All of the 540 people who attended an education convention : GMAT Problem Solving (PS)
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# All of the 540 people who attended an education convention

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

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28 Mar 2013, 11:48
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Difficulty:

35% (medium)

Question Stats:

74% (03:24) correct 26% (02:06) wrong based on 295 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?

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

Looking for a 30-sec approach.
[Reveal] Spoiler: OA

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

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28 Mar 2013, 12:15
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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:
Attachment:

Teachers.png [ 5.89 KiB | Viewed 4773 times ]
So, (Female Teachers) = 270.

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

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28 Mar 2013, 16:57
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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

Attachments

Imagen1.jpg [ 41.78 KiB | Viewed 4610 times ]

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

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28 Mar 2013, 22:34
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

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

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29 Mar 2013, 02:33
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.

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

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29 Mar 2013, 02:36
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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.
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Re: All of the 540 people who attended an education convention [#permalink]

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27 Feb 2014, 13:04
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.

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

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27 Feb 2014, 16:49
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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 [#permalink]

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08 Jul 2014, 21:59
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

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tea.jpg [ 17.76 KiB | Viewed 3179 times ]

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

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09 Dec 2015, 08:53
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 [#permalink]

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23 Dec 2015, 13:35
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.

how did you immediately get 360 and males 180?
Re: All of the 540 people who attended an education convention   [#permalink] 23 Dec 2015, 13:35
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