Year 10 Paper 5: Number, Algebra and Data
Covers number and calculation, ratio and proportion, linear algebra, sequences and graphs, geometry, and statistics and probability.
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]
Linear Equations and Inequalities
Expand and simplify 5(3x - 2) - 2(4x - 7).
Question 2 [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 3 [3 marks]
Statistics and Probability
A bag contains 6 red counters, 5 blue counters and 4 green counters.
One counter is chosen at random.
Find the probability that it is not green.
Question 4 [4 marks]
Sequences and Graphs
Find the distance between the points (-1, 2) and (5, -4), giving your answer as a simplified surd.
Question 5 [4 marks]
Number and Calculation
The HCF of two numbers is 12. The LCM of the same two numbers is 360.
One of the numbers is 72. Find the other number.
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 [5 marks]
Number and Calculation
Three warning lights flash every 18 seconds, 24 seconds and 30 seconds respectively.
All three lights flash together at 9:00 am. Find the next time, in seconds after 9:00 am, that all three lights flash together at the same moment.
Question 8 [5 marks]
Sequences and Graphs
A geometric sequence has first term 4 and common ratio 3.
Find the 6th term of the sequence, and find the sum of the first 4 terms.
Question 9 [5 marks]
Linear Equations and Inequalities
Solve the simultaneous equations
4x + 3y = 25
2x - y = 5
Question 10 [5 marks]
Angles, Polygons and Circle Theorems
ABCD is a cyclic quadrilateral. The side DC is extended to a point E, so angle BCE is an exterior angle.
Angle DAB = 112 degrees. Find angle BCD and angle BCE.
Question 11 [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 12 [5 marks]
Statistics and Probability
The grouped table shows the times in minutes taken by 50 students to complete a puzzle.
10 <= t < 15: 8 students, 15 <= t < 20: 19 students, 20 <= t < 25: 15 students, 25 <= t < 30: 8 students.
Estimate the mean time.
Question 13 [4 marks]
Linear Equations and Inequalities
Make x the subject of y = (4x + 1)/(x - 5).
Question 14 [6 marks]
Statistics and Probability
The table shows the masses, in kg, of 90 parcels.
0 <= m < 5: 18 parcels, 5 <= m < 10: 34 parcels, 10 <= m < 15: 26 parcels, 15 <= m < 20: 12 parcels.
Using linear interpolation, estimate the number of parcels with a mass greater than 12 kg.
Question 15 [6 marks]
Angles, Polygons and Circle Theorems
The interior angle of a regular polygon is 5 times the size of its exterior angle.
Find the number of sides of the polygon.
Question 16 [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 17 [6 marks]
Number and Calculation
A car travels 186 km, correct to the nearest km, in a time of 3.2 hours, correct to 1 decimal place.
Find the lower bound for the car's average speed in km/h, giving your answer to 3 significant figures.
Model solutions
| Question 1[3 marks] | |
|---|---|
| Answer or working | Marks |
| expanding to 15x - 10 - 8x + 14 | M1 |
| collecting like x terms and constants | M1 |
| 7x + 4 | A1 |
| Question 2[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 3[3 marks] | |
|---|---|
| Answer or working | Marks |
| total counters = 15 | M1 |
| not green counters = 11 | M1 |
| 11/15 | A1 |
| Question 4[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 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| using HCF x LCM = product of the two numbers | M1 |
| 12 x 360 = 4320 | M1 |
| dividing 4320 by 72 | M1 |
| 60 | 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[5 marks] | |
|---|---|
| Answer or working | Marks |
| expressing each number as a product of primes | M1 |
| 18 = 2 x 3^2, 24 = 2^3 x 3, 30 = 2 x 3 x 5 | M1 |
| identifying the highest power of each prime: 2^3, 3^2 and 5 | M1 |
| multiplying to get 8 x 9 x 5 | M1 |
| 360 seconds | A1 |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| using the nth term formula a r^(n-1) | M1 |
| 6th term = 4 x 3^5 | M1 |
| 972 | A1 |
| using the sum formula a(r^n - 1)/(r - 1) with n = 4 | M1 |
| sum of first 4 terms = 160 | A1 |
| Final answer: 6th term = 972, sum of first 4 terms = 160 | |
| Question 9[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 10[5 marks] | |
|---|---|
| Answer or working | Marks |
| using opposite angles in a cyclic quadrilateral sum to 180 degrees | M1 |
| angle BCD = 180 - 112 = 68 degrees | M1 |
| using angles BCD and BCE lie on the straight line DCE, summing to 180 degrees | M1 |
| angle BCE = 112 degrees | A1 |
| noting angle BCE equals angle DAB, the exterior angle of a cyclic quadrilateral equals the interior opposite angle | B1 |
| Final answer: angle BCD = 68 degrees, angle BCE = 112 degrees | |
| Question 11[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 12[5 marks] | |
|---|---|
| Answer or working | Marks |
| using midpoints 12.5, 17.5, 22.5 and 27.5 | M1 |
| multiplying each midpoint by its frequency | M1 |
| total fx = 990 | A1 |
| dividing by total frequency 50 | M1 |
| 19.8 minutes | A1 |
| Question 13[4 marks] | |
|---|---|
| Answer or working | Marks |
| y(x - 5) = 4x + 1 | M1 |
| yx - 5y = 4x + 1 | M1 |
| collecting x terms to get x(y - 4) = 1 + 5y | M1 |
| x = (1 + 5y)/(y - 4) | A1 |
| Question 14[6 marks] | |
|---|---|
| Answer or working | Marks |
| cumulative frequencies 18, 52, 78, 90 | M1 |
| identifying 12 kg lies in the class 10 <= m < 15 | M1 |
| using linear interpolation: cumulative frequency at 12 = 52 + (12 - 10)/5 x 26 | M1 |
| evaluating 52 + 10.4 = 62.4 | M1 |
| subtracting from the total: 90 - 62.4 = 27.6 | M1 |
| 28 parcels (nearest whole number) | A1 |
| Final answer: 28 parcels | |
| Question 15[6 marks] | |
|---|---|
| Answer or working | Marks |
| using interior angle + exterior angle = 180 degrees | M1 |
| forming the equation e + 5e = 180, where e is the exterior angle | M1 |
| solving 6e = 180 to get e = 30 degrees | M1 |
| using the exterior angles of a polygon sum to 360 degrees | M1 |
| number of sides = 360 / 30 | M1 |
| 12 sides | A1 |
| Question 16[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 17[6 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the lower bound of the distance as 185.5 km | M1 |
| identifying the upper bound of the time as 3.25 hours | M1 |
| recognising lower bound speed requires the smallest distance divided by the largest time | M1 |
| using speed = distance / time with these bounds | M1 |
| 185.5 / 3.25 = 57.0769... | M1 |
| 57.1 km/h (3 s.f.) | A1 |
| Final answer: 57.1 km/h | |