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)
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- 1.In standard form, 2,000 is written as 2 × 10ⁿ. Write down the value of n.
- 2.Solve the simultaneous equations 2x + y = 7 and x − y = 2.
- 3.Write 350 ml : 1.4 l as a ratio in its simplest form.
- 4.Which of the following is NOT one of the four basic conditions used to prove two triangles are congruent?
- 5.A charity tombola has 30 tickets, 6 of which win a prize. Priya buys a ticket at random and does not return it. Her friend Tom then buys a second ticket at random from the remaining tickets. Work out the probability that both Priya and Tom win a prize.
- 6.A pictogram shows the number of cars sold by a garage in Plymouth each month, where each whole car symbol represents 8 cars sold and a half symbol represents 4 cars. March shows 3 whole symbols and one half symbol. Work out how many cars were sold in March.
- 7.Work out 3³ + 2⁴.
- 8.Priya has a budget of £50 for a school trip. The coach costs £14 and each student ticket costs £4. Using the inequality 14 + 4s ≤ 50, work out the greatest number of student tickets, s, she can buy.
- 9.Leah puts £4000 into a savings account paying 3% compound interest each year. At the end of 2 years she takes out all of the money and spends £1500 of it on a laptop. Work out how much of the money she has left.
- 10.A map has a scale of 1 : 25000. A distance between two points measures 2 cm on the map. What is the real distance, in kilometres?
- 11.A survey of 60 readers records: 21 read only fiction, 17 read only non-fiction, 14 read both, and the rest read neither. Work out the probability that a randomly chosen reader reads exactly one of the two types, giving your answer as a fraction in its simplest form.
- 12.Work out an estimate for 89 + 52, by rounding each number to the nearest 10.
- 13.Which of these values of x is a solution of x² + 2x − 15 = 0?
- 14.A spring's extension is directly proportional to the force applied to it. A force of 5 N produces an extension of 12 mm. Work out the extension produced by a force of 20 N.
- 15.Triangle ABC is extended so that side BC is extended beyond C to a point D, forming exterior angle ACD. Angle BAC = 48° and angle ABC = 67°. Which single angle fact gives angle ACD = 115° in one step, without first working out the interior angle at C?
- 16.Work out −7 − (−3).
- 17.Make x the subject of the formula y = (x − 4)/5.
- 18.A scale drawing states '1 cm represents 25 m'. Which of these is the correct way to write this scale in the form 1 : n?
- 19.Work out √144
- 20.A car travels for 3 hours at an average speed of 80 km/h and then for 1 hour at an average speed of 40 km/h. Work out the average speed for the whole journey.
Answer key
- (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.
- (c) x = 3, y = 1 — Method: the y terms are +y and −y, so adding the two equations removes y and leaves an equation in x alone. Working: adding 2x + y = 7 and x − y = 2 gives 3x = 9, so x = 3; substituting x = 3 into x − y = 2 gives 3 − y = 2, so y = 1. Answer: x = 3, y = 1, which also satisfies 2 × 3 + 1 = 7. The distractors: x = 1, y = 3 comes from finding the two values correctly and then writing them against the wrong letters; x = 3, y = 2 comes from substituting x = 3 into 2x + y = 7 as 2 + 3 + y = 7, adding the coefficient instead of multiplying by it; x = 3, y = −1 comes from substituting into x − y = 2 as though it read x + y = 2.
- (d) 1:4 — Convert 1.4 l to millilitres: 1.4 l = 1400 ml. The ratio is 350 : 1400. Divide both parts by 350: 350 ÷ 350 = 1 and 1400 ÷ 350 = 4, giving 1 : 4. Misreading 1.4 l as 14 (moving the decimal point) gives 350 : 14, which simplifies to 25 : 1 — a very different, implausible ratio. Dividing by 175 instead of 350 gives 2 : 8, which still shares a common factor of 2, so it is not fully simplified. Swapping the order gives 4 : 1, litres to millilitres the wrong way round.
