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dMAT Sample Questions

Forty-three original questions in the exact official formats — figure sequences, equations, Latin squares, and General Academic Module passage sets. Click an answer to check it instantly. No account, no email wall.

Format-verified against g.a.s.t.'s official preparatory materials (updated July 2026). The real exam allows no note-taking — so try these in your head.

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Or work through the same questions untimed below, with the solution one click away.

1 · Figure Sequences

A symbol moves inside a grid according to a hidden rule (movement, rotation, colour change; it can bounce off walls or travel along them). Work out the rule and pick the next matrix.

Exam format: 20 series in 25 minutes. Rules can combine — e.g. a figure moves 1, then 2, then 3 fields (x + 1), while also rotating 90° each step.
Difficulty: low

The sequence below follows one rule. Which matrix comes next?

The triangle moves down one field per step in column 2. In matrix 4 it reaches the bottom row (option A). From there it would bounce off the lower boundary and move back up — the classic vertical-movement-with-bounce rule. Option B flips the triangle upside-down — nothing in the sequence rotates it. Option C shifts it to the wrong column.
Difficulty: medium

Two symbols, two rules. Which matrix comes next?

The circle moves clockwise along the outer border by 2 fields per step and alternates colour (amber → indigo). The triangle stays in place but rotates 90° to the right each step: up → right → down — so next it points left. Option C satisfies both rules; A moves the circle only 1 field, B gets both the colour and the rotation wrong.
Difficulty: high

Two symbols, two rules — one of them changes colour. Which matrix comes next?

The circle moves counter-clockwise along the top border one field per step and alternates colour (amber → indigo). The triangle stays in place but rotates 90° to the left each step (up → left → down → right). Option B satisfies both rules; A moves the circle too far, C forgets both the colour change and the rotation.
Difficulty: low

One symbol, one rule. Which matrix comes next?

The triangle moves diagonally up-right one field per step: (row 4, col 1) → (3,2) → (2,3). Next it reaches the top-right corner. From there the official rules say it would bounce and return along the same diagonal — option C. A and B break the diagonal.
Difficulty: high

The movement accelerates. Which matrix comes next?

The square moves clockwise along the border by x + 1 fields: 1 step, then 2 steps, so next it moves 3 steps — from the top-right corner three border fields clockwise lands in the bottom-right corner. This accelerating rule appears in the official materials — option B. A assumes constant 1-step movement, C overshoots by one.
Difficulty: medium

Two symbols, two rules. Which matrix comes next?

The circle travels counter-clockwise down the left border, one field per step; the square travels clockwise down the right border, one field per step. Both reach the bottom corners next. B swaps the colours (nothing in the rules changes colour); C stalls the circle.
Difficulty: high

Watch the colour carefully. Which matrix comes next?

Two rules stacked: the triangle moves down one field per step, AND its colour cycles through three states — amber → indigo → ink → back to amber. Matrix 4 needs the bottom row and the cycle restarting at amber — option C. A keeps the wrong colour phase; B forgets to move.
Difficulty: high

The rotation accelerates. Which matrix comes next?

The triangle moves right one field per step, and rotates 90° clockwise x + 1 times: once (→90°), then twice (90°+180°=270°), so next it rotates three times — 270° more, landing at 180° (pointing down) in the last column — option B. A forgets the acceleration; C forgets the movement.
Difficulty: high

Three symbols, three rules — and one of them leaves the grid. Which matrix comes next?

Watch what happens between the second and third matrix: the circle is at the right-hand edge, and instead of bouncing back it reappears on the far left of the same row. It wraps — that is a different rule from the bounce you usually see, and the sequence shows it happening so you can rely on it. The square does the same thing vertically down column 1: bottom row, then back to the top. The triangle simply turns 90° clockwise where it stands. Next: circle one more field right, square one more field down, triangle turned again (C). A keeps the triangle at its previous angle; B reads the circle as bouncing back instead of wrapping.

