In a class of 48 students, some are in chess club and some are in swim

Question

In a class of 48 students, some are in chess club and some are in swim class. If one student is selected at random, what is the probability that the student is in chess club?

(1) The probability that the student selected is in chess club or in swim class is 11/12.
(2) The probability that the student selected is in swim class is 5/6.

Bunuel · Accepted Answer

GMAT Club Official Solution:

In a class of 48 students, some are in chess club and some are in swim class. If one student is selected at random, what is the probability that the student is in chess club?

Total = Chess + Swim - Both + Neither
48 = Chess + Swim - Both + Neither
We want to find Chess/48.

(1) The probability that the student selected is in chess club or in swim class is 11/12.

This implies Chess + Swim - Both = 11/12 * 48 = 44. But this does not determine how many are in chess club. Not sufficient.

(2) The probability that the student selected is in swim class is 5/6.

This implies Swim = 5/6 * 48 = 40. But this does not determine how many are in chess club. Not sufficient.

(1)+(2) We have Chess + Swim - Both = 44 and Swim = 40. Thus, Chess + 40 - Both = 44, which gives Chess - Both = 4. Since we still do not know how many are in both swim class and chess club, we cannot determine how many are in chess club. Not sufficient.

Answer: E.

Adit_ · Answer

This is quite interesting!

Statement 1:
Firstly convert the numerator and denominator by multiplying by 4.
Thiis is done so that we can properly compare with the total of 48 students (12*4=48)

This gives us 44/48
Hmm if all students taking part in atleast 1 of them is 44 then 4 students dont take part in any!
So from statement 1 we get None=4
But we dont get anything about no.of students in chess club.
INSUFFICIENT

Statement 2:
Again after conversion we get 40/48
Not Enough 
INSUFFICIENT

Combining both:
Again this tells us that no.of swimmers are 40 and total in any 1 is 44 meaning 
4 take part in ONLY chess club
BUT
We dont know from given info how many could be part of BOTH clubs nor is it mentioned in the stem that students cant be in both the clubs.
Hence INSUFFICIENT.

Answer:  Option E  IMO

Edskore · Answer

Good question — and the existing reply gets the right answer, but let me show the full reasoning because the why is important for exam day.

Key Concept: Overlapping Sets in Data Sufficiency — the hidden free variable

Let P(C) = probability of chess, P(S) = probability of swim, P(C∩S) = probability of both.

The formula you need here: P(C∪S) = P(C) + P(S) − P(C∩S)

Statement (1) alone: P(C∪S) = 11/12
This tells us the probability of being in at least one club is 11/12. But we still have two unknowns — P(C) and P(C∩S) — and only one equation. Insufficient.

Statement (2) alone: P(S) = 5/6
This tells us swim class has 40 out of 48 students. We know nothing about chess club size. Insufficient.

Both statements together:
Substituting into the formula:
11/12 = P(C) + 5/6 − P(C∩S)
11/12 − 10/12 = P(C) − P(C∩S)
1/12 = P(C) − P(C∩S)

We have one equation and two unknowns. P(C) and P(C∩S) can take multiple valid values — for example, P(C) = 1/12 and P(C∩S) = 0, or P(C) = 3/12 and P(C∩S) = 2/12 both satisfy this. Still insufficient.

Answer: E

The common trap: The number 48 feels like it should help you pin things down, but it doesn't — the problem never says chess and swim are mutually exclusive. The moment overlap is possible, you're dealing with two unknowns that one equation can't resolve.

Takeaway: In Data Sufficiency Overlapping Sets questions, always write out the full inclusion-exclusion formula before evaluating sufficiency — if you have more unknowns than equations after combining both statements, the answer is E.

In a class of 48 students, some are in chess club and some are in swim

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GMAT Data Sufficiency (DS) Questions

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