Year 10

Year 10 Paper 3: Graphs and Geometry

Covers algebraic manipulation, simultaneous equations and inequalities, functions and their graphs, coordinate geometry, and sequences.

19 questions - 80 marks - calculator allowed

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.

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Questions

Question 1 [2 marks]

Coordinate Geometry

Find the gradient of the straight line joining the points A(2, 3) and B(6, 11).

Question 2 [2 marks]

Algebraic Manipulation

Expand and simplify (3x - 2)(x + 4).

Question 3 [3 marks]

Simultaneous Equations and Inequalities

Solve the simultaneous equations 3x + 2y = 16 and x - y = 2.

Question 4 [3 marks]

Algebraic Manipulation

Factorise fully 4x^2 - 20x + 24.

Question 5 [4 marks]

Sequences

An arithmetic sequence has first term 8 and common difference 5.

Find the number of terms needed for the sequence to first exceed 100.

Question 6 [4 marks]

Functions and their Graphs

The function f is defined by f(x) = x^2 - 3x + 1.

Find f(2) and f(-1).

Question 7 [4 marks]

Algebraic Manipulation

Factorise fully 6x^3 - 24x.

Question 8 [4 marks]

Simultaneous Equations and Inequalities

Solve the quadratic inequality x^2 - 5x - 14 <= 0, giving your answer as a single inequality.

Question 9 [4 marks]

Algebraic Manipulation

Write as a single fraction in its simplest form: 3/(x + 1) + 2/(x - 2).

Question 10 [4 marks]

Algebraic Manipulation

Factorise fully 5x^3 - 45x.

Question 11 [4 marks]

Coordinate Geometry

A circle has equation (x + 1)^2 + (y - 4)^2 = 20.

State the centre and radius of the circle, then determine whether the point (4, 5) lies inside, outside, or on the circle.

Question 12 [4 marks]

Algebraic Manipulation

Write as a single fraction in its simplest form: 4/(x - 3) - 1/(x + 2).

Question 13 [5 marks]

Sequences

The nth term of a sequence is given by Un = an^2 + b, where a and b are constants.

Given that U2 = 11 and U4 = 35, find the values of a and b.

Question 14 [5 marks]

Algebraic Manipulation

Simplify fully (3x^2 + 5x - 2)/(9x^2 - 1).

Question 15 [5 marks]

Sequences

The sum of the first n terms of an arithmetic series is given by Sn = n(3n + 7).

Find the first term and the common difference of the series.

Hence find the 20th term of the series.

Question 16 [5 marks]

Algebraic Manipulation

Solve the equation 3/(x - 1) + 2/(x + 1) = 2.

Show that your equation reduces to a quadratic before solving it, and give both solutions.

Question 17 [6 marks]

Sequences

The sum of the first n terms of an arithmetic series is given by Sn = (n/2)(3n - 1).

Find the first term and the common difference of the series, and find the value of n for which Sn = 330.

Question 18 [6 marks]

Algebraic Manipulation

Solve the equation (2x - 1)/(x + 3) = x - 3.

Show each step of your working and give both solutions.

Question 19 [6 marks]

Sequences

The first three terms of a geometric sequence are (k + 1), (k + 4), and (3k + 2), where k is a positive constant.

Find the value of k, and find the common ratio.

Model solutions

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or working	Marks
using gradient = (y2 - y1)/(x2 - x1)	M1
gradient = 2	A1

Mark scheme for Question 2 [2 marks]
Question 2[2 marks]
Answer or working	Marks
expanding to give four terms with correct signs	M1
3x^2 + 10x - 8	A1

Mark scheme for Question 3 [3 marks]
Question 3[3 marks]
Answer or working	Marks
rearranging x - y = 2 to x = y + 2	M1
substituting into 3x + 2y = 16	M1
x = 4 and y = 2	A1
Final answer: x = 4, y = 2

Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or working	Marks
taking out the common factor of 4	M1
factorising x^2 - 5x + 6 into two brackets	M1
4(x - 2)(x - 3)	A1

Mark scheme for Question 5 [4 marks]
Question 5[4 marks]
Answer or working	Marks
using Un = a + (n - 1)d to give Un = 5n + 3	M1
setting up the inequality 5n + 3 > 100	M1
solving to give n > 19.4	M1
n = 20 (the 20th term is the first to exceed 100)	A1

Mark scheme for Question 6 [4 marks]
Question 6[4 marks]
Answer or working	Marks
substituting x = 2 into f(x) = x^2 - 3x + 1	M1
f(2) = -1	A1
substituting x = -1 into f(x) = x^2 - 3x + 1	M1
f(-1) = 5	A1
Final answer: f(2) = -1, f(-1) = 5

Mark scheme for Question 7 [4 marks]
Question 7[4 marks]
Answer or working	Marks
taking out the common factor 6x	M1
6x(x^2 - 4)	A1
recognising x^2 - 4 as the difference of two squares	M1
6x(x - 2)(x + 2)	A1

