Grade 10 Math (MPM2D) Practice Quiz, With Worked Answers

The 20 questions at a glance

MPM2D has three strands. This table shows what each one asks of a student, in plain terms, and which questions below practise it.

Quadratic relations gets the most questions. It has four overall expectations in the curriculum, and the other two strands have three each.

Is MPM2D still the Grade 10 course in 2026-27?

Yes. Grade 9 math was de-streamed in 2021 and became one course, MTH1W. Grade 10 was not. The ministry's course page, updated in August 2026, still lists two Grade 10 courses: Principles of Mathematics (Academic, MPM2D) and Foundations of Mathematics (Applied, MFM2P). It says that "The 2005 Mathematics curriculum for Grade 10 and the 2007 Mathematics curriculum for Grades 11–12 remain in effect."

The ministry's version history for secondary math shows no new Grade 10 course after 2022, and we found no announcement of one as of 8 October 2026.

The one change since 2005 is a short addendum, issued in Winter 2022 and taught from September 2022. The ministry issued it to support students in their move from the de-streamed Grade 9 course, MTH1W, into MPM2D. It adds three new specific expectations under an existing analytic geometry expectation: the slopes of parallel and perpendicular lines, using the slope formula to find the equation of a line, and writing a line's equation in different forms. Question 14 covers all three.

The prerequisite for MPM2D is "Grade 9 Mathematics, De-streamed (2021), or Grade 9 Principles of Mathematics, Academic (2005)". MPM2D is the course that opens Grade 11 Functions (MCR3U). Our guide to Ontario's senior math streams sets out where each Grade 11 and 12 course leads.

How the Grade 10 math exam counts from 2026-27

The rules changed this school year. The ministry's Growing Success page, updated on 21 August 2026, says: "Beginning in the 2026-2027 school year, most secondary courses will require written exams as a percentage of the final course mark."

For Grades 9 and 10, the final mark now has three parts:

The page says this replaces the earlier evaluation policy for courses in Grades 9 to 12. The ministry's Grade 9 and 10 table puts Grade 10 math at a 20% written exam, held in the designated exam period.

The ministry page sets the weight and the timing. It does not describe a provincial Grade 10 math exam. The only EQAO math assessment it names is the Grade 9 one, which counts as the written exam for MTH1W. There is no EQAO math test in Grade 10.

No ministry document we found says whether a Grade 10 math exam comes with a formula sheet or which calculator is allowed. That is up to the school, the board and the teacher, so ask before the exam rather than assume. Our page on Ontario's new exam and attendance rules covers the rest of the change.

How to use this quiz

Do all 20 before you look at a single answer. Work on paper, with a scientific calculator for the trigonometry, and without a timer.

Twenty questions cannot predict an exam mark. What they can show is which strand is short, and which step inside it. The answer key names the skill behind every question for that reason.

Every question below is original and was written for this article. None is copied from a textbook, a school's exam or a review package.

Search for an MPM2D practice test and most of the first page, checked on 8 October 2026, is teachers' class pages, PDF review packages and a video playlist. One of the PDFs is a past exam from 2013, written for the course as it stood before the 2022 addendum. Old exams are still useful practice.

Quadratic relations: questions 1 to 9

Question 1. Expand and simplify (3x - 2)(x + 5).

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

Question 3. Factor fully.

• (a) 4x² - 12x
• (b) x² - x - 20

Question 4. Factor fully.

• (a) 6x² + 7x - 3
• (b) 25x² - 49

Question 5. For the relation y = -2(x - 3)² + 8, state the vertex, the equation of the axis of symmetry, the direction of opening and the maximum or minimum value.

Question 6. Describe the transformations that take the graph of y = x² to the graph of y = ½(x + 4)² - 5. Then state the vertex.

Question 7. Write y = 2x² - 12x + 7 in the form y = a(x - h)² + k by completing the square. State the vertex and the minimum value.

Question 8. Solve each equation.

• (a) x² + 2x - 15 = 0, by factoring
• (b) 2x² - 3x - 4 = 0, using the quadratic formula. Round to two decimal places.

Question 9. A ball is thrown upward from the top of a 25 m cliff. Its height above the water, h metres, after t seconds is h = -5t² + 20t + 25.

