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

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vasurajiv
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At a party, there were five times as many females as males [#permalink] At a party, there were five times as many females as males   18 Sep 2003, 16:21
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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
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B
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Bunuel
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Re: At a party, there were five times as many females as males [#permalink] At a party, there were five times as many females as males   24 Mar 2013, 22:00
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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] At a party, there were five times as many females as males   24 Mar 2013, 14:44
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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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At a party, there were five times as many females as males [#permalink] At a party, there were five times as many females as males   11 Oct 2015, 11:55
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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:
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B


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Re: At a party, there were five times as many females as males. There were [#permalink] At a party, there were five times as many females as males   10 Jan 2016, 08:57
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Bunuel wrote:
At a party, there were five times as many females as males. There were 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



A good QUestion.
we have to work on the info and understand what that info takes us to...

1) there were five times as many females as males..
this means F=5M.. so total number=6M..

2) There were three times as many adults as children..
A=3C...
or total = 4C..

so the total is 4C or 6M..
since C and M have to be positive integers, we can say for sure that Total is a multiple of 4 and 6..
or Total is multiple of LCM of 4,6..
LCM of 4,6= 12..
lets see which number is not multiple of 12..

A. 384...12*32
B. 258... no
C. 216....12*18
D. 120....12*10
E. 72......12*6..

so B is our answer..
Good Q..
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Re: At a party, there were five times as many females as males [#permalink] At a party, there were five times as many females as males   18 Oct 2016, 09:59
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] At a party, there were five times as many females as males   08 Dec 2018, 15:27
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.


Sure) Children are also people )
So the answer has to be (4x, 6y) the multiple of 6 and of 4.

B is not the multiple of 4.
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Re: At a party, there were 5 times as many females as males, and [#permalink] At a party, there were five times as many females as males   14 Aug 2022, 04:47
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Re: At a party, there were 5 times as many females as males, and [#permalink]
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