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# At a certain university, the ratio of the number of teaching

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Manager
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At a certain university, the ratio of the number of teaching [#permalink]  04 Jun 2009, 20:11
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At a certain university, the ratio of the number of teaching assistants to the number of students in any course must always be greater than 3:80. At this university , what is the maximum number of students possible in a course that has 5 teaching assistants?

A. 130
B. 131
C. 132
D. 133
E. 134
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Mar 2012, 23:35, edited 1 time in total.
Edited the question and added the OA
Manager
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Re: problem solving question on ratios [#permalink]  05 Jun 2009, 00:46
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Not sure whether this is the best possible way but just the way how I solve it.

Teaching Assistants = TA
Students = S

Let assume the ratio of TA/S = $$3/80$$ (Just putting aside the requirement it must be greater)

Let say x be the maximum no of students possible with 5 teaching assistants = $$3/80 = 5/x$$

$$x = 400/3 = 133.33$$. Now for ratio to be greater than $$3/80$$ reduce the denominator. So just rounded it to lowest integer as number of student can't be in decimal. The new ratio is $$5/133$$, which is less than $$3/80$$ thus, 133 is the maximum number of students possible.
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Re: problem solving question on ratios [#permalink]  17 Jul 2009, 18:34
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$$\frac{TA}{S} > \frac{3}{80}$$

$$\frac{5}{x} > \frac{3}{80}$$

400 > 3x where x has to be maximum.

Substituting the values, if x= 133, 3x=399.

hence,D:)
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Re: problem solving question on ratios [#permalink]  23 Oct 2009, 23:29
TA : S > 3 : 80

5 : S > 3 : 80

S < 400 / 3 = 133 (MAX)
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Re: problem solving question on ratios [#permalink]  16 Dec 2010, 13:36
can someone explain in further detail the relationship between the teaching assistants to the number of students in any course must always be greater than 3:80 and how to reason through this portion? I understand how to solve for x. Once I was at this point I think was stumped on which number to select and inevitably chose to round up. My rational being .33 of a student is not possible therefore it must represent the position of an entire student. Thoughts? Help?
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Re: problem solving question on ratios [#permalink]  16 Dec 2010, 13:47
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spyguy wrote:
can someone explain in further detail the relationship between the teaching assistants to the number of students in any course must always be greater than 3:80 and how to reason through this portion? I understand how to solve for x. Once I was at this point I think was stumped on which number to select and inevitably chose to round up. My rational being .33 of a student is not possible therefore it must represent the position of an entire student. Thoughts? Help?

At a certain university, the ratio of the number of teaching assistants to the number of students in any course must always be greater than 3:80. At this university, what is the maximum number of students possible in a course that has 5 teaching assistants?
A. 130
B. 131
C. 132
D. 133
E. 134

Given: $$\frac{assistants}{students}>\frac{3}{80}$$ --> $$assistants=5$$, so $$\frac{5}{s}>\frac{3}{80}$$ --> $$s_{max}=?$$

$$\frac{5}{s}>\frac{3}{80}$$ --> $$s<\frac{5*80}{3}\approx{133.3}$$ --> so $$s_{max}=133$$.

$$\frac{assistants}{students}>\frac{3}{80}$$ relationship means that if for example # of assistants is 3 then in order $$\frac{assistants}{students}>\frac{3}{80}$$ to be true then # of students must be less than 80 (so there must be less than 80 students per 3 assistants) on the other hand if # of students is for example 80 then the # of assistants must be more than 3 (so there must be more than 3 assistants per 80 students).

Hope it's clear.
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Re: problem solving question on ratios [#permalink]  16 Dec 2010, 13:57
Bunuel,

That is very clear. Thanks for breaking it down like that as it is more clear in order to solve future problems.
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Chefs to burgers [#permalink]  12 Mar 2012, 21:45
At a certain restaurant, the ratio of the number of chefs to the number of burgers on any day must always be greater than 3:80. At this restaurant, what is the maximum number of burgers possible on a day that has 5 chefs.

A) 130
B) 131
C) 132
D) 133
E) 134

[EDIT] The same problem has been solved elsewhere:
problem-solving-question-on-ratios-79240.html

Sorry, I couldn't delete this post!
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Last edited by budablasta on 12 Mar 2012, 22:50, edited 1 time in total.
Senior Manager
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Kudos [?]: 133 [1] , given: 16

Re: Chefs to burgers [#permalink]  12 Mar 2012, 22:42
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hi,

I solved it this way, any suggestions always welcome

c/b > 3/80 ( from question)

5/b > 3 / 80
(80 x 5 / 3) > b

This reduces to

133.3333 > b

So the number of burgers have to be less than 133.33 & as u dont get 0.33 burger in Mc Donalds Max burgers is 133

Give me a Big Kudoos Meal Combo if this helps
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Last edited by boomtangboy on 12 Mar 2012, 22:55, edited 1 time in total.
Manager
Joined: 23 Feb 2012
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Kudos [?]: 40 [0], given: 22

Re: Chefs to burgers [#permalink]  12 Mar 2012, 22:49
boomtangboy wrote:
hi,

Give me a Big Kudoos Meal Combo if this helps

_________________

If you like it, Kudo it!

"There is no alternative to hard work. If you don't do it now, you'll probably have to do it later. If you didn't need it now, you probably did it earlier. But there is no escaping it."

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Senior Manager
Status: May The Force Be With Me (D-DAY 15 May 2012)
Joined: 06 Jan 2012
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Kudos [?]: 133 [0], given: 16

Re: Chefs to burgers [#permalink]  12 Mar 2012, 23:10
budablasta wrote:
boomtangboy wrote:
hi,

Give me a Big Kudoos Meal Combo if this helps

Happy to Help
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Re: Chefs to burgers [#permalink]  12 Mar 2012, 23:40
Expert's post
budablasta wrote:
At a certain restaurant, the ratio of the number of chefs to the number of burgers on any day must always be greater than 3:80. At this restaurant, what is the maximum number of burgers possible on a day that has 5 chefs.

A) 130
B) 131
C) 132
D) 133
E) 134

[EDIT] The same problem has been solved elsewhere:
problem-solving-question-on-ratios-79240.html

Sorry, I couldn't delete this post!

Merging similar topics.
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Re: At a certain university, the ratio of the number of teaching [#permalink]  05 Apr 2015, 05:25
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