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)
MathsUKwww.geekhero.co.uk
- 1.Four weather stations each recorded their lowest overnight temperature. Which of these readings is the lowest?
- 2.Tom buys 2 notebooks and 6 pencils on Monday, priced at p pence each for a notebook and q pence each for a pencil. On Tuesday he buys 3 more notebooks and 1 more pencil. Write a simplified expression, in pence, for the total amount Tom has spent on notebooks and pencils.
- 3.Which of these ratios is equivalent to 5 : 4?
- 4.A tent is shaped like a triangular prism: two triangular end faces and three rectangular faces. One of the rectangular faces lies flat on the ground as the groundsheet. How many of the tent's faces are above the ground, not touching the ground?
- 5.A machine makes 4000 light bulbs a day and runs 5 days a week. Two inspectors test bulbs from this machine. Inspector A tests 40 bulbs and finds 4 faulty. Inspector B tests 500 bulbs and finds 30 faulty. Using the better of the two estimates, work out how many faulty bulbs the machine is expected to make in one week.
- 6.Four books have 120, 200, 160 and 80 pages. Work out the mean number of pages.
- 7.A sponsored walk is 36 km long. Aisha has completed 8/12 of the walk. Work out how far she has walked.
- 8.Factorise fully 56x − 24
- 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.Postcard P is similar to postcard Q. The 6 cm side of P corresponds to the 15 cm side of Q, and the 8 cm side of P corresponds to a matching side of Q. Work out the length of that matching side of Q.
- 11.A jar contains counters that are pink, gold or grey only. The probability that a counter taken at random is pink is 0.2 and the probability that it is gold is 0.45. The jar contains 60 counters in total. Work out the number of grey counters.
- 12.Grace works out 7 × 99 by writing 99 as 100 − 1. Use her method to work out 7 × 99.
- 13.A straight line has equation y = 2x − 7. Write down the coordinates of the point where the line crosses the y-axis.y = 2x − 7
- 14.In a box of pens, 3/7 of the pens are blue and the rest are black. Write down the ratio of the number of blue pens to the number of black pens, in its simplest form.
- 15.A triangle has vertices A(1, 1), B(5, 1) and C(5, 4). Work out the lengths of AB, BC and CA, and write down whether triangle ABC is right-angled.
- 16.Work out (2/3)² + (1/2)². Give your answer as a fraction.
- 17.A geometric sequence has first term 3 and common ratio 2. Work out the 5th term of the sequence.
- 18.A stall sells bottles of juice. Two bottles cost £3.00, five bottles cost £7.50 and eight bottles cost £12.00. The cost, C pounds, is in a fixed ratio to the number of bottles, n. Write down a formula for C in terms of n.
- 19.Two-digit numbers are formed using the digits 2, 5, 7 and 8, and each digit may be used only once in a number. Work out how many of these two-digit numbers are even.
- 20.A map has a scale of 1 : 50 000. Two villages are 4 cm apart on the map. Work out the real distance between the villages, in kilometres.
Answer key
- (a) −15 °C — Method: the lowest temperature is the reading furthest to the left on a number line, and among negative readings that is the one furthest below zero. Working: the three negative readings are 1 degree, 7 degrees and 15 degrees below zero, and 0 °C is not below zero at all. Fifteen degrees below zero is the furthest below. Answer: −15 °C. The distractors: −1 °C comes from ordering the negatives by the size of their digits, so that the smallest digit is taken as the lowest temperature; −7 °C comes from comparing 15 and 7 by their leading digits, deciding that 1 is less than 7 and so that −7 lies below −15; 0 °C comes from treating zero as the smallest number there is.
- (a) 5p + 7q — Tom's total spend combines like terms: (2p + 6q) + (3p + q) = 5p + 7q, adding the notebook terms (2p + 3p = 5p) and the pencil terms (6q + q = 7q) separately. Answering 12pq adds every coefficient together (2 + 6 + 3 + 1 = 12) and multiplies the letters, combining unlike terms as though notebooks and pencils were the same item. Answering 5p + 6q correctly combines the notebook terms but forgets to add Tuesday's extra pencil to the 6q. Answering 2p + 7q correctly combines the pencil terms but forgets to add Tuesday's 3 extra notebooks to the 2p. The total amount Tom has spent is 5p + 7q pence.
- (a) 15 : 12 — Multiply both parts of the ratio 5 : 4 by the same number, 3, to get an equivalent ratio: 5 × 3 = 15 and 4 × 3 = 12, giving 15 : 12. Giving 15 : 16 multiplies the two parts by different scale factors (×3 and ×4), which changes the ratio. Giving 9 : 8 adds 4 to each part instead of multiplying, which also changes the ratio. Giving 4 : 5 swaps the order of the two parts.
