Sample paper · GCSE Higher · grades 4–9
GCSE Higher 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 Higher qualification — number 15%, algebra 30%, ratio, proportion and rates of change 20%, geometry and measures 20%, probability 7.5%, statistics 7.5% — with a calculator allowed. AO1 / AO2 / AO3 at this tier: 40% / 30% / 30%.
CalculatorGCSE Higher
GCSE Higher sample Paper 2 (calculator)
MathsUKwww.geekhero.co.uk
- 1.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.
- 2.A Fibonacci-type sequence is defined by aₙ = aₙ₋₁ + aₙ₋₂ for n ≥ 3. Given that a₁ = 5 and a₃ = 17, work out a₂.
- 3.A vehicle is worth £18000. Its value decreases by 15% each year. Using the recurrence V_{n+1} = 0.85V_n, with V_0 = 18000, find the first whole number of years after which the vehicle's value drops below £10000.
- 4.A tap fills a tank at a rate of 18 litres per minute. The tank holds 0.45 m³. Work out how long the tap takes to fill the tank.
- 5.A biased spinner is spun 200 times. It lands on red 70 times, on blue 50 times, and on green 80 times. Using these results, work out the expected number of times the spinner does NOT land on red, in 500 spins of the same spinner.
- 6.In a random sample of 40 pupils at a school, 6 are left-handed. The school has 900 pupils. Work out an estimate for the number of left-handed pupils in the school.
- 7.Write these numbers in order, starting with the smallest: −1.4, 5/4, −6/5, 1.3, 0
- 8.A theatre has rows of seats arranged so that row 1 has 12 seats, row 2 has 17 seats, row 3 has 22 seats and row 4 has 27 seats, with each row having 5 more seats than the row before. Work out an expression, in terms of n, for the number of seats in row n.
- 9.The height of water in a tank, h metres, t minutes after a tap is opened is modelled by h = 0.02t² + 0.5. Estimate the instantaneous rate of change of the height at t = 10, using the gradient of the chord joining t = 9 and t = 11.
- 10.Points P and Q have coordinates P(0, 0, 0) and Q(5, 7, 9) in a three-dimensional coordinate system, with all lengths in centimetres. Work out the distance PQ. Give your answer correct to 1 decimal place.
- 11.A survey found that, of 250 shoppers questioned, 68% said they had used a self-checkout in the past month. A different store expects 1400 shoppers this week. Using this relative frequency, work out how many of the 1400 shoppers would be expected to have used a self-checkout in the past month.
- 12.A number, x, is truncated (not rounded) to 1 decimal place and the result is 6.2. Write down the error interval for x.
- 13.A circle has centre (0, 0) and passes through the point (7, 24). Work out the equation of the circle.
- 14.There are 2500 electric cars registered in a town. The number is predicted to increase by 6% each year. Work out the predicted number of electric cars after 3 years, to the nearest whole number.
- 15.A shop sells rice in packs of 250 g. Aisha buys 4 packs. Work out the total mass of rice she buys, in kilograms.
- 16.Which of these equations represents the graph of y = 2ˣ translated by 3 units in the positive y-direction?
- 17.A tap fills a tank at a rate of 15 litres per minute. Given that 1 litre = 1000 cm³, work out the rate at which the tank fills in cm³ per second.
- 18.A student says the reverse of the translation vector is . Write down the correct reverse translation vector.
- 19.Which expression means 'y squared, multiplied by 3'?
- 20.The equation x² − 3x − 7 = 0 can be solved using the iterative formula xₙ₊₁ = √(3xₙ + 7). The starting value is x₀ = 4, so x₁ is the value after the formula has been used once. Work out x₃ correct to 3 decimal places.
Answer key
- (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) 12 — Since a₃ = a₂ + a₁, rearranging gives a₂ = a₃ − a₁ = 17 − 5 = 12. 22 comes from adding instead of subtracting: a₃ + a₁ = 17 + 5 = 22. 11 comes from treating the sequence as if it were arithmetic and averaging the two given terms: (5 + 17) ÷ 2 = 11 — but a Fibonacci-type sequence isn't arithmetic, so this doesn't apply. −12 comes from subtracting in the wrong order: a₁ − a₃ = 5 − 17 = −12.
- (d) 4 years — Apply the recurrence repeatedly. V_1 = 0.85 × 18000 = 15300. V_2 = 0.85 × 15300 = 13005. V_3 = 0.85 × 13005 = 11054.25. V_4 = 0.85 × 11054.25 = 9396.1125. V_3 = £11054.25 is still above £10000, but V_4 = £9396.11 has dropped below it, so the answer is 4 years. Stopping at V_3 and calling it '3 years' misreads £11054.25 as already below £10000, or comes from wrongly modelling the fall as a flat £2700 a year (15% of the original value each time, without compounding), which crosses £10000 a year too early. Continuing one extra step to V_5 = 0.85 × 9396.1125 = 7986.70 and calling it '5 years' overshoots, since the value had already dropped below £10000 at V_4. Doubling the percentage decrease to 30% by mistake gives V_1 = 0.7 × 18000 = 12600, then V_2 = 0.7 × 12600 = 8820, which is already below £10000 after only 2 years — the wrong rate crosses the threshold too fast.
