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)
MathsUKwww.geekhero.co.uk
- 1.Light travels at 2.998 × 10⁸ metres per second. A distant object in space is 3.1 × 10¹⁵ metres from Earth. Work out an estimate for the number of seconds light takes to travel from the object to Earth, by rounding each number to 1 significant figure.
- 2.Which expression is equivalent to 7x − 3(2x − 6)?
- 3.Isla says that the ratio 4:6 is equivalent to the ratio 6:9. Is she correct? Give a reason for your answer.
- 4.A cylinder has a radius of 3 cm and a height of 10 cm. Using π = 3.14, work out the volume of the cylinder.
- 5.A card is taken at random from an ordinary pack of 52 playing cards. The pack contains 4 aces and 4 kings. Work out the probability that the card is an ace or a king.
- 6.A pie chart shows how 180 shoppers at a supermarket paid for their shopping. The sector for card payments has an angle of 90° at the centre. Work out how many of the 180 shoppers paid by card.
- 7.3/5 of the students in a year group walk to school. 90 students walk to school. Work out the total number of students in the year group.
- 8.A car park charges a £4 fixed fee plus £3 for each hour. Kofi has exactly £25 to spend on parking. Using the inequality 4 + 3h ≤ 25, work out the greatest number of whole hours, h, he can park for.
- 9.The number of calories burned, E, by a runner is plotted against the time, m minutes, on a straight-line graph. The line passes through the origin and the point (10, 80). Work out the gradient of the line.
- 10.The bearing of a campsite B from a walker's position A is 070°. What is the bearing of A from B?
- 11.A spinner is divided into 8 equal sections. 3 of the sections are red and the rest are not red. The spinner is spun once. Work out the probability that it does not land on red.
- 12.Hannah works out 3.1 × 19.6 on her calculator and writes down 6.076. Work out an estimate for 3.1 × 19.6, by rounding each number to 1 significant figure.
- 13.A graph has equation y = x² − 6x + 5. A student says its turning point has x-coordinate 6, because that's the coefficient of x. Which statement corrects the student's mistake?y = x² − 6x + 5
- 14.In a class, 50% of the students study French, 30% study Spanish and the rest study German. Write down the ratio of French : Spanish : German students, in its simplest form.
- 15.Two similar signs have lengths in the ratio 9 : 4. The larger sign has a side of length 14.4 cm. Work out the length of the corresponding side of the smaller sign.
- 16.An allotment is divided into two plots in the ratio 2:3. The larger plot has an area of 18 m². Work out the fraction of the total area taken up by the smaller plot.
- 17.A charity's fundraising total, T pounds, over d days follows T = (d − 3)(30 − d) for 3 ≤ d ≤ 30, where T = 0 marks the start and end of the campaign. Work out how many days the campaign runs for, from start to end.
- 18.A cyclist travels 45 km in 3 hours at a constant speed. Work out the cyclist's average speed, in km/h.
- 19.A tin of paint has a mass of 5.672 kg. Round this mass to 1 decimal place.
- 20.A coach journey is 372 miles in total. After a stop, the coach has travelled 217 miles. Write the distance still to travel as a fraction of the total journey. Give your answer in its simplest form.
Answer key
- (d) 1 × 10⁷ seconds — Method: the time for a journey is the distance divided by the speed, so round each number to 1 significant figure and then divide; dividing numbers in standard form means dividing the coefficients and subtracting the indices. Working: 3.1 × 10¹⁵ rounds to 3 × 10¹⁵ and 2.998 × 10⁸ rounds to 3 × 10⁸; 3 ÷ 3 = 1 for the coefficients, and 15 − 8 = 7 for the indices. Answer: about 1 × 10⁷ seconds. The distractors: 3 × 10⁷ seconds comes from subtracting the indices correctly but leaving the coefficient as 3 instead of dividing 3 by 3; 1 × 10⁻⁷ seconds comes from dividing the speed by the distance instead of the distance by the speed; 9 × 10²³ seconds comes from multiplying the two quantities instead of dividing them, since 3 × 3 = 9 and 15 + 8 = 23.
- (c) x + 18 — Expand −3(2x − 6) by multiplying both terms by −3: −3 × 2x = −6x and −3 × (−6) = 18, giving 7x − 6x + 18 = x + 18. Writing x − 18 comes from not flipping the sign of the −6 inside the bracket, so −3 × (−6) is treated as −18 instead of +18. Writing x + 6 comes from forgetting to multiply the −6 by 3, only carrying its sign. Writing 13x − 18 comes from treating the whole bracket as being added rather than subtracted, so 3(2x − 6) = 6x − 18 is added to 7x.
