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All of the 540 people who attended an education convention were either

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

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New post 28 Mar 2013, 12:48
4
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A
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Question Stats:

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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?

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

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

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New post 28 Mar 2013, 13:15
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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:

Image

So, (Female Teachers) = 270.

Answer: D.

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

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New post 28 Mar 2013, 17:57
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Maybe this helps to achieve the 30 secs you want:

Just multiply 66% x 75% = 50% (SEE ATTACHED FIGURE)

The solution is 50% of 540 ---> 270

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

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New post 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.

Female teachers = 2x-2y/3 = 360-90=270
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Re: All of the 540 people who attended an education convention were either  [#permalink]

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New post 29 Mar 2013, 03:33
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.

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

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New post 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.

:wink:


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

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New post 27 Feb 2014, 14:04
Hey there,

Yes. Grid works best but I its hard to get it all by 30 seconds unless you're really fast on your calculations.

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

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New post 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)
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Re: All of the 540 people who attended an education convention were either  [#permalink]

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New post 08 Jul 2014, 22:59
Females are twice than that of males & total = 540

So, Males = 180; Females = 360

Prepare a chart as shown in diagram below:

We require to find the value of "x" (Female - Teachers)

There are 3 times as many teachers as administrators

x+y = 3 (180+360-x-y)

4(x+y) = 3 * 540

x+y = 405 ............ (1)

1/3rd of the administrators were males

\(\frac{180-y+360-x}{3} = 180-y\)

x = 2y

\(y = \frac{x}{2}\)

Placing value of y in equation (1)

\(\frac{3x}{2} = 405\)

x = 270

Answer = D
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Re: All of the 540 people who attended an education convention were either  [#permalink]

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New post 09 Dec 2015, 09:53
I think the fastest way would be

A+T=540,
A=3T so T=135 and A=405

Male=A/3, so Male=135 and the remaining 270 will be females.
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Re: All of the 540 people who attended an education convention were either  [#permalink]

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New post 29 Jan 2018, 15:34
Hi All,

While this question is 'step-heavy', it's really just about doing the necessary arithmetic/algebra and taking organized notes.

We're told that there are 540 total people and that there were TWICE as many females as males....

M = number of Males
F = number of Females

M + F = 540

F = 2M now we'll substitute this into the first equation....

M + 2M = 540
3M = 540
M = 180

So there are 180 males and 360 females

We're also told that there are 3 times as many teachers as administrators (the math works exactly the same as what we just did above, just with different variables). You would end up with...

135 administrators and 405 teachers

Finally, we're told that 1/3 of the administrators were males...

135 total administrators
(1/3)(135) = 45 male administrators
135 - 45 = 90 female administrators

Using this information, with the original deductions (from above), we have...

360 total females
90 female administrators
360 - 90 = 270 female teachers

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

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Re: All of the 540 people who attended an education convention were either   [#permalink] 14 Feb 2019, 05:31
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