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by mukgera » Tue Sep 06, 2011 6:28 am
A certain one-day seminar consisted of a morning session and an afternoon session. If each of the 128 people attending the seminar attended at least one of the two sessions, how many of the people attended the morning session only?

1) 3/4 of the people attended both sessions
2) 7/8 of the people attended the afternoon sessions.

OA after some discussion.
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Source: — Data Sufficiency |

by gmatclubmember » Tue Sep 06, 2011 6:54 am
mukgera wrote:A certain one-day seminar consisted of a morning session and an afternoon session. If each of the 128 people attending the seminar attended at least one of the two sessions, how many of the people attended the morning session only?

1) 3/4 of the people attended both sessions
2) 7/8 of the people attended the afternoon sessions.

OA after some discussion.
-----------
Let x+a attended morning session and y+a attended afternoon sessions. 'a' being the participants who attended both sessions. x+y+a = 128.

1. a = 3/4*128 = 96. But this doesnt gives us x and y.
2. y+a = 7/8*128=102. But this doesnt gives us x or a.
if we use both conditions we get a=96,y=6 and x = 128-96-6=26.
So if we use both statements we can get the answer.

C
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by mukgera » Tue Sep 06, 2011 8:58 am
Hi GmatClubMember,
There is small mistake in the below mentioned solution...
The question asked ..."how many of the people attended the morning session only?" So we need to find x only(as per your approach)

1) is not enough that is very much right.
2) you have y+a = 102...now x + y + a = 128 so x = 128 - 102 = 26.

So B is sufficient enough.

and the OA is also B.

Hope that helps!!

gmatclubmember wrote:
mukgera wrote:A certain one-day seminar consisted of a morning session and an afternoon session. If each of the 128 people attending the seminar attended at least one of the two sessions, how many of the people attended the morning session only?

1) 3/4 of the people attended both sessions
2) 7/8 of the people attended the afternoon sessions.

OA after some discussion.
-----------
Let x+a attended morning session and y+a attended afternoon sessions. 'a' being the participants who attended both sessions. x+y+a = 128.

1. a = 3/4*128 = 96. But this doesnt gives us x and y.
2. y+a = 7/8*128=102. But this doesnt gives us x or a.
if we use both conditions we get a=96,y=6 and x = 128-96-6=26.
So if we use both statements we can get the answer.

C
Join the discussion