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At a picnic there were 3 times as many adults as children

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At a picnic there were 3 times as many adults as children [#permalink]

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New post Updated on: 24 Apr 2016, 22:31
3
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A
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D
E

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  55% (hard)

Question Stats:

67% (00:57) correct 33% (01:10) wrong based on 472 sessions

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At a picnic there were 3 times as many adults as children and twice as many women as men. If there was a total of x men, women, and children at the picnic, how many men were there, in terms of x?

A. x/2
B. x/3
C. x/4
D. x/5
E. x/6

Originally posted by saxenarahul021 on 26 Nov 2012, 22:56.
Last edited by Bunuel on 24 Apr 2016, 22:31, edited 2 times in total.
Renamed the topic.
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Re: At a picnic there were 3 times as many adults as children [#permalink]

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New post 27 Nov 2012, 00:53
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2
saxenarahul021 wrote:
At a picnic there were 3 times as many adults as children and twice as many women as men. If there was a total of x men, women, and children at the picnic, how many men were there, in terms of x?

A. x/2
B. x/3
C. x/4
D. x/5
E. x/6


Hii.
I am assuming that the question is "what fraction of men were there?".
Answer must be C.
Pick a smart number such as:
\(Adults-->18\)
Since the children are three times as many as children, hence \(Children-->6\)
Further, the adults comprise of Men and women.
The given ratio of \(\frac{#women}{#men} = \frac{2}{1}\),
therefore \(#women=12, # men=6\).
Since the total is Adults+children, i.e. 24.
Therefore fraction of men will be \(\frac{6}{24} or \frac{x}{4}\)
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Re: At a picnic there were 3 times as many adults as children [#permalink]

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New post 27 Nov 2012, 02:38
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saxenarahul021 wrote:
At a picnic there were 3 times as many adults as children and twice as many women as men. If there was a total of x men, women, and children at the picnic, how many men were there, in terms of x?

A. x/2
B. x/3
C. x/4
D. x/5
E. x/6


Say there are 2 women and 1 man (twice as many women as men).

Thus there are 2+1=3 adults and 1 child (3 times as many adults as children) --> x=3+1=4 (total # of people).

Hence, the number of men in terms of x is x/4.

Answer: C.
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Re: At a picnic there were 3 times as many adults as children [#permalink]

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New post 29 May 2014, 06:56
1
2
A=M+W=3C
W=2M

Then M=C

M+W+C=X

Then 2M+W=X
4M=X

M=X/4

Answer: C

Hope this clarifies
Cheers!
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Re: At a picnic there were 3 times as many adults as children [#permalink]

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New post 24 Apr 2016, 20:15
Bunuel, I understand your solution, but still am a little confused. The problem says three times as many adults as kids and two times as many women as men, so I had the following:

Adults: 18
Kids: 6

Women: 16 (2/3 * 24 above)
Men: 8 (1/3 * 24 above)

so: Men / (Men + Women + Kids) --> 8/30

How should I know Men and Women don't include the kids, because they 'boys vs. girls'?

Thanks! I appreciate all you do the forum, you're super helpful.

Michael
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Re: At a picnic there were 3 times as many adults as children [#permalink]

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New post 25 Apr 2016, 07:38
2
mdacosta wrote:
Bunuel, I understand your solution, but still am a little confused. The problem says three times as many adults as kids and two times as many women as men, so I had the following:

Adults: 18
Kids: 6

Women: 16 (2/3 * 24 above)
Men: 8 (1/3 * 24 above)

so: Men / (Men + Women + Kids) --> 8/30

How should I know Men and Women don't include the kids, because they 'boys vs. girls'?

Thanks! I appreciate all you do the forum, you're super helpful.