- (c) SSA — Method: recall the four basic congruence criteria for triangles and compare them with each option. Working: the four basic conditions are SSS, SAS, ASA and RHS. SSA (two sides and a non-included angle) is not one of them, because knowing two sides and an angle that is not between them does not always fix a unique triangle. Options: SSS, ASA and RHS are each genuine congruence conditions, so none of them is the correct answer to 'which is NOT'. Answer: SSA.
- (a) 1/29 — The probability that Priya's ticket wins is 6/30. Since her ticket is not returned, there are now only 5 winning tickets left out of 29 tickets in total, so the probability that Tom's ticket also wins is 5/29. Multiplying these, 6/30 × 5/29 = 30/870 = 1/29. A candidate who answers 1/25 has treated Priya's ticket as returned, using 6/30 twice. A candidate who answers 1/30 has correctly reduced the winning tickets to 5 for Tom but forgotten to reduce the total number of tickets, using 5/30 instead of 5/29. A candidate who answers 11/59 has added the numerators and added the denominators, (6+5)/(30+29), instead of multiplying.
- (c) 28 — Method: multiply the number of whole symbols by the value of one symbol, then add the value of any half symbol shown. Working: 3 whole symbols represent 3 × 8 = 24 cars. The half symbol represents 4 cars. Total cars sold in March = 24 + 4 = 28. Leaving out the half symbol, 3 × 8 = 24, undercounts by exactly the value of that half symbol. Treating the half symbol as if it were a full symbol, 4 × 8 = 32, overcounts because it doubles the value the half symbol is worth. Giving 3.5 reports the number of symbols shown, not the number of cars they represent — the key still needs to be applied. Always apply the key to every symbol shown, including a half symbol, rather than reading off the symbol count itself.
- (c) 43 — Method: work out each power separately before adding. Working: 3³ = 27 and 2⁴ = 16, so 3³ + 2⁴ = 27 + 16 = 43. Answer: 43. (25 comes from using 3² instead of 3³, giving 9 + 16. 35 comes from working out 2⁴ as 2 × 4 = 8 instead of 2 × 2 × 2 × 2, giving 27 + 8. 432 comes from multiplying the two powers together instead of adding them.)
- (b) 9 — Subtract 14 from both sides: 4s ≤ 36. Divide both sides by 4: s ≤ 9, so the greatest number of tickets is 9. A candidate who forgets the £14 coach cost solves 4s ≤ 50, getting s ≤ 12.5, rounded down to 12. A candidate who adds the £14 instead of subtracting it solves 4s ≤ 64, getting s = 16. A candidate who miscalculates 50 − 14 as 32 solves 4s ≤ 32, getting s = 8.
- (c) £2743.60 — Each year the balance is multiplied by 1.03. After the first year: 4000 × 1.03 = 4120. After the second year: 4120 × 1.03 = 4243.60, so that is what Leah takes out. She then spends £1500 of it, which leaves 4243.60 − 1500 = 2743.60. She has £2743.60 left.
- (c) 0.5 — The real distance is 2 × 25000 = 50000 cm. Converting to metres, by dividing by 100, gives 500 m, and converting to kilometres, by dividing by 1000, gives 0.5 km. A candidate who divides the 50000 cm by 1000 in one go, applying the metres-to-kilometres factor straight to the centimetres, gets 50 km. A candidate who divides by 100 twice, treating 100 m as 1 km, gets 5 km. A candidate who slips one extra decimal place when converting 500 m to kilometres gets 0.05 km. The real distance is 0.5 km.
- (d) 19/30 — Exactly one means only fiction or only non-fiction, not both: 21 + 17 = 38 out of the 60 readers, which simplifies to 19/30. Including the 14 who read both as well gives 21 + 17 + 14 = 52, so 52/60 = 13/15 — that is at least one, not exactly one. Using only the both-count, 14, as the numerator gives 14/60 = 7/30, the probability of reading both, not exactly one. Using 52, the number who read at least one type, as the denominator instead of the full 60 readers surveyed gives 38/52 = 19/26.
- (d) 140 — Method: round each number to the nearest 10, then add the rounded values. Working: 89 rounds to 90 (nearest 10) and 52 rounds to 50 (nearest 10). 90 + 50 = 140. Answer: 140. 141 is the exact value of 89 + 52, found without rounding first, so it is not an estimate. 130 comes from rounding 89 down to 80 instead of up to the nearest 10, 90. 150 comes from rounding 52 up to 60 instead of down to the nearest 10, 50.
