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The attendees at a certain convention purchased a total of 15,000 book [#permalink]
23 Jul 2006, 15:59

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

77% (01:50) correct
23% (00:44) wrong based on 66 sessions

The attendees at a certain convention purchased a total of 15,000 books. How many of these attendees were female?

(1) There was a total of 4,000 attendees at the convention. (2) The male attendees purchased an average (arithmetic mean) of 3 books each, and the female attendees purchased an average of 5 books each.

The attendees at a certain convention purchased a total of 15,000 [#permalink]
27 Jan 2007, 23:44

2

This post was BOOKMARKED

The attendees at a certain convention purchased a total of 15,000 books. How many of these attendees were female?

(1) There was a total of 4,000 attendees at the convention. (2) The male attendees purchased an average (arithmetic mean) of 3 books each, and the female attendees purchased an average of 5 books each.

Last edited by Bunuel on 28 Sep 2014, 10:33, edited 1 time in total.

Renamed the topic, edited the question and added the OA.

Re: The attendees at a certain convention purchased a total of 15,000 [#permalink]
28 Jan 2007, 22:57

1

This post received KUDOS

successstory wrote:

Total attendees at a convention purchased 15000 books. How many of these were female?

1. 4000 attended the convention 2. On average men bought 3 books, females bought 5 books

Answer C:

My solution: 5/8 * 4000 = 2500 women. Is this correct? try to check by back solving females bought 2500*5=12500 books hence 2500 books was bought by men 2500/3=not integer and obviously can't be the number of men))) (even though we don't have to solve in DS). Thanks!

agree with yezz 1st and 2 statements insuff alone
taking both together
M+F=4000 -->M=4000-F
M*3+F*5=15000-->
3*(4000-F)+5F=15000
12000-3F+5F=15000
2F=3000
F=1500
M=2500 _________________

Re: The attendees at a certain convention purchased a total of 15,000 [#permalink]
28 Sep 2014, 09:57

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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink]
19 Oct 2014, 15:50

can someone tell me why this reasoning is wrong.

For statement B - say you take it to mean for every 8 books, 5 are bought for 1 female. So for the 15,000 books 9375 are bought by females. Average out 5 books to one female and you have 1875 females.

Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink]
19 Oct 2014, 19:47

Expert's post

angelfire213 wrote:

can someone tell me why this reasoning is wrong.

For statement B - say you take it to mean for every 8 books, 5 are bought for 1 female. So for the 15,000 books 9375 are bought by females. Average out 5 books to one female and you have 1875 females.

Where is my logic going astray?

You are assuming that number of males = number of females which is not given. You say that one male and one female buy a total of 8 books so total number of pairs = 15000/8 = 1875. So we get that there are 1875 males and 1875 females.

But isn't this possible - there are 3 females who buy 15 books and rest 14985 books are bought by 4995 males? This will give us a total of 15000 books bought such that on average males buy 3 books per person and females buy 5 books per person.

We know w1/w2 = (A2 - Aavg)/(Aavg - A1) We need to find the fraction w1/w2 = Number of males/Number of females and the total number of people to get the number of females.

i) Total attendees are 4000 This gives us Aavg = 15000/4000 We also get that w1 + w2 = 4000. But we don't have A1, A2.

ii) Males purchased an average of 3 books each & females purchased an average of 5 books each. This gives us A1 = 3 and A2 = 5 We don't have Aavg.

Using both, we have Aavg, A1 and A2. So we can find w1/w2 and we also know that total number of people is 4000. This is sufficient to answer the question.

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