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

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26 Nov 2012, 22:56

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68% (01:01) correct
32% (01:15) wrong based on 307 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?

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

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27 Nov 2012, 00:53

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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

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 \(\frac{6}{24} or \frac{x}{4}\)
_________________

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).

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

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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:

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:

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

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10 Aug 2017, 10:29

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?

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

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13 Oct 2017, 20:39

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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.

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

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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!