vertically opposite angles are equal
(b)corresponding angles are equal
(c)base angles of an isosceles triangle are equal
(d)angles in a triangle add up to 180°
5.A quality inspector examines a sample of 60 items from a production line and finds that 8 are faulty. Using this sample's proportion, work out how many faulty items would be expected in a new batch of 750 items.
(a)1
(b)100
(c)8
(d)75
6.The table shows the times, t minutes, that 30 pupils took to walk to school. 0 < t ≤ 10: 8 pupils. 10 < t ≤ 20: 12 pupils. 20 < t ≤ 30: 6 pupils. 30 < t ≤ 40: 4 pupils. Write down which average can be given exactly from this table, and give a reason for your answer.
(a)The mean, as every one of the 30 times is in the table
(b)The range, as the lowest and highest times are shown
(c)The median, as the middle time is shown in the table
(d)The modal class, as the class with most pupils is shown
7.A charity trek covers 830 miles over roughly 19 days. By rounding each number to 1 significant figure, work out an estimate for the number of miles walked per day.
(a)4 miles
(b)830 miles
(c)41.5 miles
(d)40 miles
8.Work out the coordinates of the midpoint of the line segment joining (−3, 5) and (7, 9).
(a)(2, 7)
(b)(4, 14)
(c)(5, 2)
(d)(2, 9)
9.A toy manufacturer makes a model aircraft that is mathematically similar to the real aircraft, at a scale of 1 : 48. The wingspan of the model is 15 cm. Work out the wingspan of the real aircraft, giving your answer in metres.
(a)72 m
(b)7.2 m
(c)0.72 m
(d)0.31 m
10.A solid cube has a volume of 8 cm³. Work out the length of one edge of the cube.
(a)4 cm
(b)2 cm
(c)2.83 cm
(d)8 cm
11.A survey of 160 employees at a company recorded whether they cycle to work, using a frequency tree. The first branch splits them into 90 who work full-time and 70 who work part-time. Of the full-time employees, 27 cycle to work. Of the part-time employees, 14 cycle to work. Work out the probability that an employee, chosen at random from the 160, cycles to work. Give your answer as a fraction in its simplest form.
(a)27/160
(b)41/160
(c)1/5
(d)41/90
12.Write 0.005 in standard form.
(a)5 × 10³
(b)5 × 10⁻²
(c)5 × 10⁻³
(d)0.5 × 10⁻²
13.A sequence begins at 7. Each term after the first is found by adding 6 to the term before it. Work out the 6th term of the sequence.
(a)43
(b)37
(c)25
(d)31
14.A shop's takings were £4500 in May. In June the takings fell by £900. Write the June takings as a fraction of the May takings, in its simplest form.
(a)4/5
(b)1/5
(c)5/4
(d)4/9
15.Triangle LMN has a right angle at M, with hypotenuse LN = 15 cm and LM = 9 cm. Triangle PQR has a right angle at Q, with hypotenuse PR = 15 cm and PQ = 9 cm. Which condition proves the two triangles are congruent?
(a)SSS
(b)SAS
(c)RHS
(d)ASA
16.An allotment plot is divided into vegetables and flowers in the ratio 9 : 4. A third of the vegetable section is used for potatoes. What fraction of the whole plot is potatoes?
(a)3/13
(b)4/39
(c)1/3
(d)9/13
17.A plumber charges a call-out fee plus an hourly rate. The total cost, y in pounds, of a job lasting x hours is given by y = 45x + 60. Work out the total cost of a job that lasts 3 hours.
y = 45x + 60
(a)£315
(b)£108
(c)£135
(d)£195
18.A ribbon of length 90 cm is cut into two pieces in the ratio 4:5. Work out the length of the shorter piece.
(a)50 cm
(b)10 cm
(c)40 cm
(d)45 cm
19.During one night on a mountain, four hikers each recorded the temperature they felt: Ben −6 °C, Priya −2 °C, Sam −9 °C, Alex 1 °C. Work out the difference between the coldest and the warmest of these four temperatures.
(a)10 °C
(b)3 °C
(c)8 °C
(d)7 °C
20.Ollie says that the ratio 6 : 15 is equivalent to the ratio 2 : 5. Is he correct? Give a reason for your answer.
(a)Yes — dividing both parts of 6 : 15 by 3 gives 2 : 5
(b)No — ratios must start with an even number to simplify
(c)No — 15 ÷ 6 isn't whole, so it can't simplify
(d)No — 6 : 15 simplifies to 3 : 5, not 2 : 5

Answer key

(b) £37 — One box costs £4 + £3 = £7. Five boxes cost 5 × £7 = £35. Adding the single £2 delivery fee gives £35 + £2 = £37. A candidate who added the £2 delivery fee to each box instead of once for the whole order worked out 5 × (£7 + £2) = 5 × £9 = £45. A candidate who forgot the £3 markup and used the shop's buying price worked out 5 × £4 + £2 = £22. A candidate who added the £3 markup only once, after multiplying the buying price by 5, worked out 5 × £4 + £3 + £2 = £25.
