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M01-15

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M01-15  [#permalink]

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New post 16 Sep 2014, 00:15
25
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

65% (01:59) correct 35% (01:55) wrong based on 241 sessions

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Re M01-15  [#permalink]

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New post 16 Sep 2014, 00:15
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Official Solution:

At Daifu university, 40% of all students are members of both a chess club and a swim team. If 20% of members of the swim team are not members of the chess club, what percentage of all Daifu students are members of the swim team?

A. 20%
B. 30%
C. 40%
D. 50%
E. 60%


Assume there are total of 100 students. 40 students are members of both clubs. We are told that: "20% of members of the swim team are not members of the chess club", thus if S is a # of members of the swim team then 0.2S is # of members of ONLY the swim team:

\(40+0.2S=S\), so \(S=50\).

Or another way: since "20% of members of the swim team are not members of the chess club" then the rest 80% of members of the swim team (S) ARE members of the chess club, so members of both clubs: \(0.8*S=40\), so \(S=50\).


Answer: D
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Re: M01-15  [#permalink]

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New post 10 Feb 2015, 09:16
I think the answer should be E. The question does not say that ALL swim team members are included in the 40%, but rather that 40% of all students are members of both. If we assume that there are 100 total students, there could be 40 members of the swim team, or 100 members of the swim team. All we know is that 40% of these students are members of both.

I set up a table:

Chess
Yes No Total
Swim Yes .40x .20x .60x
No n/a n/a .40x
Total 1.0x


Sorry if the above table is garbled, but I hope it presents properly. If you add all the members of the swim team (40% who are both Swim and Chess and 20% who are not on the Chess team) then you get 60%.
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Re: M01-15  [#permalink]

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New post 10 Feb 2015, 09:25
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brinckma wrote:
I think the answer should be E. The question does not say that ALL swim team members are included in the 40%, but rather that 40% of all students are members of both. If we assume that there are 100 total students, there could be 40 members of the swim team, or 100 members of the swim team. All we know is that 40% of these students are members of both.

I set up a table:

Chess
Yes No Total
Swim Yes .40x .20x .60x
No n/a n/a .40x
Total 1.0x


Sorry if the above table is garbled, but I hope it presents properly. If you add all the members of the swim team (40% who are both Swim and Chess and 20% who are not on the Chess team) then you get 60%.


No, the answer is D, not E. Here is the correct table:
Attachment:
Untitled.png
40 + 0.2S = S --> S = 50.

Hope it helps.
>> !!!

You do not have the required permissions to view the files attached to this post.


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Re: M01-15  [#permalink]

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New post 19 Jan 2016, 04:59
EMPOWERgmatRichC

Please, can you help me here? Should not the tic tac toe board works here? I dont know what Im doing wrong...

Regards

Bunuel wrote:
brinckma wrote:
I think the answer should be E. The question does not say that ALL swim team members are included in the 40%, but rather that 40% of all students are members of both. If we assume that there are 100 total students, there could be 40 members of the swim team, or 100 members of the swim team. All we know is that 40% of these students are members of both.

I set up a table:

Chess
Yes No Total
Swim Yes .40x .20x .60x
No n/a n/a .40x
Total 1.0x


Sorry if the above table is garbled, but I hope it presents properly. If you add all the members of the swim team (40% who are both Swim and Chess and 20% who are not on the Chess team) then you get 60%.


No, the answer is D, not E. Here is the correct table:
Attachment:
Untitled.png
40 + 0.2S = S --> S = 50.

Hope it helps.
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M01-15  [#permalink]

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New post 21 Jan 2016, 16:28
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Hi mestrec,

Yes, the tic-tac-toe board will work on this question - you just have to be careful about how you're filling in the grid. Bunuel's solution (directly above your post) shows how to do it.

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Re M01-15  [#permalink]

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New post 03 Sep 2016, 23:28
I think this is a poor-quality question and I agree with explanation. The question should have been 'what percentage of all Daifu students are members of the swim team and not members of chess club?' The answer would then be 50. For the question as given, the answer would be 50+40 = 90%
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Re: M01-15  [#permalink]

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New post 11 May 2017, 16:34
Bunuel wrote:

No, the answer is D, not E. Here is the correct table:
Attachment:
Untitled.png
40 + 0.2S = S --> S = 50.

Hope it helps.


Perfect table. I was trying to make this table when I was trying to do the problem but couldn't think of it. Venn diagram was not helping. Once the table is made the answer is easy to find
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Re: M01-15  [#permalink]

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New post 07 Nov 2017, 20:18
The question does not mention if there are students who are not a part of either clubs. I feel its a poor question.
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Re: M01-15  [#permalink]

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New post 07 Nov 2017, 21:06
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New post 09 Jul 2018, 08:26
Hi,

I solved it the following way, please let me know where I went wrong-

Given - 40% of total students are members of both the clubs.
So, we are left with 60% of students who are members of either of the 2 clubs but not both the clubs

Equation formed - Let's say total X students are there
.4X + .2(.6X) = S (total swim club members)

I have assumed - all students are members of either or both clubs, so for each club - students in both clubs + students only in that club will give us the exact no. of students for that club.
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Re: M01-15  [#permalink]

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New post 04 Oct 2018, 05:45
Bunuel Please help to understand where I did mistake ? I can get answer of 30%.

Ground work :
----------------------------
Total Students = T
Only Swim Club = S
Only Chess Club = C
Both Swim and Chess Club = B= 40% of T = 0.4 T
Also C= 0.8 S
----------------------------
Equations :
S + C+ B = T
S + 0.8S + 0.4T = T
1.8S = 0.6T
So 1.8/0.6 = 30% i.e. 30 % percentage of all Daifu students are members of the swim team
--------------------
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Re: M01-15  [#permalink]

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New post 24 Oct 2018, 10:33
Bunuel though i got the right answer i still had to assume that all the members are in atleast i cllub

your explanation : Or another way: since "20% of members of the swim team are not members of the chess club" then the rest 80% of members of the swim team (S) ARE members of the chess club, so members of both clubs: 0.8∗S=400.8∗S=40, so S=50S=50.

how did u assume that the rest 80% are member of chess club?

according to me the same statement could have been : the rest 80% are NOT members of swim ... this would include both member of chess and memebers of NEITHER

Though i agree that "NEITHER" part plays no role here ... ( the answer will anyway be same)... but dont you think that the question , to maintain the clarity, would be much standard?
Pardon me if I'm wrong somewhere... Thankyou
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Re: M01-15  [#permalink]

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New post 26 Feb 2019, 05:37
I'm still not clear with this Q because the wording is ambiguous, correct me where I'm wrong.

When it says, 20% members of swim team are not members of Chess club.
How do i know whether this 20% is ONLY swim team or Swim team + common.
Because the common ones are also members of swim team.
I feel in the explanation, we have assumed that it is ONLY swim team.

If not, How do I recognize what's what?
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Re: M01-15  [#permalink]

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New post 08 Mar 2019, 03:35
This question is accurate!
Let's say the number of all students at Daifu university is X, and the number of students, who are members of Swim team, is S.
We are told that 40% of all students at the university are members of both a chess club and a swim team. i.e 0.4X and 20% of members of the swim team are not members of the chess club( i.e are members of Swim team only) i.e 0.2S
therefore S = 0.4X + 0.2S
0.8S =0.4X
S =0.5X as shown, it is 50% of all students at Daifu university
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Re: M01-15   [#permalink] 08 Mar 2019, 03:35
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