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

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?

Bunuel,

That is very clear. Thanks for breaking it down like that as it is more clear in order to solve future problems.

A ratio of Teacher/Student > 3/80

In words;
3 Teachers teach < 80 students
1 teacher teaches < 80/3 students
5 teachers teach < (80/3)*5 students
5 teachers teach < 400/3 students
5 teachers teach < 133.33 students

Students can only be integers
5 teachers teach < 134 students

or a maximum of 133 Students.

Ans: "D"

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

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

\(\frac{Assistant}{Student} > \frac{3}{80}\)

\(\frac{5}{Student} > \frac{3}{80}\)

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

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

i.e. \(Student < 133.33\)

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

Answer: option D

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!!!!

We are given that the ratio of the number of teaching assistants to the number of students in any course must always be greater than 3:80, and we need to determine the maximum number of students possible in a course that has 5 teaching assistants. Let’s use the following formula, in which t = 5 = the number of teaching assistants and s = the number of students.

t/s> 3/80

5/s > 3/80

400 > 3s

400/3 > s

133.33 > s

Since s must be a whole number, the largest possible value of s is 133.

Answer: D

Hello Bunuel,

As per your explanation here, then max should be A: 130 to be greater than 3/80. Is my thinking correct.

Hi Aaron,

Bunuel's explanation is, of course, correct. The answer is D.

Think of it this way:
If you have a ratio \(\frac{a}{b}\), and you want to increase the value of that ratio, there are two ways you can do that. You can either increase the numerator or decrease the denominator. In our case, we want to choose the denominator so that the resulting ratio is less than the given value \(\frac{3}{80}\). To do that we will have to increase the denominator as much as we can so that the ratio \(\frac{assistants}{students}\) is still greater than \(\frac{3}{80}\). But if we increase the denominator too much the ratio of assistants to students will drop below our limit of \(\frac{3}{80}\).

Imagine a simpler scenario where the target ratio wasn't \(\frac{3}{80}\), but instead \(\frac{5}{80}\). In that case the greatest number of students that would be allowed and still maintain the ratio of assistants to students greater than \(\frac{5}{80}\) would be 79. One more student and the ratio would be equal to, not greater than \(\frac{5}{80}\).

In our problem, we want the maximum number of students such that the ratio \(\frac{assistants}{students}>\frac{3}{80}\). You can solve this algebraically, as has been done above to find that students must be less than 133.3. The largest integer less than 133.3 is 133. If we decrease the students down to 130, then the ratio will indeed be larger than the target of \(\frac{3}{80}\), but we still have space to add more students and still maintain the ratio greater than \(\frac{3}{80}\).

I hope it helps!

I got this incorrect only because I think I rounded before I needed to. Can someone let me know if this is the reason why?

I got to the point where 133.33 > x but then I just rounded down and assumed 133 > x, which I think was wrong. By doing that, I flipped it so that x < 133 and then I chose 132 but I really should've phrased it such that x < 133.33 and chose 133. Ugh so dumb.

Yes, you have identified your problem. It should be x < 133.33 and since you need an integer value, x = 133.

Ratio of \(\frac{T}{S} > \frac{3}{80}\)

What this means is that if you have 3 TAs, then you must have less than 80 students. Max number of students you can have then is 79.
If you have 80 students, then you must have more than 3 TAs. Hence, you must have at least 4 TAs in this case.

In terms of 5 TAs, what is this ratio equivalent to?

\(\frac{3}{80} = \frac{5}{S}\)
S = 133.33

Ratio of \(\frac{T}{S} > \frac{5}{133.33}\)

So if you have 5 TAs, then you must have less than 133.33 students.

Hence, the maximum number of students you can have with 5 TAs is 133.

Answer (D)

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