Mark scheme for Question 8 [4 marks]
Question 8[4 marks]
Answer or working	Marks
factorising x^2 - 5x - 14 as (x - 7)(x + 2)	M1
identifying the critical values x = -2 and x = 7	M1
reasoning about the sign of the quadratic between the roots	M1
-2 <= x <= 7	A1

Mark scheme for Question 9 [4 marks]
Question 9[4 marks]
Answer or working	Marks
using the common denominator (x + 1)(x - 2)	M1
writing 3(x - 2) + 2(x + 1) as the numerator	M1
simplifying the numerator to 5x - 4	A1
(5x - 4)/((x + 1)(x - 2))	A1

Mark scheme for Question 10 [4 marks]
Question 10[4 marks]
Answer or working	Marks
taking out the common factor 5x	M1
5x(x^2 - 9)	A1
recognising x^2 - 9 as the difference of two squares	M1
5x(x - 3)(x + 3)	A1

Mark scheme for Question 11 [4 marks]
Question 11[4 marks]
Answer or working	Marks
centre (-1, 4)	B1
radius sqrt(20), i.e. 2sqrt(5)	B1
calculating the distance from (4, 5) to the centre as sqrt(5^2 + 1^2)	M1
concluding the point lies outside the circle, since sqrt(26) > sqrt(20)	A1
Final answer: Centre (-1, 4), radius 2sqrt(5); the point (4, 5) lies outside the circle

Mark scheme for Question 12 [4 marks]
Question 12[4 marks]
Answer or working	Marks
using the common denominator (x - 3)(x + 2)	M1
writing 4(x + 2) - (x - 3) as the numerator	M1
simplifying the numerator to 3x + 11	A1
(3x + 11)/((x - 3)(x + 2))	A1

Mark scheme for Question 13 [5 marks]
Question 13[5 marks]
Answer or working	Marks
forming the equation 4a + b = 11 from U2	M1
forming the equation 16a + b = 35 from U4	M1
subtracting the equations to eliminate b, giving 12a = 24	M1
a = 2	A1
b = 3	A1
Final answer: a = 2, b = 3 (so Un = 2n^2 + 3)

Mark scheme for Question 14 [5 marks]
Question 14[5 marks]
Answer or working	Marks
factorising the numerator as (3x - 1)(x + 2)	M1
the correct factorisation (3x - 1)(x + 2)	A1
factorising the denominator as the difference of two squares (3x - 1)(3x + 1)	M1
cancelling the common factor (3x - 1)	M1
(x + 2)/(3x + 1)	A1

Mark scheme for Question 15 [5 marks]
Question 15[5 marks]
Answer or working	Marks
using S1 = a to find the first term a = 10	M1
finding S2 and using T2 = S2 - S1	M1
common difference d = 6	A1
using T20 = a + 19d	M1
T20 = 124	A1
Final answer: first term = 10; common difference = 6; 20th term = 124

Mark scheme for Question 16 [5 marks]
Question 16[5 marks]
Answer or working	Marks
multiplying both sides by (x - 1)(x + 1)	M1
expanding to 3(x + 1) + 2(x - 1) = 2(x - 1)(x + 1)	M1
rearranging to 2x^2 - 5x - 3 = 0	M1
solving the quadratic, e.g. by factorising as (2x + 1)(x - 3) = 0	M1
x = 3 or x = -0.5	A1

Mark scheme for Question 17 [6 marks]
Question 17[6 marks]
Answer or working	Marks
using S1 = a to find a = 1	M1
finding S2 and using T2 = S2 - S1 to find the common difference d = 3	M1
a = 1 and d = 3	A1
forming the equation (n/2)(3n - 1) = 330	M1
rearranging to 3n^2 - n - 660 = 0 and solving, e.g. using the quadratic formula	M1
n = 15	A1
Final answer: first term = 1; common difference = 3; n = 15

Mark scheme for Question 18 [6 marks]
Question 18[6 marks]
Answer or working	Marks
cross-multiplying to give 2x - 1 = (x - 3)(x + 3)	M1
expanding the right side to x^2 - 9	M1
rearranging to x^2 - 2x - 8 = 0	M1
factorising as (x - 4)(x + 2) = 0	M1
x = 4	A1
x = -2	A1
Final answer: x = 4 or x = -2

Mark scheme for Question 19 [6 marks]
Question 19[6 marks]
Answer or working	Marks
using the geometric sequence property (k + 4)^2 = (k + 1)(3k + 2)	M1
expanding the left side to k^2 + 8k + 16	M1
expanding the right side to 3k^2 + 5k + 2	M1
rearranging to 2k^2 - 3k - 14 = 0 and solving, rejecting k = -2 since k is positive	M1
k = 3.5	A1
common ratio = 5/3	A1
Final answer: k = 3.5; common ratio = 5/3