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At a certain university, the ratio of the number of teaching [#permalink]
 04 Jun 2009, 21:11
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57% (01:44) correct
42% (00:53) wrong based on 80 sessions
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
Last edited by Bunuel on 13 Mar 2012, 00:35, edited 1 time in total.
Edited the question and added the OA
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Re: problem solving question on ratios [#permalink]
05 Jun 2009, 01:46
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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, 19:34
\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]
24 Oct 2009, 00:29
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, 14:36
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, 14:47
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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]
16 Dec 2010, 14:57
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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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, 23:50, edited 1 time in total.
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Re: Chefs to burgers [#permalink]
12 Mar 2012, 23:42
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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, 23:55, edited 1 time in total.
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Re: Chefs to burgers [#permalink]
12 Mar 2012, 23:49
boomtangboy wrote: hi, Give me a Big Kudoos Meal Combo if this helps Hi BoomTang, great answer! +1
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Re: Chefs to burgers [#permalink]
13 Mar 2012, 00:10
budablasta wrote: boomtangboy wrote: hi, Give me a Big Kudoos Meal Combo if this helps Hi BoomTang, great answer! +1 Happy to Help 
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Re: Chefs to burgers [#permalink]
13 Mar 2012, 00:40
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Re: Chefs to burgers
[#permalink]
13 Mar 2012, 00:40
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