A certain one-day seminar consisted of a morning session and

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 after noon session

(1) 3/4 of 128 is 96. So we know that 96 people attended both seminars. If you draw Venn diagram you'll see that 96 is the intersection of the people who attended morning session and the evening session. This only tells that minimum # of people for the morning session is 96. Morning only could be from 0 to 32. Not sufficient.

(2) 7/8 of the people attended the after noon session=112 people. This directly tells that 112 attended after noon session, as each of the 128 people attending the seminar attended at least one of the two sessions hence rest of the people 16, must have been attending the morning session and not attending the evening, which IS only morning session attendant #.

Again on Venn diagram 112 would be: only afternoon + intersection of morning and afternoon, so the third part which is morning only would be: total-(Afternoon and intersection)=128-112=16.

Answer: B.

P.S. You can attach the file directly to the question, so that everyone see it on the page.
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Actually, just the reverse would be correct. Please see the attached picture.



Morning Session Only = Grey Color = M
Both (Morning Session+Afternoon) = Orange Color = M&A
Afternoon Session Only = Yellow Color = A

Q: If M+M&A+A=128
M=?

St1:
M&A=96
Not Sufficient.

St2:
M&A+A=112
Now, we know M&A+A+M=128
M&A+A=112
So, M=16(Morning Session only)
Sufficient.

Ans: "B"

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Now, if the question were: How many people attended morning session, we are asked to find M+M&A

St1:
M&A=96
Not Sufficient.

St2:
M&A+A=112
Now, we can't find "M&A+M" using this information and the stem.
Not Sufficient.

Together:
We can find "A" and eventually "M&A+M".
Sufficient.

Ans: "C"

Dear Fluke
were "only" not there
then B would have been a straight away answer (why would we need St A)

M only = M - both ?

1. Not sufficient

both = (3/4)*128 = 96

Total = M + A - both

M-both = 128- A

we still need to know either M or A to find M only.

2. Sufficient

A = (7/8)*128 = 112

Total = M + A - both

=>M-both = Total - A = 128-112 = 16.

Answer is B.
128 =

Bunuel,
Statement 2 says that 7/8 attended the afternoon session (i.e. 112 ppl). From these, we don't know how many attended afternoon only or morning + afternoon, hence how can we say for sure that 16 attended morning only?

Consider this: we are asked to find how many of the people attended the morning session only. So, we need to find {Morning} - {Both}. Now:

{Total} = {Morning} + {Afternoon} - {Both};
{128} = {Morning} + 112 - {Both};
{Morning} - {Both} = 16.

Does this make sense?
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I tried solving this prob. by using double set matrix. Is that a wrong approach for this problem? I got (C) by doing that.

Hi arkle,

The question asks us to find the number of people who attended the morning session only i.e. who attended the morning session but did not attend the afternoon session.

Even when using the double set matrix you will get option B as the answer. Please refer the diagram of the matrix below for statement-II:



We are given that 7/8 of the people attended the afternoon session i.e. number of people who attended afternoon session irrespective of whether they attended morning session or not is 112. Since the total number of people who attended the session is 128, we get the number of people who did not attend the afternoon session as 128 - 112 = 16.

As everyone attended at least one session there were 0 number of people who did not attend any of the session. That leaves us with 16 number of people who did not attend the afternoon session but attended only the morning session which is what the question asks us .

Hence st-II is sufficient to answer the question.

Hope this helps

Regards
Harsh
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C-trap problem.

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
3/4 x 128 = 96 attended both but we don't know how many attended the morning session but NOT the afternoon session as well.
Insufficient

(2) 7/8 of the people attended the afternoon session
7/8 x 128 = 128
This tells us that the number of people who DID NOT attend the afternoon session is 16.

Sufficient.
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Answer: Option B

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Hi avigutman - my question is on the MECE groups in this question.

I eventually got the table right, which is below



My question is the following however :

a) would you agree - morning and afternoon are MECE groups ?

What I mean is -- I thought the grouping instead was this instead


I know that my 2nd table is NOT RIGHT but my question still stands :

-- Afternoon and Morning can be MECE groups, can they not ?
--

Sure, jabhatta2. If you know that it’s impossible to attend both morning AND afternoon sessions, and if you know that everyone did attend one of them, then by definition they are MECE. If at least one of those conditions is false, then they’re not MECE.

Posted from my mobile device
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Thanks so Much avigutman ! I think you are saying – my 2nd table above is possible.

Attached is an example of my 2nd table with morning/afternoon vs with boys/girls





Regarding the yellow – I think you are referring to morning/afternoon specifically in my 2nd table above

Everyone attends either morning/afternoon
Its impossible to attend BOTH morning/afternoon

Both of these are true in regards to table 2 specifically (where morning /afternoon are on the vertical line)

I ask because in the original question -- you can attend morning and afternoon AT the same time.

Right, jabhatta2, so in the original question morning and afternoon are not MECE.

Posted from my mobile device
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thanks so much avigutman - interesting

afternoon/morning can be MECE / may not be MECE (Depending on the question)

girls/boys I think will always be MECE

That's way too politically charged for me to weigh in on, jabhatta2.
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Hi avigutman - should have clarified.

I meant in the context of the GMAT - girls/boys are always MECE (in the context of the GMAT)

Obviously, based in Canada - that is not the case.

jabhatta2 Any published official GMAC problems that mention boys and girls do require that we assume that those categories are MECE, yes.
However, I recall hearing around 5 years ago that current and future questions on the test will no longer require that assumption.
So I don't think we'll face that issue going forward. They'll probably just avoid using gender as a category for overlapping sets questions.
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