Free dMAT mathematical equations practice — no login.

01How this mock test works

Generated fresh, proven to have one solution.

Every system is built in your browser the moment you reach it, so no two runs are the same. Before a task is shown, the generator checks by exhaustive search that exactly one combination of values in 1–20 satisfies all the equations — so there is always a single correct answer and a clean chain of substitutions to reach it. That chain is the working you read afterwards.

02Format and timing

What the mathematical equations subtest actually asks.

03Worked examples

Four systems, solved in dependency order.

Fixed examples, shown with the full solution path rather than just the answer. No system here hands you a letter outright: each one names the letter it rewrites everything else in terms of, and the order the substitutions run in — which is the part that transfers to the next system. They run from two letters to four.

1. 01

   Write one letter in terms of the other

   low

   Neither equation gives a value on its own — that is the point. Pick the equation that already reads as one letter equalling something built from the other, and push that expression into the second equation. Two unknowns become one.

   A + B = 9

   2 × B − 9 = A

   Find A.

   A = 3 (B = 6, A = 3)

   How you get there

   1. 1No equation gives a letter on its own, so rewrite every letter in terms of B.
   2. 22 × B − 9 = A already puts A in terms of B: A = 2 × B − 9.
   3. 3Substituting all of that into A + B = 9 leaves 3 × B − 9 = 9, so B = 6.
   4. 4Reading back up the chain: A = 3.
2. 02

   Choose the letter everything else hangs off

   medium

   With three letters, one of them is the hub: the others are each defined against it, directly or through a short chain. Find that letter first, rewrite the rest in terms of it, and only then use the equation that carries a plain number.

   A = 2 × B + 7

   28 = B + C + A

   B = C + 3

   Find A.

   A = 19 (C = 3, B = 6, A = 19)

   How you get there

   1. 1No equation gives a letter on its own, so rewrite every letter in terms of C.
   2. 2B = C + 3 already puts B in terms of C: B = C + 3.
   3. 3A = 2 × B + 7, with B rewritten, gives A = 2 × C + 13.
   4. 4Substituting all of that into 28 = B + C + A leaves 4 × C + 16 = 28, so C = 3.
   5. 5Reading back up the chain: B = 6, A = 19.
3. 03

   The combination equation is the only way in

   high

   Every relation here ties two letters together, so none of them can be solved alone. The equation holding a bare number across several letters is the one that closes the system — but it is the last step, not the first. Reach it with everything already rewritten.

   B = A + 3

   A + 11 = C

   17 = B + D + C − A

   A + 1 = D

   Find B.

   B = 4 (A = 1, C = 12, B = 4, D = 2)

   How you get there

   1. 1No equation gives a letter on its own, so rewrite every letter in terms of A.
   2. 2A + 11 = C already puts C in terms of A: C = A + 11.
   3. 3B = A + 3 already puts B in terms of A: B = A + 3.
   4. 4A + 1 = D already puts D in terms of A: D = A + 1.
   5. 5Substituting all of that into 17 = B + D + C − A leaves 2 × A + 15 = 17, so A = 1.
   6. 6Reading back up the chain: C = 12, B = 4, D = 2.
4. 04

   Chains of two: substitute through the middle letter

   high

   Some letters are not defined against the base directly but against a letter that is. Compose the two steps before substituting — get each letter down to a multiple of the base plus an offset — and the closing equation collapses in one line.

   A = 2 × C + 8

   C + 5 = D

   A − D + B = 20

   B = 3 × D − 6

   Find D.

   D = 7 (C = 2, A = 12, D = 7, B = 15)

   How you get there

   1. 1No equation gives a letter on its own, so rewrite every letter in terms of C.
   2. 2A = 2 × C + 8 already puts A in terms of C: A = 2 × C + 8.
   3. 3C + 5 = D already puts D in terms of C: D = C + 5.
   4. 4B = 3 × D − 6, with D rewritten, gives B = 3 × C + 9.
   5. 5Substituting all of that into A − D + B = 20 leaves 4 × C + 12 = 20, so C = 2.
   6. 6Reading back up the chain: A = 12, D = 7, B = 15.