- (b) 4 — A triangular prism has 5 faces in total: 2 triangular ends and 3 rectangular faces. One rectangular face is the groundsheet, lying on the ground, so the other 5 − 1 = 4 faces are above the ground. A candidate who forgets one of the triangular ends when counting the remaining faces answers 3. A candidate who forgets to subtract the groundsheet at all answers 5. A candidate who only counts the two sloped rectangular faces, forgetting the two triangular ends, answers 2. The number of faces above the ground is 4.
- (a) 1200 — Method: take the estimate from the larger sample, because an unbiased relative frequency tends towards the true probability as the sample grows, then multiply by the number of bulbs made in a week. Working: Inspector B tested 500 bulbs, far more than Inspector A's 40, so use B's relative frequency: 30 ÷ 500 = 0.06. A week's production is 4000 × 5 = 20000 bulbs. The expected number of faulty bulbs is 20000 × 0.06 = 1200. Answer: about 1200 faulty bulbs a week. The distractors: 2000 uses Inspector A's estimate, 4 ÷ 40 = 0.1, giving 20000 × 0.1 = 2000, and so rests on a sample of only 40 bulbs; 1600 comes from averaging the two estimates of 0.1 and 0.06 to get 0.08, and 20000 × 0.08 = 1600, which gives the small sample equal weight with the large one; 240 uses the right estimate but stops at a single day, 4000 × 0.06 = 240.
- (c) 140 — Method: the mean shares the total equally between the items, so add the values and then divide by how many there are. Working: the total is 120 + 200 + 160 + 80 = 560 pages and there are 4 books, so the mean is 560 ÷ 4 = 140 pages. Answer: 140. The distractors: 560 comes from stopping at the total number of pages and never dividing by 4; 280 comes from dividing the total by 2 instead of by the 4 books; 120 comes from working out the range, 200 − 80, which measures spread instead of centre.
- (d) 24 km — Method: a fraction acts as an operator, so finding 8/12 of a distance means dividing by the denominator and multiplying by the numerator. Working: 36 ÷ 12 = 3, so one twelfth of the walk is 3 km, and eight twelfths is 3 × 8 = 24 km. Answer: 24 km. The distractors: 12 km comes from working out the part of the walk still left, the other four twelfths, instead of the part already completed; 288 km comes from multiplying by the numerator without dividing by the denominator, giving 36 × 8 = 288; 4.5 km comes from dividing by the numerator instead of multiplying by it, giving 36 ÷ 8 = 4.5.
- (d) 8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8.
- (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.
- (d) 20 cm — The scale factor from P to Q is 15 ÷ 6 = 2.5. Apply the same scale factor to the other side: 8 × 2.5 = 20 cm. A pupil who divides instead of multiplying by the scale factor gets 8 ÷ 2.5 = 3.2 cm. A pupil who adds the difference between the two known sides, 15 − 6 = 9, to 8 instead of scaling gets 8 + 9 = 17 cm. A pupil who rounds the scale factor 2.5 down to 2 gets 8 × 2 = 16 cm. The correct length is 20 cm.
- (d) 21 — Pink, gold and grey cover every counter, so their probabilities sum to 1: P(grey) = 1 − 0.2 − 0.45 = 0.35. Number of grey counters = 0.35 × 60 = 21. Multiplying the pink probability by 60 gives 12, the number of pink counters, not grey. Multiplying the gold probability by 60 gives 27, the number of gold counters. Adding 0.2 and 0.45 without subtracting from 1 gives 0.65, and 0.65 × 60 = 39 counts the pink and gold counters together, not the grey ones.
- (c) 693 — Method: multiplying a bracket by a number multiplies every term inside it, so 7 × (100 − 1) = 7 × 100 − 7 × 1. Working: 7 × 100 = 700 and 7 × 1 = 7, so the calculation becomes 700 − 7 = 693. Answer: 693. The distractors: 699 comes from subtracting the 1 itself rather than 7 lots of it, giving 700 − 1 = 699; 707 comes from adding the second product instead of subtracting it, giving 700 + 7 = 707; 700 comes from rounding 99 up to 100 and then offering the estimate 7 × 100 as an exact value.
- (d) (0, −7) — At the y-axis, x = 0, so y = 2(0) − 7 = −7, giving the point (0, −7). A candidate who drops the negative sign on the constant term would write (0, 7). A candidate who swaps the coordinates, confusing the y-intercept with an x-intercept, would write (−7, 0). A candidate who uses the gradient, 2, as the x-coordinate instead of 0 would write (2, −7).