2 · Mathematical Equations

Solve the system so every equation holds. Each letter is an integer between 1 and 20, and there is exactly one solution.

Exam format: 20 systems in 25 minutes, solved mentally — no rough paper.
Difficulty: low

What is B?

6 + A = 13
B − A = 5
10
13
11
12
First equation: A = 13 − 6 = 7. Substitute into the second: B − 7 = 5 → B = 12.
Difficulty: medium

What is B?

3 × C = A
A + C = 16
B = A + C − 6
12
10
8
16
Replace A with 3C in the second equation: 3C + C = 16 → C = 4, so A = 12. Then B = 12 + 4 − 6 = 10.
Difficulty: high

What is D?

A = 5 × B
C = B + 9
A − B + C − D = 12
D = 3 × B + 1
9
5
7
11
Substitute everything into the third equation: 5B − B + (B + 9) − (3B + 1) = 12 → 2B + 8 = 12 → B = 2. Then D = 3 × 2 + 1 = 7. (And A = 10, C = 11 — all within 1–20.)
Difficulty: low

What is A?

B ÷ 2 = A
A + B = 18
4
9
12
6
Multiply the first equation by 2: B = 2A. Substitute: A + 2A = 18 → A = 6 (and B = 12).
Difficulty: high

What is B?

A + 3 = B
D = 2 × A
C = D − B
C = 4
7
10
14
13
Since C = 4, the third equation gives D − B = 4. Substitute D = 2A and B = A + 3: 2A − A − 3 = 4 → A = 7. So B = 10 (and D = 14). Notice the habit: substitute everything into the one equation with a known value.
Difficulty: low

What is B?

A − 4 = B
A + B = 14
5
9
4
7
From the first equation A = B + 4, so (B + 4) + B = 14 → B = 5 (and A = 9).
Difficulty: medium

What is A?

C = A + B
A = 2 × B
C = 9
3
9
6
4
Substitute A = 2B into the first equation: 2B + B = 9 → B = 3, so A = 6.
Difficulty: medium

What is B?

B ÷ A = 4
B − A = 12
16
12
8
20
B = 4A, so 4A − A = 12 → A = 4 and B = 16.
Difficulty: medium

What is C?

A + B + C = 22
B = A + 2
C = 3 × A
6
16
10
12
Substitute everything into the first: A + (A + 2) + 3A = 22 → 5A = 20 → A = 4, so C = 12 (and B = 6).
Difficulty: high

What is A?

D − C = 5
C = 2 × B
A = B + C + D
B = 2
13
17
15
11
Start from the known value: B = 2 gives C = 4, then D = 9, so A = 2 + 4 + 9 = 15. Always anchor on the equation that already contains a number.
Difficulty: high

What is C?

C − B + D − A = 13
8 × A = C
3 × A = D
A + 5 = B
16
24
12
18
There is no equation here that hands you a number outright — the official materials' hardest equation exercise is built exactly this way, and the technique is always the same: pick the letter every other equation depends on and write everything in terms of it. Here that letter is A. C = 8A, D = 3A, B = A + 5. Substitute into the long equation: 8A − (A + 5) + 3A − A = 13, which is 9A − 5 = 13, so A = 2. Then C = 16 (and B = 7, D = 6 — all inside 1–20, as the format guarantees). 24 is 8 × 3 — multiplying the two coefficients instead of solving; 12 answers for D + wrong A; 18 comes from dropping the −5 when you collect terms.

3 · Latin Squares

Each of the letters A–E appears exactly once per row and once per column of the 5×5 grid. Which letter belongs in the ? field? Sometimes you must fill other cells in your head first.

Exam format: 20 grids in 25 minutes. Only elimination is needed — never guesswork.
Difficulty: low

Which letter replaces the question mark?

?CD
EAB
ABCED
DC
CA
A
B
C
D
E
Look at column 1: it already contains E, A, D and C. The only letter missing is B.
Difficulty: medium

Which letter replaces the question mark?

