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
GCSE Foundation sample Paper 1 (non-calculator)
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- 1.A number, n, is equal to 3.7 when rounded to 1 decimal place. Write down the error interval for n.
- 2.Which of these is a formula, rather than an expression, an equation, or an identity?
- 3.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.
- 4.A diameter divides a circle into two equal halves. Write down the special name given to each of these two halves.
- 5.A small pack contains 5 cards, numbered 1 to 5. One card is drawn at random from the pack, and a fair coin is flipped. Every outcome is listed as a pair, such as (3, H). Work out how many equally likely outcomes there are in total.
- 6.A school has 1,500 pupils. The head teacher takes a random sample of 150 of them from the school register and asks how long they spend on homework. Rory says the sample is too small for the result to mean anything. Is Rory right? Give a reason for your answer.
- 7.Work out (4 × 10⁻³) × (2 × 10⁵). Give your answer in standard form.
- 8.Solve x² − 5x + 6 = 0 by factorising.
- 9.Two mathematically similar rectangles have lengths in the ratio 3 : 5. The area of the smaller rectangle is 27 cm². Work out the area of the larger rectangle.
- 10.Point A is at (2, 3). It is translated by the vector (4, −1) to give point A′. Work out the coordinates of A′.
- 11.Ellie drops a bottle top and records whether it lands open end up. In her first 20 drops it lands open end up 13 times. She carries on, and after 200 drops in total it has landed open end up 84 times. Work out the best estimate of the probability that the bottle top lands open end up.
- 12.Work out an estimate for 312 × 19, by rounding each number to 1 significant figure.
- 13.A cyclist's journey is shown on a distance–time graph. The graph is a straight line from (0 minutes, 0 km) to (40 minutes, 12 km). Work out her speed in kilometres per hour.
- 14.A box contains 12 red counters and 18 blue counters. Write the number of red counters as a fraction of the number of blue counters. Give your answer in its simplest form.
- 15.a is the column vector with top number 3 and bottom number 2. b is the column vector with top number −1 and bottom number 4. Work out a + b, giving your answer as a column vector in the form (top, bottom).
- 16.A shop assistant says that 7.2 × 3.9 = 56.16. Work out an estimate for 7.2 × 3.9, by rounding each number to the nearest whole number, to show that the assistant’s answer cannot be correct.
- 17.p = 4. Work out the value of 2p³.
- 18.In a science lesson Priya has 10 litres of a solution that is 30% salt. She adds water to make a solution that is 20% salt. Work out how many litres of water she adds.
- 19.Write these numbers in order, starting with the smallest: −1.4, 5/4, −6/5, 1.3, 0
- 20.A shade of paint is made by mixing blue paint and white paint. To make 5 litres of the shade, 2 litres of blue paint is used and the rest is white paint. Write the ratio of blue paint to white paint in its simplest form.
Answer key
- (b) 3.65 ≤ n < 3.75 — Rounding to 1 decimal place means n can be up to half of one decimal place, 0.05, below or above 3.7 before it would round to a different value. This gives a lower bound of 3.7 − 0.05 = 3.65 and an upper bound of 3.7 + 0.05 = 3.75. A value exactly at 3.75 would round up to 3.8, not 3.7, so the upper bound is excluded while the lower bound, 3.65, does still round to 3.7. Writing 3.65 ≤ n ≤ 3.75 wrongly includes 3.75. Writing 3.6 ≤ n < 3.8 uses a whole decimal place, 0.1, either side instead of half of one, 0.05. Writing 3.65 < n < 3.75 wrongly excludes 3.65, which does round to 3.7.
- (d) V = lwh — V = lwh is a formula: it relates one quantity, the volume V, to others, the length, width and height, in a way that is true generally. A candidate who picks lwh has chosen the expression, not a full statement relating two quantities. A candidate who picks lwh = 60 has chosen an equation, since it is only true for particular values of l, w and h that multiply to 60. A candidate who picks 2(l + w) ≡ 2l + 2w has chosen an identity, mistaking the identity symbol ≡ for a sign that makes a statement a formula.
- (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.
- (b) Semicircle — Method: recall the special name for exactly half of a circle. Working: a diameter splits the circle into two identical halves, and each half has its own specific name. A student who answers segment has confused this special case with the general region cut off by any chord. A student who answers sector has confused it with the pie-slice shape made by two radii. A student who answers arc has only named the curved edge, not the whole two-dimensional region. Answer: semicircle.
- (c) 10 — There are 5 possible cards and 2 possible coin results, so listing every pair gives 5 × 2 = 10 equally likely outcomes. Choosing 5 comes from listing only the card outcomes and forgetting the coin flip adds a second stage to each one. Choosing 7 comes from adding the two stages instead of combining them, 5 + 2 = 7, rather than pairing every card with every coin result. Choosing 20 comes from counting each coin result twice for every card, 5 × 2 × 2 = 20, effectively pairing every card with the coin twice over.
