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Tricky Math Question - Error in Question (pls do not refer)

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sandhyash
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Tricky Math Question - Error in Question (pls do not refer) [#permalink] New post   Updated on: 13 Feb 2011, 07:39
00:00
A
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C
D
E

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Question Stats:

50% (01:26) correct 50% (01:41) wrong based on 2 sessions
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Q- In the junior basketball league there are 18 teams, 2/3 of them are bad and 1/2 are rich. What can't be the number of teams that are rich and bad?
a)4 b)6 c) 10 d)13 e)15

The Answer is (C)

The given explanation is :
12 teams are bad and 9 are rich.
The no. of teams which are rich and that are bad must be between 9 and (9+12-18=3)

The only answer, which is not in that range is (C)

I did not understand this explanation. Please help me !! Thanks in advance.



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C

Originally posted by sandhyash on 13 Feb 2011, 01:48.
Last edited by sandhyash on 13 Feb 2011, 07:39, edited 1 time in total.
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Re: Tricky Math Question - Error in Question (pls do not refer) [#permalink] New post   13 Feb 2011, 03:32
They can be both bad and rich.

Think of it as a Venn-diagram, left circle bad and right circle rich.

If 10 are both bad and rich, that implies that at least 10 are bad, but only 9 are bad so it cannot be true.
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Re: Tricky Math Question - Error in Question (pls do not refer) [#permalink] New post   13 Feb 2011, 03:58
Expert Reply
Moved to PS subforum.

sandhyash wrote:
Q- In the junior basketball league there are 18 teams, 2/3 of them are bad and 1/2 are rich. What can't be the number of teams that are rich and bad?

a) 4
b) 6
c) 10
d) 13
e) 15

The Answer is (C)

The given explanation is :
12 teams are bad and 9 are rich.
The no. of teams which are rich and that are bad must be between 9 and (9+12-18=3)

The only answer, which is not in that range is (C)

I did not understand this explanation. Please help me !! Thanks in advance.


How can the number of teams that are BOTH rich and bad be more than the teams that are rich (9)? So 9 is the upper limit for the number of teams that are BOTH rich and bad. The lower limit for the the number of teams that are BOTH rich and bad is 3: TOTAL=Bad+Rich-Both+Neither --> 18=12+9-Both+Neither --> Both=3+Neither, so if there are no teams that are neither bad or rich then Both=3.

So there are 3 values among answer choices which can't be the number of teams that are both rich and bad: C, D, and E.

P.S.

!
Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.
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Re: Tricky Math Question - Error in Question (pls do not refer) [#permalink] New post   13 Feb 2011, 06:16
Yes! C,D& E all can't be both bad and rich. And yet the OA states that only C is correct. How is this so?
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Re: Tricky Math Question - Error in Question (pls do not refer) [#permalink] New post   13 Feb 2011, 07:37
thanks for your prompt response. Sorry for posting my question here, I did not know where to post my query on the website.

As for the solution, even this was not making sense to me. I was practising from a coaching class material from an academy. I had a feeling that there must be an error in question but I did not want to be over confident either. I dont wish to confuse more people with this question hence I am deleting this thread.
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Re: Tricky Math Question - Error in Question (pls do not refer) [#permalink] New post   14 Feb 2011, 12:31
sandhyash wrote:
Q- In the junior basketball league there are 18 teams, 2/3 of them are bad and 1/2 are rich. What can't be the number of teams that are rich and bad?
a)4 b)6 c) 10 d)13 e)15

The Answer is (C)

The given explanation is :
12 teams are bad and 9 are rich.
The no. of teams which are rich and that are bad must be between 9 and (9+12-18=3)

The only answer, which is not in that range is (C)

I did not understand this explanation. Please help me !! Thanks in advance.



I Used the manhattan gmat technique for word translation problems for such questions. make a table of rich/not rich /total along columns and

Bad/not bad/total along rows. Is there an easier way to solve these- provided the answer was right?



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Re: Tricky Math Question - Error in Question (pls do not refer) [#permalink]
14 Feb 2011, 12:31
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