Michael


Women/men refers to adults. If it were otherwise it would be female/male instead.
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Re: At a picnic there were 3 times as many adults as children [#permalink]

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New post 10 Aug 2017, 10:29
1
This question is brutal. I don't know how we should know to assume that the terms "men" and "women" are excluding children. This actually brings me to an important question: what are common verbal traps set in the GMAT quant section similar to this? Is there a reasonable list of indicators that a question may be trying to mislead you verbally?
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Re: At a picnic there were 3 times as many adults as children [#permalink]

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New post 13 Oct 2017, 11:28
I am getting x/6. Can someone explain where I am going wrong?

Given:
3A=C and 2W=M (a=adults, C= children, w=women and m=men)
x=m+w+c

therefore,

A=w+m
3(w+m)=c
3(m+m/2)=c
9m=2c ------eq 1

x=m+w+c ------multiply both side with 2

2x=2m+2w+2c

substitute eq 1

2x=2m+m+9m (2w=m given)
2x=12m
m=x/6

help!
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Re: At a picnic there were 3 times as many adults as children [#permalink]

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New post 13 Oct 2017, 20:39
2
pra1785 wrote:
I am getting x/6. Can someone explain where I am going wrong?

Given:
3A=C and 2W=M (a=adults, C= children, w=women and m=men)
x=m+w+c

therefore,

A=w+m
3(w+m)=c
3(m+m/2)=c
9m=2c ------eq 1

x=m+w+c ------multiply both side with 2

2x=2m+2w+2c

substitute eq 1

2x=2m+m+9m (2w=m given)
2x=12m
m=x/6

help!


Hi Pra1987,

I believe you are getting mixed up right in the beginning. A=3C and W=2M, not the other way around (Remember the women are 2x are many men and the children are 3x as many adults). When you write 3A=C I think you might be looking at it like 3 adults for every child. That wording is ok, but the math is not. If you are thinking ratios, you need to first say A/C = 3/1 and then solve for one or the other. I solved for A=3/1C or just A=3C.

I hope this helps!
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Re: At a picnic there were 3 times as many adults as children [#permalink]

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New post 15 Oct 2017, 04:10
zflodeen wrote:
pra1785 wrote:
I am getting x/6. Can someone explain where I am going wrong?

Given:
3A=C and 2W=M (a=adults, C= children, w=women and m=men)
x=m+w+c

therefore,

A=w+m
3(w+m)=c
3(m+m/2)=c
9m=2c ------eq 1

x=m+w+c ------multiply both side with 2

2x=2m+2w+2c

substitute eq 1

2x=2m+m+9m (2w=m given)
2x=12m
m=x/6

help!


Hi Pra1987,

I believe you are getting mixed up right in the beginning. A=3C and W=2M, not the other way around (Remember the women are 2x are many men and the children are 3x as many adults). When you write 3A=C I think you might be looking at it like 3 adults for every child. That wording is ok, but the math is not. If you are thinking ratios, you need to first say A/C = 3/1 and then solve for one or the other. I solved for A=3/1C or just A=3C.

I hope this helps!


You're right. I solved it with a=3c and w=2m and I got the right answer. I think the wording of the question confused me. thank you for the clarification!
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Re: At a picnic there were 3 times as many adults as children [#permalink]

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New post 16 Oct 2017, 07:33
Pra1785, I'm happy to help!
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Re: At a picnic there were 3 times as many adults as children [#permalink]

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New post 21 Apr 2018, 09:34
pra1785 wrote:
I am getting x/6. Can someone explain where I am going wrong?

Given:
3A=C and 2W=M (a=adults, C= children, w=women and m=men)
x=m+w+c

therefore,

A=w+m
3(w+m)=c
3(m+m/2)=c
9m=2c ------eq 1

x=m+w+c ------multiply both side with 2

2x=2m+2w+2c

substitute eq 1

2x=2m+m+9m (2w=m given)
2x=12m
m=x/6

help!


let's Adult = A, Children = C, Men = M, Women = W
we find, A=3C, W= 2M
(M+W)/3+M+W=x
from the equation we find, M = x/4
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Re: At a picnic there were 3 times as many adults as children   [#permalink] 21 Apr 2018, 09:34
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