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matrikabp
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Post subject: Word Translation:Chapter 7: Q No 8 Posted: Mon Aug 15, 2011 11:26 am |
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There are 26 students who have read a total of 56 books among them. The only books they have read, though, are Aye, Bee, Cod, and Dee. If 10 students have only read Aye, and 8 students have read only Cod and Dee, what is the smallest number of books any of the remaining students could have read?
I was Convinced with the following explanation on the book: 2 books
Students|Books ---------------------------------- 10 | 10 8 | 16 ---------------------------------- 18 | 26
Remaining (26-18)= 8 Students, (56-26)=30 books Smallest No. of books read by any of the remaining students = 7x4 + 1x2 So, 2 books
But, one of the person in other forums posted following question and I got confused:
"I am not too convinced with the explanation; why can't 1 person read 23 of the 30 remaining books and let the rest 7 read 7 different books, 1 each. In that case; the smallest number of books that any of the remaining students reads is going to be 1. And that 1 is just not read by one person; but 7. There are no mandates that a student must read different books. Am I misinterpreting something?
_________________ ~fluke "
Any Manhattan Staffs over there please clarify!
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mithunsam
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Post subject: Re: Word Translation:Chapter 7: Q No 8 Posted: Tue Aug 16, 2011 7:33 pm |
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Posts: 76
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Question states that students read only 4 books. That means, there are only 4 books available - Aye, Bee, Cod, and Dee.
So, how can someone read 23 books? Maximum number of books that can be read by one person is 4 (Aye, Bee, Cod, and Dee)
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matrikabp
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Post subject: Re: Word Translation:Chapter 7: Q No 8 Posted: Fri Aug 19, 2011 3:40 am |
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jnelson0612
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Post subject: Re: Word Translation:Chapter 7: Q No 8 Posted: Sat Sep 17, 2011 10:03 pm |
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Posts: 1857
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Thanks!
_________________ Jamie Nelson ManhattanGMAT Instructor
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