Year 10 Paper 2: Graphs and Sequences
Covers linear algebra, graphs and coordinates, sequences, number and calculation, ratio and proportion, and angles and geometrical reasoning.
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 [2 marks]
Sequences
Here are the first five terms of a sequence. 2, 5, 7, 12, 19
Explain how each term of the sequence, after the first two, is found from the previous two terms.
Work out the next term in the sequence.
Question 2 [2 marks]
Angles and Geometrical Reasoning
Shape P has vertices at (1, 1), (1, 3) and (3, 1). Shape Q has vertices at (1, -1), (1, -3) and (3, -1).
Question 3 [2 marks]
Number and Calculation
Round 15.976 to 3 significant figures.
Question 4 [3 marks]
Linear Algebra
Each part below shows a different function machine with input n. Select the expression that represents the output.
input n --> [ x 2 ] --> [ +5 ] --> output. Which expression represents the output?
input n --> [ +3 ] --> [ x 4 ] --> output. Which expression represents the output?
input n --> [ x 5 ] --> [ -2 ] --> output. Which expression represents the output?
Question 5 [3 marks]
Ratio and Proportion
A football squad has 24 players. The ratio of forwards to midfielders to defenders is 3:5:4. Work out the percentage of the squad who are defenders.
Question 6 [3 marks]
Linear Algebra
x = -2 and y = 1/4. Work out the value of 1/x + 1/y.
Question 7 [4 marks]
Graphs and Coordinates
A sequence of points is generated on a grid. The first point is (1, -2). Each new point is formed by adding 3 to the x-coordinate and adding 2 to the y-coordinate of the previous point.
Write down the coordinates of the second point in the sequence.
Write down the coordinates of the third point in the sequence.
One of the points in the sequence has coordinates (16, y). Work out the value of y.
Question 8 [1 marks]
Number and Calculation
Work out 9^2 - 12.
Question 9 [2 marks]
Ratio and Proportion
Share £63 in the ratio 1:2. Work out how much the smaller share is.
Question 10 [2 marks]
Angles and Geometrical Reasoning
The exterior angle of a regular polygon is 15 degrees. Work out the number of sides of the polygon.
Question 11 [2 marks]
Sequences
The sequence is 5, 9, 13, 17, ... Find the value of n for the term 29.
Question 12 [2 marks]
Number and Calculation
Show that 41472 is divisible by 8.
Question 13 [2 marks]
Angles and Geometrical Reasoning
v is the column vector (9, 12). Calculate the magnitude of v, |v|.
Question 14 [3 marks]
Number and Calculation
A jacket normally costs £45. In a sale, the price is reduced by 1/5. Work out the sale price of the jacket.
Question 15 [3 marks]
Linear Algebra
The formula for the cost C (in pounds) of m apples and n bananas is C = 0.30m + 0.20n. Work out C when m = 4 and n = 6. Give your answer to the nearest penny.
Question 16 [3 marks]
Number and Calculation
Use long division to work out 9876 ÷ 12.
Question 17 [4 marks]
Ratio and Proportion
Sara walks 3.6 km in 45 minutes, at a constant speed. Calculate her average speed in m/s. Give your answer to 3 significant figures.
Question 18 [4 marks]
Linear Algebra
Form and solve an equation. "Sarah has £2.50. She buys a drink costing d pounds and has £1.10 left. How much did the drink cost?"
Form an equation in d to represent the situation.
Solve the equation and give the cost of the drink.
Question 19 [6 marks]
Number and Calculation
The integers a, b and c are given by a = -4, b = 7 and c = -3. This question has four parts.
Work out a + b - c.
Work out a x c.
Hence or otherwise, work out (a x c) / b, giving your answer as a decimal rounded to 1 decimal place.
Given that d is an integer such that b + d = a, work out the value of d.
Question 20 [5 marks]
Sequences
Here are the first four terms of a geometric sequence. 5, 15, 45, 135
Find the common ratio of the sequence.
Find the 6th term of the sequence.
Explain whether 3000 is a term of the sequence.
