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

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At a certain university, the ratio of the number of teaching [#permalink] New post 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
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Re: problem solving question on ratios [#permalink] New post 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] New post 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] New post 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] New post 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] New post 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 \(s_{max}=133\).

Answer: D.

\(\frac{assistants}{students}>\frac{3}{80}\) 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] New post 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] New post 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

Please help. The phrase "must always be greater than" is completely throwing me off.

[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.
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Re: Chefs to burgers [#permalink] New post 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 :-P
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Last edited by boomtangboy on 12 Mar 2012, 22:55, edited 1 time in total.
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Re: Chefs to burgers [#permalink] New post 12 Mar 2012, 22:49
boomtangboy wrote:
hi,

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Hi BoomTang, great answer! +1
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Re: Chefs to burgers [#permalink] New post 12 Mar 2012, 23:10
budablasta wrote:
boomtangboy wrote:
hi,

Give me a Big Kudoos Meal Combo if this helps :-P


Hi BoomTang, great answer! +1


Happy to Help :-D
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Re: Chefs to burgers [#permalink] New post 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

Please help. The phrase "must always be greater than" is completely throwing me off.

[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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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) ; 12. Tricky questions from previous years.

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: At a certain university, the ratio of the number of teaching [#permalink] New post 05 Apr 2015, 05:25
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