- (c) 3 : 4 — Method: a fraction compares a part with the whole, while this ratio compares one part with the other part, so find the fraction that is black before writing the ratio. Working: if 3/7 are blue then the black pens make up 7/7 − 3/7 = 4/7 of the box, so out of every 7 pens 3 are blue and 4 are black, and blue : black = 3 : 4. Answer: 3 : 4. The distractors: 3 : 7 comes from reading the numerator and the denominator of 3/7 straight off as the two parts, which compares the blue pens with the whole box rather than with the black pens; 4 : 3 comes from writing the black pens before the blue pens, reversing the order asked for; 4 : 7 is the same numerator-and-denominator reading applied to the black fraction 4/7, again comparing a part with the whole box.
- (d) Yes, since 3² + 4² = 5² — AB is horizontal with length 5 − 1 = 4, BC is vertical with length 4 − 1 = 3, and CA = √(4² + 3²) = √25 = 5. Since the two shorter sides satisfy 3² + 4² = 5², the triangle is right-angled, with the right angle at B. "No, since 3 + 4 ≠ 5" wrongly tests Pythagoras' theorem by adding the sides instead of squaring them first. "No, since AB, BC and CA are not all equal" confuses a right-angled triangle with an equilateral one — a triangle does not need equal sides to have a right angle. "Yes, since 4² + 5² = 3²" reaches the correct conclusion but puts the longest side, 5, on the wrong side of the equation, as if it were one of the two shorter sides instead of the hypotenuse.
- (c) 25/36 — Method: square a fraction by squaring its numerator and its denominator separately, then add the two results over a common denominator. Working: (2/3)² = 4/9 and (1/2)² = 1/4; the lowest common denominator of 9 and 4 is 36, so 4/9 = 16/36 and 1/4 = 9/36, and 16 + 9 = 25. Answer: 25/36. The distractors: 49/36 comes from adding the two fractions first and squaring the total, giving (7/6)²; 5/13 comes from squaring correctly but then adding the numerators and the denominators, as (4 + 1)/(9 + 4); 7/3 comes from doubling each fraction instead of squaring it, giving 4/3 + 1.
- (d) 48 — Each term is found by multiplying the previous term by the common ratio, 2: 3, 6, 12, 24, 48 — the 5th term is 48. A candidate who mistakes the common ratio for a common difference, and adds 2 four times, would reach 3+4×2=11. A candidate who works out the multiplier 2⁴=16 but forgets to multiply it by the first term would give 16. A candidate who multiplies one time too many (finding the 6th term instead of the 5th) would reach 3×2⁵=96.
- (c) C = 1.5n — Method: a fixed ratio between C and n means C is always the same multiple of n, and that multiple is the cost of one bottle. Working: 3.00 ÷ 2 = 1.5, 7.50 ÷ 5 = 1.5 and 12.00 ÷ 8 = 1.5, so every bottle costs £1.50 and C = 1.5n. Answer: C = 1.5n. The distractors: C = n + 1 comes from subtracting on the first row, 3 − 2 = 1, and adding that difference instead of multiplying; it fits the first row and fails the other two, which is why three rows are given; C = 3n reads the £3.00 as the price of one bottle when it is the price of two; C = n/1.5 divides the number of bottles by the price of one bottle, which works out how many bottles a pound buys instead of what n bottles cost.
- (c) 6 — The units digit must be even, so it can be 2 or 8, giving 2 choices. The tens digit can then be any of the remaining 3 digits, since one digit has been used for the units. Multiply: 2 × 3 = 6. 12 comes from working out how many two-digit numbers can be made in total, 4 × 3 = 12, ignoring the requirement that the number is even. 8 comes from choosing the units digit from 2 options and then wrongly allowing any of the 4 digits again for the tens digit, 2 × 4 = 8, which lets a digit repeat. 2 comes from counting only the choices for the units digit and forgetting the tens digit.
- (a) 2 km — Method: multiply the map distance by the scale to get the real distance in centimetres, then convert centimetres to kilometres using 100 cm = 1 m and 1000 m = 1 km. Working: 4 × 50 000 = 200 000 cm; 200 000 ÷ 100 = 2000 m; 2000 ÷ 1000 = 2. Answer: 2 km. The distractors: 200 km comes from dividing the 200 000 cm by 1000 in a single step, as if a kilometre were 1000 cm rather than the 100 000 cm it is; 20 km comes from converting to metres correctly, 200 000 ÷ 100 = 2000 m, and then dividing those metres by 100 instead of by 1000; 0.2 km comes from dividing by 1000 to reach metres, as if a metre were 1000 cm, and then dividing by 1000 again, so 200 000 is divided by 1 000 000 altogether.
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.