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%.

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GCSE Foundation sample Paper 1 (non-calculator)

Maths · GCSE Foundation · Paper 1 · Non-calculator · 20 questions

MathsUKwww.geekhero.co.uk

1.Work out (−3) × 4 + 2 × (−5)
(a)−22
(b)50
(c)−2
(d)22
2.The formula for the perimeter of a square is P = 4s, where s is the side length. Make s the subject of the formula.
(a)s = P − 4
(b)s = P + 4
(c)s = P / 4
(d)s = 4P
3.A jar contains 24 green sweets and 16 orange sweets. Write the ratio of green sweets to orange sweets in its simplest form.
(a)3:2
(b)3:5
(c)2:3
(d)6:4
4.Point B is at (3, 5). It is reflected in the line y = 2. Work out the coordinates of the image of point B.
(a)(3, −4)
(b)(3, −1)
(c)(3, −5)
(d)(−3, 5)
5.120 members of a sports club were asked whether they play football or netball; each member plays exactly one of the two sports. 70 of the members are male. 45 of the male members play football. 38 of the female members play netball. Work out how many female members play football.
(a)45
(b)38
(c)25
(d)12
6.The lowest temperature in a garden in Sheffield was recorded on each of seven days: 3 °C, 5 °C, 2 °C, 7 °C, 4 °C, 6 °C and 3 °C. Write down what the range of these seven temperatures describes.
(a)The temperature that occurred most often in the week
(b)The temperature that was typical of the seven days
(c)How many different temperatures were recorded
(d)How much the temperatures varied over the seven days
7.Insert one pair of brackets into 2 + 3 × 5 − 1 so that the calculation is equal to 24. Which calculation is correct?
(a)(2 + 3 × 5) − 1
(b)(2 + 3) × 5 − 1
(c)2 + 3 × (5 − 1)
(d)(2 + 3) × (5 − 1)
8.Solve the simultaneous equations x + y = 1 and 2x + y = 5.
(a)x = 4, y = −3
(b)x = −2, y = 3
(c)x = 6, y = −5
(d)x = 4, y = 5
9.A straight-line graph shows the cost, C pounds, of buying n litres of paint. The line passes through the origin and through the point (5, 40). Work out the gradient of the line.
(a)35
(b)8
(c)0.125
(d)40
10.Point A and point B are 5 cm apart. How many points are exactly 4 cm from A and exactly 3 cm from B?
(a)4
(b)0
(c)2
(d)1
11.The numbers 1 to 10 are each written on a card. Set A is the set of multiples of 3 from 1 to 10 and set B is the set of even numbers from 1 to 10. Work out how many of the numbers from 1 to 10 are in set A but not in set B.
(a)3
(b)7
(c)2
(d)4
12.Work out ∛64.
(a)21.3
(b)4
(c)192
(d)8
13.A stack of firewood has 3 logs in the top layer. Each layer below has 4 more logs than the layer above it. Work out the number of logs in the 6th layer from the top.
(a)23
(b)27
(c)24
(d)20
14.1 metre = 100 cm. Work out how many square centimetres there are in 1 square metre.
(a)100 cm²
(b)10000 cm²
(c)1000000 cm²
(d)200 cm²
15.A church tower is due east of a village. What is the three-figure bearing of the church tower from the village?
(a)180
(b)270
(c)090
(d)009
16.Work out 3/7 × 14/9. Give your answer as a fraction in its simplest form.
(a)17/16
(b)2/21
(c)2/3
(d)27/98
17.A number machine adds 1 to its input. Isla says that an input of 5 gives an output of 6. Which statement is correct?
(a)Isla is wrong, because 5 − 1 = 4
(b)Isla is wrong, because 5 × 1 = 5
(c)Isla is right, because the machine adds 1 to its output
(d)Isla is right, because 5 + 1 = 6
18.Work out 50% of 60.
(a)6
(b)3000
(c)120
(d)30
19.Work out √25 + 4² − 12 ÷ 3
(a)9
(b)17
(c)3
(d)37
20.A shop sells fruit juice concentrate that must be diluted with water in the ratio 2 : 5 (concentrate : water) to make squash. Priya has 8 litres of concentrate. Work out how much water she needs to keep the same ratio.
(a)40 litres
(b)3.2 litres
(c)11 litres
(d)20 litres

