Study the following information carefully and answer the question given below

Eight persons from A to H has different degree i.e., BBA, MBA and MCA but not necessarily in the same order. At least two but not more than three persons have the same degree. Consecutive alphabetically named persons do not have the same degree.

D does not have an MBA degree. Only C and F have the same degree but not a BBA degree. Both A and H have the same degree but not as D have. E does not have an MBA degree. B does not have a BBA degree.

Which of the following degree does G have?

Correct Answer: MCA

Step-by-Step Explanation:

Let's systematically analyze the given information to determine the degree of each person from A to H.

1. Basic Rules and Constraints:

- There are 8 persons: A, B, C, D, E, F, G, H.

- There are 3 degrees: BBA, MBA, MCA.

- At least 2 and at most 3 persons have the same degree.

- Consecutive alphabetically named persons do NOT have the same degree (e.g., A and B cannot have the same degree, B and C cannot, etc.).

2. Analyzing Specific Clues:

- "Only C and F have the same degree but not a BBA degree."

This is a crucial clue. It means exactly two persons have this particular degree, and those two persons are strictly C and F. Since it is not BBA, this degree must be either MBA or MCA.

- "D does not have an MBA degree." and "E does not have an MBA degree."

Neither D nor E can be in MBA.

- "Both A and H have the same degree but not as D have."

A and H share a degree, but D has a different degree.

- "B does not have a BBA degree."

3. Determining the Degree for C and F:

Since 8 persons are divided among 3 degrees with group sizes between 2 and 3, the group size distribution must be 3, 3, and 2.

Since the degree of C and F contains ONLY C and F, the degree of C and F has exactly 2 persons.

Therefore, the other two degrees must each have exactly 3 persons.

Now, let's check which degree belongs to C and F:

Could C and F have MBA? If C and F have MBA, then only C and F have MBA (2 persons). But we know E does not have MBA, D does not have MBA, B does not have BBA (so B must be MBA or MCA). Let's test the possibilities carefully.

Let's test if C and F have MCA:

If MCA = {C, F} (size 2), then BBA and MBA must each have 3 persons.

Since D does not have MBA and C/F are the ONLY ones with MCA, D MUST have BBA.

Similarly, E does not have MBA and C/F are the ONLY ones with MCA, so E MUST have BBA.

So far: MCA = {C, F}, BBA has D and E.

We know A and H have the same degree, but NOT the same as D (which is BBA). Therefore, A and H must have MBA (since MCA is strictly only C and F).

So, MBA has A and H.

Now, B cannot have BBA (given in clue). Since MCA is strictly {C, F}, B MUST have MBA.

So, MBA = {A, B, H} (3 persons - full!).

Now we have BBA needing one more person to reach 3 persons. The remaining person is G.

So, BBA = {D, E, G} (3 persons).

Let's check the consecutive alphabetical constraint for this configuration:

- MCA: C, F

- MBA: A, B, H

- BBA: D, E, G

Notice that A and B are both in MBA! But consecutive alphabetical named persons CANNOT have the same degree. Also D and E are both in BBA! So this configuration violates the rule.

Therefore, C and F MUST have MBA!

- MBA = {C, F} (exactly 2 persons).

- BBA must have 3 persons, and MCA must have 3 persons.

Let's place the remaining persons into BBA and MCA:

- D does not have MBA. So D is in BBA or MCA.

- E does not have MBA. So E is in BBA or MCA.

- B does not have BBA, and MBA is strictly {C, F}, so B MUST have MCA!

Thus, MCA has B.

Now, C has MBA, so B (MCA) and C (MBA) are different, which satisfies the consecutive rule.

D and E must be assigned to BBA and MCA such that consecutive letters don't match:

- C is MBA.

- If D is MCA, then D and B are both MCA (non-consecutive, fine). But then E cannot be MCA (as D and E would be consecutive in MCA). So if D is MCA, E must be BBA.

- If D is BBA, then E must be MCA (so D and E are not both BBA).

Let's check A and H:

A and H have the same degree, which cannot be MBA (since MBA is only {C, F}).

So A and H have either BBA or MCA, but not as D have.

Case 1: D has BBA

- Then A and H must have MCA (since they have a different degree from D).

- So MCA has {B, A, H} (3 persons - full!).

- Since B, A, H are in MCA, and C, F are in MBA, the remaining persons D, E, G must be in BBA.

- But if D, E, G are in BBA, then D and E have the same degree (BBA), which violates the consecutive rule (D and E are alphabetical consecutive)! So Case 1 is invalid.

Case 2: D has MCA

- Then A and H must have BBA (since they have a different degree from D).

- So BBA has {A, H}.

- MCA has {B, D}.

- We need 3 persons in BBA and 3 persons in MCA.

- The remaining two persons to assign are E and G.

- Since D has MCA, E CANNOT have MCA (consecutive rule for D and E). Therefore, E MUST have BBA!

- This fills BBA with {A, H, E} (3 persons - full!).

- Therefore, G MUST have MCA to complete MCA with {B, D, G} (3 persons - full!).

4. Verification of the Final Distribution:

- MBA: C, F (2 persons)

- BBA: A, E, H (3 persons)

- MCA: B, D, G (3 persons)

Let's check all conditions:

1. At least 2, at most 3 per degree: MBA=2, BBA=3, MCA=3. (Satisfied)

2. Only C and F have MBA (not BBA): Yes, MBA = {C, F}. (Satisfied)

3. D does not have MBA: D has MCA. (Satisfied)

4. E does not have MBA: E has BBA. (Satisfied)

5. B does not have BBA: B has MCA. (Satisfied)

6. A and H have the same degree, different from D: A and H have BBA, D has MCA. (Satisfied)

7. No consecutive alphabetical pairs have the same degree:

- A (BBA) and B (MCA) - Different

- B (MCA) and C (MBA) - Different

- C (MBA) and D (MCA) - Different

- D (MCA) and E (BBA) - Different

- E (BBA) and F (MBA) - Different

- F (MBA) and G (MCA) - Different

- G (MCA) and H (BBA) - Different

All conditions are perfectly satisfied!

Hence, G has an MCA degree.