(c) x = 2 and x = 3 — x² − 5x + 6 factorises as (x − 2)(x − 3), since −2 and −3 multiply to give 6 and add to give −5. The graph crosses the x-axis where each bracket equals zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. The option x = −2 and x = −3 comes from reading the signs inside the brackets directly instead of solving x − 2 = 0 and x − 3 = 0. The option x = 2 and x = −3 mixes up the sign of only one root. The option x = −1 and x = −6 comes from picking the wrong pair of factors of 6 (1 and 6 instead of 2 and 3) and then reading their signs directly from the brackets.
(d) 24 — Method: split 60 into 3 + 7 = 10 equal parts, find the value of one part, then use the difference in ratio parts. Working: 60 ÷ 10 = 6, so the numbers are 3 × 6 = 18 and 7 × 6 = 42, and their difference is 42 − 18 = 24. Answer: 24. 4 comes from finding the difference between the ratio numbers, 7 − 3, but forgetting to multiply by the value of one part. 60 comes from adding the two numbers back together instead of subtracting, which just repeats the given sum. 80 comes from dividing 60 by the first ratio number, 3, instead of by the total number of parts, 10, giving a part value of 20 and a difference of 7 × 20 − 3 × 20 = 80.
(c) base angles of an isosceles triangle are equal — Because AB = AC, triangle ABC is isosceles, and the base angles of an isosceles triangle — the two angles opposite the equal sides — are always equal, which is why angle C equals angle B, 70°. Angles in a triangle adding up to 180° is a true fact about the triangle as a whole, but it is not the reason two specific angles are equal to each other. Corresponding angles are equal is a fact about parallel lines cut by a transversal, which does not apply inside a single triangle like this. Vertically opposite angles are equal is a fact about two lines crossing, not about a triangle's base angles.
(b) 100 — The sample shows a proportion of 8/60 = 2/15 faulty. Apply that proportion to the new batch of 750: 750 × 2/15 = 100. Flipping the ratio, calculating 8/750 × 60 instead of 8/60 × 750, gives 0.64, which rounds to about 1. Assuming the same number of faulty items applies to the new batch, without scaling for its larger size, just repeats the sample's count of 8. Rounding the proportion 8/60 = 0.1333... down to 0.1 before multiplying gives 750 × 0.1 = 75.
(d) The modal class, as the class with most pupils is shown — Method: a grouped frequency table records how many values fall into each class, but not the values themselves, so any average that needs the individual times can only be estimated from it. Working: the four frequencies are 8, 12, 6 and 4, and 8 + 12 + 6 + 4 = 30, so every pupil is counted. The largest frequency is 12, which belongs to the class 10 < t ≤ 20, and that class can be written down exactly, because finding it needs nothing but the counts the table already gives. Answer: the modal class, as the class with most pupils is shown. The distractors: the mean is said to use all 30 times, but the table does not hold them; the usual method replaces each class by its midpoint, 5, 15, 25 and 35, which gives an estimate of the mean and not its true value; the median is said to be shown, but the table locates only the class holding the 15th and 16th times, which is 10 < t ≤ 20, without saying what either time was; the range is said to be shown, but 0 and 40 are the boundaries of the first and last classes, not the fastest and slowest times actually recorded.
(d) 40 miles — Method: round each number to 1 significant figure first, then divide to estimate the daily distance. Working: 830 rounds to 800, and 19 rounds to 20, and 800 ÷ 20 = 40, so the estimate is 40 miles per day. 41.5 miles comes from rounding only the number of days and working out 830 ÷ 20 = 41.5, without rounding the distance too. 830 miles is the total distance for the whole trek, given as the answer without dividing by the number of days at all. 4 miles comes from working out 80 ÷ 20 = 4, misplacing a digit in the rounded distance. Answer: 40 miles.
(a) (2, 7) — Midpoint = ((x1+x2)/2, (y1+y2)/2) = ((−3+7)/2, (5+9)/2) = (4/2, 14/2) = (2, 7). (4, 14) comes from adding the coordinates correctly but forgetting to divide by 2. (2, 9) comes from correctly averaging the x-coordinates but simply copying the y-coordinate of the second point instead of averaging the y-coordinates. (5, 2) comes from subtracting the coordinates instead of adding them before halving: ((7−(−3))/2, (9−5)/2) = (5, 2).
(b) 7.2 m — Multiply the model wingspan by the scale factor: 15 × 48 = 720. This is in centimetres, and 720 cm = 7.2 m, since 1 m = 100 cm. Giving 0.31 m divides by the scale factor instead of multiplying (15 ÷ 48 ≈ 0.31), scaling the model down rather than the real aircraft up. Giving 72 m converts centimetres to metres by dividing by 10 instead of 100. Giving 0.72 m converts by dividing by 1000 instead of 100.
(b) 2 cm — Method: the volume of a cube is edge × edge × edge, so finding the edge from the volume means undoing a cube, not a square or a halving. Working: edge × edge × edge = 8; testing whole numbers, 1 × 1 × 1 = 1 is too small and the next whole number gives 2 × 2 × 2 = 8, which matches. Answer: 2 cm. The distractors: 8 cm comes from writing the volume down as the edge length and changing only the unit; 4 cm comes from halving the volume, 8 ÷ 2, as though a cube were undone by dividing by 2; 2.83 cm is the square root of 8, from taking a square root where a cube root is needed.