• (a) Factor the expression to find when the ball hits the water.
• (b) Find the maximum height of the ball, and when it reaches it.

Analytic geometry: questions 10 to 15

Question 10. Solve by substitution.

• y = 2x - 1
• 3x + 2y = 19

Question 11. A school sells 150 tickets to a play. Adult tickets cost $12 and student tickets cost $8. The school takes in $1,480. Write a linear system and solve it by elimination to find how many of each ticket were sold.

Question 12. The endpoints of line segment AB are A(-3, 4) and B(5, -2). Find the midpoint of AB and the length of AB.

Question 13. A circle with its centre at the origin passes through the point (-5, 12).

• (a) Write the equation of the circle.
• (b) Is the point (12, -5) on the circle? Is (10, 8) on it, inside it or outside it?

Question 14. A line passes through P(2, -1) and Q(6, 7).

• (a) Find the slope of PQ.
• (b) Write the equation of the line in the form y = mx + b, and again in the form Ax + By + C = 0.
• (c) Write the equation of the line parallel to PQ that passes through (0, 3).
• (d) State the slope of any line perpendicular to PQ.

Question 15. A triangle has vertices A(-2, 1), B(2, 3) and C(4, -1). Show that it is a right isosceles triangle.

Trigonometry: questions 16 to 20

Question 16. A student 1.6 m tall casts a shadow 2.4 m long. At the same time, a tree casts a shadow 18 m long. How tall is the tree?

Question 17. A 6.0 m ladder leans against a wall and makes an angle of 72° with the level ground. How far up the wall does it reach? Round to one decimal place.

Question 18. A ramp rises 0.9 m over a horizontal distance of 10.8 m.

• (a) Find the angle the ramp makes with the ground, to one decimal place.
• (b) Find the length of the ramp's surface, to two decimal places.

Question 19. In acute triangle ABC, ∠A = 48°, ∠B = 62° and side a = 15 cm. Find side b, to one decimal place.

Question 20. Two sides of an acute triangle measure 8 cm and 11 cm, and the angle between them is 52°. Find the third side, to one decimal place.

Answer key, with the working

The working

Question 1. Multiply each term in the first bracket by each term in the second: 3x² + 15x - 2x - 10. Collect the like terms: 3x² + 13x - 10.

Question 2. Work each part separately. (x - 4)² = x² - 8x + 16, so 2(x - 4)² = 2x² - 16x + 32. (x + 3)(x - 3) is a difference of squares, x² - 9. Subtract the whole of it: 2x² - 16x + 32 - x² + 9 = x² - 16x + 41. The usual slip is writing -x² - 9 instead of -x² + 9. Brackets around the second product prevent it.

Question 3. (a) Both terms share 4x, so 4x² - 12x = 4x(x - 3). (b) Find two numbers that multiply to -20 and add to -1. They are -5 and 4, so x² - x - 20 = (x - 5)(x + 4). Expand it again to check.

Question 4. (a) Multiply a and c: 6 × (-3) = -18. Find two numbers that multiply to -18 and add to 7: 9 and -2. Split the middle term: 6x² + 9x - 2x - 3. Group: 3x(2x + 3) - 1(2x + 3) = (3x - 1)(2x + 3). (b) 25x² is (5x)² and 49 is 7², so this is a difference of squares: (5x - 7)(5x + 7).

Question 5. In y = a(x - h)² + k, the vertex is (h, k). Here h = 3 and k = 8, so the vertex is (3, 8) and the axis of symmetry is x = 3. The value of a is -2, which is negative, so the parabola opens down and the vertex is a maximum. The maximum value is 8.

Question 6. a = ½, so the graph is compressed vertically by a factor of ½. It opens up, because a is positive. The bracket is (x + 4), which is (x - (-4)), so h = -4 and the graph moves 4 units left. k = -5 moves it 5 units down. The vertex is (-4, -5). Watch the sign: (x + 4) moves the graph left, not right, because the sign inside the bracket is the opposite of the sign of h.

