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 ....
Gmat Prep PS -Ratio
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here i have a concern -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 ....
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.
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava
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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
(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
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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.
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.
- AleksandrM
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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.
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.
- Stuart@KaplanGMAT
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This method definitely works, but you could solve for: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.
5/x > 3/80
400 > 3x
400/3 > x
133.333 > x
therefore the greatest possible integer value for x is 133.
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Hey Rara:
These are the numbers when you plug:
5/133=0,03759398
3/180=0,03750000
5/134=0,03731343
These are the numbers when you plug:
5/133=0,03759398
3/180=0,03750000
5/134=0,03731343
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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
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
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