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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]  18 Sep 2003, 15:21
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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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[#permalink]  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]  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]  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]  24 Mar 2013, 13:44
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]  24 Mar 2013, 21:00
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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]  14 Aug 2014, 01:25
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Re: At a party, there were five times as many females as males [#permalink]  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.
Re: At a party, there were five times as many females as males   [#permalink] 24 Sep 2014, 06:01
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# At a party, there were five times as many females as males

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