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At a picnic there were 3 times as many adults as children [#permalink]
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Updated on: 24 Apr 2016, 22:31
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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
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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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27 Nov 2012, 00:53
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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27 Nov 2012, 02:38



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Re: At a picnic there were 3 times as many adults as children [#permalink]
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29 May 2014, 06:56
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! J



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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:
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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25 Apr 2016, 07:38



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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?



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Re: At a picnic there were 3 times as many adults as children [#permalink]
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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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13 Oct 2017, 20:39
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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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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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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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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