Year 10 Paper 4: Proportion and Linear Methods
Covers fractions, decimals and percentages, indices, surds and standard form, ratio and proportion, linear algebra, sequences and graphs, and geometry.
Year 10 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 10, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.
Questions
Question 1 [3 marks]
Angles, Polygons and Circle Theorems
A regular nonagon (9-sided polygon) is drawn.
Find the sum of its interior angles, and find the size of one interior angle.
Question 2 [4 marks]
Fractions, Decimals and Percentages
Write these in order of size, starting with the smallest: 5/8, 0.62, 59%.
Question 3 [4 marks]
Sequences and Graphs
Find the distance between the points (-1, 2) and (5, -4), giving your answer as a simplified surd.
Question 4 [5 marks]
Linear Equations and Inequalities
Tickets to a fair cost 6 pounds each, and there is a fixed 8 pound parking charge.
A family has a budget of 50 pounds.
Form an inequality and find the greatest number of tickets the family can buy without exceeding their budget.
Question 5 [5 marks]
Indices, Surds and Standard Form
A = 4.6 x 10^5 and B = 9.2 x 10^4.
Work out A - B, giving your answer in standard form.
Question 6 [5 marks]
Ratio and Proportion
y is directly proportional to the square root of x.
When x = 16, y = 20.
Find y when x = 81.
Question 7 [6 marks]
Fractions, Decimals and Percentages
The price of a laptop is reduced by 20% in a sale, then a student discount reduces the sale price by a further 10%.
The final price paid is 216 pounds. Find the original price of the laptop before any reduction.
Question 8 [5 marks]
Linear Equations and Inequalities
At a cinema, 3 adult tickets and 2 child tickets cost 34 pounds in total.
2 adult tickets and 5 child tickets cost 41 pounds in total.
Find the cost of one adult ticket and the cost of one child ticket.
Question 9 [6 marks]
Angles, Polygons and Circle Theorems
A chord AB of length 24 cm is drawn in a circle of radius 13 cm, centre O.
Find the perpendicular distance from O to AB, and find angle AOB, giving your answer to 1 decimal place.
Question 10 [5 marks]
Linear Equations and Inequalities
Solve the simultaneous equations
4x + 3y = 25
2x - y = 5
Question 11 [6 marks]
Sequences and Graphs
A quadratic sequence has nth term n^2 + 5n - 14.
Find the positive value of n for which the nth term equals 112.
Question 12 [6 marks]
Ratio and Proportion
y is directly proportional to x and inversely proportional to z.
When x = 8 and z = 4, y = 20. Find y when x = 15 and z = 6.
Model solutions
| Question 1[3 marks] | |
|---|---|
| Answer or working | Marks |
| using the sum of interior angles = (n - 2) x 180 | M1 |
| (9 - 2) x 180 = 1260 degrees | M1 |
| one interior angle = 1260 / 9 = 140 degrees | A1 |
| Final answer: sum = 1260 degrees, one interior angle = 140 degrees | |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| converting 5/8 to a decimal, 0.625 | M1 |
| converting 59% to a decimal, 0.59 | M1 |
| comparing all three as decimals: 0.59, 0.62, 0.625 | M1 |
| 59%, 0.62, 5/8 | A1 |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| finding the horizontal difference 5 - (-1) = 6 | M1 |
| finding the vertical difference -4 - 2 = -6 | M1 |
| using distance = sqrt(6^2 + 6^2) = sqrt(72) | M1 |
| 6 sqrt(2) | A1 |
| Question 4[5 marks] | |
|---|---|
| Answer or working | Marks |
| forming the inequality 6n + 8 <= 50 | M1 |
| subtracting 8 from both sides to get 6n <= 42 | M1 |
| dividing both sides by 6 | M1 |
| n <= 7 | A1 |
| the greatest number of tickets is 7 | A1 |
| Final answer: 7 tickets | |
| Question 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| converting to the same power of 10, such as A = 46 x 10^4 | M1 |
| identifying B = 9.2 x 10^4 | M1 |
| 46 x 10^4 - 9.2 x 10^4 = 36.8 x 10^4 | M1 |
| 368000 | A1 |
| 3.68 x 10^5 | A1 |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| y = k sqrt(x) | M1 |
| 20 = k x 4, since sqrt(16) = 4 | M1 |
| k = 5 | A1 |
| substituting x = 81 into y = 5 sqrt(x) | M1 |
| 45 | A1 |
| Question 7[6 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the final price is 90% of the price after the first reduction | M1 |
| 216 / 0.9 = 240 | M1 |
| recognising 240 is 80% of the original price | M1 |
| 240 / 0.8 = 300 | M1 |
| original price = 300 pounds | A1 |
| checking 300 x 0.8 x 0.9 = 216 | B1 |
| Final answer: 300 pounds | |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| forming the equations 3a + 2c = 34 and 2a + 5c = 41 | M1 |
| multiplying the first equation by 5 and the second by 2 to align the c terms | M1 |
| subtracting to eliminate c, giving 11a = 88 | M1 |
| a = 8 (adult ticket = 8 pounds) | A1 |
| c = 5 (child ticket = 5 pounds) | A1 |
| Final answer: adult ticket = 8 pounds, child ticket = 5 pounds | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the perpendicular from the centre bisects the chord, giving half the chord = 12 cm | M1 |
| using Pythagoras to find the perpendicular distance: sqrt(13^2 - 12^2) | M1 |
| the perpendicular distance = 5 cm | A1 |
| using trigonometry in the right-angled triangle formed, such as cos(AOM) = 5/13 | M1 |
| doubling the angle found to find angle AOB | M1 |
| angle AOB = 134.8 degrees (1 d.p.) | A1 |
| Final answer: perpendicular distance = 5 cm, angle AOB = 134.8 degrees | |
| Question 10[5 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging the second equation to y = 2x - 5 | M1 |
| substituting into the first equation | M1 |
| forming 10x = 40 | M1 |
| x = 4 | A1 |
| y = 3 | A1 |
| Final answer: x = 4, y = 3 | |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| forming the equation n^2 + 5n - 14 = 112 | M1 |
| rearranging to n^2 + 5n - 126 = 0 | M1 |
| finding two numbers with product -126 and sum 5 | M1 |
| identifying 14 and -9, giving factors (n + 14)(n - 9) = 0 | M1 |
| n = 9 or n = -14 | A1 |
| n = 9, since n must be a positive integer | A1 |
| Final answer: n = 9 | |
| Question 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| y = kx/z | M1 |
| 20 = 8k/4, so 20 = 2k | M1 |
| k = 10 | A1 |
| substituting x = 15 and z = 6 into y = 10x/z | M1 |
| y = 10 x 15 / 6 | M1 |
| 25 | A1 |