Year 10 Paper 4: Algebra and Sequences
Covers algebraic manipulation, surds and indices, simultaneous equations and inequalities, 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]
Surds and Indices
Simplify sqrt(50) - sqrt(18), giving your answer in the form k*sqrt(2).
Question 2 [3 marks]
Algebraic Manipulation
Factorise fully 4x^2 - 20x + 24.
Question 3 [4 marks]
Simultaneous Equations and Inequalities
Solve the simultaneous equations 4x + y = 16 and 2x - 3y = -6.
Question 4 [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 5 [4 marks]
Algebraic Manipulation
Expand and simplify (3x + 1)(x - 2) - (x - 1)^2.
Question 6 [4 marks]
Coordinate Geometry
Line L1 has equation y = -3x + 4.
Line L2 is perpendicular to L1 and passes through the point (6, 1).
Find the equation of L2, giving your answer in the form y = mx + c.
Question 7 [4 marks]
Algebraic Manipulation
Simplify fully (x^2 - 9)/(x^2 + x - 6).
Question 8 [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 9 [4 marks]
Sequences
The nth term of a sequence is given by Un = 2n^2 + 3n.
Find the 6th term of the sequence, and determine whether 200 is a term of the sequence, showing your working.
Question 10 [5 marks]
Coordinate Geometry
Find the equation of the straight line passing through the points C(-2, 5) and D(4, -7), giving your answer in the form y = mx + c.
State the coordinates of the point where the line crosses the y-axis.
Question 11 [5 marks]
Coordinate Geometry
Find the equation of the straight line passing through the points E(3, -4) and F(-1, 8), giving your answer in the form y = mx + c.
State the coordinates of the point where the line crosses the x-axis.
Question 12 [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 13 [6 marks]
Coordinate Geometry
A circle has equation (x - 3)^2 + (y + 1)^2 = 40.
Find the coordinates of the points where the circle crosses the x-axis, giving your answers in surd form.
Question 14 [6 marks]
Surds and Indices
Show that (4 + sqrt(3))/(4 - sqrt(3)) can be written in the form a + b*sqrt(3), where a and b are rational numbers to be found.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| writing sqrt(50) = 5sqrt(2) and sqrt(18) = 3sqrt(2) | M1 |
| 2sqrt(2) | A1 |
| Question 2[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 |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging 4x + y = 16 to make y the subject, y = 16 - 4x | M1 |
| substituting into 2x - 3y = -6 | M1 |
| simplifying to 14x = 42 | M1 |
| x = 3 and y = 4 | A1 |
| Final answer: x = 3, y = 4 | |
| Question 4[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 |
| Question 5[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 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the gradient of L1 as -3 | M1 |
| using the perpendicular gradient rule to find gradient 1/3 | M1 |
| using y - 1 = (1/3)(x - 6) | M1 |
| y = x/3 - 1 | A1 |
| Question 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| factorising the numerator as (x - 3)(x + 3) | M1 |
| factorising the denominator as (x + 3)(x - 2) | M1 |
| cancelling the common factor (x + 3) | M1 |
| (x - 3)/(x - 2) | A1 |
| Question 8[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 | |
| Question 9[4 marks] | |
|---|---|
| Answer or working | Marks |
| substituting n = 6 into Un = 2n^2 + 3n | M1 |
| the 6th term = 90 | A1 |
| forming the equation 2n^2 + 3n - 200 = 0 and using the discriminant | M1 |
| concluding 200 is not a term since n is not a positive integer, as the discriminant 1609 is not a perfect square | A1 |
| Final answer: 6th term = 90; 200 is not a term of the sequence | |
| Question 10[5 marks] | |
|---|---|
| Answer or working | Marks |
| using gradient = (y2 - y1)/(x2 - x1) | M1 |
| gradient = -2 | A1 |
| using y - (-7) = -2(x - 4) | M1 |
| y = -2x + 1 | A1 |
| the y-intercept (0, 1) | B1 |
| Final answer: y = -2x + 1; crosses the y-axis at (0, 1) | |
| Question 11[5 marks] | |
|---|---|
| Answer or working | Marks |
| using gradient = (y2 - y1)/(x2 - x1) | M1 |
| gradient = -3 | A1 |
| using y + 4 = -3(x - 3) | M1 |
| y = -3x + 5 | A1 |
| the x-intercept (5/3, 0) | B1 |
| Final answer: y = -3x + 5; crosses the x-axis at (5/3, 0) | |
| Question 12[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 13[6 marks] | |
|---|---|
| Answer or working | Marks |
| substituting y = 0 into the circle equation | M1 |
| simplifying to (x - 3)^2 + 1 = 40 | M1 |
| rearranging to (x - 3)^2 = 39 | M1 |
| taking the square root to get x - 3 = +/- sqrt(39) | M1 |
| x = 3 + sqrt(39) | A1 |
| x = 3 - sqrt(39) | A1 |
| Final answer: (3 + sqrt(39), 0) and (3 - sqrt(39), 0) | |
| Question 14[6 marks] | |
|---|---|
| Answer or working | Marks |
| multiplying numerator and denominator by (4 + sqrt(3)) | M1 |
| the denominator simplifying to 16 - 3 = 13 | M1 |
| expanding the numerator (4 + sqrt(3))^2 | M1 |
| the numerator simplifying to 19 + 8sqrt(3) | A1 |
| a = 19/13 | A1 |
| b = 8/13 | A1 |
| Final answer: a = 19/13, b = 8/13, so 19/13 + (8/13)sqrt(3) | |