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Test: Syllogism - 2 - Bank Exams MCQ


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10 Questions MCQ Test - Test: Syllogism - 2

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Test: Syllogism - 2 - Question 1

Statements: All green are blue. All blue are white.

Conclusions:

  1. Some blue are green.
  2. Some white are green.
  3. Some green are not white.
  4. All white are blue.

Detailed Solution for Test: Syllogism - 2 - Question 1

 


 

From the statements:

  • All green are blue (G ⊆ B).

  • All blue are white (B ⊆ W).
    Thus G ⊆ B ⊆ W, so every green is white.

Evaluate conclusions:

  1. Some blue are green — True (since all green are blue, assuming at least one green exists).

  2. Some white are green — True (greens are a subset of white).

  3. Some green are not white — False (all green are white).

  4. All white are blue — False (we only know all blue are white, not the converse).

Correct choice: A (Only 1 and 2).

So Option A is correct

Test: Syllogism - 2 - Question 2

Statements: All men are vertebrates. Some mammals are vertebrates.

Conclusions:

  1. All men are mammals.
  2. All mammals are men.
  3. Some vertebrates are mammals.
  4. All vertebrates are men.

Detailed Solution for Test: Syllogism - 2 - Question 2


 

From the statements:

  • All men are vertebrates (Men ⊆ Vertebrates).

  • Some mammals are vertebrates (Mammals ∩ Vertebrates ≠ ∅).

Check conclusions:

  1. All men are mammals — Not implied. Men could be non-mammal vertebrates.

  2. All mammals are men — Not implied.

  3. Some vertebrates are mammals — True (given directly: some mammals are vertebrates).

  4. All vertebrates are men — Not implied.

Thus, only conclusion 3 follows.

So Option C is correct

Test: Syllogism - 2 - Question 3

Statements: All the phones are scales. All the scales are calculators.

Conclusions:

  1. All the calculators are scales.
  2. All the phones are calculators
  3. All the scales are phones.
  4. Some calculators are phones.

Detailed Solution for Test: Syllogism - 2 - Question 3


 

From the statements:

  • All phones are scales (P ⊆ S).

  • All scales are calculators (S ⊆ C).
    Therefore P ⊆ S ⊆ C, so P ⊆ C and S ⊆ C.

Evaluate conclusions:

  1. All calculators are scales (C ⊆ S): Not implied (we only have S ⊆ C, not the converse).

  2. All phones are calculators (P ⊆ C): True (by transitivity).

  3. All scales are phones (S ⊆ P): Not implied.

  4. Some calculators are phones: True (since all phones are calculators, assuming at least one phone exists, those are calculators).

Hence only (2) and (4) follow.

Answer: C.

So Option C is correct

Test: Syllogism - 2 - Question 4

Statements: Some pens are books. Some books are pencils.

Conclusions:

  1. Some pens are pencils.
  2. Some pencils are pens.
  3. All pencils are pens.
  4. All books are pens.
Detailed Solution for Test: Syllogism - 2 - Question 4


 

Given:

  • Some pens are books. (P ∩ B ≠ ∅)

  • Some books are pencils. (B ∩ C ≠ ∅) [let C = pencils]

Check conclusions:

  1. Some pens are pencils — Not necessarily. From “some P are B” and “some B are C,” the overlapping parts of B with P and with C may be disjoint. Not guaranteed.

  2. Some pencils are pens — Same as (1); not guaranteed.

  3. All pencils are pens — Not implied.

  4. All books are pens — Not implied.

None of the conclusions necessarily follow.

So Option E is correct

Test: Syllogism - 2 - Question 5

Statements: All the goats are tigers. All the tigers are lions.

Conclusions:

  1. All the goats are lions.
  2. All the lions are goats.
  3. Some lions are goats.
  4. Some tigers are goats.
Detailed Solution for Test: Syllogism - 2 - Question 5


 

Let G = goats, T = tigers, L = lions.

Given:

  • All G are T (G ⊆ T).

  • All T are L (T ⊆ L).
    Therefore G ⊆ T ⊆ L, so G ⊆ L.

