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3/8 of all students at Social High are in all three of the
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21 Aug 2010, 06:07
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69% (02:26) correct 31% (02:45) wrong based on 472 sessions
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3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs? (A) 1/8 (B) 1/4 (C) 3/8 (D) 1/2 (E) 5/8
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Re: The Quest for 700: Weekly GMAT Challenge
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21 Aug 2010, 06:33
zisis wrote: 3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?
(A) 1/8 (B) 1/4 (C) 3/8 (D) 1/2 (E) 5/8 Let's # of students at Social High be 8 (I picked 8 as in this case 3/8 of total and 5/8 of total will be an integer). 3/8 of all students at Social High are in all three clubs > 3/8*8=3 people are in exactly 3 clubs; 1/2 of all students are in Albanian club > 1/2*8=4 people are in Albanian club; 5/8 of all students are in Bardic club > 5/8*8=5 people are in Bardic club; 3/4 of all students are in Checkmate club > 3/4*8=6 people are in Checkmate club; Also as every student is in at least one club then # of students in neither of clubs is 0; Total=A+B+C{# of students in exactly 2 clubs}2*{# of students in exactly 3 clubs}+{# of students in neither of clubs}; 8=4+5+6{# of students in exactly 2 clubs}2*3+0 > {# of students in exactly 2 clubs}=1, so fraction is 1/8. Answer: A. For more about the formula used check my post at: formulaefor3overlappingsets69014.html?hilit=exactly%20groups or overlappingsetsproblems87628.html?hilit=exactly%20membersHope it helps.
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Re: The Quest for 700: Weekly GMAT Challenge
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20 Sep 2010, 22:30
If you plot this data on a venn diagram, the problem would be much clearer . General formula for 3 intersecting sets of data: n(A u B u C) = n(A) + n(B) + n(C)  n(A int B)  n(B int C)  n(A int C) + n(A int B int C). Assuming all students were part of atleast 1. My solution is giving me an absurd result and it was definitely not 1/8. Please explain, I might have missed the point.



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Re: The Quest for 700: Weekly GMAT Challenge
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20 Sep 2010, 22:43
vicksikand wrote: If you plot this data on a venn diagram, the problem would be much clearer . General formula for 3 intersecting sets of data: n(A u B u C) = n(A) + n(B) + n(C)  n(A int B)  n(B int C)  n(A int C) + n(A int B int C). Assuming all students were part of atleast 1. My solution is giving me an absurd result and it was definitely not 1/8. Please explain, I might have missed the point. It's not clear how you applied the formula you wrote and what answer did you get, so it's hard to tell where you went wrong. But the problem might be in formula you apply. Actually there are 2 formulas for 3 overlapping sets and the second one is better for this particular problem. Check the solution above and also this link about the mentioned 2 formulas: formulaefor3overlappingsets69014.html?hilit=exactly%20groupsHope it helps.
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Re: The Quest for 700: Weekly GMAT Challenge
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20 Sep 2010, 22:57
1. For 3 sets A, B, and C: P(A u B u C) : P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C) Thats the one I have used all along and has worked fine for me. I would think that this formula would apply to this problem.



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Re: The Quest for 700: Weekly GMAT Challenge
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20 Sep 2010, 23:02
vicksikand wrote: 1. For 3 sets A, B, and C: P(A u B u C) : P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C) Thats the one I have used all along and has worked fine for me. I would think that this formula would apply to this problem. It depends how you apply the formula. Again it's not the only formula for 3overlapping sets, there is a second one and it fits this problem better. You can check the link I gave if you are interested to know the difference and to study how to apply either of them.
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Re: The Quest for 700: Weekly GMAT Challenge
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20 Sep 2010, 23:08
5. To determine the No of persons in two or more sets (at least 2 sets) : P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C)
3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?
It doesnt say 1/2 of students are in Albanian only  In my opinion 1/2 includes the number that are common with Bardic and Checkmate. Formula 2 is for finding no. of persons in one set. I kind of see how the author arrived at that formula, but I wouldnt apply it to our case.
We have to find (P(A n B) + P(A n C) + P(B n C))/2



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Re: The Quest for 700: Weekly GMAT Challenge
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20 Sep 2010, 23:23
vicksikand wrote: 5. To determine the No of persons in two or more sets (at least 2 sets) : P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C)
3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?
It doesnt say 1/2 of students are in Albanian only  In my opinion 1/2 includes the number that are common with Bardic and Checkmate. Formula 2 is for finding no. of persons in one set. I kind of see how the author arrived at that formula, but I wouldnt apply it to our case.
We have to find (P(A n B) + P(A n C) + P(B n C))/2 You are right it doesn't say that 1/2 of students are in Albanian only and nowhere in the solution above it's mentioned. Also this is not the formula I meant. Refer to my post at: formulaefor3overlappingsets69014.html?hilit=exactly%20groupsAnd refer to the solution above to see how the second formula is applied to this problem.
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Re: The Quest for 700: Weekly GMAT Challenge
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21 Sep 2010, 00:07
I agree. Good explanation on the proof. (a n b n c) was bothering me and you did a good job explaining why we need to subtract it twice and let that 1 remain so that it is counted at least once. I remember having seen 1 question based on this in the MGMAT test series. Thanks!



