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The attendees at a certain convention purchased a total of 15,000 book

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The attendees at a certain convention purchased a total of 15,000 book [#permalink] New post 23 Jul 2006, 15:59
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88% (02:26) correct 13% (00:24) wrong based on 9 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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Oct 2014, 07:32, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink] New post 23 Jul 2006, 16:23
Form 1. 4000 = m + f
From 2. 15000 = 3m + 5f
using both solve for f

Thanks
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink] New post 23 Jul 2006, 20:42
C it is.

Statement 1....... M + F = 4000,
Statement 2.......3M + 5F = 15000,

Solving => F = 1500.
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink] New post 23 Jul 2006, 21:20
St1:
# of male attendees = m
# of female attendees = 4000-m

But we can't work further. Insufficient.

St2:
3m + 5f = 15,000

Can't solve as there are many possibilites for (m,f) sets

Using St1 and St2:
3m + 5(4000-m) = 15,000

Can solve for m, and thus f. Sufficient.

ANS c
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink] New post 23 Jul 2006, 23:46
shehreenquayyum wrote:
Attendees at a certain convention purchased 15000 boks. How many of these attendees are females?

i) Total attendees are 4000
ii) Males purchased an average of 3 books each & females purchased an average of 5 books each.

I think we have to use the weighted average? Answer is C....


Straight C

from 1: m+f =4000
from 2: 3m+5f=15000

2 equations and 2 unknowns, 1 and 2 are insuff. but combined can give the answer.
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink] New post 25 Jul 2006, 03:40
C

1) M + F = 4000
Not Suff

2) 3M +5F = 15000
Not Suff

Together

3(4000-F) + 5F = 15000
12000 -3F +5F = 15000
2F = 3000
F = 1500
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The attendees at a certain convention purchased a total of 15,000 [#permalink] New post 27 Jan 2007, 23:44
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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.
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Re: The attendees at a certain convention purchased a total of 15,000 [#permalink] New post 28 Jan 2007, 02:01
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? (even though we don't have to solve in DS). Thanks!

ONE IS OBVIOUSLY NOT SUFF

FROM TWO .......NOT SUFF

BOTH TOGETHER

TOTAL NUMBER OF BOOKS MEN PURCHASED/ TOTAL NMBER OF MEN = 3

X/Y =3

TOTAL NMBE OF BOOKS PURCHASED BY WOMEN/ NUMBER OF WOMEN =5

15000 - X / 4000- Y = 5

WE HAVE TWO EQUATIONS FOR TWO UNKNOWNS

3Y =X

15000 - 3Y = 20000 - 5Y

2Y = 5000 IE: Y = 2500 THUS ............SUFF

THE ANSWER IS c
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Re: The attendees at a certain convention purchased a total of 15,000 [#permalink] New post 28 Jan 2007, 22:57
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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))) :wink: (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
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Re: The attendees at a certain convention purchased a total of 15,000 [#permalink] New post 28 Sep 2014, 09:57
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink] New post 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.

Where is my logic going astray?
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink] New post 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.

Similarly, there are many other cases possible.

This question can be done by weighted averages concept (discussed here: http://www.veritasprep.com/blog/2011/03 ... -averages/) in seconds.

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.

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