04Official practice exercises

The six exercises from the official materials.

These are the mathematical equations exercises published in the g.a.s.t. preparatory materials, with their model solution paths. Two each at low, medium and high difficulty. Try to solve each one in your head before opening the solution.

1. Exercise 1

   low

   7 + A = 14

   B − 3 = A

   Show the solution

   A = 7, B = 10

   1. 1The first equation makes it clear that A = 7 if you subtract 7 on both sides.
   2. 2Insert A = 7 into B − 3 = A to get B − 3 = 7, so adding 3 gives B = 10.

2. Exercise 2

   low

   B ÷ 2 = A

   B − A = 8

   Show the solution

   A = 8, B = 16

   1. 1Multiplying by 2 in the first equation gives B = 2A.
   2. 2Replacing B in the second equation gives 2A − A = 8, so A = 8.
   3. 3Substituting back, B ÷ 2 = 8, so B = 16.

3. Exercise 3

   medium

   3 × C = A

   A + C = 8

   2 × A + 2 × C = B

   Show the solution

   A = 6, B = 16, C = 2

   1. 1From the first equation A = 3C; substitute into A + C = 8 to get 4C = 8, so C = 2.
   2. 2Then A = 3 × 2 = 6.
   3. 3Finally 2 × 6 + 2 × 2 = B, so B = 16.

4. Exercise 4

   medium

   18 − B = A

   3 × A = C

   B ÷ 2 = A

   Show the solution

   A = 6, B = 12, C = 18

   1. 1The third equation gives B = 2A; substitute into 18 − B = A to get 18 − 2A = A, so 18 = 3A and A = 6.
   2. 2Then B ÷ 2 = 6, so B = 12.
   3. 3And 3 × 6 = C, so C = 18.

5. Exercise 5

   high

   A − B + C − D = 2

   10 × B = C

   5 × B = A

   11 + B = D

   Show the solution

   A = 5, B = 1, C = 10, D = 12

   1. 1Substitute A = 5B, C = 10B and D = 11 + B into the first equation: 5B − B + 10B − (11 + B) = 2.
   2. 2That simplifies to 13B − 11 = 2, so 13B = 13 and B = 1.
   3. 3Then C = 10, A = 5 and D = 12.

6. Exercise 6

   high

   C + D − A = 1

   5 × C = D

   13 − C = A

   3 × C − 1 = B

   Show the solution

   A = 11, B = 5, C = 2, D = 10

   1. 1Substitute D = 5C and A = 13 − C into the first equation: C + 5C − (13 − C) = 1.
   2. 2That simplifies to 7C − 13 = 1, so 7C = 14 and C = 2.
   3. 3Then D = 10, A = 11 and B = 3 × 2 − 1 = 5.

Exercises and solutions are taken from the official g.a.s.t. preparatory materials for the dMAT, published at d-mat.de.

05dMAT preparation

Four habits that speed you up.

• Find the letter you can solve first

  One equation almost always pins a single letter to a number on its own. Solve that letter, then substitute its value everywhere it appears — that usually unlocks the next one.

• Substitute, do not guess

  Because every system has exactly one solution, there is always a clean chain of substitutions. If you find yourself trying numbers, you have probably missed the letter that solves directly.

• Keep it mental

  No calculator or rough paper in the exam. The arithmetic here is deliberately small — the skill being tested is spotting the order to substitute, and holding two or three values in your head.

• Move on and come back

  20 systems in 25 minutes is 75 seconds each. If a system does not resolve in about 40 seconds, guess and move on — a fresh look later is faster than staring.

Installable as an app — use your browser's Install or Add to Home Screen option to practise offline.

Keep going

Next on dmatgrid

• Latin squaresThe other quick core subtest: a 5×5 grid where you name one marked cell.
• Figure sequencesContinue a series of matrices as the figures move, turn and recolour.
• Exam patternHow the modules fit together and how long each one runs.