OMPT-D Practice Questions: 6 Worked Multi-Step Problems

Every August the same email arrives: "do you have OMPT-D sample questions?" The honest answer is that omptest.org publishes none in public; the only samples sit inside the paid practice material. So I wrote my own, in the shape the syllabus describes: several steps, more than one chapter per problem, an exact answer at the end, every number checked numerically. For format and pacing, read the OMPT-D survival guide; this post is the maths.

Work each problem on paper under a nine-minute timer before reading the solution, then compare your note with the "what earns partial credit" paragraph.

How the OMPT-D scores a problem

On the D and the F a written solution is required and two graders read every test. omptest.org's blog describes a per-step rubric in the spirit of "25% of the full grade if this step is written". Your final answer must be exact unless a decimal is asked for: the help centre's own example is -48/408, which scores as -2/17 and not as -0.12. You get a scientific on-screen calculator, no formula sheet, three A4 sheets of scratch paper and no negative marking. With 21 questions, each one carries close to five points.

Problem 1: the largest rectangle under a parabola

A rectangle has its base on the x-axis and its two upper corners on the parabola below. Find the largest possible area of the rectangle, exactly, and show that it is a maximum.

Worked solution

Write the area as a function of the half-width x. The width is 2x and the height is the parabola's value at x, which must be positive, so x lives strictly between 0 and 3.

Differentiate and set the derivative to zero. Only the positive root is admissible, because x is half a width.

Confirm a maximum with the second derivative, then substitute back: the height at x = sqrt(3) is 27 - 9 = 18 and the width is 2 sqrt(3).

Typed answer: A = 36*sqrt(3). The calculator will show 62.35, and 62.35 is not what the grader wants.

What earns partial credit: the area function with its domain; the correct derivative; solving A'(x) = 0 and rejecting the negative root with a reason; a maximum test; the exact value. Setting up A(x) and differentiating it correctly, then stopping, is roughly half of what the rubric credits.

The common wrong turn: maximising the height instead of the area, or reporting x = sqrt(3) when the question asked for the area. Read the last line of the question before you type. The applications of derivatives drills are built around that "which quantity did they ask for" habit.

Problem 2: position and speed on a Lissajous figure

A point moves along the parametric curve below, a Lissajous figure, which the OMPT-D learning outcomes name explicitly. Give the position of the point at t = pi/6 and its speed at that instant, both exactly.

Worked solution

Position is substitution with special values. At t = pi/6, cos(pi/6) = sqrt(3)/2 and 2t = pi/3, so sin(pi/3) = sqrt(3)/2.

Speed is the length of the velocity vector, so differentiate each coordinate with respect to t. The chain rule brings a factor 2 out of sin 2t.

As a bonus the tangent direction falls out: the velocity vector is (-1, 1), so the tangent has slope dy/dx = 1/(-1) = -1, a third part the D likes to ask for.

Typed answer: position (sqrt(3), sqrt(3)/2), speed = sqrt(2). If the answer field is a vector, the SOWISO syntax is [sqrt(3), sqrt(3)/2].

What earns partial credit: the two exact coordinates; both derivatives with the chain-rule factor present; the velocity components at t = pi/6; the speed as the root of the sum of squares. A note reading "v = sqrt(x'^2 + y'^2)" followed by the numbers is the trail a grader wants.

The common wrong turn: differentiating sin 2t to cos 2t and losing the factor 2, which gives a speed of sqrt(5)/2 that looks plausible enough that nobody checks it. A second trap is degrees: the D expects radians throughout. Practise the special values in the trig functions drills until sin(pi/3) is a reflex.

Problem 3: the area enclosed by a parabola and a line

Find the exact area of the region enclosed between the parabola and the line below. Set up the integral yourself; the intersection points are part of the work.

Worked solution

First find where the curves meet, by setting them equal and solving the quadratic.

Decide which curve is on top between the intersections. At x = 0 the parabola gives 4 and the line gives 2, so the parabola is the upper curve on the whole interval and the integrand is parabola minus line.

