Exam Structure Set B: Higher Paper 3
An 80-mark, 90-minute calculator paper using the Edexcel 1MA1 paper structure as its reference.
Questions
Question 1 [2 marks]
Area, Volume and Measures
Rectangle A measures 5 cm by 8 cm.
Which of these rectangles is mathematically similar to rectangle A?
Write down the scale factor of enlargement from rectangle A to the rectangle you chose in part (a).
Question 2 [2 marks]
Fractions, Decimals and Percentages
Write 0.3333... (3 recurring) as a fraction in its simplest form.
Question 3 [5 marks]
Statistics and Probability
Priya Chandra manages Riverside Car Park. She recorded how many cars and vans were parked there in the morning and in the afternoon.
Question 4 [1 marks]
Angles and Geometrical Reasoning
A student is asked to enlarge a triangle by scale factor -2 from a given centre. The student draws the image on the same side of the centre as the original triangle, but twice the size. Explain the error in the student's diagram.
Question 5 [1 marks]
Graphs and Coordinates
Write down the equation of the graph obtained by reflecting y = f(x) in the y-axis.
Question 6 [1 marks]
Linear Algebra
Simplify 7y - 2y + 4y.
Question 7 [1 marks]
Indices and Standard Form
Work out the value of 16^(1/2).
Question 8 [2 marks]
Ratio and Proportion
A recipe for 6 portions uses 300 g of flour. Work out how much flour is needed for 15 portions, assuming the amount of flour is directly proportional to the number of portions.
Question 9 [2 marks]
Area, Volume and Measures
A sphere has radius 3 cm. Work out the volume of the sphere. Give your answer as a multiple of pi.
Question 10 [2 marks]
Linear Algebra
Make b the subject of the formula: a = (b + 11)/6
Question 11 [3 marks]
Number and Calculation
Find the highest common factor (HCF) of 24 and 36. You must show your method.
Question 12 [3 marks]
Quadratics
Consider the equation x^3 + 4x - 10 = 0 and the iteration x_{n+1} = (10 - x_n^3)/4. Starting with x_0 = 1.5, calculate x_1 and x_2. From these two iterations, state whether the sequence appears to be converging.
Question 13 [3 marks]
Pythagoras and Trigonometry
A cone has base radius 5 cm and perpendicular height 12 cm. Diagram: cone shown in cross-section, base radius marked 5 cm from the centre of the base to the edge, perpendicular height marked 12 cm from the centre of the base to the apex, and the slant height labelled l cm from the apex to the edge of the base. Calculate the slant height, l, of the cone.
Question 14 [4 marks]
Sequences
A sequence has nth term given by T_n = 5n - 3 for n odd, and T_n = 5n + 1 for n even. (a) Work out T_5 and T_6. (b) Explain briefly whether this sequence is arithmetic overall.
Question 15 [4 marks]
Indices and Standard Form
a = 3 x 10^5 b = 4 x 10^-2
Work out a x b. Give your answer in standard form.
Work out a / b. Give your answer in standard form.
Question 16 [4 marks]
Pythagoras and Trigonometry
An isosceles triangle has two equal sides of length 10 cm and a base of length 12 cm. A line is drawn from the apex of the triangle perpendicular to the base, meeting the base at its midpoint. Diagram: isosceles triangle with the two equal sides marked 10 cm, base marked 12 cm, and a dashed perpendicular line from the apex to the midpoint of the base labelled h cm.
Explain why the perpendicular line splits the base into two lengths of 6 cm each.
Calculate the perpendicular height, h, of the triangle.
Question 17 [4 marks]
Linear Algebra
Solve the simultaneous equations: y = 3x - 2 2x + y = 13
Question 18 [4 marks]
Number and Calculation
A rectangle has length 3 times its width. The perimeter of the rectangle is 96 cm. (a) Find the width and the length. (b) Work out the area of the rectangle.