- (d) Yes, because 4 × 9 = 6 × 6 — Method: two ratios are equal when their cross-products are equal, so multiply the first part of each ratio by the second part of the other. Working: 4 × 9 = 36 and 6 × 6 = 36; the two products match, so the ratios are equal, and simplifying both to 2:3 shows the same thing. Answer: yes, because 4 × 9 = 6 × 6. The distractors: the reason that the number 6 appears in both ratios reaches the right verdict from a surface match, since a figure shared by two ratios says nothing about equivalence — 4:6 and 6:5 share a 6 and are not equal; the reason built on 9 − 6 and 6 − 4 compares the differences inside each ratio, 3 against 2, which is additive thinking and ends at a verdict of no; the reason built on 4 × 6 and 6 × 9 multiplies the two parts of each ratio together instead of across the pair, giving 24 against 54 and again a verdict of no.
- (c) 282.6 cm³ — Volume of a cylinder = πr²h = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6 cm³. (94.2 cm³ comes from using πrh and forgetting to square the radius; 90 cm³ comes from using r²h and leaving π out altogether; 1130.4 cm³ comes from using the diameter, 6 cm, in place of the radius.)
- (b) 8/52 — Method: a card cannot be an ace and a king at the same time, so the two events are mutually exclusive and their probabilities are added, keeping the denominator the same. Working: P(ace) = 4/52 and P(king) = 4/52, so P(ace or king) = 4/52 + 4/52, and 4 + 4 = 8 fifty-seconds. Answer: 8/52. The distractors: 4/52 comes from giving the probability of just one of the two events and forgetting to add the other; 16/52 comes from multiplying the two counts, 4 × 4, instead of adding them; 1/52 comes from giving the probability of one particular named card rather than any of the eight.
- (d) 45 — Method: convert the angle into a fraction of the full circle, 360°, then apply that fraction to the total number of shoppers. Working: the card sector is 90° out of 360°, a fraction of 90 ÷ 360 = 0.25. Applying that fraction to the 180 shoppers gives 0.25 × 180 = 45 shoppers. Giving 90 states the angle itself, not a number of shoppers — the angle first has to be converted into a fraction. Using the remaining angle, 360 − 90 = 270°, and scaling that, 270 ÷ 360 × 180 = 135, finds the number who did NOT pay by card, not the number who did. Dividing 360 by 90, 360 ÷ 90 = 4, finds how many equal 90° sectors fit in the circle, a fact about the pie chart's shape, not about the shoppers at all. Always convert the angle to a fraction of 360° first, and apply that same fraction to the total number of people.
- (c) 150 — Since 90 students represent 3 of the 5 equal parts, one part is 90 ÷ 3 = 30, and the whole year group is five parts: 30 × 5 = 150. Applying the fraction forwards to 90 instead of reversing it, 90 × 3/5 = 54, treats the given number as the whole rather than as three fifths of it. Finding one part correctly as 30 but forgetting to scale up to the whole year group leaves 30 as the final answer. Treating 90 as the whole year group and adding on 2/5 of 90 for the students who do not walk, 90 + (90 × 2/5) = 126, applies the missing fraction to the wrong base amount.
- (a) 7 — Subtract 4 from both sides: 3h ≤ 21. Divide both sides by 3: h ≤ 7, so the greatest whole number of hours is 7. A candidate who forgets the £4 fixed fee solves 3h ≤ 25, getting h ≤ 8.33, rounded down to 8. A candidate who adds the £4 instead of subtracting it solves 3h ≤ 29, getting h ≤ 9.67, rounded down to 9. A candidate who reaches h ≤ 7 but wrongly assumes the boundary value cannot be used, thinking that spending the whole £25 is not allowed, answers 6.
- (b) 8 — The gradient of a line through the origin is the y-value divided by the x-value at any point on the line. Using (10, 80): gradient = 80 ÷ 10 = 8.
- (a) 250° — The back bearing (the bearing of A from B) differs from the bearing of B from A by exactly 180°. Because the given bearing, 070°, is less than 180°, add 180°: 070 + 180 = 250°, so the bearing of A from B is 250°. Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 110° comes from subtracting 180° from 070° and dropping the negative sign (070 − 180 = −110) instead of adding 180°. Choosing 160° comes from adding only 90° instead of 180° (070 + 90 = 160).
- (c) 5/8 — Method: the sections are all the same size, so every section is equally likely; count the sections that are not red and write that count over the total number of sections. Working: 8 − 3 = 5 sections are not red, and there are 8 sections altogether. Answer: 5/8, a value between 0 and 1 and a little above the halfway point of the scale. The distractors: 3/8 comes from giving the probability that the spinner does land on red; 5/11 comes from adding the 3 red sections to the 8 sections to make a total of 11 instead of using the 8 sections that exist; 1/2 comes from assuming that 'red' and 'not red' must be equally likely because there are only two possibilities.