Typed answer: A = 9/2. Not 4.5, even though it is the same number; the rule is exact form unless a decimal is requested.

What earns partial credit: the intersection points from a correctly solved quadratic; a stated reason for which curve is on top; the definite integral with the right limits and integrand; the antiderivative; the evaluated result. The integral line is the one I would weight most heavily; the official learning outcome is "draws up a definite integral and calculates this exactly".

The common wrong turn: integrating line minus parabola, getting -9/2, and silently dropping the sign. A grader credits "the area is 9/2 because I subtracted in the wrong order" more readily than an unexplained flip. The arithmetic at the lower limit, with -2 cubed and the double negatives, is the other leak. Drill the mechanics in integration basics.

Problem 4: a trig equation on an interval with exact solutions

Solve the equation below exactly for all x on the interval 0 ≤ x < 2pi. Give every solution.

Worked solution

The equation mixes cos squared and sin, so convert to one function with cos^2 x = 1 - sin^2 x. That gives a quadratic in sin x.

Now read each value off the unit circle on the given interval. Sine equals 1/2 at pi/6 and, by symmetry about pi/2, at 5pi/6; it equals -1 only at 3pi/2.

Typed answer: x = pi/6 or x = 5*pi/6 or x = 3*pi/2. Several solutions go in as a disjunction with "or", the form the SOWISO documentation shows for equations with more than one root.

What earns partial credit: the identity substitution; the quadratic in sin x; factorising it (the quadratic formula is fine too); each sine value; the full set of angles with the symmetric partner 5pi/6 included. Two solutions out of three is rarely two thirds of the marks, because a rubric credits "solves completely on the interval" as one step.

The common wrong turn: dividing by sin x + 1, which throws a solution away, or stopping at x = pi/6 and forgetting the second-quadrant partner. A quieter one is writing 30 degrees; the answer field expects pi/6. The trig equations drills are all interval problems for this reason.

Problem 5: a logarithmic equation with a candidate to reject

Solve the equation below exactly. Part of the marks are for the domain.

Worked solution

Before touching the equation, write the domain. Both arguments must be positive: x - 1 > 0 and x + 3 > 0, so x > 1 is the condition that matters.

Combine the logarithms with the product rule, then remove the log by writing the equation in exponential form.

Check both candidates against the domain. x = -7 makes both arguments negative, so it solves the quadratic and not the original equation. Substituting x = 5 back: log_2(4) + log_2(8) = 2 + 3 = 5.

Typed answer: x = 5. If a question ever needs a logarithm typed, the syntax is log(2, x) for base 2, ln(x) for the natural log and log(x) for base 10; the answer entry guide has the full table.

What earns partial credit: the domain stated up front; the product rule; the conversion to (x - 1)(x + 3) = 32; the quadratic solved; the rejection of x = -7 with the reason written. That last step is the one students skip. A note ending "x = 5 or x = -7" has done the algebra and missed the mathematics.

The common wrong turn: expanding log(x - 1) into log x - log 1, the invented log law from the algebra mistakes post. Subtler: writing the domain as x > -3 because that is the first bracket seen, and keeping -7 anyway. Take the stricter condition. Both habits are drilled in logarithms.

Problem 6: a circle, a chord and a dot product

The circle and line below meet at two points A and B. Find A and B. Let M be the centre of the circle and N the midpoint of AB. Show with a dot product that MN is perpendicular to the line, and give the exact length of AB.

Worked solution

Substitute the line into the circle. The centre is M(2, 1) and the radius is 5; write both down first.

The midpoint of AB is N(4, 0), so MN = (2, -1). A direction vector of the line y = 2x - 8 is d = (1, 2), since the slope is 2. Perpendicular means the dot product is zero.

So MN is perpendicular to the chord, as the line from a centre to a chord's midpoint always is. The length of AB comes from the coordinates.

