agrwal12 wrote:At a certain university, the ratio of 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 any course that has 5 teaching assistants ?

A) 130
b) 131
c) 132
d) 133
e) 134

How to solve this ???
I tried like this

for every 3 assistant required for = 80 students
1 assistant required for = 80/3 student
5 assistants required for = 80/3*5 = 130 students

What to do next ??
Help needed ....

here i have a concern -
5 assistants required for = 80/3*5 = 130 students -> 5 assistants required for = 80*5/3 ~ 133 students. but the ratio is greater than 3:80. so the no uf students should be 134.

The question says :
(T.A/Students) > (3/80).. i.e the ratio shud be greater than 3/80.
On substituting T.A as 5 we get somewhere arround 133.33
No if we decrease the denominator( Number of students as 132,121,130..) then the ratio eventually increases but the number of students decreases ( as per question we need to spot the Maximum number of students) And if we take 134 then the ratio falls below the requirement i.e it5/134 is less than 3/80 .Therefore the answer shud be 133 ...
Only when the number of students is 133 we get the max. number of students and the ratio is more than 3/80.

If I am wrong let meknow !
Thanks
Senthil

I had this problem on GMAT prep and got it wrong too. After going through all your replies, I tried it another way. Just need confirmation if this method is correct. According to the Question
5/x > 3/80

=> now try ans choices

=> 5/134 > 3/80
cross mutliply
=> 5 * 80 > 134 * 3
=> 400 > 402
not the right choice

let's try ans choice 133
=> 5/133 > 3/80
cross mutliply
=> 400 > 399
right choice.


I would appreciate it if someone can confirm that this method is right. Thanks.

sangeethai, I got the same answer.

I just set it up as follows 3/80 = 5/x which gives you 3x = 400. Since you cannot have partial students, at this point it is a good idea to just go and plug in the numbers (you are looking for an answer that will give you a units digit of 9). You are also looking for an answer that is the biggest possible number under 400, and 133 is just such a number. It seems right to me.

AleksandrM wrote:sangeethai, I got the same answer.

I just set it up as follows 3/80 = 5/x which gives you 3x = 400. Since you cannot have partial students, at this point it is a good idea to just go and plug in the numbers (you are looking for an answer that will give you a units digit of 9). You are also looking for an answer that is the biggest possible number under 400, and 133 is just such a number. It seems right to me.

This method definitely works, but you could solve for:

5/x > 3/80

400 > 3x

400/3 > x

133.333 > x

therefore the greatest possible integer value for x is 133.

Hello All,

I went through the explanations but still did not understand why the answer was 133 instead of 134. I tried the plug in method and the only answer that fit the 5/x > 3/80 was 134. Help!

Hey Rara:

These are the numbers when you plug:

5/133=0,03759398
3/180=0,03750000
5/134=0,03731343

just make a ratio table

T S Total
3 80 83
----------------------
x 5/8 5/8 5/8
----------------------
= 5 133.4
----------------------

Hence 133.4 ~ 133

here's how it made sense to me when I re-looked at it (btw got it wrong the first time - I chose 134)

If the number of TA to students must be > 3:80 means we need MINIMUM 0.375 TA per student

Now between the 2 options 133 and 134:
3/133 = 0.3759
3/134 = 0.3731 ---> we don't have the minimum number of TA per students here. There are less TA / student in this option.

So means 133 is correct. because it complies with the information in the stem.

Hope it helps