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 circle has a radius of 3 cm. Which of these is the exact area of the circle?
- 2.Here are the first four terms of an arithmetic sequence: 5, 9, 13, 17. Work out the 5th term.
- 3.A coach travels 210 miles at an average speed of 70 mph. Work out how long the journey takes.
- 4.Triangle ABC has AB = 9 cm, BC = 6 cm and angle A = 35°. Triangle DEF has DE = 9 cm, EF = 6 cm and angle D = 35°. Which of the following correctly identifies the condition shown here?
- 5.180 people entered a fun run. A frequency tree splits them into 108 adults and the rest children. It then splits each of those groups into those who finished and those who did not. 99 of the adults finished, and 153 of the 180 people finished altogether. Work out how many children did not finish.
- 6.A café owner in Brighton records the midday temperature, x °C, and the number of hot chocolates sold, y, on 12 days. The temperatures recorded run from 4 °C to 18 °C, and the line of best fit is y = −3x + 74. The forecast for tomorrow gives a midday temperature of 12 °C. Work out the number the line of best fit predicts, and write down how much confidence the owner can have in it.y = -3x + 74
- 7.Tickets to a theme park cost £38.50 for an adult and £19.75 for a child. A family estimates the total cost for 4 adults and 3 children, by rounding each ticket price to the nearest £5. Work out their estimate for the total cost.
- 8.The formula for the area of a trapezium is A = (a + b)h / 2, where a and b are the parallel side lengths and h is the height. Make h the subject of the formula.
- 9.A laptop bag has a mass of 2.4 kg. A school bag has a mass of 3.6 kg. Write the mass of the laptop bag as a fraction of the mass of the school bag, giving your answer in its simplest form.
- 10.A written description of a diagram states: 'AB is marked with a single arrow, CD is marked with a single arrow, and EF is marked with a double arrow.' Using the standard convention for marking parallel lines, what can be concluded?
- 11.A two-way table records how 130 pupils travel to school. 70 of the pupils are girls and the rest are boys. 42 of the girls walk to school and the rest of the girls cycle. 33 of the boys walk to school. Work out what fraction of the girls walk to school.
- 12.Two fair six-sided dice are rolled together. One die is red and the other is blue. Work out how many different outcomes are possible.
- 13.Solve x² − 2x − 24 = 0.
- 14.300 g of a salt solution of concentration 10% is poured into 200 g of a salt solution of concentration 25%. Work out the concentration of salt in the mixture.
- 15.An angle is measured with a protractor as 47°. What type of angle is this?
- 16.Round 0.0759 to 2 decimal places.
- 17.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.
- 18.Two mathematically similar flags are made in different sizes. The length scale factor from the smaller flag to the larger flag is 5. Work out the area scale factor from the smaller flag to the larger flag.
- 19.Simplify x⁶ ÷ x²
- 20.A tin of beans has a mass of 400 g. A tin of soup has a mass of 500 g. Write the mass of the tin of beans as a fraction of the mass of the tin of soup, giving your answer in its simplest form.
Answer key
- (c) 9π cm² — The area of a circle is π × r². With a radius of 3 cm this is π × 3² = 9π cm², and this is exact because π has not been replaced by any approximation. Writing 28.3 cm² replaces π with a rounded decimal value, 3.14, and then rounds the result again, so it is only an approximation. Writing 28.26 cm² uses π ≈ 3.14 without a final rounding step, but this is still only an approximation of 9π, not the exact value. Writing 27 cm² comes from replacing π with the rough approximation 3, which is even further from the true value.
- (b) 21 — Method: find the gap between neighbouring terms, then add one gap to the last term that is known. Working: 9 − 5 = 4, 13 − 9 = 4 and 17 − 13 = 4, so 4 is added each time; the 5th term is one step on from the 4th term, so it is 17 + 4. Answer: 21. The distractors: 25 comes from adding the 4 twice and landing on the 6th term; 20 comes from multiplying the position by the common difference, 5 × 4, and ignoring the fact that the sequence starts at 5 rather than at 4; 22 comes from adding the first term, 5, to 17 instead of adding the common difference.
- (c) 3 hours — Time = distance ÷ speed, so 210 ÷ 70 = 3 hours. Working out 210 − 70 = 140 subtracts the speed from the distance instead of dividing. Working out 210 + 70 = 280 adds the two values together, which does not give a time. Working out 210 × 70 = 14700 multiplies distance and speed together instead of dividing. The journey takes 3 hours.