Cross-check, since it costs thirty seconds: the distance from M to the line 2x - y - 8 = 0 is |4 - 1 - 8| / sqrt(5) = sqrt(5), so the half-chord is sqrt(25 - 5) = 2 sqrt(5) and the chord is 4 sqrt(5).

Typed answer: A = (2, -4) and B = (6, 4); |AB| = 4*sqrt(5). Not sqrt(80), which is the same number unsimplified; the grading engine can reject an unsimplified root.

What earns partial credit: the substitution and the quadratic; both points; the midpoint; the two vectors named; the dot product computed and the conclusion stated in words; the exact chord length. If time runs out after the points, a bare list "N = (4, 0), MN = (2, -1), d = (1, 2), dot product = 0" is worth more than a blank.

The common wrong turn: using the normal vector (2, -1) of the line as its direction, which gives a dot product of 5 and a panicked conclusion that the geometry is broken. A direction vector of y = mx + c is (1, m); the normal is (m, -1). The other slip is losing the 36x middle term in (2x - 9)^2. Quadratic mechanics live in quadratic equations.

The six problems against the official OMPT-D weights

The 30% block is the largest and the least glamorous: equations, inequalities, domains, asymptotes, functions with a parameter. Problem 5 is the only one here that sits in it; on the real test expect six or seven questions from it. And "geometry" on the D is coordinate geometry with vectors, not the triangle geometry of the OMPT-C. Run the free D mock to see which block leaks, then feed the result into the study planner.

How to write an OMPT-D note

The note field is plain text next to the answer field, read by two graders, and it is where partial credit lives. Write it as you go, in this order.

1. State what you are computing and from what, in one line: "Area of rectangle, A(x) = 2x(27 - 3x^2), 0 < x < 3." Domains and conditions go here, before any algebra.
2. Put each transformation on its own line with the operation named: "differentiate", "set equal to zero", "substitute y = 2x - 8". A grader crediting the step needs to find it.
3. Show the substitution of numbers once, before simplifying: "A(sqrt(3)) = 2 sqrt(3)(27 - 9)". Jumping from formula to final number hides the step being credited.
4. Reject candidates in writing, with the reason: "x = -7 rejected, x - 1 < 0". A silent rejection looks like a guess.
5. Verify when it is cheap, by substituting back or by a second route as in Problem 6, and say that you did.
6. End with the exact answer in the form you will type, then type it. If the two differ, the grader sees a contradiction.
7. If stuck, write what you have and the step you cannot do. Partial knowledge on the page is scored; partial knowledge in your head is not.

Two students, two ways to lose the same points

Student one is strong: two years of calculus, all six problems correct on paper, and she typed 62.35 for Problem 1 and 4.5 for Problem 3 because the calculator showed those numbers. Her notes were three lines each. On a mock she scored in the low sixties against a 75% target. The fix took a week: exact form always, every step written, and a last pass at minute 160 comparing each typed answer with the last line of its note. Her next mock was in the eighties. Nothing mathematical changed.

Student two came from a curriculum with almost no vectors and no parametric curves, applying to a programme that wants 60% on the D (Maastricht's DSAI and Computer Science programmes, at two attempts; the Maastricht page has the detail). He could do Problems 3, 4 and 5 cleanly and could not start 2 or 6. That is a geometry-block gap, learnable in about two weeks because the underlying algebra is easy; the unfamiliarity is the whole difficulty. He spent those weeks on vectors, circles and parametric derivatives, wrote full notes on every problem, and left differentiation alone. His mocks climbed from the forties to the mid-sixties, and the note habit carried the borderline questions.

If your diagnostic mock is under 40%, the problem is the prerequisite chapters; rebuild OMPT-B-level algebra first. Between 40% and 60%, take the two weakest blocks from the breakdown and write full notes for two weeks. Above 60% with a 75% target, the gap is almost always exact form and note discipline. The sample questions walkthrough shows the same discipline on the shorter variants.