EDCA
?A
BD
AEC
E
A
B
C
D
E
Column 1 already has E and B, so the ? can only be A, C or D. Look at row 4 (A, E, C given): it still needs B and D, and its column-1 cell can't be B (column 1 already has one) — so that cell takes D. Now row 2: it already contains A, ruling A out for the ?. Only C remains.
Difficulty: high

Which letter replaces the question mark?

D
AB
ABCE
?A
AE
A
B
C
D
E
Column 3 already shows D, A and C — only B and E are missing, split between the ? and the bottom cell. Row 5 already contains E, so the bottom cell of column 3 must take B. That forces the ? to be E. Two steps, no guessing.
Difficulty: low

Which letter replaces the question mark?

CD
EA?B
BC
BCEA
CDE
A
B
C
D
E
Row 2 already contains E, A and B; column 4 adds E and D. Four letters excluded — only C fits.
Difficulty: low

Which letter replaces the question mark?

DACB
C
EA
CDB
BAE?
A
B
C
D
E
Column 5 contains B, C and A; row 5 adds B, A and E. The union rules out A, B, C and E — only D fits.
Difficulty: medium

Which letter replaces the question mark?

DE
ADBE
BD
E?AB
A
A
B
C
D
E
Row 4 still needs C and D, split between the ? and its column-3 cell. Column 3 already has a D (row 2), so the column-3 cell must take C — leaving D for the ?.
Difficulty: high

Which letter replaces the question mark?

EDBA
D?E
D
EAC
B
A
B
C
D
E
Row 2 and column 2 together narrow the ? to A or C. Now test A: following the forced cells through columns 4 and 5 runs into a dead end — the grid can't complete. Only C works (verified unique by exhaustive check). On hard items, eliminate with the row-plus-column union first, then test the survivors.
Difficulty: high

Which letter replaces the question mark?

DC
A
C
BD
?C
A
B
C
D
E
Row 5 and column 1 between them exclude only B and C, so three letters are still possible at the ? — you cannot read this one off. Turn it round and ask where a letter is still allowed. Look down column 1: C has exactly one legal free field there, so place it. That fills a cell in row 2, and now D has exactly one legal free field left in row 2 — place that too. Finally run the same test along row 5: every other free field in that row sits in a column that already contains a D, so the ? is the only place left for it. Answer: D. This is the “hidden single” move the official materials demonstrate — what you reach for when simple row-and-column elimination stalls, and the difference between a square you can solve and one you end up guessing.

4 · General Academic Module

You read a short academic text, then answer single-choice questions that apply it — 4 options each, one correct. 90 minutes for the whole module. This format is now confirmed in g.a.s.t.'s official GAM preparatory materials (published 16 July 2026), with topics spanning mathematics, natural sciences, engineering, business, economics, and social sciences. Warm up with five standalone questions, then try three passage sets in the official style below.

Applied reasoning

A standardised test reports results as percentile ranks: the percentage of all test takers whose score was equal to or lower than yours. In one sitting, 1,600 candidates take the test. Priya's score is higher than 1,120 of them and equal to none. What is Priya's percentile rank?

30
70
48
75
1,120 of 1,600 scored lower → 1,120 ÷ 1,600 = 0.70. Percentile rank = 70: she did as well as or better than 70% of all candidates. This is exactly how your dMAT certificate reports results (plus a 0–200 score per module).
Applied reasoning

A learning study reports: of 240 students who used spaced practice, 168 passed their exam; of 160 students who crammed, 96 passed. Which statement follows from this data?

Spaced practice caused the higher pass rate
Most students who passed had crammed
Fewer than half of the crammers passed
The pass rate was higher with spaced practice
168 ÷ 240 = 70% vs 96 ÷ 160 = 60% — the spaced-practice pass rate is higher, and that's all the data supports. "Caused" is the classic trap: the data shows association, not causation (maybe stronger students choose spaced practice). Passers: 168 > 96, so most passers did NOT cram; and 60% of crammers passing is more than half. Distinguishing what follows from what merely sounds plausible is the core General Academic Module skill.
Difficulty: Applied reasoning

A dMAT test city has 480 available seats across its centres, booked first-come-first-served. If about 5,000 candidates prefer that city, roughly what share of them will have to choose another city?

