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BrentGMATPrepNow
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Solved this in another way, though I must confess I don't know whether that's correct or not, but let me present it:

Imo, this is a conditional probability problem.
Let R be "people that play rugby" and C, "people that play chess".

We know that:

1) P(C and R) = 0,4
2) P(not C | R) = 0,4

Looking at 2), we know that P(not C | R) = 1 - P(C|R), thus: P(C|R) = 0,6

Now, we know that P(C|R) = P(C and R) / P(R):

0,6 = 0,4 / P(R) -> P(R) = 0,4/0,6 = 2/3 (B)
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At a certain school, 40 percent of the students play both rugby and play chess. If 40 percent of the students who play rugby do not play chess, what percent of the students play rugby?

A) 60
B) 66 2/3
C) 72
D) 75
E) 80

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Let total be 100.

A : Only Rugby
B : Both Rugby and Chess
C : Only Chess

A+B+C =100

Also, B = 40 (Given)

We are given 40% of (A+B) = A

or 2 (A+40) = 3A

or A = 80/3

Now, We need to find (A+B)/Total *100 = 66 2/3. Hence, B is the answer.



Aren't we forgetting people who are not going to play both chess and rugby.
without mentioning the fact that all the student will play one of the game, this question can't be solved.
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There are 2 possibilities for students who play Rugby:

(1) play rugby AND play chess

(2) play rugby, do NOT play chess


100% of rugby players = (1) + (2)

Call the No. of Rugby players R

R = (1) + (2)

Assume there are 100 total students


“40% of the students play both.

(1) = 40 students

“40% of the students play rugby but not chess”

(2) = 40%(R)


R = 40 + 40%(R)

60%(R) = 40

R = 400/6 = 200/3

As a percentage of 100 students, (200/3) is:

66 and 2/3 %

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