Test: Syllogism - 2 - CUET Commerce MCQ

​
 

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

​
 

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

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

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

​
 

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

​
 

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

So Option A is correct

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. 

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

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).