About 90%
About 10%
About 48%
Cannot be determined
480 of 5,000 preferences can be satisfied: 480 ÷ 5,000 = 9.6% get seats, so roughly 90% must book elsewhere. 'Cannot be determined' tempts because real demand is unknown — but the question says 'if about 5,000', making the calculation valid under its own assumption. (This is also why you register early.)
Difficulty: Applied reasoning

The dMAT score scale runs 0–200 with a fixed mean of 100 per module. Priya scores 118 on the core module and 96 on the subject module. Which statement follows?

She answered 118 core questions correctly
Her overall percentile must be above 50
Her core result is above average; her subject result is below average
Her subject performance failed the exam
The 0–200 scale has mean 100, so 118 is above and 96 below average — that is all the scale tells you. The score is a conversion, not a raw count (rules out B); the percentile depends on the whole distribution (C); and the dMAT has no pass/fail (D).
Difficulty: Applied reasoning

In a survey, 500 test takers who used official preparatory materials averaged 112, while 300 who didn't averaged 104. Which conclusion follows from the data alone?

On average, users of the official materials scored 8 points higher
Using the official materials raises a candidate's score by 8 points
Every user of the materials outscored every non-user
Most of the 800 candidates scored above 104
Only the difference in group averages follows from the data. 'Raises' claims causation (motivated students may self-select into preparing); averages say nothing about every individual (C); and a group mean doesn't locate the median (D).

5 · Passage sets — official GAM style

In the real subject module, each task is a text plus several questions. Read the passage once (no notes allowed in the exam!), then answer every question under it. One set each from business, natural sciences, and research methods — the topic areas named in the official materials.

PASSAGE A · BUSINESS ADMINISTRATION

Break-even analysis

A firm's fixed costs (rent, salaries, machinery) stay the same regardless of output, while variable costs are incurred per unit produced. If each unit sells at price p and costs v to make, the difference p − v is the contribution margin: what each sale contributes toward covering fixed costs F. The break-even quantity is the output at which total revenue exactly equals total cost: Q* = F ÷ (p − v). Below Q* the firm makes a loss; above it, a profit.

Difficulty: Passage · apply

A firm has fixed costs of ₹60,000 per month. Each unit sells for ₹50 and costs ₹30 to produce. What is the monthly break-even quantity?

1,200 units
2,000 units
4,000 units
3,000 units
Contribution margin = 50 − 30 = ₹20 per unit. Q* = 60,000 ÷ 20 = 3,000 units. The trap answers divide by the price alone (60,000 ÷ 50 = 1,200) or by the variable cost (60,000 ÷ 30 = 2,000) — both ignore that only the margin covers fixed costs.
Difficulty: Passage · concept

Which change lowers a firm's break-even quantity?

A rise in fixed costs
A rise in variable cost per unit
A rise in the selling price
A fall in the selling price
Q* = F ÷ (p − v). A higher selling price widens the contribution margin (the denominator), so fewer units are needed to cover fixed costs. Higher F raises Q*; higher v or lower p shrink the margin and raise Q* too.
Difficulty: Passage · transfer

Price and variable cost per unit both increase by exactly ₹5. What happens to the break-even quantity?

It rises
It stays the same
It falls
It cannot be determined
The contribution margin is (p + 5) − (v + 5) = p − v — unchanged. Since F is also unchanged, Q* is identical. The question tests whether you reason from the formula's structure rather than recalculating blindly — exactly the transfer skill the module measures.
PASSAGE B · NATURAL SCIENCES

Density and flotation

Density is mass per volume: ρ = m ÷ V. An object placed in a fluid floats if its average density is lower than the fluid's, and sinks if it is higher. A floating object displaces exactly enough fluid to support its weight, so the fraction of its volume below the surface equals the ratio of the densities: object ÷ fluid. Fresh water has a density of 1.0 g/cm³.