- (b) 25 minutes — Method: a rate in litres per minute can only be used on a volume measured in litres, so convert the tank first and then divide. Working: 1 m³ = 1000 litres, so the tank holds 0.45 × 1000 = 450 litres, and the time is 450 ÷ 18 = 25. Answer: 25 minutes. Using 1 m³ = 100 litres gives 45 ÷ 18 = 2.5 minutes. Using 1 m³ = 1 000 000 litres, which is the factor that turns cubic metres into cubic centimetres, gives 450 000 ÷ 18 = 25 000 minutes. Multiplying by the rate instead of dividing by it gives 450 × 18 = 8100.
- (c) 325 — Method: first find the relative frequency of NOT landing on red from the 200 spins, then scale that up to 500 spins. Working: non-red results = 50 + 80 = 130, out of 200 spins, so P(not red) = 130 ÷ 200 = 0.65. Expected non-red results in 500 spins = 500 × 0.65 = 325. Answer: 325. Watch out: writing down 175 finds the expected number of RED results instead, 70 ÷ 200 × 500 = 175, answering the opposite of what was asked. Writing down 250 assumes landing red or not landing red must be a fair 50-50 split, but the spinner is biased and the actual results do not split evenly. And writing down 130 stops after finding how many of the 200 spins were non-red and forgets to scale that figure up to the 500 spins asked for.
- (b) 135 — Method: use the sample to find the PROPORTION of left-handed pupils, then apply that same proportion to the whole school population. Working: in the sample, 6 out of 40 pupils are left-handed, a proportion of 6 ÷ 40 = 0.15. Applying that proportion to the school's 900 pupils gives an estimate of 0.15 × 900 = 135 pupils. Giving 6 simply repeats the number of left-handed pupils IN THE SAMPLE, without scaling up to the whole school at all. Multiplying the population by the number of left-handed pupils in the sample without first dividing by the sample size, 900 × 6 = 5400, badly overestimates — that is more pupils than the whole school has. Dividing the population by the sample size but forgetting to multiply by the number of left-handed pupils found, 900 ÷ 40 = 22.5, finds the scale factor but stops one step short of using it. Always find the proportion in the sample first, then scale that same proportion up to the population.
- (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) 5n + 7 — Method: find the common difference between the rows, then find the constant by adjusting the first row's total. Working: each row has 5 more seats than the last, so the coefficient of n is 5. The constant is the first row's total minus the common difference: 12 − 5 = 7. Answer: the nth term is 5n + 7. 5n + 12 comes from using row 1's total, 12, as the constant without subtracting the common difference. 5n + 2 comes from a slip in working out the constant, subtracting the common difference twice (12 − 5 − 5 = 2) instead of once. 7n + 5 comes from swapping the common difference and the constant.
- (d) 0.40 m/min — To estimate an instantaneous rate of change at a point without a diagram, use the gradient of a chord joining two points close to it, one on each side. At t = 9: 0.02 × 81 = 1.62, so h = 1.62 + 0.5 = 2.12. At t = 11: 0.02 × 121 = 2.42, so h = 2.42 + 0.5 = 2.92. The change in height is 2.92 − 2.12 = 0.80 and the change in time is 11 − 9 = 2, so the gradient of the chord is 0.80 ÷ 2 = 0.40. Reporting the change in height, 0.80, on its own is not a rate, because it has not been divided by the 2 minutes over which it happened. Using the chord from t = 0 (where h = 0.5) to t = 11 instead gives 2.92 − 0.5 = 2.42, and 2.42 ÷ 11 = 0.22, which is the average gradient over the whole 11 minutes, not the instantaneous rate at t = 10. Substituting t = 10 into the formula gives 0.02 × 100 + 0.5 = 2.50, which is the height of the water at that moment, not the rate at which the height is rising. The estimated instantaneous rate of change at t = 10 is 0.40 m/min.
- (c) 12.4 cm — In three dimensions, the distance between two points extends Pythagoras' theorem to three squared terms: PQ² = 5² + 7² + 9² = 25 + 49 + 81 = 155. Taking the square root, PQ = √155 = 12.4 cm (1 d.p.). Using only the x- and y-coordinates, and ignoring the third dimension entirely, gives PQ = √(5² + 7²) = √74 = 8.6 cm (1 d.p.), which is not the distance in 3D space. Adding the three coordinates directly, 5 + 7 + 9 = 21 cm, treats the coordinates as if they were lengths along a single straight path rather than the sides of a right-angled arrangement. Squaring and adding all three coordinates but forgetting to take the square root leaves 155 cm, the squared distance rather than the distance itself.
- (a) 952 — 68% = 0.68. The relative frequency from the survey applies to the new group of 1400 shoppers, so the expected number is 0.68 × 1400 = 952. Working out 1 − 0.68 = 0.32 and applying that instead, 0.32 × 1400 = 448, finds the number who have NOT used a self-checkout, not the number who have. Applying 68% to the original sample size of 250 instead of the new total of 1400 gives 0.68 × 250 = 170. Applying the complement percentage to the original sample size, 0.32 × 250 = 80, compounds both mistakes.