- (a) x = 3 — Method: factorise x² + 2x − 15 as (x + 5)(x − 3), since 5 × (−3) = −15 and 5 + (−3) = 2. Setting each bracket equal to zero gives x + 5 = 0 or x − 3 = 0, so x = −5 or x = 3. Only x = 3 is offered here. Distractor origins: x = −3 reverses the sign of the factor pair, treating the bracket (x − 3) as giving x = −3 instead of x = 3; x = 5 takes the number from the other factor, (x + 5), but with the wrong sign, giving x = 5 instead of x = −5; x = 15 takes the constant term of the original expression as if it were a root, without factorising at all.
- (c) 48 mm — Method: extension = k × force, where k = extension ÷ force. Working: k = 12 ÷ 5 = 2.4 mm per N. At 20 N: extension = 2.4 × 20 = 48 mm. Wrong options: 32 mm comes from adding the extension and force numbers instead of scaling (12 + 20); 3 mm comes from treating the relationship as inverse proportion (12 × 5 ÷ 20); 36 mm comes from using an incorrect scale factor of 3 between the forces instead of the correct factor of 4 (20 ÷ 5).
- (c) exterior angle = sum of the two interior opposite angles — Angle ACD is the exterior angle at C, and the exterior angle of a triangle is always equal to the sum of the two interior angles at the other two vertices — here, angle BAC and angle ABC — so angle ACD = 48° + 67° = 115°. "angles in a triangle add up to 180°" is a true fact, but on its own it only gives the INTERIOR angle at C (180 − 48 − 67 = 65°), not the exterior angle 115° — reaching 115° that way still needs a further step. "angles on a straight line add up to 180°" is the fact that links the interior and exterior angles at C to each other, not the fact that gives 115° directly from the two OTHER angles. "exterior angle is always double the smallest angle" is not a genuine angle fact: double the smallest angle here is 48° × 2 = 96°, not 115°.
- (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.
- (d) x = 5y + 4 — To make x the subject of y = (x − 4)/5, first multiply both sides by 5 to clear the fraction: 5y = x − 4, then add 4 to both sides: x = 5y + 4. Writing x = 5y − 4 multiplies correctly but keeps the minus sign on the 4 instead of changing it to a plus when moving it across. Writing x = y/5 + 4 divides by 5 instead of multiplying, the wrong inverse of the fraction. Writing x = 5(y + 4) adds 4 before multiplying by 5, reversing the correct order of the two steps. The correct rearrangement is x = 5y + 4.
- (d) 1 : 2500 — Method: convert the real-world measurement to the same unit as the drawing (centimetres) before writing the ratio. Working: 25 m = 2500 cm, so the scale is 1 : 2500. Wrong options: 1 : 25 comes from not converting metres to centimetres at all; 1 : 250 comes from converting metres to centimetres using ×10 instead of ×100; 1 : 2.5 comes from converting in the wrong direction (treating 25 m as 2.5 cm).
- (c) 12 — Method: a square root asks which positive number multiplied by itself gives the number under the root sign. Working: 11 × 11 = 121, which is too small, and 12 × 12 = 144. Answer: 12. The distractors: 14 comes from recalling the wrong square fact, since 14 × 14 = 196; 36 comes from dividing 144 by 4 instead of taking a square root; 72 comes from halving 144, treating a square root as a halving.
- (a) 70 km/h — Method: average speed for a whole journey is the total distance divided by the total time, not the mean of the separate speeds. Working: 3 × 80 = 240 km and 1 × 40 = 40 km, giving 280 km in 3 + 1 = 4 hours, so 280 ÷ 4 = 70. Answer: 70 km/h. The distractors: 60 km/h comes from taking the mean of 80 and 40, which would only be right if equal times were spent at each speed; 280 km/h is the total distance written with a speed unit, from forgetting to divide by the total time; 80 km/h comes from quoting the speed of the longer leg as the average for the whole journey.
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.