- (a) SSA - not sufficient to prove congruence — The angle given, angle A, is not the angle between sides AB and BC — it is not the included angle — so this data is SSA (side, side, angle), which is not sufficient to prove congruence on its own; two triangles can share this SSA information without being congruent. SAS is wrong because the given angle is not the one included between the two given sides. ASA is wrong because only one angle (A) is given, not two. AAS is wrong for the same reason — only one angle is given, not two.
- (d) 18 — Method: a frequency tree can be filled from either end, because a pair of branches adds up to the number above it and the matching end branches add up to the total for that outcome. Working: the children number 180 − 108 = 72. Of the 153 finishers, 99 were adults, so the children who finished number 153 − 99 = 54. Along the children's branch, 72 − 54 = 18 children did not finish. Answer: 18 children did not finish. The distractors: 27 is 180 − 153 and counts everyone who did not finish, adults included; 9 is 108 − 99 and counts only the adults who did not finish; 54 is the number of children who did finish, which is the branch next to the one asked for.
- (d) 38, and fairly confident, as 12 °C is inside the range — Method: substitute the forecast temperature into the equation of the line of best fit, then judge the prediction by where that temperature sits among the data the line was drawn from. Working: putting x = 12 into y = −3x + 74 gives −3 × 12 + 74 = 38, so the line predicts 38 hot chocolates. The recorded temperatures run from 4 °C to 18 °C, and 12 °C lies inside that interval, so this is interpolation, the safer kind of prediction. Answer: 38, and fairly confident, as 12 °C is inside the range; the owner should still expect the true figure to differ a little, since the points only lie near the line and not on it. The distractors: being completely certain treats a line of best fit as a rule that fixes each day's sales, when it describes a trend that individual days depart from; saying 12 °C is outside the range misreads the interval 4 °C to 18 °C, and the wrong warning would be attached to a sound prediction; 110 comes from −3 × 12 being taken as +36, giving 36 + 74 = 110, which loses the negative gradient and so predicts that a warm day sells more hot chocolate than a cold one.
- (b) £220 — Method: round each ticket price to the nearest £5, multiply each rounded price by the number of tickets, then add the two totals. Working: the adult price £38.50 rounds to £40, and 4 × £40 = £160; the child price £19.75 rounds to £20, and 3 × £20 = £60; £160 + £60 = £220. Answer: £220. £213.25 is the exact total cost, found without rounding the prices first, so it is not an estimate. £160 comes from including the cost of the adult tickets only and forgetting the three children's tickets. £200 comes from swapping the two ticket quantities, using 3 adults and 4 children instead of 4 adults and 3 children.
- (d) h = 2A / (a + b) — Method: undo the multiplication by 1/2 by multiplying both sides by 2, then undo the multiplication by (a + b) by dividing both sides by it. Working: A = (a + b)h / 2, so multiplying both sides by 2 gives 2A = (a + b)h, then dividing both sides by (a + b) gives h = 2A / (a + b). The value h = A / (a + b) comes from forgetting to multiply by 2 to clear the 1/2 first. The value h = 2A / a + b comes from dividing only by a and leaving b outside the fraction, instead of dividing by the whole bracket (a + b). The value h = (a + b) / (2A) comes from inverting the fraction.
- (a) 2/3 — Put the laptop bag's mass over the school bag's mass: 2.4/3.6. Multiply both numbers by 10 to clear the decimals: 24/36. Divide both by their highest common factor, 12: 24÷12 = 2, 36÷12 = 3, giving 2/3. (3/2 comes from writing the masses the wrong way round. 1/3 comes from finding the difference in the masses, 3.6 − 2.4 = 1.2 kg, and writing it as a fraction of the school bag's mass, 1.2/3.6. 2/5 comes from comparing the laptop bag's mass to the total mass of both bags, 2.4/6.)
- (c) AB ∥ CD only; EF not confirmed — By convention, lines marked with the same number of arrows are parallel to each other, but lines marked with a different number of arrows belong to a different, unrelated family of parallel lines. AB and CD both have a single arrow, so AB is parallel to CD. EF has a double arrow, showing it is not part of the same family as AB and CD; it may be parallel to some other line marked with a double arrow, but nothing here confirms it is parallel to AB or CD, so 'AB ∥ CD only; EF not confirmed' is correct. 'AB, CD and EF are all parallel' and 'EF is parallel to AB' both wrongly treat every arrow mark as showing the same relationship. 'None of the lines are parallel' wrongly assumes a different arrow count rules out any parallel relationship at all, when it actually just signals a different pairing.
