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All of the 540 people who attended an education convention [#permalink]

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28 Mar 2013, 12:48

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A

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D

E

Difficulty:

35% (medium)

Question Stats:

74% (03:19) correct
26% (02:11) wrong based on 243 sessions

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All of the 540 people who attended an education convention were either teachers or administrators. There were twice as many females as males at the convention and 3 times as many teachers as administrators. If 1/3 of the administrators were males, how many of the females were teachers?

Re: All of the 540 people who attended an education convention [#permalink]

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28 Mar 2013, 13:15

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Expert's post

megafan wrote:

All of the 540 people who attended an education convention were either teachers or administrators. There were twice as many females as males at the convention and 3 times as many teachers as administrators. If 1/3 of the administrators were males, how many of the females were teachers?

(A) 90 (B) 135 (C) 180 (D) 270 (E) 405

Looking for a 30-sec approach.

Total = 540 people.

Twice as many females as males --> (Female) = 360 and (Males) = 180.

3 times as many teachers as administrators --> (Teachers) = 405 and (Administrators) = 135.

1/3 of the administrators were males --> (Male Administrators) = 135*1/3 = 45.

Re: All of the 540 people who attended an education convention [#permalink]

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28 Mar 2013, 23:34

megafan wrote:

All of the 540 people who attended an education convention were either teachers or administrators. There were twice as many females as males at the convention and 3 times as many teachers as administrators. If 1/3 of the administrators were males, how many of the females were teachers?

(A) 90 (B) 135 (C) 180 (D) 270 (E) 405

Looking for a 30-sec approach.

Given:

1. No.of people = 540 2. No of males =x, no of females = 2x 3. No of administrators=y, no of teachers = 3y 4. y/3 = male administrators and 2y/3 = female administrators

Question

No of Female teachers?

= Number of Females - Number of Female administrators

= 2x-2y/3

Deductions

1. x + 2x = 540 or x=180. 2. y + 3y = 540 or y=135.

Re: All of the 540 people who attended an education convention [#permalink]

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29 Mar 2013, 03:36

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johnwesley wrote:

SravnaTestPrep, maybe it's much faster doing the grid, isn't it? The grid helps to avoid doing calculations that you don't need and gain seconds.

Yes johnwesley, the grid method is faster. But if it is possible to adopt a more or less uniform approach to problems, you may gain on the average. _________________

Re: All of the 540 people who attended an education convention [#permalink]

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27 Feb 2014, 17:49

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one way to solve in 30 secs is if you can make the below connection very quickly:

Females comprise 2/3 of conference Females comprise 2/3 of admin, so they must also comprise 2/3 of teachers (so female teachers comprise 2/3*3/4 = 1/2 of conference) _________________

Re: All of the 540 people who attended an education convention [#permalink]

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23 Dec 2015, 14:35

Bunuel wrote:

megafan wrote:

All of the 540 people who attended an education convention were either teachers or administrators. There were twice as many females as males at the convention and 3 times as many teachers as administrators. If 1/3 of the administrators were males, how many of the females were teachers?

(A) 90 (B) 135 (C) 180 (D) 270 (E) 405

Looking for a 30-sec approach.

Total = 540 people.

Twice as many females as males --> (Female) = 360 and (Males) = 180.

3 times as many teachers as administrators --> (Teachers) = 405 and (Administrators) = 135.

1/3 of the administrators were males --> (Male Administrators) = 135*1/3 = 45.

Make a matrix:

Attachment:

Teachers.png

So, (Female Teachers) = 270.

Answer: D.

how did you immediately get 360 and males 180?

gmatclubot

Re: All of the 540 people who attended an education convention
[#permalink]
23 Dec 2015, 14:35

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