- (a) No, 150 pupils are a tenth of the school, chosen at random — Method: judge a sample on two things, whether every member of the population had the same chance of being chosen, and whether the sample is large enough to carry a pattern. Working: the 150 pupils were drawn from the register of every pupil in the school, so no year group or set is shut out and no pupil chooses to take part; and 150 ÷ 1,500 = 0.1, so one pupil in ten has been asked. A random sample of that share is ample for an estimate of how long the school's pupils spend on homework. Answer: no, because 150 pupils are a tenth of the school and were chosen at random. The distractors: saying a random sample always gives the exact school figure reaches the same verdict for a reason that is false, since a second random sample of 150 would give a slightly different mean; saying 150 pupils cannot be picked at random from 1,500 treats randomness as something only a whole population can have, when drawing names from the register is exactly how a random sample is taken; saying that only asking all 1,500 could show anything rejects sampling altogether, which would leave no way to study any population too large to count.
- (a) 8 × 10² — Method: to multiply numbers written in standard form, multiply the coefficients and add the indices. Working: 4 × 2 = 8 for the coefficients, and −3 + 5 = 2 for the indices; 8 already lies between 1 and 10, so no adjustment is needed. Answer: 8 × 10². The distractors: 6 × 10² comes from adding the coefficients, 4 + 2, instead of multiplying them; 8 × 10⁸ comes from ignoring the minus sign and adding 3 + 5; 8 × 10⁻¹⁵ comes from multiplying the indices, −3 × 5, instead of adding them.
- (c) x = 2 or x = 3 — Method: factorise into two brackets whose numbers multiply to the constant term and add to the coefficient of x, then set each bracket equal to zero. Working: two numbers that multiply to 6 and add to −5 are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3) = 0; then x − 2 = 0 gives x = 2 and x − 3 = 0 gives x = 3. Answer: x = 2 or x = 3. The distractors: x = −2 or x = −3 comes from reading the numbers inside the brackets as the solutions instead of changing their signs; x = 1 or x = 6 comes from taking the first factor pair of 6 without checking that the pair adds to −5; x = 5 or x = 6 comes from reading the solutions straight off the 5 and the 6 in the equation.
- (b) 75 cm² — Areas of similar shapes are in the ratio of the squares of their lengths. Squaring both parts of 3 : 5 gives an area ratio of 9 : 25, so the larger area is 25/9 of the smaller one. Working with the smaller area: 27 ÷ 9 = 3, and 3 × 25 = 75. The area of the larger rectangle is 75 cm².
- (c) (6, 2) — Method: add the vector's components to the point's coordinates, x-component to x, y-component to y. Working: A′ = (2 + 4, 3 + (−1)) = (6, 2). Options: (6, 4) comes from adding the size of the y-component, 1, instead of subtracting it; (1, 7) comes from swapping the components, adding the y-component (−1) to the x-coordinate and the x-component (4) to the y-coordinate; (−2, 4) comes from reversing the sign of both components, using the vector (−4, 1) instead of (4, −1). Answer: (6, 2).
- (d) 0.420 — Method: use the relative frequency worked out from the larger number of trials as the best estimate of the probability, since a bigger sample tends to sit closer to the true probability. Working: 84 out of 200 drops land open end up, so the relative frequency from the total is 84 ÷ 200 = 0.420. Answer: 0.420. Watch out: writing down 0.650 comes from 13 ÷ 20, using only the first, much smaller sample instead of the total. Writing down 0.535 comes from averaging 0.650 and 0.420, treating the 20-drop run and the 200-drop run as equally reliable instead of using the larger sample on its own. And writing down 0.580 comes from 1 − 0.420, working out the probability that the bottle top lands the other way up instead of open end up.
- (c) 6000 — Method: round each number to 1 significant figure, then multiply the rounded values. Working: 312 rounds to 300 (1 s.f.) and 19 rounds to 20 (1 s.f.). 300 × 20 = 6000. Answer: 6000. 5928 is the exact value of 312 × 19, found by multiplying without rounding first, which is not an estimate. 600 comes from rounding 19 down to 2 instead of to 20, losing a zero from its place value. 6200 comes from rounding 312 to the nearest 10, 310, instead of to 1 significant figure, 300, then multiplying by the correctly rounded 20.
- (c) 18 km/h — A straight line on a distance–time graph means a steady speed, so the speed is the distance gained divided by the time taken. First convert 40 minutes to hours: 40 ÷ 60 = 2/3 of an hour. Speed is distance ÷ time, so speed = 12 ÷ (2/3) = 12 × 3/2 = 18 km/h. The option 8 km/h comes from multiplying the distance by the time in hours instead of dividing: 12 × 2/3 = 8. The option 30 km/h comes from scaling the distance-per-minute rate by 100 instead of 60: 12 ÷ 40 = 0.3, then 0.3 × 100 = 30. The option 16 km/h comes from mistakenly treating 40 minutes as three-quarters of an hour instead of two-thirds: 12 ÷ 0.75 = 16.
