Year 10 Paper 1: Core Skills
Covers the full Year 10 further maths topic list: algebraic manipulation, surds and indices, simultaneous equations and inequalities, functions and their graphs, coordinate geometry, and sequences.
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
An arithmetic sequence has first term 5 and common difference 4.
Find the 12th term of the sequence.
Question 2 [4 marks]
Algebraic Manipulation
Expand and simplify (3x + 1)(x - 2) - (x - 1)^2.
Question 3 [4 marks]
Simultaneous Equations and Inequalities
Solve the simultaneous equations 2x + 3y = 13 and 3x - y = 3.
Question 4 [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 5 [4 marks]
Simultaneous Equations and Inequalities
Solve algebraically the simultaneous equations y = 2x - 3 and x^2 + y^2 = 18.
Question 6 [5 marks]
Coordinate Geometry
A circle has equation x^2 + y^2 + 8x - 2y + 8 = 0.
Find the centre and radius of the circle by completing the square.
Question 7 [4 marks]
Surds and Indices
Simplify (2x^3y^-2)^3 / (4x^-1y), giving your answer in the form kx^ay^b.
Question 8 [4 marks]
Coordinate Geometry
Line L1 has equation y = 2x - 1.
Line L2 is perpendicular to L1 and passes through the point (4, 5).
Find the equation of L2, giving your answer in the form y = mx + c.
Question 9 [5 marks]
Surds and Indices
Simplify (2x^3y^-1)^3 x (x^-4y^2), giving your answer in the form kx^ay^b.
Question 10 [4 marks]
Sequences
The first three terms of an arithmetic sequence are 3k - 1, 5k + 2, and 9k - 3, where k is a constant.
Find the value of k, and find the common difference of the sequence.
Question 11 [5 marks]
Surds and Indices
Show that (sqrt(7) - 1)/(sqrt(7) + 1) can be written in the form a + b*sqrt(7), where a and b are rational numbers to be found.
Question 12 [5 marks]
Coordinate Geometry
A circle has centre (-2, 3) and passes through the point (2, 6).
Find the equation of the circle.
Then find the equation of the tangent to the circle at the point (2, 6), giving your answer in the form y = mx + c.
Question 13 [5 marks]
Algebraic Manipulation
Solve the equation 4/(x - 1) - 2/(x + 2) = 1.
Show that your equation reduces to a quadratic before solving it, and give both solutions.
Question 14 [5 marks]
Coordinate Geometry
A circle has centre (1, 2) and passes through the point (4, 6).
Find the equation of the circle.
Then find the equation of the tangent to the circle at the point (4, 6), giving your answer in the form y = mx + c.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| using the formula a + (n - 1)d with n = 12 | M1 |
| T12 = 49 | A1 |
| Final answer: 49 | |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| expanding (3x + 1)(x - 2) to give 3x^2 - 5x - 2 | M1 |
| expanding (x - 1)^2 to give x^2 - 2x + 1 | M1 |
| combining like terms | M1 |
| 2x^2 - 3x - 3 | A1 |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging 3x - y = 3 to make y the subject, y = 3x - 3 | M1 |
| substituting into 2x + 3y = 13 | M1 |
| simplifying to 11x = 22 | M1 |
| x = 2 and y = 3 | A1 |
| Final answer: x = 2, y = 3 | |
| Question 4[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 | |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| substituting y = 2x - 3 into the second equation | M1 |
| expanding and simplifying to 5x^2 - 12x - 9 = 0 | M1 |
| solving the quadratic (e.g. using the quadratic formula) to find x = 3 or x = -0.6 | M1 |
| (x, y) = (3, 3) and (-0.6, -4.2) | A1 |
| Final answer: (x, y) = (3, 3) or (-0.6, -4.2) | |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| completing the square on the x terms: (x + 4)^2 - 16 | M1 |
| completing the square on the y terms: (y - 1)^2 - 1 | M1 |
| rearranging to (x + 4)^2 + (y - 1)^2 = 9 | M1 |
| centre (-4, 1) | A1 |
| radius 3 | A1 |
| Final answer: Centre (-4, 1), radius 3 | |
| Question 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| cubing to give 8x^9y^-6 | M1 |
| dividing the coefficients to give 2 | M1 |
| subtracting the indices to give x^10 and y^-7 | M1 |
| 2x^10y^-7 | A1 |
| Final answer: 2x^10y^-7 (equivalently 2x^10/y^7) | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the gradient of L1 as 2 | M1 |
| using the perpendicular gradient rule to find gradient -1/2 | M1 |
| using y - 5 = -0.5(x - 4) | M1 |
| y = -0.5x + 7 | A1 |
| Question 9[5 marks] | |
|---|---|
| Answer or working | Marks |
| cubing to give 8x^9y^-3 | M1 |
| multiplying coefficients: 8 x 1 = 8 | M1 |
| adding the indices of x: 9 + (-4) = 5 | M1 |
| adding the indices of y: -3 + 2 = -1 | M1 |
| 8x^5y^-1 | A1 |
| Final answer: 8x^5y^-1 (equivalently 8x^5/y) | |
| Question 10[4 marks] | |
|---|---|
| Answer or working | Marks |
| setting up the equation (T2 - T1) = (T3 - T2) | M1 |
| forming 2k + 3 = 4k - 5 | M1 |
| k = 4 | A1 |
| common difference = 11 | A1 |
| Final answer: k = 4; common difference = 11 | |
| Question 11[5 marks] | |
|---|---|
| Answer or working | Marks |
| multiplying numerator and denominator by (sqrt(7) - 1) | M1 |
| the denominator simplifying to 7 - 1 = 6 | M1 |
| expanding the numerator (sqrt(7) - 1)^2 | M1 |
| the numerator simplifying to 8 - 2sqrt(7) | A1 |
| a = 4/3 and b = -1/3 | A1 |
| Final answer: a = 4/3, b = -1/3, so 4/3 - (1/3)sqrt(7) | |
| Question 12[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding the radius as sqrt(4^2 + 3^2) = 5 | M1 |
| the circle equation (x + 2)^2 + (y - 3)^2 = 25 | A1 |
| finding the gradient of the radius to (2, 6) as 3/4 | M1 |
| using the perpendicular gradient -4/3 with the point (2, 6) | M1 |
| the tangent y = -(4/3)x + 26/3 | A1 |
| Final answer: Circle: (x + 2)^2 + (y - 3)^2 = 25; tangent: y = -(4/3)x + 26/3 | |
| Question 13[5 marks] | |
|---|---|
| Answer or working | Marks |
| multiplying both sides by (x - 1)(x + 2) | M1 |
| expanding to 4(x + 2) - 2(x - 1) = (x - 1)(x + 2) | M1 |
| rearranging to x^2 - x - 12 = 0 | M1 |
| factorising as (x - 4)(x + 3) = 0 | M1 |
| x = 4 or x = -3 | A1 |
| Question 14[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding the radius as sqrt(3^2 + 4^2) = 5 | M1 |
| the circle equation (x - 1)^2 + (y - 2)^2 = 25 | A1 |
| finding the gradient of the radius to (4, 6) as 4/3 | M1 |
| using the perpendicular gradient -3/4 with the point (4, 6) | M1 |
| the tangent y = -0.75x + 9 | A1 |
| Final answer: Circle: (x - 1)^2 + (y - 2)^2 = 25; tangent: y = -0.75x + 9 | |