Question 19 [4 marks]
Fractions, Decimals and Percentages
The table shows the number of apples picked at three orchards and the fraction of each orchard's apples that were sold at a farmers' market. Orchard | Apples picked | Fraction sold at market Ashcroft Farm | 96 | 3/8 Bramble Hill | 105 | 2/5 Cedar Grove | 90 | 7/9 Work out which orchard sold the most apples at the market. You must show your working to support your answer.
Question 20 [7 marks]
Statistics and Probability
Amelia recorded the number of text messages she sent on each of 13 days: 2, 5, 6, 6, 7, 8, 9, 9, 10, 12, 14, 15, 20
Find the median number of texts sent.
Find the lower quartile and the upper quartile of the number of texts sent.
Hence find the interquartile range of the number of texts sent.
Draw a box plot to represent Amelia's data on the grid provided.
Question 21 [1 marks]
Quadratics
An iteration formula is x_(n+1) = 3/(x_n - 2). Which of these values of x0 would make it impossible to calculate x1?
Question 22 [2 marks]
Statistics and Probability
In a capture-recapture study, n1 = 40 animals are marked and released. In the second sample, m = 5 of the recaptured animals are marked, and this gives a Lincoln-Petersen population estimate of N = 240. Find n2, the number of animals caught in the second sample.
Question 23 [3 marks]
Indices and Standard Form
Expand and simplify (3 + sqrt(5))(5 - sqrt(5)).
Question 24 [5 marks]
Area, Volume and Measures
A rectangular lawn ABCD has AB = 12 m and BC = 7 m. An isosceles triangular flower bed is attached to side AB, outside the rectangle, with the two equal slant sides of the triangle each measuring 7.5 m. Work out the total area of the lawn and the flower bed together.
Question 25 [3 marks]
Statistics and Probability
A histogram represents the finishing times, in minutes, of 200 athletes in a marathon, grouped into three classes: 180 <= t < 190 (width 10), 190 <= t < 210 (width 20), and 210 <= t < 240 (width 30). The heights of the three bars are in the ratio 3 : 2 : 1 respectively.
Question 26 [4 marks]
Graphs and Coordinates
A bus leaves a stop and accelerates uniformly to a speed of 12 m/s in 15 seconds. It then travels at 12 m/s for 40 seconds, before decelerating uniformly to rest in 10 seconds. The bus then waits at a second stop for 20 seconds, before accelerating uniformly back up to 12 m/s over the next 12 seconds.
Question 27 [3 marks]
Statistics and Probability
Using the cumulative frequency graph for the fun run times (60 runners, shown in Question 3), the organisers decide to award a medal to the fastest 20% of runners. Estimate the completion time below which a runner receives a medal.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| B) 15 cm by 24 cm | B1 |
| 3 cao ft their (a) | B1 |
| Final answer: a) B b) 3 | |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| 3/9 seen or equivalent unsimplified fraction with denominator 9 | M1 |
| 1/3 | A1cao |
| Question 3[5 marks] | |
|---|---|
| Answer or working | Marks |
| Afternoon column Total = 60 | B1cao |
| Vans Morning = 15 | B1cao |
| Cars Afternoon = 35 | B1cao |
| Cars Total = 80 | B1cao |
| Vans Total = 40 | B1cao |
| Final answer: Afternoon Total = 60, Vans Morning = 15, Cars Afternoon = 35, Cars Total = 80, Vans Total = 40 | |
| Question 4[1 mark] | |
|---|---|
| Answer or working | Marks |
| the image should be drawn on the opposite side of the centre of enlargement (and inverted), because the scale factor is negative | B1oe |