- (b) 6.2 ≤ x < 6.3 — Truncating simply cuts off the digits after the required decimal place instead of rounding them, so every value from 6.2 up to (but not reaching) 6.3 truncates to 6.2. This gives the error interval 6.2 ≤ x < 6.3, with no allowance made on the lower side because truncation never rounds a smaller value up into this interval. Using 6.15 ≤ x < 6.25 applies the rounding rule of going half a unit either side, which does not apply to truncation. Writing 6.1 < x ≤ 6.2 puts the interval below 6.2 instead of above it. Writing 6.2 ≤ x ≤ 6.3 wrongly includes 6.3, which truncates down to itself, not to 6.2.
- (b) x² + y² = 625 — Since (7, 24) lies on the circle, x² + y² = 7² + 24² = 49 + 576 = 625, so the equation is x² + y² = 625. Choosing x² + y² = 31 adds the coordinates 7 and 24 directly instead of squaring them first. Choosing x² + y² = 576 uses only 24² and forgets to add 7². Choosing x² + y² = 49 uses only 7² and forgets to add 24².
- (a) 2978 — A rise of 6% is a multiplier of 1.06, applied once for each year. After year 1: 2500 × 1.06 = 2650. After year 2: 2650 × 1.06 = 2809. After year 3: 2809 × 1.06 = 2977.54, which is 2978 to the nearest whole number. Multiplying by 1.18 in one go would be wrong, because the second and third years grow from larger numbers than the first.
- (d) 1 kg — First find the total mass in grams: 4 × 250 = 1000 g. Then convert to kilograms by dividing by 1000: 1000 ÷ 1000 = 1 kg. Finding the correct total in grams but forgetting to divide by 1000 gives 1000 kg. Adding the number of packs to the pack mass instead of multiplying, 4 + 250 = 254 g, gives 0.254 kg. Dividing the pack mass by the number of packs instead of multiplying, 250 ÷ 4 = 62.5 g, gives 0.0625 kg.
- (a) y = 2ˣ + 3 — A translation of 3 units in the positive y-direction shifts the whole graph up, which means adding to the output: y = f(x) + k with k = 3, so the image is y = 2ˣ + 3. Adding the 3 inside the power instead of outside it, which translates the graph horizontally instead of vertically, gives y = 2ˣ⁺³. Using a negative 3, which moves the graph down instead of up, gives y = 2ˣ − 3. Mistaking 2ˣ for the linear expression 2x and adding 3 inside brackets gives y = 2(x + 3).
- (d) 250 cm³/s — Method: first change litres per minute into cm³ per minute, then change per minute into per second. Working: 15 × 1000 = 15000 cm³ per minute, then 15000 ÷ 60 = 250 cm³ per second. So the tank fills at 250 cm³ per second. Distractor 15000 cm³/s comes from stopping after the first step and forgetting to change minutes into seconds. Distractor 900000 cm³/s comes from multiplying by 60 instead of dividing. Distractor 2500 cm³/s comes from dividing by 6 instead of 60.
- (c) $\binom{-4}{9}$ — The reverse of a translation negates both components, so the reverse of $\binom{4}{-9}$ is $\binom{-4}{9}$. $\binom{4}{9}$ is the student's mistake — only the bottom number has been negated, and the top number was left unchanged. $\binom{-4}{-9}$ makes the opposite error, negating only the top number. $\binom{4}{-9}$ is simply the original vector, unchanged.
- (a) 3y² — 'y squared, multiplied by 3' means the square is applied to y only, and the result is then multiplied by 3, written as 3y². Writing y³ mistakes the multiplication by 3 for an extra factor of y, adding to the power instead of using a coefficient. Writing (3y)² squares the whole of 3y, including the 3, which gives 9y² rather than 3y² — the square should apply to y alone. Writing 3 + y² adds the 3 instead of multiplying by it. The expression for 'y squared, multiplied by 3' is 3y².
- (d) 4.521 — x₁ = √(3 × 4 + 7) = √19 = 4.358898944. x₂ = √(3 × 4.358898944 + 7) = √20.076696833 = 4.480702716. x₃ = √(3 × 4.480702716 + 7) = √20.442108148 = 4.521294964, which rounds to 4.521. Mislabelling the starting value x₀ as x₁, so that the working stops one iteration too early, reports the true x₂ = 4.480702716, which rounds to 4.481. Working out one iteration too many reports the true x₄ = √(3 × 4.521294964 + 7) = 4.534741987, which rounds to 4.535. Applying the formula in the wrong order, calculating √(3xₙ) + 7 at every step instead of √(3xₙ + 7), gives, from x₀ = 4: √12 + 7 = 10.464101615, then √(3 × 10.464101615) + 7 = 12.602883619, then √(3 × 12.602883619) + 7 = 13.148873950, which rounds to 13.149.
How the 20 questions are shared out
- Number — 3 questions (15% of the qualification)
- Algebra — 6 questions (30% of the qualification)
- Ratio, proportion and rates of change — 4 questions (20% of the qualification)
- Geometry and measures — 4 questions (20% 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.