- (d) 60 — Method: round each number to 1 significant figure and multiply; the estimate then shows whether the calculator answer is sensible. Working: 3.1 rounds to 3 and 19.6 rounds to 20, so the estimate is 3 × 20 = 60. Answer: 60. Hannah's 6.076 is about ten times too small, which is what happens when 19.6 is keyed in as 1.96. The distractors: 62 comes from rounding 19.6 only and leaving 3.1 as it stands, giving 3.1 × 20 = 62; 6 comes from trusting the calculator display rather than checking it against an estimate; 600 comes from rounding 19.6 to 200 instead of to 20, a place-value slip, giving 3 × 200 = 600.
- (d) The roots of x² − 6x + 5 = 0 are x = 1 and x = 5 (since it factorises to (x − 1)(x − 5)), so by symmetry the turning point has x-coordinate 3, not 6. — Factorising, x² − 6x + 5 = (x − 1)(x − 5), so the roots are x = 1 and x = 5. The turning point lies midway between the roots by symmetry: (1 + 5) ÷ 2 = 3. The coefficient of x has no direct role in locating the turning point this way. The option giving −6 makes an arbitrary sign change with no mathematical basis. The option giving 5 wrongly takes just one of the two roots instead of their midpoint.
- (d) 5:3:2 — German = 100% − 50% − 30% = 20%. The ratio 50 : 30 : 20 simplifies by dividing every part by 10 to give 5 : 3 : 2.
- (d) 6.4 cm — The scale factor from the larger sign to the smaller sign is 4 ÷ 9, so the smaller side is 14.4 × 4 ÷ 9 = 6.4 cm. The distractor 32.4 cm comes from using the ratio the wrong way round, 14.4 × 9 ÷ 4 = 32.4. The distractor 9.4 cm comes from subtracting the difference between the ratio numbers (9 − 4 = 5) from the given length, 14.4 − 5 = 9.4. The distractor 3.6 cm comes from dividing the given length by 4 only, 14.4 ÷ 4 = 3.6, without also using the other ratio number.
- (a) 2/5 — The ratio 2:3 has 2 + 3 = 5 parts in total, and the larger plot is 3 of those parts. Since the larger plot is 18 m², each part is 18 ÷ 3 = 6 m², so the total area is 5 × 6 = 30 m² and the smaller plot is 2 × 6 = 12 m². The fraction of the total area taken up by the smaller plot is 12/30, which simplifies to 2/5. Giving the fraction for the larger plot instead of the smaller one gives 3/5. Comparing the smaller plot to the larger plot instead of to the total area gives 2/3. Assuming the two plots split the area evenly, ignoring the given ratio altogether, gives 1/2.
- (c) 27 days — The campaign starts at d = 3 and ends at d = 30, so it runs for 30 − 3 = 27 days. Getting 33 days comes from adding the two values, 3 + 30 = 33, instead of subtracting them. Getting 30 days uses only the end day and ignores that the campaign did not start at day 0. Getting 24 days comes from subtracting the start day twice, 30 − 3 − 3 = 24, instead of once.
- (a) 15 km/h — Speed = distance ÷ time, so 45 ÷ 3 = 15 km/h. Working out 45 × 3 = 135 multiplies distance and time together instead of dividing. Working out 45 + 3 = 48 and 45 − 3 = 42 both combine the two values by addition or subtraction, which does not give a speed at all. The cyclist's average speed is 15 km/h.
- (c) 5.7 kg — Method: to round to 1 decimal place, keep one digit after the decimal point and let the digit in the second decimal place decide whether that digit stays as it is or goes up. Working: 5.672 has 6 in the first decimal place and 7 in the second decimal place; 7 is 5 or more, so the 6 goes up to 7 and the digits beyond the first decimal place are dropped. Answer: 5.7 kg. The distractors: 5.6 kg comes from chopping the digits after the first decimal place off instead of rounding them, which is truncation rather than rounding; 6.0 kg comes from rounding to the nearest whole kilogram instead of to 1 decimal place; 5.0 kg comes from chopping everything after the decimal point off, so the mass is both truncated and given to the wrong degree of accuracy.
- (a) 5/12 — Work out the distance still to travel: 372 − 217 = 155 miles. Form the fraction 155/372; both numbers share a factor of 31, so 155 ÷ 31 = 5 and 372 ÷ 31 = 12, giving 5/12. 7/12 comes from writing the distance already travelled as the fraction of the journey (217/372 = 7/12), instead of the distance still to travel. 145/372 comes from miscalculating 372 − 217 as 145 instead of 155. 5/7 comes from comparing the remaining distance with the distance already travelled (155/217 = 5/7), instead of with the total 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.