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At a party, there were five times as many females as males

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At a party, there were five times as many females as males [#permalink]

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At a party, there were five times as many females as males and three times as many adults as children. Which of the following could NOT be the number of people at the party?

A. 384
B. 258
C. 216
D. 120
E. 72
[Reveal] Spoiler: OA

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New post 19 Sep 2003, 04:51
B. the number must be a multiple of both 6 and 4.

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Re: At a party, there were five times as many females as males [#permalink]

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New post 24 Mar 2013, 09:02
We can see that the total number of people at the party must be divisible by 6 since the ratio is 5:1(add 5+1 to get 6).
From the second sentence,the total must be divisible by both 6 and 4,which means that the number would be divisible by 12.
The only number that is not divisible by 12 is 258.

So option b.

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Re: At a party, there were five times as many females as males [#permalink]

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New post 24 Mar 2013, 13:33
rajrj17 wrote:
We can see that the total number of people at the party must be divisible by 6 since the ratio is 5:1(add 5+1 to get 6).
From the second sentence,the total must be divisible by both 6 and 4,which means that the number would be divisible by 12.
The only number that is not divisible by 12 is 258.

So option b.


You forgot to count children in people. The total shouls be divisible by 8. Answer remains the same.

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Re: At a party, there were five times as many females as males [#permalink]

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New post 24 Mar 2013, 13:44
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It's very simple, we are looking for a number that is both multiple of \(6\)(\(F=5M\) so the number must be divisible by \(5+1=6\)) and of \(4\) (\(A=3C\) so \(3+1=4\))

Only B is not multiple of 4

\(\frac{258}{4}=64.5\)
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Re: At a party, there were five times as many females as males [#permalink]

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vasurajiv wrote:
At a party, there were five times as many females as males and three times as many adults as children. Which of the following could NOT be the number of people at the party?

A. 384
B. 258
C. 216
D. 120
E. 72


Five times as many females as males --> F = 5M.
Three times as many adults as children --> (F + M) = 3C.

The number of people at the party = F + M + C = 3C + C = 4C.

The number of people at the party must be a multiple of 4. The only answer choice which is NOT a multiple of 4 is B.

Answer: B.
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Re: At a party, there were five times as many females as males [#permalink]

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New post 14 Aug 2014, 01:25
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Re: At a party, there were five times as many females as males [#permalink]

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New post 24 Sep 2014, 06:01
Bunuel wrote:
vasurajiv wrote:
At a party, there were five times as many females as males and three times as many adults as children. Which of the following could NOT be the number of people at the party?

A. 384
B. 258
C. 216
D. 120
E. 72


Five times as many females as males --> F = 5M.
Three times as many adults as children --> (F + M) = 3C.

The number of people at the party = F + M + C = 3C + C = 4C.

The number of people at the party must be a multiple of 4. The only answer choice which is NOT a multiple of 4 is B.

Answer: B.


Here adults is assumed to be M+F. However, even children can also be categorized into Male/Female?

Adults is a categorization based on AGE so children could be classified as a male or female.

So i find it a bit awkward to classify total no. of people (T) = M + F + Children?

T= M+F (gender basis) and T = A+C (Age basis)

kindly correct me if i am wrong.
Thank you.

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Re: At a party, there were five times as many females as males [#permalink]

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New post 11 Oct 2015, 08:25
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Re: At a party, there were five times as many females as males [#permalink]

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Hi earnit,

While your post in this thread is over a year old, I think that you're still an active member of this Forum, so I'll answer your questions.

Yes, the 'children' in this question would also be categorized as male or female, so if you were to create all of the potential 'categories of people', they would be...

Adult Male, Adult Female, Child Male and Child Female

That level of detail isn't required in this question though. The information in the prompt clues us in to the idea that we're dealing with MULTIPLES....

There were 5 TIMES as many Females as Males. So F = 5M, and the total number of people can be written as M + F = M + 5M = 6M. Thus, the TOTAL must be a multiple of 6.

Also, there were 3 TIMES as many Adults as Children. So A = 3C and the total number of people can also be written as A + C = 3C + C = 4C. Thus, the TOTAL must also be a multiple of 4.

There's only one answer that does NOT fit BOTH of those multiples...

Final Answer:
[Reveal] Spoiler:
B


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Re: At a party, there were five times as many females as males [#permalink]

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New post 11 Oct 2015, 11:35
EMPOWERgmatRichC wrote:
Hi earnit,

While your post in this thread is over a year old, I think that you're still an active member of this Forum, so I'll answer your questions.

Yes, the 'children' in this question would also be categorized as male or female, so if you were to create all of the potential 'categories of people', they would be...

Adult Male, Adult Female, Child Male and Child Female

That level of detail isn't required in this question though. The information in the prompt clues us in to the idea that we're dealing with MULTIPLES....

There were 5 TIMES as many Females as Males. So F = 5M, and the total number of people can be written as M + F = M + 5M = 6M. Thus, the TOTAL must be a multiple of 6.

Also, there were 3 TIMES as many Adults as Children. So A = 3C and the total number of people can also be written as A + C = 3C + C = 4C. Thus, the TOTAL must also be a multiple of 4.

There's only one answer that does NOT fit BOTH of those multiples...

Final Answer:
[Reveal] Spoiler:
B


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Rich



Hi Rich,

I appreciate you taking time out to clarify this part..

As a matter of fact, i just tried this Qs again and used the same approach as mentioned already by you ie. The answer should be a common multiple of BOTH 4 and 6.

However, looking back at the query i posted here, it was in regard to the solution by Bunuel where his answer relies on the fact that the answer should be a multiple of 4 only since he assumed M+F = Adults, and children a separate category.

P.S. Please do check your inbox for i have mailed you a certain query.

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New post 15 Oct 2016, 13:00
Hello from the GMAT Club BumpBot!

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Re: At a party, there were five times as many females as males [#permalink]

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New post 18 Oct 2016, 08:59
We know from the prompt that:

5F = 1M & 3A = 1C

M+F = A --> Thus, we have 6A = 2C

Together there are 8 people, so we need to look for a number that is not a multiple of 8 (i.e. doesn't contain 8 as a factor). B does not, thus it is the correct answer. :)

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Re: At a party, there were five times as many females as males   [#permalink] 18 Oct 2016, 08:59
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