Difficulty: Passage · apply

A solid block has a mass of 400 g and a volume of 500 cm³. What fraction of the block is below the surface when it floats in fresh water?

20%
80%
50%
100%
Density = 400 ÷ 500 = 0.8 g/cm³. Submerged fraction = 0.8 ÷ 1.0 = 80%. The 20% option is the part above water — read what the question asks. 100% would mean the block barely sinks, which needs density ≥ 1.0.
Difficulty: Passage · concept

The same block is moved from fresh water into denser seawater (1.05 g/cm³). What happens?

It sinks deeper into the water
Exactly the same fraction is submerged
It sinks completely
It floats higher — a smaller fraction is submerged
Submerged fraction = ρobject ÷ ρfluid = 0.8 ÷ 1.05 ≈ 76%, less than the 80% in fresh water. A denser fluid supports the same weight with less displaced volume — the reason ships ride higher in the sea than in a river.
Difficulty: Passage · transfer

Objects A and B have equal mass, but A has twice the volume of B. Which statement follows?

A's density is twice B's
Their densities are equal
A's density is half of B's
It cannot be determined from the given information
ρ = m ÷ V: same numerator, double the denominator → half the density. No numbers are needed — the relationship alone answers it. "Cannot be determined" tempts because no masses are given, but the ratio is fully determined.
PASSAGE C · RESEARCH METHODS

Sampling and study design

Conclusions from a study are only as good as its design. A random sample gives every member of the population an equal chance of inclusion, so results can be generalised. A convenience or self-selected sample — whoever is nearby or chooses to respond — may differ systematically from the population. Observational data can show that two things are associated, but a causal claim requires an experiment in which researchers assign participants to conditions at random, so that no hidden third factor explains the difference.

Difficulty: Passage · apply

A university surveys students found in the library at 9 a.m. about campus-wide study habits. What is the main weakness?

The sample is too small to compute percentages
Library visitors at 9 a.m. may differ systematically from the student body
The question wording was leading
It proves nothing because correlation implies causation
This is a convenience sample: early-morning library users are plausibly more studious than average, so generalising to all students is unsafe. Nothing in the stem tells us the sample size or wording — don't import weaknesses the text doesn't state.
Difficulty: Passage · concept

What distinguishes a randomised experiment from an observational study?

Treatment is assigned at random by the researchers, which supports causal conclusions
It always includes more participants
It always runs for a longer period
It does not need a control group
Random assignment balances hidden factors across groups, so a difference in outcomes can be attributed to the treatment. Sample size and duration are separate design choices, and experiments typically rely on control groups rather than dropping them.
Difficulty: Passage · transfer

A fitness app finds that members who log meals lose more weight. What would be needed to support the claim that logging meals causes weight loss?

A much larger sample of app members
Random assignment of members to logging and non-logging groups
Repeating the same survey a year later
More precise measurement of weight
The association could reflect motivation: disciplined members both log meals and diet harder. Only random assignment removes that confound. A bigger sample, a repeat survey, or better scales make the association more precise — not causal.
Difficulty: Passage · transfer

A newspaper invites readers to vote online on a policy; 78% of the 100,000 votes oppose it. Why can't this result be generalised to the whole population?

The sample is too small
Percentages cannot be calculated from online votes
No reason — a large sample generalises automatically
Readers who choose to vote may differ systematically from the general population
100,000 is plenty — the flaw is self-selection: people with strong opinions (and this newspaper's readership) choose to vote. A biased sampling method isn't cured by size; a small random sample beats a huge self-selected one.

All questions on this page are original PrepDMAT items written in the official dMAT task formats documented by g.a.s.t. — nothing here is reproduced from copyrighted exam materials. Official sample tasks and introduction videos: d-mat.de.

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