MPM2D Mixed Practice Test: Linear Systems, Quadratics and Trigonometry

A useful MPM2D practice test mixes topics so you must choose a method rather than follow the previous question’s pattern. Define variables, show algebraic steps, label units and check each answer against the original question.

This original fifteen-question set samples linear systems, coordinate geometry, quadratics and trigonometry. It is not an official exam or a complete assessment of every MPM2D expectation. Use your teacher’s outline for the full assessed scope and calculator rules.

How to use the practice test

Attempt the questions before reading the solutions. If you are stuck, record the first step you cannot choose and move on. Afterwards, separate errors into setup, algebra, interpretation and checking. A total score alone does not show what to study next.

The questions deliberately switch topics. This gives you practice recognizing whether a situation calls for a system, a quadratic relationship or a triangle model. Unless otherwise stated, triangle angles are in degrees.

Questions

• 1. Solve y = 2x + 1 and y = −x + 7.
• 2. Solve 2x + y = 11 and x − y = 1.
• 3. A school sells 40 tickets at $8 for adults and $5 for students. Revenue is $260. How many of each ticket were sold?
• 4. Find the midpoint of A(−2, 3) and B(4, 7).
• 5. Find the distance between the points in Question 4.
• 6. Find the equation of the line through (0, 4) and (2, 0).
• 7. Factor x² − 9x + 20.
• 8. Solve 2x² − 5x − 3 = 0.
• 9. Write y = x² − 6x + 5 in vertex form and identify the vertex.
• 10. How many real solutions does 2x² + 4x + 5 = 0 have?
• 11. A height model is h(t) = −5t² + 20t + 1, with height in metres and time in seconds. Find the maximum height and when it occurs.
• 12. A rectangle has perimeter 30 cm and width x cm. Express its area as a function of x and find the greatest possible area.
• 13. In a right triangle, an acute angle is 35° and the hypotenuse is 12 cm. Find the opposite side.
• 14. A ramp rises 1.2 m over a horizontal run of 4.0 m. Find its angle to the horizontal.
• 15. Two sides of a non-right triangle are 6 cm and 8 cm with included angle 60°. Find the third side.

Worked solutions

1. Equate the two expressions for y

2x + 1 = −x + 7 gives 3x = 6, so x = 2. Then y = 5. The solution is (2, 5), which satisfies both equations.

2. Eliminate one variable

Add the equations: 3x = 12, so x = 4. From x − y = 1, y = 3. The solution is (4, 3). Addition works because the y-coefficients are opposites.

3. Build separate count and revenue equations

Let a and s be adult and student ticket counts. Then a + s = 40 and 8a + 5s = 260. Subtract five times the count equation from the revenue equation: 3a = 60. Therefore 20 adult and 20 student tickets were sold. Check: 160 + 100 = 260.

4. Average corresponding coordinates

The midpoint is ((−2 + 4)/2, (3 + 7)/2) = (1, 5). Do not average an x-coordinate with a y-coordinate.

5. Use the coordinate differences

Distance = √[(4 − (−2))² + (7 − 3)²] = √(36 + 16) = √52 = 2√13, approximately 7.21 coordinate units.

6. Find slope and intercept

m = (0 − 4)/(2 − 0) = −2. Since the line passes through (0, 4), its y-intercept is 4. The equation is y = −2x + 4.

7. Factor the expression

Two numbers multiply to 20 and add to −9: −4 and −5. Therefore x² − 9x + 20 = (x − 4)(x − 5). The question asks for a factorization, not solutions to an equation.

8. Apply the zero-product property

2x² − 5x − 3 = (2x + 1)(x − 3). Since the product equals zero, x = −1/2 or x = 3. Substitute either value into the original equation to check.

9. Complete the square

x² − 6x + 5 = (x² − 6x + 9) − 9 + 5 = (x − 3)² − 4. The vertex is (3, −4). The added and subtracted 9 preserve the expression’s value.

10. Interpret the discriminant

D = 4² − 4(2)(5) = 16 − 40 = −24. Because D is negative, there are no real solutions. This does not mean the associated parabola has no points; it means it has no x-intercepts.

11. Find the vertex of the height model

Completing the square gives h(t) = −5(t − 2)² + 21. The maximum is 21 m at t = 2 s because the squared term is multiplied by a negative coefficient. Interpret the result within the stated model.

12. Model area with a domain

If width is x, length is 15 − x because twice their sum is 30. Thus A(x) = x(15 − x) = −(x − 7.5)² + 56.25, with 0 < x < 15 for positive dimensions. The greatest area is 56.25 cm², at dimensions 7.5 cm by 7.5 cm.

13. Select sine

sin 35° = opposite/12, so the opposite side is 12 sin 35° ≈ 6.88 cm. It is shorter than the hypotenuse, as required.

14. Select tangent and then its inverse

tan θ = 1.2/4.0 = 0.3. Therefore θ ≈ 16.7°. The horizontal run is not the length of the ramp surface.

15. Use the cosine law

c² = 6² + 8² − 2(6)(8)cos 60° = 36 + 64 − 48 = 52. Therefore c = 2√13 cm ≈ 7.21 cm. The included angle lets the cosine law connect the two known sides to the opposite unknown side.

Decide what to revise next

Errors in Questions 1–3 suggest revisiting equation setup and elimination. Questions 4–6 check coordinate relationships. Questions 7–12 connect quadratic algebra, graphs and context. Questions 13–15 test triangle-model selection.

Choose one question you could not start. Explain why the method applies, then solve a fresh version with different numbers. Repeating only the original answer can test memory rather than independent method choice.

Find the next resource

Use our linear relations guide, quadratic equations guide and Grade 10 tutoring page. Book a free one-hour assessment for a personalised learning roadmap.

Curriculum reference: Ontario Grades 9–10 Mathematics.