End-of-Unit Quiz 15 MC + 3 Short Answer Covers Lessons 1–20

Unit Quiz

Algebraic Techniques & Linear Relationships. A comprehensive assessment of everything you have learned in Unit 2.

Instructions: This quiz covers all topics from Unit 2. For multiple choice, select one answer and click Check. For short answer, show all working. Your responses are saved automatically.

Section A — Multiple Choice

1 mark each. Select the best answer.

1 mark Simplify $3x + 2y - x + 4y$. L1

$2x + 6y$ $4x + 6y$ $2x + 2y$ $3x + 6y$

1 mark Expand and simplify $2(x + 3) + 3(x - 2)$. L2

$5x$ $5x + 12$ $5x - 12$ $5x$

1 mark Factorise $x^2 - 5x + 6$ fully. L4

$(x - 2)(x - 3)$ $(x + 2)(x + 3)$ $(x - 1)(x - 6)$ $(x + 1)(x - 6)$

1 mark Solve $\dfrac{x}{3} + 2 = 5$. L9

$x = 1$ $x = 9$ $x = 21$ $x = 7$

1 mark The solution to $2(3x - 1) = 4x + 10$ is: L8

$x = 2$ $x = 3$ $x = 4$ $x = 6$

1 mark If $3x - 5 > 7$, then: L11

$x > 4$ $x < 4$ $x > \dfrac{2}{3}$ $x > -4$

1 mark Solve simultaneously: $x + y = 5$ and $x - y = 1$. L13

$x = 2, y = 3$ $x = 3, y = 2$ $x = 4, y = 1$ $x = 1, y = 4$

1 mark The distance between $(1, 2)$ and $(5, 5)$ is: L15

$3$ $4$ $5$ $\sqrt{41}$

1 mark The gradient of the line through $(0, 3)$ and $(4, 7)$ is: L16

$1$ $\dfrac{4}{3}$ $\dfrac{5}{2}$ $2$

1 mark The equation of the line with gradient $-3$ passing through $(2, 4)$ is: L18

$y = -3x + 10$ $y = -3x - 2$ $y = 3x + 10$ $y = -3x + 4$

1 mark A line perpendicular to $y = 2x + 1$ has gradient: L19

$2$ $-2$ $-\dfrac{1}{2}$ $\dfrac{1}{2}$

1 mark In the model $C = 25t + 50$, the value $25$ represents: L20

The total cost The fixed cost The cost per unit time The time taken

1 mark Make $x$ the subject of $y = mx + c$: L10

$x = \dfrac{y + c}{m}$ $x = \dfrac{y - c}{m}$ $x = y - m - c$ $x = \dfrac{c - y}{m}$

1 mark The midpoint of $(-2, 5)$ and $(6, -1)$ is: L15

$(2, 2)$ $(4, 4)$ $(2, 3)$ $(-4, 6)$

1 mark Using data from $t = 0$ to $t = 10$ to predict at $t = 8$ is: L20

Extrapolation Interpolation Modelling Regression

Section B — Short Answer

Show all working.

Question 16

5 marks Apply / Analyse

Expand and simplify $(2x - 3)(x + 4) - (x - 1)^2$.

Marking Criteria

2 marks: Correctly expand $(2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12$
2 marks: Correctly expand $(x - 1)^2 = x^2 - 2x + 1$
1 mark: Correct final simplification: $2x^2 + 5x - 12 - x^2 + 2x - 1 = x^2 + 7x - 13$

Question 17

5 marks Analyse / Evaluate

A rectangle has vertices $A(1, 2)$, $B(5, 2)$, $C(5, 6)$, and $D(1, 6)$.

(a) Find the length of diagonal $AC$. (2 marks)

(b) Find the equation of the line through the midpoint of $AB$ that is perpendicular to $AB$. (3 marks)

Marking Criteria

1 mark: Set up distance formula for $AC$
1 mark: Correct answer: $\sqrt{32} = 4\sqrt{2}$ units
1 mark: Find midpoint of $AB$: $(3, 2)$
1 mark: Identify perpendicular gradient (undefined/vertical since $AB$ is horizontal)
1 mark: Correct equation: $x = 3$

Question 18

5 marks Evaluate / Create

Solve the following system of equations and interpret your solution in context.

$2x + 3y = 12$

$3x - 2y = 5$

Context: At a café, 2 coffees and 3 muffins cost $12. At the same café, 3 coffees and 2 muffins cost $13. Let $x$ be the price of a coffee and $y$ be the price of a muffin.

(a) Solve the system to find the price of each item. (3 marks)

(b) Verify your solution satisfies both equations. (1 mark)

Marking Criteria

1 mark: Set up elimination (e.g., multiply first by 3 and second by 2)
1 mark: Correctly eliminate one variable and solve
1 mark: Find both values: coffee $x = 3$, muffin $y = 2$
1 mark: Substitute back to verify both equations
1 mark: Explain that coefficients are small integers making elimination straightforward

I have completed the Unit Quiz and checked my answers.