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VettamedHUMAT PREPARATION
Free calibration

Worked examples

Three public questions in the style of Scientific Thinking, each reviewed in full: the step that decides it, a reason for every option, and the method to keep for the next question of its kind.

Example 1

Symbols in place of variables.

A mathematical-thinking question from the free calibration. Three equations, one value to find, and four ways to report the wrong one.

Mathematical thinking · symbols in place of variables

The symbols ◆, ● and ▲ each stand for an integer. They satisfy:

◆ + ● + ▲ = 5
2◆ − ● + ▲ = −3
◆ − ● + ▲ = −1

What is the value of ●?

  1. A−2
  2. B4
  3. C−3
  4. D2
  5. E3

The decisive step

Equations 1 and 3 differ only in ●. Subtract one from the other and ◆ and ▲ cancel: 2● = 6, so ● = 3. The full solution is ◆ = −2, ● = 3, ▲ = 4.

Why each option stands or falls

  1. A−2Ruled outThis is the value of ◆. Solving the system correctly and then reporting the wrong symbol is the easiest way to lose this mark.
  2. B4Ruled outThis is the value of ▲: the same trap, a correct solution with the wrong symbol reported.
  3. C−3Ruled outA sign slip. Equation 3 minus equation 1 gives −2● = −6; dropping one minus sign leaves ● = −3, and equation 1 no longer holds.
  4. D2Ruled outThe size of ◆ with its sign lost. No value of ● other than 3 satisfies equations 1 and 3 together.
  5. E3KeyEquation 1 minus equation 3 cancels ◆ and ▲ and leaves 2● = 6.

Method that transfers. Before solving a system, look for two equations that differ in a single term. Subtracting them isolates that term in one step.

Example 2

Collecting elements from different sets.

A procedural-thinking question. Four rules, five candidate sets, and one rule that settles the rest.

Procedural thinking · collecting elements from different sets

Five samples, A to E, may be included in a set. The set must satisfy four rules:

  • Exactly one of A and B is included.
  • D is included.
  • If C is included, D is not.
  • Exactly one of C and E is included.

Which set satisfies every rule?

  1. AA, C and D
  2. BA, D and E
  3. CB and D
  4. DA, B, D and E
  5. EB, C, D and E

The decisive step

D is forced in. C would force D out, so C is out, and then E must be in. Only A, D and E keeps D and E with exactly one of A and B.

Why each option stands or falls

  1. AA, C and DRuled outC appears together with D, which the third rule forbids.
  2. BA, D and EKeyD and E are in, C is out, and exactly one of A and B is in: every rule holds.
  3. CB and DRuled outNeither C nor E is included, but exactly one of them must be.
  4. DA, B, D and ERuled outA and B are both included; exactly one of them is allowed.
  5. EB, C, D and ERuled outC appears with D, and C and E are both in: two rules fail.

Method that transfers. Start from the rule that fixes something outright, follow its consequences, and only then read the options.

Example 3

Following a protocol.

A procedural-thinking question. A three-step procedure, and a property that removes two options before any arithmetic.

Procedural thinking · following a protocol

A sorting protocol gives each sample a code. Start with the sample number n.

  1. If n is even, divide it by 2. If n is odd, add 5.
  2. If the result is greater than 10, subtract 4. Otherwise, multiply it by 3.
  3. The code is the remainder when the result is divided by 4.

Which sample number receives code 3?

  1. A6
  2. B9
  3. C10
  4. D11
  5. E14

The decisive step

Work backwards. An odd n becomes even at step 1, and both branches of step 2 keep it even, so its code is 0 or 2: 9 and 11 are out at once. Of 6, 10 and 14, step 2 gives 9, 15 and 21; only 15 leaves remainder 3.

Why each option stands or falls

  1. A6Ruled out6 → 3 → 9, and 9 leaves remainder 1: code 1.
  2. B9Ruled out9 is odd, so step 1 gives 14. 14 → 10, which leaves remainder 2: code 2.
  3. C10Key10 → 5 → 15, and 15 = 3 × 4 + 3: code 3.
  4. D11Ruled out11 → 16 → 12: code 0. Like every odd number, it can only reach code 0 or 2.
  5. E14Ruled out14 → 7 → 21, and 21 leaves remainder 1: code 1. It takes the same route as 10 and lands on the wrong remainder.

Method that transfers. When a procedure ends in a property, ask which inputs can produce that property at all. Here parity removes two options before any arithmetic.

Try eight more, against the clock.

The free calibration is eight timed Scientific Thinking questions at exam pace, each with a worked solution. 18 minutes, no account needed.

Start the free calibrationHow review works