Evaluate conclusions:

  1. All goats are lions — True (G ⊆ L).

  2. All lions are goats — Not implied.

  3. Some lions are goats — True (if goats exist, they are lions).

  4. Some tigers are goats — True (since all goats are tigers, assuming goats exist).

Hence, only (1), (3), and (4) follow.

So Option B is correct

Test: Syllogism - 2 - Question 6

Statements: All the books are pencils. No pencil is eraser.

Conclusions:

  1. All the pencils are books.
  2. Some erasers are books.
  3. No book is eraser.
  4. Some books are erasers.
Detailed Solution for Test: Syllogism - 2 - Question 6


 

Let B = books, P = pencils, E = erasers.

Given:

  • All books are pencils: B ⊆ P.

  • No pencil is an eraser: P ∩ E = ∅.

Evaluate conclusions:

  1. All pencils are books (P ⊆ B): Not implied; we only know B ⊆ P.

  2. Some erasers are books: False; since B ⊆ P and P has no overlap with E, books cannot be erasers.

  3. No book is an eraser: True; B ⊆ P and P ∩ E = ∅ ⇒ B ∩ E = ∅.

  4. Some books are erasers: False (same reason as 2).

Hence only conclusion (3) follows.

 

So Option A is correct

Test: Syllogism - 2 - Question 7

Statements: All the research scholars are psychologists. Some psychologists are scientists.

Conclusions:

  1. All the research scholars are scientists.
  2. Some research scholars are scientists.
  3. Some scientists are psychologists.
  4. Some psychologists are research scholars.
Detailed Solution for Test: Syllogism - 2 - Question 7

So Option A is correct

Test: Syllogism - 2 - Question 8

Statements:
All apples are fruits.
All fruits are healthy.

Conclusions:
1. Some apples are healthy.
2. All apples are healthy.
3. Some healthy things are fruits.
4. Some fruits are apples.

Detailed Solution for Test: Syllogism - 2 - Question 8

All apples are fruits, all fruits are healthy ⇒ All apples are healthy ⇒ (1) and (2) follow.
Some healthy things are fruits (true by basic logic) ⇒ (3) follows.
(4) must hold as "all apples are fruits" implies "some fruits are apples" (assuming apples exist).

So the correct answer is Option D. 

Test: Syllogism - 2 - Question 9

Statements:
Some fruits are mangos.
All mangos are guavas.
No guava is a banana.
Conclusions:
I. All guavas are fruits.
II. Some guavas are fruits

Detailed Solution for Test: Syllogism - 2 - Question 9

Some fruits are mangos.
All mangos are guavas.
I + A => I - type of conclusion
"Some fruits are guavas".
Conclusion II is Converse of it.
All mangos are guavas
No guava is a banana.
A + E => E - type of conclusion
"No mango is a banana

So Option B is correct

Test: Syllogism - 2 - Question 10

Statements:
No tree is an animal.
Some animals are humans.

Conclusions:
1. Some trees are humans.
2. No animal is a tree.
3. Some humans are not trees.
4. All animals are humans.

Detailed Solution for Test: Syllogism - 2 - Question 10

Conclusion 1: “Some trees are humans.”

  • We know trees ∩ animals = ∅.

  • We know animals ∩ humans is nonempty, but nothing tells us that humans extend beyond the animal set.

  • It is possible (in one model) that every human is also an animal—then trees (which are disjoint from animals) cannot overlap humans.

  • Or humans could include non-animals, letting trees overlap those—but that’s only a possibility, not a necessity.
    ⇒ Conclusion 1 does not logically follow.

Conclusion 2: “No animal is a tree.”

  • “No tree is an animal” is equivalent (by symmetry) to “No animal is a tree.”

  • This holds in every case.
    ⇒ Conclusion 2 does follow.

Conclusion 3: “Some humans are not trees.”

  • From “Some animals are humans,” pick any one of those animal-humans.

  • Since that picked animal cannot be a tree (trees and animals are disjoint), that human is not a tree.

  • Thus there exists at least one human who is not a tree.
    ⇒ Conclusion 3 does follow.

Conclusion 4: “All animals are humans.”

  • We only know “some animals are humans,” not that every animal is human.
    ⇒ Conclusion 4 does not follow.

Final answer: Conclusions 2 and 3 follow.
Choice (c).

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