Could you please help with the following problem from MBA practice test 1.
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
Ratio 5/S> 3/80, therefore S>400/5 or S>133.3. My answer is 134, but the correct one is 133.
Could you please help me understand the correct way of solving this problem?
Thanks!
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- GregS&P
Here's my approach:
3/80 is the ratio we begin with (TAs / Students). We know that our ultimate ratio of TAs to students has to exceed this ratio.
We're told the course in question has 5 TAs ---> 5/x (TAs / Unknown Number of Students)
Solving for the unknown number of students can be done by setting up:
3/80 = 5/x ---> x = 133.333(333,etc.).
So, when x = 133.333(333,etc.), 5/x = 3/80. We want x such that 5/x > 3/80. Using an x < 133.333(333,etc.) would help us attain this. More specifically, we want the highest integer value (since we can't have .333(333,etc.) of a student...). This value is 133.
3/80 is the ratio we begin with (TAs / Students). We know that our ultimate ratio of TAs to students has to exceed this ratio.
We're told the course in question has 5 TAs ---> 5/x (TAs / Unknown Number of Students)
Solving for the unknown number of students can be done by setting up:
3/80 = 5/x ---> x = 133.333(333,etc.).
So, when x = 133.333(333,etc.), 5/x = 3/80. We want x such that 5/x > 3/80. Using an x < 133.333(333,etc.) would help us attain this. More specifically, we want the highest integer value (since we can't have .333(333,etc.) of a student...). This value is 133.
- RonPurewal
- Students
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- Joined: Tue Aug 14, 2007 8:23 am
Gphil,
The problem is that you've accidentally reversed the inequality sign. There are two ways to see that this is true:
(1) Take the original equation that you wrote (5/S > 3/80), which is CORRECT, and multiply both sides of the equation by 80S. Since 80S is positive, the sign doesn't reverse, so you get 400 > 3S. Therefore, 133.333333 > S, meaning that S is less than that number, not greater.
(2) Just think about the problem. The problem is asking you for the maximum number of students, and you know from the problem statement that the number of students is not allowed to exceed a certain value to maintain the desired ratios. Based on these observations, it's clear that the final inequality HAS to be something of the form S < number, not S > number. So, since you got S > number, something's wrong, and you should just switch the sign.
The rest of your math is fine. Incidentally, you wrote 400/5 where you should have written 400/3.
The problem is that you've accidentally reversed the inequality sign. There are two ways to see that this is true:
(1) Take the original equation that you wrote (5/S > 3/80), which is CORRECT, and multiply both sides of the equation by 80S. Since 80S is positive, the sign doesn't reverse, so you get 400 > 3S. Therefore, 133.333333 > S, meaning that S is less than that number, not greater.
(2) Just think about the problem. The problem is asking you for the maximum number of students, and you know from the problem statement that the number of students is not allowed to exceed a certain value to maintain the desired ratios. Based on these observations, it's clear that the final inequality HAS to be something of the form S < number, not S > number. So, since you got S > number, something's wrong, and you should just switch the sign.
The rest of your math is fine. Incidentally, you wrote 400/5 where you should have written 400/3.
3 posts Page 1 of 1