| Final answer: The image should be on the opposite side of the centre of enlargement (and inverted), because the scale factor is negative. | |
| Question 5[1 mark] | |
|---|---|
| Answer or working | Marks |
| y = f(-x) | B1cao |
| Question 6[1 mark] | |
|---|---|
| Answer or working | Marks |
| 9y | B1cao |
| Final answer: 9y (cao) | |
| Question 7[1 mark] | |
|---|---|
| Answer or working | Marks |
| 4 | B1cao |
| Question 8[2 marks] | |
|---|---|
| Answer or working | Marks |
| use direct proportion: 300/6 * 15 or equivalent method | M1 |
| 750 g | A1cao |
| Final answer: 750 g (grams). This is 300/6 * 15 = 50 * 15 = 750 g. Working check: 6 portions -> 300 g, so per portion 50 g; 15 portions -> 750 g. | |
| Question 9[2 marks] | |
|---|---|
| Answer or working | Marks |
| correct substitution (4/3) x pi x 3^3 | M1oe |
| 36 x pi cm^3 (oe, cao - exact value required, not a decimal) | A1 |
| Final answer: 36 x pi cm^3 (exact) | |
| Question 10[2 marks] | |
|---|---|
| Answer or working | Marks |
| multiplies both sides by 6, e.g. 6a = b + 11 | M1 |
| b = 6a - 11 oe | A1cao |
| Final answer: b = 6a - 11 | |
| Question 11[3 marks] | |
|---|---|
| Answer or working | Marks |
| factors of 24 listed: 1, 2, 3, 4, 6, 8, 12, 24 (or use of prime factorisation) | M1 |
| factors of 36 listed: 1, 2, 3, 4, 6, 9, 12, 18, 36 and common factors identified | M1 |
| 12 | A1cao |
| Question 12[3 marks] | |
|---|---|
| Answer or working | Marks |
| calculate x_1 = (10 - 1.5^3)/4 = 1.65625 | M1 |
| calculate x_2 = (10 - x_1^3)/4 = 1.36415863 | M1awrt |
| conclude that the sequence does not appear to be converging (it oscillates) | B1oe |
| Final answer: x_1 = (10 - 1.5^3)/4 = 6.625/4 = 1.65625; x_2 = (10 - 1.65625^3)/4 approx 1.36415863. The values jump (1.5 -> 1.65625 -> 1.36416), so the sequence appears to oscillate and not settle (not converging). | |
| Question 13[3 marks] | |
|---|---|
| Answer or working | Marks |
| recognises right-angled triangle formed by the radius, height and slant height | M1oe |
| 5^2 + 12^2 oe (= 169) | M1 |
| 13 cm | A1cao |
| Question 14[4 marks] | |
|---|---|
| Answer or working | Marks |
| substitute n = 5 and n = 6 into correct formula depending on parity | M1 |
| 22 and 31 | A1cao |
| compare differences between consecutive terms and note they are not constant | M1 |
| statement: not arithmetic because differences are not constant (example given) | A1 |
| Final answer: T_5 = 22, T_6 = 31 | Not arithmetic, because consecutive differences are not constant (they alternate). | |
| Question 15[4 marks] | |
|---|---|
| Answer or working | Marks |
| 3 x 4 = 12 and 10^5 x 10^-2 = 10^3 seen, or 12 x 10^3 oe unsimplified | M1 |
| 1.2 x 10^4 | A1cao |
| 3 / 4 = 0.75 and 10^5 / 10^-2 = 10^7 seen, or 0.75 x 10^7 oe unsimplified | M1 |
| 7.5 x 10^6 | A1cao |
| Final answer: 1.2 x 10^4 | 7.5 x 10^6 | |
| Question 16[4 marks] | |
|---|---|
| Answer or working | Marks |
| correct reason, e.g. the perpendicular from the apex of an isosceles triangle bisects the base (by symmetry), so each half is 12 / 2 = 6 cm | B1oe |
| 10^2 - 6^2 oe (= 64) | M1 |
| sqrt(their 64) | M1 |
| 8 cm | A1cao |
| Final answer: The perpendicular from the apex of an isosceles triangle bisects the base by symmetry, so each half is 12 divided by 2 = 6 cm. | 8 cm | |
| Question 17[4 marks] | |
|---|---|
| Answer or working | Marks |
