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At a certain school, 40 percent of the students play rugby

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At a certain school, 40 percent of the students play rugby  [#permalink]

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New post 20 Sep 2016, 13:49
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Question Stats:

68% (02:27) correct 32% (02:03) wrong based on 207 sessions

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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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Re: At a certain school, 40 percent of the students play rugby  [#permalink]

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New post 21 Sep 2016, 01:21
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GMATPrepNow wrote:
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

Kudos for all correct solutions




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.
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Re: At a certain school, 40 percent of the students play rugby  [#permalink]

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New post 21 Sep 2016, 12:28
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Explanation:

40% of Total Students play Rugby and Chess.

40% of Rugby players only play Rugby ---> 60% of rugby players play rugby and chess ---> 60%R = 40% Total Students

Rugby = 40%/60% x Total Students = 66 2/3% Total Students
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Re: At a certain school, 40 percent of the students play rugby  [#permalink]

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New post 05 Sep 2017, 19:09
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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Re: At a certain school, 40 percent of the students play rugby  [#permalink]

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New post 29 Sep 2017, 20:43
abhimahna wrote:
GMATPrepNow wrote:
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

Kudos for all correct solutions






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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Re: At a certain school, 40 percent of the students play rugby  [#permalink]

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New post 27 Oct 2018, 07:42
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Re: At a certain school, 40 percent of the students play rugby   [#permalink] 27 Oct 2018, 07:42
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