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Attendees at a certain convention purchased 15000 boks. How

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Intern
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Attendees at a certain convention purchased 15000 boks. How [#permalink] New post 23 Jul 2006, 15:59
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....
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 [#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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 [#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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 [#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: Average problem [#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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 [#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
  [#permalink] 25 Jul 2006, 03:40
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