Answer key

(a) −22 — Method: both multiplications are carried out before the addition, and a positive multiplied by a negative is negative. Working: (−3) × 4 = −12 and 2 × (−5) = −10, so the calculation becomes −12 + (−10) = −22. Answer: −22. The distractors: 22 comes from ignoring the minus signs and working out 3 × 4 + 2 × 5 = 22; 50 comes from working from left to right with no priority at all, giving −12 + 2 = −10 and then −10 × (−5) = 50; −2 comes from taking 2 × (−5) as +10, so that −12 + 10 = −2.
(c) s = P / 4 — Method: undo the multiplication by 4 by dividing both sides by 4. Working: P = 4s, so dividing both sides by 4 gives s = P / 4. The value s = 4P comes from multiplying by 4 instead of dividing — using the wrong inverse operation. The value s = P + 4 comes from adding 4 instead of dividing. The value s = P − 4 comes from subtracting 4 instead of dividing.
(a) 3:2 — There are 24 green sweets and 16 orange sweets. The highest common factor of 24 and 16 is 8. Divide both numbers by 8: 24 ÷ 8 = 3 and 16 ÷ 8 = 2, so the ratio is 3 : 2. Dividing by 4 instead of 8 gives 6 : 4, which still has a common factor of 2, so it is not fully simplified. Writing green sweets to the total number of sweets, 24 : 40, simplifies to 3 : 5 — that compares green to everything, not green to orange, so it answers a different question. Swapping the order gives 2 : 3, green and orange the wrong way round.
(b) (3, −1) — Method: find the distance from the point to the mirror line, then place the image the same distance on the other side of the line. Working: B is 5 − 2 = 3 units above the line y = 2, so its image is 3 units below the line, at y = 2 − 3 = −1, giving (3, −1); the x-coordinate does not change, since the mirror line is horizontal. Options: (3, −5) comes from reflecting in the x-axis (y = 0) instead of the line y = 2; (−3, 5) comes from reflecting in the y-axis instead of a horizontal line; (3, −4) comes from doubling the distance from the line instead of reflecting it, giving 2 − 2×3 = −4. Answer: (3, −1).
(d) 12 — There are 120 − 70 = 50 female members. 38 of them play netball, so the rest play football: 50 − 38 = 12. Writing 45 is wrong because that is the number of MALE members who play football, not female. Writing 38 again is wrong because that is the female netball total, not the female football total — the question needs the total minus 38, not 38 itself. Writing 25 is wrong because that comes from the male branch (70 − 45 = 25 male netball players), not the female branch. 12 female members play football.
(d) How much the temperatures varied over the seven days — Method: the range of a set of values is the largest value take away the smallest, so it is built from two values only and it measures the gap they leave between them. Working: the largest of the seven readings is 7 °C and the smallest is 2 °C, so the range is 7 − 2 = 5 °C. That figure says the week's readings covered a band 5 °C wide; it names no particular day and no particular reading. Answer: the range describes how much the temperatures varied over the seven days. The distractors: the temperature that occurred most often is the mode, which here is 3 °C, and a mode counts repeats instead of measuring a gap; the temperature typical of the week is an average, and the range is not an average, since it throws away every value lying between the two extremes; the number of different temperatures recorded is 6, a count of how many distinct values appear, while the range is a difference between two of them.