Question 7. Take out the common factor from the first two terms: y = 2(x² - 6x) + 7. Half of -6 is -3, and (-3)² = 9. Add and subtract 9 inside the bracket: y = 2(x² - 6x + 9 - 9) + 7. Move the -9 out, remembering it is multiplied by 2: y = 2(x - 3)² - 18 + 7 = 2(x - 3)² - 11. The vertex is (3, -11). a = 2 is positive, so the minimum value is -11. Check: at x = 0 the original gives 7, and 2(9) - 11 = 7.

Question 8. (a) Find two numbers that multiply to -15 and add to 2: 5 and -3. So (x + 5)(x - 3) = 0, and x = -5 or x = 3. (b) Here a = 2, b = -3 and c = -4. The quadratic formula is x = (-b ± √(b² - 4ac)) / 2a. The part under the root is (-3)² - 4(2)(-4) = 9 + 32 = 41. So x = (3 ± √41) / 4. √41 ≈ 6.403, so x ≈ 9.403 / 4 ≈ 2.35, or x ≈ -3.403 / 4 ≈ -0.85.

Question 9. (a) Take out -5: h = -5(t² - 4t - 5) = -5(t - 5)(t + 1). The zeros are t = 5 and t = -1. Time cannot be negative, so the ball hits the water at t = 5 seconds. (b) The axis of symmetry sits halfway between the zeros: t = (5 + (-1)) / 2 = 2. At t = 2, h = -5(4) + 20(2) + 25 = -20 + 40 + 25 = 45. The ball reaches 45 m above the water after 2 seconds.

Question 10. The first equation already gives y. Substitute it into the second: 3x + 2(2x - 1) = 19, so 3x + 4x - 2 = 19, 7x = 21 and x = 3. Then y = 2(3) - 1 = 5. Check in the second equation: 3(3) + 2(5) = 19. The solution is (3, 5).

Question 11. Let a be the number of adult tickets and s the number of student tickets. Then a + s = 150 and 12a + 8s = 1480. Multiply the first equation by 8: 8a + 8s = 1200. Subtract it from the second: 4a = 280, so a = 70. Then s = 150 - 70 = 80. Check: 12(70) + 8(80) = 840 + 640 = 1480.

Question 12. The midpoint is the average of the coordinates: ((-3 + 5) / 2, (4 + (-2)) / 2) = (1, 1). The length is √((5 - (-3))² + (-2 - 4)²) = √(64 + 36) = √100 = 10.

Question 13. (a) The radius is the distance from (0, 0) to (-5, 12): √(25 + 144) = √169 = 13. A circle centred at the origin with radius r is x² + y² = r², so the equation is x² + y² = 169. (b) For (12, -5): 144 + 25 = 169, so it is on the circle. For (10, 8): 100 + 64 = 164, which is less than 169, so the point is inside.

Question 14. (a) Slope = (7 - (-1)) / (6 - 2) = 8 / 4 = 2. (b) Put m = 2 and the point (2, -1) into y = mx + b: -1 = 2(2) + b, so b = -5 and y = 2x - 5. Move every term to one side for the other form: 2x - y - 5 = 0. (c) Parallel lines have equal slopes, so the slope is 2, and the line crosses the y-axis at 3: y = 2x + 3. (d) The slopes of perpendicular lines are negative reciprocals, so the slope is -½. Check: 2 × (-½) = -1.

Question 15. Find two side lengths. AB = √((2 - (-2))² + (3 - 1)²) = √(16 + 4) = √20. BC = √((4 - 2)² + (-1 - 3)²) = √(4 + 16) = √20. Two equal sides make it isosceles. Now the slopes: AB has slope 2 / 4 = ½ and BC has slope -4 / 2 = -2. Their product is -1, so AB is perpendicular to BC and the angle at B is 90°. The triangle is right isosceles. As a check, AC = √(36 + 4) = √40, and 20 + 20 = 40, which fits the Pythagorean theorem.

Question 16. The sun's rays meet the ground at the same angle for both, so the two triangles are similar and their sides are in proportion. Height / shadow is the same for both: h / 18 = 1.6 / 2.4. So h = 18 × 1.6 / 2.4 = 12. The tree is 12 m tall.