Question 21 [2 marks]
Number and Calculation
Work out (-8) + 15 - 6.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| each term is the sum of the two previous terms | B1oe |
| 31 cao, ft from their rule in part (a) | B1 |
| Final answer: Each term is found by adding the two previous terms together. | 31 | |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| reflection (oe 'reflected') | B1 |
| mirror line x-axis, y = 0 | B1oe |
| Final answer: Reflection in the x-axis (y = 0) | |
| Question 3[2 marks] | |
|---|---|
| Answer or working | Marks |
| identifies the first 3 significant figures as 1, 5, 9 and the next digit as 7 | M1oe |
| 16.0 | A1cao |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| A | B1cao |
| B | B1cao |
| A | B1cao |
| Final answer: A, B, A | |
| Question 5[3 marks] | |
|---|---|
| Answer or working | Marks |
| 24 / 12 = 2 oe (one share found) | M1 |
| defenders = 4 x 2 = 8 oe, then 8/24 x 100 | M1dep |
| awrt 33.3% (exact value 100/3%) | A1 |
| Final answer: 33.3% (1 dp); exact value 100/3% | |
| Question 6[3 marks] | |
|---|---|
| Answer or working | Marks |
| 1 / (-2) (= -0.5) seen | M1oe |
| 1 / (1/4) (= 4) seen | M1oe |
| 3.5 oe (7/2) | A1 |
| Final answer: 3.5 (or 7/2) | |
| Question 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| (4, 0) | B1cao |
| (7, 2) | B1cao |
| correct method to find which term has x = 16, e.g. 1 + 3n = 16 so n = 5 | M1oe |
| y = 8 | A1cao |
| Final answer: (4, 0) | (7, 2) | y = 8 | |
| Question 8[1 mark] | |
|---|---|
| Answer or working | Marks |
| 69 | B1cao |
| Question 9[2 marks] | |
|---|---|
| Answer or working | Marks |
| divide 63 by 3 to find one part (63/3 = 21) or equivalent method | M1 |
| £21 | A1cao |
| Question 10[2 marks] | |
|---|---|
| Answer or working | Marks |
| 360 / 15 | M1 |
| 24 | A1cao |
| Final answer: 24 sides | |
| Question 11[2 marks] | |
|---|---|
| Answer or working | Marks |
| use nth term formula 4n + 1 = 29 or show 5 + (n-1)4 = 29 | M1oe |
| 7 | A1cao |
| Question 12[2 marks] | |
|---|---|
| Answer or working | Marks |
| Uses the divisibility test for 8: a number is divisible by 8 if its last three digits are divisible by 8, and identifies the last three digits as 472 | M1 |
| 472 = 8 x 59, so 41472 is divisible by 8 | A1cso |
| Final answer: 41472 is divisible by 8, since its last three digits (472) equal 8 x 59. | |
| Question 13[2 marks] | |
|---|---|
| Answer or working | Marks |
| sqrt(9^2 + 12^2) | M1oe |
| 15 | A1cao |
| Question 14[3 marks] | |
|---|---|
| Answer or working | Marks |
| 45 / 5 (= 9) seen | M1 |
| 45 - their 9 (dependent method) | M1 |
| £36 | A1cao |
| Question 15[3 marks] | |
|---|---|
| Answer or working | Marks |
| substitute m = 4 and n = 6 into 0.30m + 0.20n and evaluate | M1 |
| add totals for apples and bananas correctly | M1 |
| £2.40 | A1cao |
| Final answer: 2.40 | |
| Question 16[3 marks] | |
|---|---|
| Answer or working | Marks |
| correct stages of long division shown (e.g. subtract 9600 leaving 276 or equivalent) | M1 |
| correct subsequent division step shown (e.g. 276 ÷ 12 = 23) | M1 |
| 823 | A1cao |
| Question 17[4 marks] | |
|---|---|
| Answer or working | Marks |
| 3.6 km = 3600 m | M1oe |
| 45 minutes = 2700 seconds | M1oe |
| 3600 / 2700 oe (ft their distance in m and time in s) | M1 |
| 1.33 (m/s) | A1awrt |
| Final answer: 1.33 m/s (3 s.f.) | |
| Question 18[4 marks] | |
|---|---|
| Answer or working | Marks |
| equation formed, e.g. 2.50 - d = 1.10 | M1oe |
| subtract 1.10 from both sides or rearrange (2.50 - 1.10 = d seen) | M1 |
| calculate correctly 2.50 - 1.10 = 1.40 | M1 |
| d = 1.40 | A1cao |
| Final answer: 2.50 - d = 1.10, d = 1.40 | |
| Question 19[6 marks] | |
|---|---|
| Answer or working | Marks |
| 6 | B1cao |
| 12 | B1cao |
| 12 / 7 seen, or correct division set up | M1oe |
| 1.7 | A1awrt |
| correct rearrangement: d = a - b | M1oe |
| -11 | A1cao |
| Final answer: 6 | 12 | 1.7 (awrt) | -11 | |
| Question 20[5 marks] | |
|---|---|
| Answer or working | Marks |
| 3 | B1cao |
| 5 x 3^5 oe, or lists terms up to the 6th | M1 |
| 1215 | A1cao |
| identifies the two terms either side of 3000 (1215 and 3645), or sets up 5 x 3^(n-1) = 3000 | M1oe |
| correct conclusion: no, 3000 is not a term, e.g. it lies strictly between 1215 and 3645 | A1oe |
| Final answer: 3 | 1215 | No, 3000 is not a term of the sequence | |
| Question 21[2 marks] | |
|---|---|
| Answer or working | Marks |
| -8 + 15 (= 7) (oe), or equivalent complete method | M1 |
| 1 | A1cao |