(b) (2 + 3) × 5 − 1 — 2 + 3 = 5, then 5 × 5 = 25, then 25 − 1 = 24, so the brackets belong around 2 + 3. Placing them around 5 − 1 instead gives 5 − 1 = 4, then 3 × 4 = 12, then 2 + 12 = 14. Leaving the multiplication bracketed instead changes nothing, because it already had priority: 3 × 5 = 15, then 2 + 15 = 17, then 17 − 1 = 16. Bracketing both 2 + 3 and 5 − 1 uses two pairs instead of the one asked for: 2 + 3 = 5, 5 − 1 = 4, then 5 × 4 = 20.
(a) x = 4, y = −3 — Method: both equations contain +y with the same coefficient, so subtracting one equation from the other removes y. Working: (2x + y) − (x + y) = 5 − 1 gives x = 4; substituting x = 4 into x + y = 1 gives 4 + y = 1, so y = −3. Answer: x = 4, y = −3, which also satisfies 8 − 3 = 5. The distractors: x = 4, y = 5 comes from rearranging x + y = 1 as y = 1 + x; x = −2, y = 3 comes from eliminating x by doubling the first equation and then reading −y = 3 as y = 3; x = 6, y = −5 comes from adding the constants instead of subtracting them while eliminating y, taking x as 5 + 1.
(b) 8 — Method: the gradient of a straight line is the change in the vertical value divided by the change in the horizontal value between two points on the line. Working: from the origin (0, 0) to (5, 40) the vertical change is 40 − 0 = 40 and the horizontal change is 5 − 0 = 5, so the gradient is 40 ÷ 5 = 8, which here means a cost of £8 for each litre. Answer: 8. The distractors: 0.125 comes from dividing the horizontal change by the vertical change, 5 ÷ 40, which gives litres per pound instead of the gradient; 40 comes from reading off the vertical value of the point and calling it the gradient, ignoring the 5 litres it took to reach that cost; 35 comes from subtracting the two coordinates, 40 − 5, instead of dividing them.
(c) 2 — A point 4 cm from A lies on a circle of radius 4 cm centred at A; a point 3 cm from B lies on a circle of radius 3 cm centred at B. Since AB = 5 cm, and 4 + 3 = 7 is greater than 5 while 4 − 3 = 1 is less than 5, the two circles genuinely cross each other, at two separate points, one on each side of line AB. "1" comes from wrongly assuming the circles only touch rather than cross, which would need 4 + 3 to equal exactly 5. "0" comes from wrongly assuming the circles miss each other completely. "4" comes from counting where each circle crosses the line AB itself (two points each) instead of counting where the two circles cross each other.
(c) 2 — Set A, the multiples of 3, is {3, 6, 9}. Removing the numbers that are also in set B, the even numbers {2, 4, 6, 8, 10}, leaves {3, 9} — 2 numbers are in A but not B. Giving 3, the whole size of set A, forgets to remove the number 6, which is also even. Giving 7 counts every number in A or B combined, {2, 3, 4, 6, 8, 9, 10}, rather than only those in A but not B. Giving 4 counts the numbers in set B but not set A, {2, 4, 8, 10}, the wrong way round.
(b) 4 — Method: the cube root of a number is the value that multiplies by itself three times to give that number. Working: 4 × 4 × 4 = 64, so ∛64 = 4. Answer: 4. (8 comes from finding the square root of 64 instead of the cube root. 192 comes from multiplying 64 by 3 instead of finding the number that cubes to 64. 21.3 comes from dividing 64 by 3 instead of finding its cube root.)
(a) 23 — The number of logs increases by 4 for each layer down, starting from 3 in the top layer, so the nth layer has 3+(n−1)×4 logs. For the 6th layer: 3+5×4=3+20=23. A candidate who adds the difference of 4 six times instead of five, treating the top layer as needing an addition too, would compute 3+6×4=27. A candidate who uses 4n instead of 4n−1, omitting the adjustment for the first layer, would compute 4×6=24. A candidate who subtracts the common difference, 4, instead of 1 when adjusting the multiplier would compute 4×6−4=20.
