GFree OMPT-G mock exam
An OMPT-G mock exam — with the answers.
The OMPT-G is the science variant of the OMPT: a maths core plus physics and chemistry chapters, including mechanics, sat by applicants whose programme asks for introductory science alongside maths. On this page: the exact OMPT-G format, six sample questions with fully worked answers — free to read, no signup — and a free full-length mock at the OMPT-G's real length and real clock.
The sample questions below need no account at all. The full mock needs a free one — takes a minute, no card.
Sample OMPT-G questions — with answers.
Six questions, one per official OMPT-G syllabus group, each tagged with the group's published share of the exam. Every solution is fully worked and visible right here — no signup, no PDF to hunt down. Try each one on paper first.
Question 1 · Functions, graphs & equations 45% of the OMPT-G
Solve for x: log2(x) + log2(x − 6) = 4.
Worked solution
- Domain first: both logarithms must have positive arguments, so x > 6.
- Combine the logs: log2(x(x − 6)) = 4, so x(x − 6) = 24 = 16.
- Expand and rearrange: x2 − 6x − 16 = 0, which factors as (x − 8)(x + 2) = 0.
- The candidates are x = 8 and x = −2. Only x = 8 satisfies x > 6.
- Check: log2(8) + log2(2) = 3 + 1 = 4. Correct.
Answer to enter: 8. The rejected root is the classic trap — an answer of −2 scores zero, because the original logarithms don't exist there.
Question 2 · Sequences & series 10% of the OMPT-G
An arithmetic sequence has third term a3 = 11 and seventh term a7 = 23. Find the first term, the common difference, and the sum of the first 20 terms.
Worked solution
- From a3 to a7 is 4 steps of the common difference d: 23 − 11 = 4d, so d = 3.
- Walk back two steps from a3: a1 = 11 − 2 × 3 = 5.
- The 20th term is a20 = 5 + 19 × 3 = 62.
- Sum of the first 20 terms: S20 = 20 × (a1 + a20) / 2 = 20 × (5 + 62) / 2 = 10 × 67 = 670.
Answers to enter: a1 = 5, d = 3, S20 = 670.
Question 3 · Trigonometric functions 15% of the OMPT-G
Solve sin(2x) = cos(x) for 0 ≤ x ≤ π.
Worked solution
- Use the double-angle identity: sin(2x) = 2 sin(x) cos(x), so the equation becomes 2 sin(x) cos(x) − cos(x) = 0.
- Factor — never divide by cos(x), you'd lose solutions: cos(x) · (2 sin(x) − 1) = 0.
- cos(x) = 0 gives x = π/2 on this interval.
- sin(x) = 1/2 gives x = π/6 and x = 5π/6 on this interval.
- Check x = π/6: sin(π/3) = √3/2 and cos(π/6) = √3/2. Correct.
Answer to enter: x = π/6, π/2, 5π/6. Dividing both sides by cos(x) at step 2 silently discards x = π/2 — a full lost point on the real exam.
Question 4 · Geometry & coordinates 10% of the OMPT-G
A circle has equation x2 + y2 − 6x + 4y − 12 = 0. Find its centre and radius, and determine whether the point (7, 1) lies on the circle.
Worked solution
- Group and complete the square in each variable: (x2 − 6x) + (y2 + 4y) = 12.
- x2 − 6x = (x − 3)2 − 9 and y2 + 4y = (y + 2)2 − 4.
- So (x − 3)2 + (y + 2)2 − 13 = 12, i.e. (x − 3)2 + (y + 2)2 = 25.
- Centre (3, −2), radius √25 = 5.
- Test (7, 1): (7 − 3)2 + (1 + 2)2 = 16 + 9 = 25. The point lies on the circle.
Answers to enter: centre (3, -2), radius 5, and yes — (7, 1) is on the circle.
Question 5 · Mechanics 10% of the OMPT-G
A ball is thrown vertically upward with an initial speed of 19.6 m/s. Take g = 9.8 m/s2 and neglect air resistance. (a) How long does the ball take to reach its highest point? (b) What is the maximum height above the launch point? (c) How long after launch does it return to the launch height?
Worked solution
- (a) At the top, the velocity is zero: v = v0 − gt gives 0 = 19.6 − 9.8t, so t = 2.0 s.
- (b) With v2 = v02 − 2gh at the top: h = v02 / (2g) = (19.6)2 / (2 × 9.8) = 384.16 / 19.6 = 19.6 m.
- (c) By symmetry (no air resistance), the descent mirrors the ascent: total time = 2 × 2.0 = 4.0 s.
Answers to enter: t = 2.0 s, h = 19.6 m, t_total = 4.0 s.
Question 6 · Chemistry 10% of the OMPT-G
Hydrogen burns in oxygen according to 2 H2 + O2 → 2 H2O. What mass of water is produced when 8.0 g of hydrogen gas reacts completely? (Molar masses: H = 1.0 g/mol, O = 16.0 g/mol.)
Worked solution
- Moles of H2: molar mass of H2 is 2.0 g/mol, so 8.0 / 2.0 = 4.0 mol.
- The equation gives a 2 : 2 ratio of H2 to H2O — that is, 1 : 1 — so 4.0 mol of water forms.
- Molar mass of H2O: 2 × 1.0 + 16.0 = 18.0 g/mol.
- Mass of water: 4.0 × 18.0 = 72 g.
- Sanity check by conservation of mass: the 4.0 mol of H2 consumes 2.0 mol of O2 = 64.0 g, and 8.0 g + 64.0 g = 72 g. Correct.
Answer to enter: 72 g.
How to read your result: six questions can show you the question style, not your level. If any solution above surprised you, that chapter is where your first hours should go — the
OMPT-G guide maps every chapter and its weight. To find out where you actually stand, sit the
free 10-question test or the full-length mock below.