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Re: 3/8 of all students at Social High are in all three of the
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11 Jun 2013, 08:23
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Re: 3/8 of all students at Social High are in all three of the
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30 Oct 2014, 09:58
Why did you multiply {# of students in exactly 3 clubs} by 2?



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Re: 3/8 of all students at Social High are in all three of the
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30 Oct 2014, 10:01
mmcneilly wrote: Why did you multiply {# of students in exactly 3 clubs} by 2? This is explained in details here ADVANCED OVERLAPPING SETS PROBLEMSHope it helps.
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3/8 of all students at Social High are in all three of the
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01 Nov 2014, 22:40
Basic formula is:
Total=A+B+C  exactly two  2*all three, so
Let pick 16 as total number of students
16=8+10+12  exactly two  2*6 16=30x12 x=2
2/16=1/8
with fractions:
8/8=1/2+5/8+3/4  x  3/4 8/8=15/8x3/4 x=1/8
A



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Re: 3/8 of all students at Social High are in all three of the
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05 Jul 2018, 16:45
zisis wrote: 3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?
(A) 1/8 (B) 1/4 (C) 3/8 (D) 1/2 (E) 5/8 We can let the total number of students = 8, and thus: Since 3/8 of all students are in all 3 clubs, 3 students are in all 3 clubs Since 1/2 of all students are in Albanian, 4 students are in Albanian Since 5/8 of all students are in Bardic, 5 students are in Bardic Since 3/4 of all students are in Checkmate, 6 students are in Checkmate. From the given information, we know that 0 students are in none Letting t = the number of students in exactly two clubs, we can now use the equation: Total = A + B + C  (exactly two clubs) 2(all three clubs) + none 8 = 4 + 5 + 6  t  2(3) + 0 8 = 15  t  6 t = 1 So, 1/8 are in exactly two clubs. Answer: A
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Re: 3/8 of all students at Social High are in all three of the
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11 Jul 2018, 22:19
ScottTargetTestPrep wrote: zisis wrote: 3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?
(A) 1/8 (B) 1/4 (C) 3/8 (D) 1/2 (E) 5/8 We can let the total number of students = 8, and thus: Since 3/8 of all students are in all 3 clubs, 3 students are in all 3 clubs Since 1/2 of all students are in Albanian, 4 students are in Albanian Since 5/8 of all students are in Bardic, 5 students are in Bardic Since 3/4 of all students are in Checkmate, 6 students are in Checkmate. From the given information, we know that 0 students are in none Letting t = the number of students in exactly two clubs, we can now use the equation: Total = A + B + C  (exactly two clubs) 2(all three clubs) + none 8 = 4 + 5 + 6  t  2(3) + 0 8 = 15  t  6 t = 1 So, 1/8 are in exactly two clubs. Answer: A



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Re: 3/8 of all students at Social High are in all three of the
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11 Jul 2018, 22:26
Let the total number of students be 96, a multiple of 8. Then P(A)=48, P(B)=60, P(C)=72 and P(An BnC)= 36,APPLYING FORMULA P(AUBUC)= P(A)+P(B)+P(C)P(An b)P(BnC)P(AnC)+ P(AnBnC) We get P(AnB) +P(BnC)+P(AnC)= 120 So, number of students in exactly two clubs will be 120 3*36= 12, which is 1/8th of 96. Answer: A



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Re: 3/8 of all students at Social High are in all three of the
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30 Jan 2019, 13:27
zisis wrote: 3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?
(A) 1/8 (B) 1/4 (C) 3/8 (D) 1/2 (E) 5/8 When they ask "what fraction of some population is category x?" Pick an easy to use number. 8, 4, and 2 are divisible by 64, so let the total number of students = 64. then the number of students in all three groups = 3/8 * 64 = 24 The number of students in Albanian = 32, and the number in Bardic = 40, and the number in checkmake = 48 The formula for 3 set overlaps is Group A + Group B + Group C  (sum of 2 group overlaps)  2*(3 group overlaps)  neither group = total Since every student is in one of the clubs, we don’t have to worry about the neither group So then by the formula 32+40 +48(2overlap)  2* 24=64 Then 120  (2 group overlap) 48 = 64 and 72  2 group overlap  64 so 2 group overlap =8. Since we chose 64 to equal the total student body, then 8/64 = 1/8 of the student body is in exactly 2 clubs.



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Re: 3/8 of all students at Social High are in all three of the
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05 Feb 2019, 12:26
zisis wrote: 3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?
(A) 1/8 (B) 1/4 (C) 3/8 (D) 1/2 (E) 5/8 My reasoning: \(\frac{1}{2} + \frac{5}{8} + \frac{3}{4}  BOTH  2*\frac{3}{8} = 1\) BOTH = \(\frac{1}{8}\)




Re: 3/8 of all students at Social High are in all three of the
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