(b) 41/160 — In total, 27 + 14 = 41 of the 160 employees cycle to work, so the probability is 41/160 (41 and 160 share no common factor, so this is already in its simplest form). Writing 27/160 is wrong because it only counts the full-time cyclists and leaves out the 14 part-time cyclists. Writing 41/90 is wrong because it uses the full-time total (90) as the denominator instead of the whole survey (160). Writing 1/5 is wrong because it only uses the part-time branch, simplifying 14/70 to 1/5 and ignoring the full-time cyclists completely. The probability is 41/160.
(c) 5 × 10⁻³ — Method: in standard form the coefficient must be at least 1 and less than 10, and for a number smaller than 1 the index is negative and counts the places the decimal point moves to the right. Working: the only significant digit is 5, so the coefficient is 5; moving the decimal point in 0.005 three places to the right gives 5, so the index is −3. Answer: 5 × 10⁻³. The distractors: 0.5 × 10⁻² is the same value written the wrong way, because 0.5 is smaller than 1 and so is not an allowed coefficient; 5 × 10⁻² comes from counting the two zeros after the decimal point instead of the three places the point moves; 5 × 10³ comes from taking the index as positive, which describes a number in the thousands rather than one smaller than 1.
(b) 37 — The terms are 7, 13, 19, 25, 31, 37 — each found by adding 6 to the term before, so the 6th term is 37. Adding 6 six times to the first term instead of five times, 7 + 6 × 6 = 43, treats the first term as if it were before the sequence starts. Stopping one term early gives the 5th term, 31. Stopping two terms early gives the 4th term, 25.
(a) 4/5 — Method: find the June takings first, then write them over the May takings and cancel. Working: the takings fell by £900, so June is £4500 − £900 = £3600; the fraction is 3600/4500, and dividing the numerator and the denominator by 900 gives 4/5. Answer: 4/5 of the May takings. The distractors: 1/5 comes from writing the fall over the May takings, 900/4500, which answers how far the takings dropped rather than what June's takings are compared with May's; 5/4 comes from writing May over June, 4500/3600, reversing the order the question asks for; 4/9 comes from writing June over the two months added together, 3600/8100, a part-to-whole fraction when the comparison asked for is with May alone.
(c) RHS — Method: check which basic congruence condition matches the facts given — a right angle, the hypotenuse, and one other side, in both triangles. Working: both triangles have a right angle (at M and Q), the hypotenuse is given for both (LN = PR = 15 cm), and one other side is given for both (LM = PQ = 9 cm) — this is exactly Right angle, Hypotenuse, Side. Options: SAS would need the given angle to sit between the two given sides, but the right angle at M is not between LM and LN, since LN is the hypotenuse, opposite the right angle; SSS would need three sides given in each triangle, but only two sides are known here; ASA would need two angles and the side between them, but only one angle is given. Answer: RHS.
(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.
(d) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
(c) 40 cm — Method: split the total length into the number of parts shown by the ratio, then find the value of the shorter share. Working: the ratio 4:5 has 4 + 5 = 9 parts, so one part is 90 ÷ 9 = 10 cm, and the shorter piece is 4 × 10 = 40 cm. So the shorter piece is 40 cm. Distractor 50 cm is the length of the LONGER piece, not the shorter one. Distractor 45 cm comes from splitting the ribbon into two equal halves, ignoring the ratio. Distractor 10 cm is the value of one part, found correctly but never multiplied by 4.
(a) 10 °C — Method: work out the coldest and warmest of the four temperatures, then subtract to find the difference. Working: the coldest temperature is Sam's, −9 °C, and the warmest is Alex's, 1 °C. The difference is 1 − (−9) = 1 + 9 = 10. Answer: 10 °C. 8 °C comes from working out 1 − 9 = −8 and reporting 8, dropping the negative sign on −9 instead of turning the subtraction into an addition. 3 °C comes from comparing the wrong pair, Sam's −9 °C and Ben's −6 °C, instead of the coldest and the warmest: −6 − (−9) = 3. 7 °C comes from comparing Ben's −6 °C with Alex's 1 °C, mistakenly treating Ben's reading as the coldest instead of Sam's.
(a) Yes — dividing both parts of 6 : 15 by 3 gives 2 : 5 — Method: divide both parts of the ratio by their highest common factor and compare. Working: the highest common factor of 6 and 15 is 3. 6 ÷ 3 = 2 and 15 ÷ 3 = 5, giving 2 : 5, so Ollie is correct. Wrong options: '6 : 15 simplifies to 3 : 5' divides incorrectly, giving the wrong simplified ratio; 'cannot simplify a ratio that does not start with an even number' states a false rule — any ratio can be simplified if its parts share a common factor; '15 ÷ 6 is not a whole number, so it cannot be simplified' wrongly tries to divide one part by the other instead of finding a common factor.

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.