- (c) 2/3 — Method: 'as a fraction of' puts the quantity named after the word 'of' on the bottom, so the red counters go on top and the blue counters go on the bottom. Working: the fraction is 12/18; the highest common factor of 12 and 18 is 6, and 12 ÷ 6 = 2 while 18 ÷ 6 = 3. Answer: 2/3. The distractors: 2/5 comes from comparing the red counters with the total number of counters, 12/30, a part-to-whole fraction when the question asks for one part compared with the other part; 3/2 comes from writing the blue counters over the red counters, 18/12, reversing the order the question gives; 1/2 comes from simplifying 12/18 by subtracting the common factor instead of dividing by it, 12 − 6 = 6 and 18 − 6 = 12, giving 6/12, which then cancels to 1/2.
- (b) (2, 6) — To add column vectors, add the top numbers together and add the bottom numbers together, separately: top = 3 + (−1) = 2, bottom = 2 + 4 = 6, giving (2, 6). A candidate who writes the top and bottom numbers of the correct answer in the wrong order gives (6, 2). A candidate who subtracts instead of adds, working out 3 − (−1) = 4 and 2 − 4 = −2, gives (4, −2). A candidate who adds the top numbers correctly but subtracts the bottom numbers by mistake, working out 2 − 4 = −2, gives (2, −2). The correct answer is (2, 6).
- (d) 28 — Method: round each number to the nearest whole number, then multiply the rounded numbers to get an estimate that can be compared with the assistant's answer. Working: 7.2 rounds to 7, and 3.9 rounds to 4, so the estimate is 7 × 4 = 28. Since 28 is much smaller than 56.16, the assistant's answer cannot be correct. 56 comes from rounding the assistant's answer to the nearest whole number, instead of rounding the two numbers being multiplied and then multiplying them. 35 comes from rounding both numbers correctly but then slipping in the seven times table, writing 7 × 5 = 35 in place of 7 × 4 = 28. 21 comes from rounding 3.9 down to 3 instead of 4, giving 7 × 3 = 21. Answer: 28.
- (d) 128 — In 2p³ the index belongs to p only, so cube p first and multiply by the coefficient afterwards. Cubing gives 4 × 4 × 4 = 64, and then 2 × 64 = 128. Cubing the coefficient as well would mean working out (2 × 4)³, which is 512. Reading the index as an instruction to multiply by 3 gives 2 × 4 × 3 = 24, and ignoring the coefficient altogether leaves 64.
- (b) 5 litres — Method: adding water changes the total volume but adds no salt, so work out the volume of salt, then the total volume that makes that salt 20% of the mixture, then the extra water. Working: 30% of 10 litres is 0.3 × 10 = 3 litres of salt. For the same 3 litres to be 20% of the new mixture, the new total volume is 3 ÷ 0.2 = 15 litres. The water added is the extra volume, 15 − 10 = 5 litres. Answer: 5 litres. The distractors: 3 litres is the volume of salt in the solution, which is the first step and not what the question asks for; 15 litres is the total volume of the new mixture, which counts the 10 litres already in the container as water that was poured in; 2 litres comes from taking 20% of the original 10 litres, applying the new percentage to the old volume instead of to the new one.
- (d) −1.4, −6/5, 0, 5/4, 1.3 — Method: convert the fractions 5/4 and −6/5 to decimals so every number is written the same way, then compare all five decimals. Working: 5/4 = 1.25 and −6/5 = −1.2. Comparing −1.4, −1.2, 0, 1.25 and 1.3 in size gives the order −1.4, −1.2, 0, 1.25, 1.3. Answer: −1.4, −6/5, 0, 5/4, 1.3. −6/5, −1.4, 0, 5/4, 1.3 swaps the two negative numbers, treating −6/5 as more negative than −1.4 even though −1.2 is closer to zero than −1.4. 1.3, 5/4, 0, −6/5, −1.4 lists the numbers from largest to smallest instead of smallest to largest. −1.4, −6/5, 0, 1.3, 5/4 swaps 5/4 and 1.3, comparing the numerator 5 directly with 1.3 instead of converting 5/4 to the decimal 1.25 first.
- (c) 2:3 — The white paint is 5 − 2 = 3 litres. The ratio of blue paint to white paint is 2 : 3, which has no common factor, so it is already in simplest form. Getting 2 : 5 compares the blue paint to the total amount of shade instead of to the white paint. Getting 3 : 2 has the two parts the wrong way round. Getting 5 : 3 uses the total amount of shade instead of the blue paint as the first part.
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.