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In college M the average

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sandy
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In college M the average [#permalink] New post   09 May 2015, 11:22
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In college M the average (arithmetic mean) number of students per course is 30 and the ratio of the number of students to the number of faculty is 20 to 1.


Quantity A
Quantity B
The total number of students in College M
600


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

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Re: GRE Math Challenge #103 [#permalink] New post   12 May 2015, 08:40
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sandy wrote:
In college M the average (arithmetic mean) number of students per course is 30 and the ratio of the number of students to the number of faculty is 20 to 1.

Quantity A: The total number of students in College M
Quantity B: 600


Each piece of information is insufficient to determine the total number of students.

The average (arithmetic mean) number of students per course is 30
To determine the number of students, we need to know the number of courses.

The ratio of the number of students to the number of faculty is 20 to 1
To determine the number of students, we need to know the number of faculty.

Consider these two possible cases:

Case #1: There are 3 faculty members, who teach a total of 2 courses. This means there are 60 students in College M, in which case Quantity B is greater.
Case #2: There are 300 faculty members, who teach a total of 200 courses. This means there are 6000 students in College M, in which case Quantity A is greater.

Answer:
Show: ::
D


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Re: GRE Math Challenge #103-In college M the average [#permalink] New post   13 Dec 2016, 12:26
i cant understand this solution, can anybody explain this in easy process please?
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Re: In college M the average [#permalink] New post   13 Dec 2019, 10:27
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Re: In college M the average [#permalink] New post   12 Feb 2021, 20:46
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Students/Courses = 30/1 == students s: courses c = s : c = 30 : 1 == c : s = 1: 30

Students / Faculty = 20/1 == s : f = 20:1

c:s = 1:30 == 2:60

similarly, s:f = 20:1 == 60:3

therefore c:s:f = 2:60:3

As you can see the by writing it in a ratio, there is no definitive answer to to total number of students as long as this ratio is maintained.
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Re: In college M the average [#permalink] New post   12 Feb 2021, 22:48
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malihanajia wrote:
i cant understand this solution, can anybody explain this in easy process please?


Let the number of courses be N
So, \(\frac{X1 + X2 + X3 + ........ + Xn}{N} = 30\)

Also, \(\frac{30N}{F} = \frac{20}{1}\)
i.e. 30N = 20F or Total Students = 20F

We have 2 variable (N and F) and a single equation, therefore we cannot answer this question.

Hence, D
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Re: In college M the average [#permalink] New post   16 Aug 2022, 04:59
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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