| substitutes y = 3x - 2 into the second equation | M1 |
| forms and simplifies a single linear equation in x, e.g. 5x - 2 = 13 | M1 |
| x = 3 | A1 |
| y = 7 | A1cao |
| Final answer: x = 3, y = 7 | |
| Question 18[4 marks] | |
|---|---|
| Answer or working | Marks |
| forms equation 2(3w + w) = 96 or 8w = 96 | M1 |
| width = 12 cm and length = 36 cm | A1cao |
| uses area = length × width | M1 |
| 432 cm^2 | A1cao |
| Final answer: width 12 cm, length 36 cm | 432 cm^2 | |
| Question 19[4 marks] | |
|---|---|
| Answer or working | Marks |
| 96 / 8 = 12, 12 x 3 = 36 oe (Ashcroft Farm's apples sold) | M1 |
| 105 / 5 = 21, 21 x 2 = 42 oe (Bramble Hill's apples sold) | M1 |
| 90 / 9 = 10, 10 x 7 = 70 oe (Cedar Grove's apples sold) | M1 |
| cao, Cedar Grove sold the most, since 70 is greater than 42 and 36 | A1 |
| Final answer: Cedar Grove sold the most, 70 apples (Ashcroft Farm = 36, Bramble Hill = 42, Cedar Grove = 70). | |
| Question 20[7 marks] | |
|---|---|
| Answer or working | Marks |
| 9 | B1cao |
| splitting the data into a lower half (2,5,6,6,7,8) and an upper half (9,10,12,14,15,20) and finding the median of each half | M1 |
| LQ = 6 and UQ = 13, both | A1cao |
| ft: 7 | B1cao |
| ft: whiskers drawn correctly from the minimum (2) to the box, and from the box to the maximum (20) | B1 |
| ft: box drawn correctly from the lower quartile (6) to the upper quartile (13) | B1 |
| ft: median (9) correctly marked inside the box | B1 |
| Final answer: 9 texts | LQ = 6 texts, UQ = 13 texts | 7 texts | Box plot with min 2, LQ 6, median 9, UQ 13, max 20 | |
| Question 21[1 mark] | |
|---|---|
| Answer or working | Marks |
| C | B1cao |
| Question 22[2 marks] | |
|---|---|
| Answer or working | Marks |
| rearrange N = n1 * n2 / m to make n2 the subject: n2 = N * m / n1 | M1oe |
| 30 | A1cao |
| Question 23[3 marks] | |
|---|---|
| Answer or working | Marks |
| at least 3 of the 4 terms 15, -3sqrt(5), 5sqrt(5), -5 correct | M1 |
| correct unsimplified expansion 15 - 3sqrt(5) + 5sqrt(5) - 5 (all 4 terms right) | A1 |
| 10 + 2sqrt(5) oe | A1cao |
| Final answer: 10 + 2sqrt(5) | |
| Question 24[5 marks] | |
|---|---|
| Answer or working | Marks |
| half base = 6 m identified | M1 |
| height^2 = 7.5^2 - 6^2 (Pythagoras) | M1 |
| height = 4.5 m | A1cao |
| triangle area = 1/2 x 12 x 4.5 (=27) and rectangle area = 12 x 7 (=84) | M1 |
| 111 m^2 | A1cao |
| Question 25[3 marks] | |
|---|---|
| Answer or working | Marks |
| expressing frequencies in terms of k: 10 x 3k = 30k, 20 x 2k = 40k, 30 x k = 30k | M1 |
| 30k + 40k + 30k = 200, so 100k = 200, k = 2 | M1 |
| frequency for 190 <= t < 210 = 40k = 80 | A1cao |
| Final answer: 80 athletes | |
| Question 26[4 marks] | |
|---|---|
| Answer or working | Marks |
| correct straight line from (0,0) rising to (15,12) | B1 |
| correct horizontal line from (15,12) to (55,12) | B1 |
| correct straight line falling from (55,12) to (65,0), then horizontal at 0 from (65,0) to (85,0) | B1 |
| correct straight line rising from (85,0) to (97,12), with key time and velocity values labelled on the axes | B1 |
| Final answer: Velocity-time graph with vertices at (0,0), (15,12), (55,12), (65,0), (85,0), (97,12). | |
| Question 27[3 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the required position as the 12th value (20% of 60 = 12) | M1 |
| correct interpolation method between (40,4) and (50,13) | M1 |
| awrt 49 minutes (accept 47-51) | A1 |
| Final answer: awrt 49 minutes | |