At a picnic there were 3 times as many adults as children : Problem Solving (PS)

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

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 $\frac{6}{24} or \frac{x}{4}$

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

24 Apr 2016, 20:15

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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]

25 Apr 2016, 07:38

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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]

10 Aug 2017, 10:29

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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]

13 Oct 2017, 11:28

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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]

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.

I hope this helps!

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

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]

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]

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]

06 Nov 2018, 17:55

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spence11 wrote:

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?

I just coul not agree more.. Any expert may want to share an opinion?

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

15 Jan 2020, 03:20

spence11 wrote:

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?

I was caught in the same trap as you. Hence, I couldn't figure out the answer. I assumed women and men included children as well.

If any expert could share more trap warnings with us, it would be very helpful.

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

18 Feb 2020, 06:59

Hello Bunuel

I tried setting up equation here as -

A=3C=M+W

W=2M

M+W+C=X

But am stuck in creating relation between C and M here; can C be considered as M or will it be W+M too?

Please help

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

18 Feb 2020, 07:18

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Anurag06 wrote:

Hello Bunuel

I tried setting up equation here as -

A=3C=M+W

W=2M

M+W+C=X

But am stuck in creating relation between C and M here; can C be considered as M or will it be W+M too?

Please help

M + W + C = X
Since C = (M + W)/3 and W = 2M, then M + (2M) + (M + 2M)/3 = X --> M = X/4.

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

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

04 Aug 2023, 21:19

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