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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]

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New post Updated on: 20 Oct 2014, 07:32
3
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Question Stats:

77% (00:57) correct 23% (00:43) wrong based on 332 sessions

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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.

Originally posted by shehreenquayyum on 23 Jul 2006, 15:59.
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]

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New post 23 Jul 2006, 16:23
1
2
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]

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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]

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New post 23 Jul 2006, 21:20
1
1
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]

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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]

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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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Re: The attendees at a certain convention purchased a total of 15,000  [#permalink]

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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]

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New post 28 Jan 2007, 22:57
1
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 book  [#permalink]

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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]

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New post 19 Oct 2014, 19:47
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]

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New post 01 May 2017, 18:24
Hoping someone can help me to understand what I am doing wrong:

For statement 1, I understood that M + F = 4000

But for statement 2, when I read "average" this is the type of algebraic statement I came up with:
For males, where number of total books bought by males I noted as "Bm" and overall total of males M, so equation is Bm/M =3, and similarly for females, Bf/F = 5. I then tried to set up Bm/M + BF/f = 15,000.

How come in the solution we reduce it to very simple terms as being 3M + 5F = 15,000? I'm not sure I'm quite understanding...thanks!
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Re: The attendees at a certain convention purchased a total of 15,000 book  [#permalink]

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New post 01 May 2017, 19:38
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1
infinitemac wrote:
Hoping someone can help me to understand what I am doing wrong:

For statement 1, I understood that M + F = 4000

But for statement 2, when I read "average" this is the type of algebraic statement I came up with:
For males, where number of total books bought by males I noted as "Bm" and overall total of males M, so equation is Bm/M =3, and similarly for females, Bf/F = 5. I then tried to set up Bm/M + BF/f = 15,000.

How come in the solution we reduce it to very simple terms as being 3M + 5F = 15,000? I'm not sure I'm quite understanding...thanks!


What is 15000? The sum of the total number of books purchased.

Average number of books purchased by males = 3
Average number of books purchased by females = 5

Say there are M males and F females.

Average = Number of books purchased by males/Number of males

3 * M = Number of books purchased by males

Average = Number of books purchased by females/Number of females

5*F = Number of books purchased by females

Total number of books purchased = Number of books purchased by males + Number of books purchased by females = 3M + 5F

15000 = 3M + 5F


Note that Bm/M + Bf/F = 15,000 is wrong. You have already established that Bm/M = 3 and Bf/F = 5. These are the averages. When you add these two, you get 3 + 5 = 8.
15000 is the sum of the number of books purchased.
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Re: The attendees at a certain convention purchased a total of 15,000 book  [#permalink]

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New post 02 May 2017, 08:57
VeritasPrepKarishma wrote:
infinitemac wrote:
Hoping someone can help me to understand what I am doing wrong:

For statement 1, I understood that M + F = 4000

But for statement 2, when I read "average" this is the type of algebraic statement I came up with:
For males, where number of total books bought by males I noted as "Bm" and overall total of males M, so equation is Bm/M =3, and similarly for females, Bf/F = 5. I then tried to set up Bm/M + BF/f = 15,000.

How come in the solution we reduce it to very simple terms as being 3M + 5F = 15,000? I'm not sure I'm quite understanding...thanks!


What is 15000? The sum of the total number of books purchased.

Average number of books purchased by males = 3
Average number of books purchased by females = 5

Say there are M males and F females.

Average = Number of books purchased by males/Number of males

3 * M = Number of books purchased by males

Average = Number of books purchased by females/Number of females

5*F = Number of books purchased by females

Total number of books purchased = Number of books purchased by males + Number of books purchased by females = 3M + 5F

15000 = 3M + 5F


Note that Bm/M + Bf/F = 15,000 is wrong. You have already established that Bm/M = 3 and Bf/F = 5. These are the averages. When you add these two, you get 3 + 5 = 8.
15000 is the sum of the number of books purchased.



Thank you so much Karishma - seems like I missed out on a simple concept but you have explained it very clearly! Thanks!
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Re: The attendees at a certain convention purchased a total of 15,000 book  [#permalink]

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New post 24 Jun 2017, 19:06
shehreenquayyum wrote:
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 goal of the problem is to find F, the number of female attendees at the convention.

Statement 1) F+M = 4000.

Insufficient. We don't know the breakdown or distribution of males or females.

Statement 2) Books bought by males / M = 3.

Books bought by females / F = 5.

5F + 3M = books bought by males and books bought by females = 15000

Insufficient because we don't know the total number of males or females.

Statements 1+2) Sufficient.

3M + 5F = 15000

M + F = 4000
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Re: The attendees at a certain convention purchased a total of 15,000 book  [#permalink]

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New post 26 Jun 2017, 13:08
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.

Total attendees \(= 4,000\)

This does not give us any information on the number of female attendees.

Hence, (1) =====> is NOT SUFFICIENT

(2) The male attendees purchased an average (arithmetic mean) of 3 books each, and the female attendees purchased an average of 5 books each

This gives us average of Male and Female purchase of books

\(3m + 5f = 15000\)

As we are not aware of total count of the participants

Hence, (2) =====> is NOT SUFFICIENT.

Now lets combine (1) & (2)

\(m + f = 4,000\)

\(3m + 5f = 15,000\)

As you can see that with the help of these two equations we can find the total number of female attendees.

Please note that you need not do the actual calculation, only the information that we can arrive at a number should be sufficient for us.

Hence, Answer is C
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Re: The attendees at a certain convention purchased a total of 15,000  [#permalink]

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New post 09 Jul 2017, 07:58
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Ans = C
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Re: The attendees at a certain convention purchased a total of 15,000 book  [#permalink]

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