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# Guests at a recent party ate a total of fifteen hamburgers.

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Manager
Joined: 04 Jan 2008
Posts: 116
Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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16 Sep 2008, 07:26
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Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?

(1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians.

(2) 30% of the guests were vegetarian non-students.

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VP
Joined: 21 Jul 2006
Posts: 1457
Re: Zumit DS 027 [#permalink]

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16 Sep 2008, 09:29
dancinggeometry wrote:
Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?

(1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians.

(2) 30% of the guests were vegetarian non-students.

I think the answer should be A. I'm not too sure, but will give it a try:

Denote the following:

s = students
v = vegetarians
b = both
n = none
t = total

our equation here is $$t = s+v+n-b$$, we already know from the question that none is 15 because only none ate hamburgers and each ate exactly 1, and the total hamburgers eaten were 15. So none is 15.

Also, half of the guests were vegetarians, so:$$v = (1/2)t$$

so, the other half of the guests aren't vegetarians, so: $$s-b= (1/2)t$$

our established equation is now: $$t=s+v+15-b$$ ----> question is asking us to look for $$t$$

(1) So $$2$$ vegetarians students and $$3$$ non-student vegetarians, making a total of $$5$$ vegetarians. Since this is half of the non-vegetarians, which means student minus both, therefore $$s-b=10$$. So look again at our established equation, we already have the value of $$s-b$$ and we already know that $$v=(1/2)t$$. Plug in what we got so far into the equation:

$$t= 10 +(1/2)t+15$$, therefore: $$(1/2)t=25$$, finally: $$t=50$$ ----->Suff.

(2) $$(3/10)t = v-b$$

Our established equation will become $$t = s+(3/10)t+15$$ or $$(7/10)t = s+15$$------> 1 equation with 2 variables...Not. Suff.

What's the OA?

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Re: Zumit DS 027   [#permalink] 16 Sep 2008, 09:29
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# Guests at a recent party ate a total of fifteen hamburgers.

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