Question 17. The ladder is the hypotenuse, and the height on the wall is opposite the 72° angle. sin 72° = height / 6.0, so the height is 6.0 × sin 72° ≈ 6.0 × 0.9511 ≈ 5.7 m. A calculator set to radians gives a wrong answer here. Check that it is in degree mode first.

Question 18. (a) The rise is opposite the angle and the horizontal distance is adjacent to it, so tan θ = 0.9 / 10.8 ≈ 0.0833. θ = tan⁻¹(0.0833) ≈ 4.8°. (b) The surface is the hypotenuse: √(10.8² + 0.9²) = √(116.64 + 0.81) = √117.45 ≈ 10.84 m.

Question 19. The sine law pairs each side with the angle opposite it: b / sin B = a / sin A. So b = 15 × sin 62° / sin 48° ≈ 15 × 0.8829 / 0.7431 ≈ 17.8 cm. The third angle is 180° - 48° - 62° = 70°, so all three angles are acute and the sine law applies without complications.

Question 20. Two sides and the angle between them call for the cosine law: c² = a² + b² - 2ab cos C. c² = 8² + 11² - 2(8)(11) cos 52° = 185 - 176 × 0.6157 ≈ 185 - 108.36 = 76.64. So c ≈ √76.64 ≈ 8.8 cm. A common slip is to subtract before multiplying, working out (185 - 176) × cos 52°. The order of operations matters.

What to do with the result

Count the wrong answers by strand, not in total.

A few wrong in each strand. The topics are probably fine. Go back to the questions that were missed and look for the step that went wrong: a sign, a bracket, degree mode on the calculator.

One strand clearly weaker. That is the useful result, and it is what the quiz is for. Take that strand back to the unit notes from class.

Quadratics weak, especially questions 4, 7 and 8. Check factoring first. Completing the square and solving both lean on it, so a factoring gap shows up three times.

Question 14 wrong, and the rest of analytic geometry right. Ask the teacher when the class covers the slopes of parallel and perpendicular lines and the forms of a line's equation. They are the expectations the ministry added in 2022 for students moving up from the de-streamed Grade 9 course, and Question 14 is where all three meet. Once the class has taught them, practise them from the unit notes.

Nearly everything wrong. Stop practising and find out what is going on underneath. More practice tests teach very little when the skills under them are missing, and they can make a student feel worse. If Grade 9 left gaps, our guide to Grade 9 math in Ontario covers what MPM2D assumes on day one. If Grade 9 was recent, our EQAO Grade 9 math practice quiz is a quick way to check the algebra MPM2D builds on. If the problem is freezing on tests rather than not knowing the math, math anxiety in children explains how to tell the two apart.

Where the course goes next

For students who started Grade 9 in 2024-25 or later, the ministry's diploma requirements list "3 credits in mathematics (Grade 9, Grade 10 and 1 credit in Grade 11 or 12)". So a student in Grade 10 this year must earn a Grade 10 math credit, and either Grade 10 course, MPM2D or MFM2P, earns it.

MPM2D is the listed prerequisite for Grade 11 Functions (MCR3U): the ministry's course table gives "Grade 10 Principles of Mathematics, Academic". Grade 11 Functions and Applications (MCF3M) accepts either Grade 10 course. Our page on Ontario's senior math streams shows which Grade 12 courses each path opens.

When a teacher helps more than another practice test

A practice quiz shows which strand is short. It cannot show why, and the why decides what to do next.

A student who misses Question 7 might not know how to complete the square. Or they might be fine with that and lose the sign when they move the -9 out of the bracket. Those look the same on a mark sheet and need different teaching.

A teacher who reads the working can see which of the two it was. On Genie, the math page says a teacher can work through "practice questions, past unit tests, and provincial-test style problems using the student's own course material". Genie works only with Canadian Certified Teachers, licensed by a provincial governing body such as the Ontario College of Teachers. Before a teacher sees a student profile, their certificate number and issuing body are checked against the provincial registry, with a live government-ID match. Genie calls this the Trust Engine. Students are matched by course, pace and personality, and parents get a per-lesson record of what was covered and what their child understood.

A certified math teacher can also chart which courses a student needs to reach a target programme, then focus the teaching on the subject areas that get them there.