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VettamedHUMAT HAZIRLIK
Ücretsiz kalibrasyon

Çözümlü örnekler

Bilimsel Düşünme tarzında, herkese açık üç soru; her biri baştan sona incelendi: soruyu çözen adım, her seçeneğin gerekçesi ve aynı türden bir sonraki soru için akılda tutulacak yöntem. Sorular ve çözümleri, HUMAT’ta olduğu gibi İngilizcedir.

Örnek 1

Değişkenlerin yerinde semboller.

Ücretsiz kalibrasyondan bir matematiksel düşünme sorusu. Üç denklem, bulunacak tek bir değer ve yanlış değeri vermenin dört yolu.

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

Soruyu çözen adım

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.

Her seçeneğin gerekçesi

  1. A−2ElendiThis is the value of ◆. Solving the system correctly and then reporting the wrong symbol is the easiest way to lose this mark.
  2. B4ElendiThis is the value of ▲: the same trap, a correct solution with the wrong symbol reported.
  3. C−3ElendiA sign slip. Equation 3 minus equation 1 gives −2● = −6; dropping one minus sign leaves ● = −3, and equation 1 no longer holds.
  4. D2ElendiThe size of ◆ with its sign lost. No value of ● other than 3 satisfies equations 1 and 3 together.
  5. E3Doğru cevapEquation 1 minus equation 3 cancels ◆ and ▲ and leaves 2● = 6.

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

Örnek 2

Farklı kümelerden öğe toplamak.

Bir işlemsel düşünme sorusu. Dört kural, beş aday küme ve geri kalanını belirleyen tek bir kural.

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

Soruyu çözen adım

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.

Her seçeneğin gerekçesi

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

Akılda tutulacak yöntem. Start from the rule that fixes something outright, follow its consequences, and only then read the options.

Örnek 3

Bir prosedürü izlemek.

Bir işlemsel düşünme sorusu. Üç adımlı bir prosedür ve hiç işlem yapmadan iki seçeneği eleyen bir özellik.

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

Soruyu çözen adım

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.

Her seçeneğin gerekçesi

  1. A6Elendi6 → 3 → 9, and 9 leaves remainder 1: code 1.
  2. B9Elendi9 is odd, so step 1 gives 14. 14 → 10, which leaves remainder 2: code 2.
  3. C10Doğru cevap10 → 5 → 15, and 15 = 3 × 4 + 3: code 3.
  4. D11Elendi11 → 16 → 12: code 0. Like every odd number, it can only reach code 0 or 2.
  5. E14Elendi14 → 7 → 21, and 21 leaves remainder 1: code 1. It takes the same route as 10 and lands on the wrong remainder.

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

Sekiz soru daha, süreyle.

Ücretsiz kalibrasyon, sınav temposunda süreli sekiz Bilimsel Düşünme sorusudur; her birinin çözümlü açıklaması vardır. 18 dakika, hesap gerekmez.

Ücretsiz kalibrasyonu başlatınİnceleme nasıl işler