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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]

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04 Jun 2009, 21:11
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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 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]

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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]

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17 Jul 2009, 19:34
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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]

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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]

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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]

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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 $$s_{max}=133$$.

$$\frac{assistants}{students}>\frac{3}{80}$$ 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]

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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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12 Mar 2012, 22:45
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, 23:50, edited 1 time in total.

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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: At a certain university, the ratio of the number of teaching [#permalink]

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26 Jan 2016, 10:18
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!

same for me
Please help. The phrase "must always be greater than" states it has to be 134 & not 133
what is the catch here?

I thought I was good at ratios!

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Re: At a certain university, the ratio of the number of teaching [#permalink]

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26 Jan 2016, 14:53
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The question states that the ratio must always be greater than 3:80, not the number of students (or burgers). So when you calculate the ratio $$\frac{5}{x}>\frac{3}{80}$$, increasing the value of $$x$$ will decrease the ratio $$\frac{5}{x}$$, and decreasing the value of $$x$$ will increase the ratio $$\frac{5}{x}$$.

If you calculate the number of burgers to be 133.3, then decide whether to round up or down, understand what will happen to the ratio of $$\frac{5}{x}$$.

If $$\frac{5}{133.33}=\frac{3}{80}$$, and that is the minimum (because $$\frac{5}{x}$$ must always be greater than $$\frac{3}{80}$$), what happens if you round $$x$$ up to 134? Is $$\frac{5}{134}$$ > or < $$\frac{3}{80}$$?

As explained above, if you increase $$x$$ to 134, then the ratio $$\frac{5}{x}$$ is decreased, and it will be less than the minimum of $$\frac{3}{80}$$. If you round $$x$$ down to 133, then the ratio $$\frac{5}{x}$$ will increase, and you will not violate the condition that it must always be greater than $$\frac{3}{80}$$.

Looking at it another way, if we know that the ratio of assistants to students must always be greater than 3:80, then we know that for any given number of assistants, there is a maximum number of students allowed. For every assistant, a maximum of 26.66 students are allowed (80/3). So if there is 1 assistant and 27 students, that is too many. 26 is the maximum number of students allowed if there is only 1 assistant in order to keep the ratio greater than 3:80. Using the same logic, if there are 5 assistants, then the maximum number of students allowed is 133.33. If there were 134 students that would be more than the maximum, therefore the maximum number of students allowed is 133.

Does that help?

Cheers
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Re: At a certain university, the ratio of the number of teaching [#permalink]

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27 Jan 2016, 06:29
Yes, davedekoos
understand what will happen to the ratio that solved it 50%

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Re: At a certain university, the ratio of the number of teaching [#permalink]

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27 Jan 2016, 06:59
scorpio7 wrote:
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

$$\frac{Assistant}{Student} > \frac{3}{80}$$

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

$$Student < \frac{(5*80)}{3}$$

$$Student < \frac{(400)}{3}$$

i.e. $$Student < 133.33$$

i.e. Maximum value of No. of students = 133

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At a certain university, the ratio of the number of teaching [#permalink]

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25 Feb 2016, 12:56
Brute Force Method:

$$\frac{3}{80}$$ As we are looking to a similar ratio for 5 assistants instead of 3, convert the both numerator (3) and denominator (80) to multiple of 5 by multiplying with 5

$$\frac{3*5}{80*5}$$equivalent to

$$\frac{15}{400}$$

Now as we need ratio for 05 assistants; again divide both numerator and denominator with 3. Pay attention to denominator which we need to answer:

$$\frac{5}{133.33}$$ (Post division of both numerator and denominator with 03)

DONE

Maximum students could be 133 because if students 134 ratio would be less.

Hope it helps!!!!

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Re: At a certain university, the ratio of the number of teaching [#permalink]

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