- (b) 3/5 — The question asks about the girls only, so use the girls' total of 70 as the denominator: 42 out of 70 girls walk, giving 42/70 = 3/5. Choosing 21/65 comes from using the whole survey of 130 pupils as the denominator instead of just the 70 girls, 42/130 = 21/65. Choosing 33/70 comes from using the boys' walking count, 33, over the girls' total of 70, mixing up the two rows of the table. Choosing 2/5 comes from using the number of girls who CYCLE, 70 − 42 = 28, instead of the number who walk, giving 28/70 = 2/5.
- (c) 36 — Method: set the outcomes out in a grid with one die along the top and the other down the side, so that every cell of the grid is one outcome, and count the cells. Working: the red die can land in 6 ways, so the grid has 6 columns, and the blue die can also land in 6 ways, so the grid has 6 rows; the number of cells is 6 × 6 = 36. Answer: 36. The distractors: 12 comes from adding 6 and 6 instead of multiplying them; 6 comes from counting the outcomes of a single die and forgetting that the second die also has to land; 21 comes from treating the two dice as indistinguishable, so that a red 2 with a blue 3 and a red 3 with a blue 2 are counted as one outcome.
- (a) x = 6 or x = −4 — We need two numbers that multiply to −24 and add to −2: these are −6 and 4, since −6 × 4 = −24 and −6 + 4 = −2. So x² − 2x − 24 = (x − 6)(x + 4) = 0, giving x = 6 or x = −4. A candidate who swaps the signs, using 6 and −4 the wrong way round in the brackets, gets x = −6 or x = 4. A candidate who picks the wrong factor pair, 8 and −3 (which multiply to −24 but add to +5, not −2), gets (x + 8)(x − 3) = 0 and answers x = −8 or x = 3. A candidate who makes both factors negative gets x = −6 or x = −4, which would require the constant term to be +24, not −24.
- (c) 16% — Method: scale each concentration to its own mass to find the salt it contains, add the two masses of salt, then write the ratio of salt to mixture per 100 g. Working: 10:100 = x:300 gives 30 g of salt, and 25:100 = y:200 gives 50 g of salt; the mixture holds 30 + 50 = 80 g of salt in 300 + 200 = 500 g of solution; 80:500 = 16:100. Answer: 16%. The distractors: 17.5% is the mean of 10% and 25%, which ignores that there is more of the weaker solution than of the stronger one; 19% comes from swapping the two concentrations over, working out (300 × 25% + 200 × 10%) ÷ 500; 26.7% comes from dividing the 80 g of salt by the 300 g of the first solution rather than by the 500 g of mixture.
- (b) Acute — An acute angle is any angle less than 90°. Since 47° is less than 90°, it is acute. An obtuse angle is between 90° and 180°, which 47° is not. A reflex angle is greater than 180°, which is far too big for 47°. A right angle is exactly 90°, not 47°. The angle is acute.
- (a) 0.08 — Method: decimal places are counted from the decimal point, including any zeros straight after it, so rounding to 2 decimal places is decided by the digit in the third decimal place. Working: 0.0759 has 7 in the second decimal place and 5 in the third, and 5 counts as rounding up, so the 7 goes up to 8. Answer: 0.08. The distractors: 0.07 comes from chopping the digits after the second decimal place off instead of rounding them; 0.076 is 0.0759 correct to 2 significant figures rather than 2 decimal places, because the zeros in front of the 7 are not significant figures; 0.1 is 0.0759 rounded to 1 decimal place, a coarser degree of accuracy than the question asks for.
- (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) 25 — Method: for similar shapes, the area scale factor is the length scale factor squared. Working: 5² = 25. Answer: 25. 5 comes from using the length scale factor itself as if it were the area scale factor, without squaring it. 10 comes from doubling the length scale factor instead of squaring it. 125 comes from cubing the length scale factor, which would give the volume scale factor, not the area scale factor.
- (d) x⁴ — Method: dividing two powers of the same letter subtracts the index of the divisor from the index of the term being divided. Working: six factors of x on the top and two on the bottom cancel in pairs, leaving 6 − 2 = 4 factors of x. Answer: x⁴. The distractors: x³ comes from dividing the indices, 6 ÷ 2, instead of subtracting them; x⁸ comes from adding the indices, 6 + 2, as though the powers were being multiplied; x¹² comes from multiplying the indices, 6 × 2, as though a power were being raised to a power.
- (b) 4/5 — Write the mass of the beans over the mass of the soup: 400/500. Divide the top and bottom by 100 to get 4/5. Choosing 5/4 comes from writing the soup's mass over the beans' mass, the wrong way round. Choosing 1/5 comes from finding the difference in mass (500 − 400 = 100) and writing that over the mass of the soup, instead of using the mass of the beans. Choosing 5/9 comes from writing the mass of the soup over the total mass of both tins (500 out of 900), instead of over the mass of the beans.
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.