(b) 10000 cm² — Method: an area conversion factor is the square of the length conversion factor, because both sides of the square are scaled. Working: a square metre is a square of side 100 cm, so its area is 100 × 100 = 10000 cm². Answer: 10000 cm². The distractors: 100 cm² comes from using the length factor without squaring it; 200 cm² comes from doubling the length factor instead of squaring it; 1000000 cm² comes from cubing the factor, which is the conversion for a volume, not an area.
(c) 090 — Method: bearings are measured clockwise from north; east is a quarter turn clockwise. Working: a quarter turn is 90°, written with a leading zero as three figures. A student who answers 009 has kept the correct digits but put the leading zero in the wrong place. A student who answers 180 has confused east with south. A student who answers 270 has confused east with west. Answer: 090.
(c) 2/3 — Method: multiply the numerators together and the denominators together, then divide both parts of the result by their highest common factor. Working: 3 × 14 = 42 and 7 × 9 = 63, giving 42/63; the highest common factor of 42 and 63 is 21, and 42 ÷ 21 = 2 with 63 ÷ 21 = 3. Answer: 2/3. The distractors: 17/16 comes from adding the numerators and adding the denominators, giving (3 + 14)/(7 + 9); 27/98 comes from turning the second fraction upside down and multiplying, which divides instead of multiplying and gives 3/7 × 9/14; 2/21 comes from cancelling the 7 into the 14 in the numerator but leaving the 7 in the denominator, giving 6/63.
(d) Isla is right, because 5 + 1 = 6 — Method: put the input through the machine yourself, then compare what comes out with the output claimed, and check that the reason given with the verdict is itself true. Working: the machine adds 1 to whatever is put in, so an input of 5 gives 5 + 1 = 6, which is exactly the output claimed. Answer: Isla is right, and the reason is that 5 + 1 = 6. The distractors: the statement that the machine adds 1 to its output describes a different machine — this one adds 1 to what goes in — so the verdict is right but the reason is false; 5 − 1 = 4 comes from running the machine backwards and subtracting instead of adding; 5 × 1 = 5 comes from reading ‘adds 1’ as ‘multiplies by 1’, which leaves the input unchanged.
(d) 30 — Method: 50% is one half, so 50% of a quantity is the quantity divided by 2. Working: 60 ÷ 2 = 30. Answer: 30. The distractors: 120 comes from multiplying by 2 instead of dividing; 3000 comes from multiplying by 50 without dividing by 100; 6 comes from finding 10% instead of 50%.
(b) 17 — Roots and powers are worked out first: √25 = 5 and 4² = 16. Division comes next: 12 ÷ 3 = 4. Then addition and subtraction, left to right: 5 + 16 − 4 = 17. A candidate who treated 4² as 4 × 2 = 8, multiplying the base by the exponent instead of squaring it, worked out 5 + 8 − 4 = 9. A candidate who did not evaluate the root and used 25 itself worked out 25 + 16 − 4 = 37. A candidate who ignored the priority of division and worked through 5 + 16 − 12 ÷ 3 strictly left to right got 5 + 16 = 21, then 21 − 12 = 9, then 9 ÷ 3 = 3.
(d) 20 litres — The ratio of concentrate to water is 2 : 5, so water = concentrate × 5 ÷ 2. 8 × 5 ÷ 2 = 20, so Priya needs 20 litres of water. Giving 40 litres multiplies by 5 but forgets to divide by 2 (8 × 5 = 40). Giving 3.2 litres uses the ratio inverted, multiplying by 2 ÷ 5 instead of 5 ÷ 2 (8 × 2 ÷ 5 = 3.2). Giving 11 litres uses additive reasoning instead of multiplicative: it adds the difference between the ratio parts, 5 − 2 = 3, onto the amount of concentrate (8 + 3 = 11), but ratios scale by multiplying, not by adding a fixed amount.

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.