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A Survey of 100 people revealed that 72 of them had eaten at

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A Survey of 100 people revealed that 72 of them had eaten at [#permalink] New post 11 Feb 2004, 16:17
A Survey of 100 people revealed that 72 of them had eaten at restaurant P and that 52 of them had eaten at restaurant Q .Which of the following could not be the number of persons in the surveryed group who had eaten at both P and Q?

a) 20
b)24
c)30
d)50
e)52
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shubhangi

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 [#permalink] New post 11 Feb 2004, 16:25
Is the question right..
Which of the following could not be the number of persons in the surveryed group who had eaten at both P and Q?

If its could be...then its 24.If its could not be ....then Im lost

Vivek.
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 [#permalink] New post 11 Feb 2004, 16:42
Good Question !

Is it A) 20
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 [#permalink] New post 11 Feb 2004, 16:46
Answer is A.

Total = P+Q+neither-Both
Both = P+Q+Neither-Total
72+52+Neither-100 = 24+neither.
Even if Neither=0 Both will be 24. SO minimum no of people eating at botht the restaurants is 24.
A is the choice because A <24
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 [#permalink] New post 11 Feb 2004, 16:48
anandnk wrote:
Answer is A.

Total = P+Q+neither-Both
Both = P+Q+Neither-Total
72+52+Neither-100 = 24+neither.
Even if Neither=0 Both will be 24. SO minimum no of people eating at botht the restaurants is 24.
A is the choice because A <24


I need to learn to reason like you!
good explanation!!
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 [#permalink] New post 11 Feb 2004, 16:49
hi kpadma,

how did you choose A then? I am curious.

Thanks,
Anand.
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 [#permalink] New post 11 Feb 2004, 17:13
Whoopsiedaisy..I interpreted the question in the wrong way..my bad.

Vivek.
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 [#permalink] New post 11 Feb 2004, 19:04
Yes ,it is A. i could not figure it out at first.. did you guys.??
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 [#permalink] New post 14 Feb 2004, 03:18
anandnk wrote:
hi kpadma,

how did you choose A then? I am curious.

Thanks,
Anand.


This is a very easy question to solve logically. There are 5 answers ranging from 20 to 52. By the type of question asked, the answer must be at one end of the range or another (think about this). Well, the number CAN clearly be 52 so the answer must be at the other extreme 20.

Another way to think of it is "what is the minimum number that can be both"? and the anwers can be found using a Venn Diagram technique but it will take a little longer.
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AkamaiBrah
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Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

  [#permalink] 14 Feb 2004, 03:18
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