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At certain university ,the ratio of teachers to students for

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At certain university ,the ratio of teachers to students for [#permalink] New post 17 Oct 2006, 02:09
At certain university ,the ratio of teachers to students for evry course is always greater than 3:80. At this universty ,what is te maximum number of students possible in a course that has 5 teachers?
a.>130
b.>131
c.>132
d.>133
e.>134

I will appreciate answers with explanations.
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 [#permalink] New post 17 Oct 2006, 03:30
D...

t/s>3/80
5/s>3/80
3s>400
s>133
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 [#permalink] New post 17 Oct 2006, 09:10
The OA is d. I am trying to figure how is it d???
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 [#permalink] New post 17 Oct 2006, 09:19
ishtmeet wrote:
The OA is d. I am trying to figure how is it d???


The ratio is given: teachers/students>3/80

If there 5 teachers in the course simply write 5 instead of teachers in the equation:

teachers/students>3/80.... include 5 instead of teachers
5/students>3/80...cross multiply
3students>400
students>133... this is the answer....
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 [#permalink] New post 17 Oct 2006, 09:23
ratio is 3:80 minimum

if we have 5, we have increased the number of teachers by 2/3 (the equivalent of multiplying it by 5/3), therefore we need to increase the number of students by 2/3 (times by 5/3) also:

80*5/3 = 400/3
400/3 = 133.33333
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 [#permalink] New post 17 Oct 2006, 09:39
Since the question is saying ratio is greater than 3/80 and we get 133.33 so should'nt it be 134. I got this question in GMATPREP in the morning and OA is d i.e 134.
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 [#permalink] New post 17 Oct 2006, 10:09
yes u are right we cannot say > 133 as it is 133.3333, so we must say > 134
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 [#permalink] New post 17 Oct 2006, 10:45
ishtmeet wrote:
Since the question is saying ratio is greater than 3/80 and we get 133.33 so should'nt it be 134. I got this question in GMATPREP in the morning and OA is d i.e 134.


In your first post you say D >133 and here you are saying D is >134 - which one is it? Please clarify

I think it should be >133 as you cannot have 1/3 of a student so >134 is out.
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 [#permalink] New post 17 Oct 2006, 10:55
At certain university ,the ratio of teachers to students for evry course is always greater than 3:80. At this universty ,what is te maximum number of students possible in a course that has 5 teachers?

a.>130 b.>131 c.>132 d.>133 e.>134


3x = 5 thus x = 5/3

80x = 80*5/3= 133.3 the max is less than 133.3 = 133 my answer is D
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 [#permalink] New post 17 Oct 2006, 11:02
The answer must surely be 134.
Because we get s>133.3... it is E...
Got bit confused with answer choices indicated as >133 etc....

Because choice E as indicated here >134 is incorrect.....

The answer choices should be:

A) 130
B) 131
etc....

Or maybe it was intentional? I mean the answer choices, >133, >134
Then D is correct... Someone confirm the answer choices....
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 [#permalink] New post 17 Oct 2006, 11:08
3:80 = 0.0375
5:133 = 0.03759

5:134 = 0.0373

133 IS THE RIGHT ANSWER

Last edited by yezz on 17 Oct 2006, 11:14, edited 1 time in total.
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 [#permalink] New post 17 Oct 2006, 11:12
if we assumed we have 3 teachers / 80 students the ratio can increase but never decrease ( THE TEACHERS NUMBER WONT BE ENOUGH TO TEACH TO A BIGGER NUMBER OF STUDENTS

so if 5 teachers can teach to 133.3 students

then they can teach to less for sure but not for more

my answer is D 133
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 [#permalink] New post 17 Oct 2006, 11:15
Please confirm the answer choices. The > is throwing me off. I agree that it could be 134 if this means S is LESS THAN 134.
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 [#permalink] New post 17 Oct 2006, 11:31
I too had the same doubt ..SAMq has changed the inequality while cross multiplying..
thus answer should be s<133 and then D makes sense...

Please confirm the answer choices..
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 [#permalink] New post 17 Oct 2006, 11:32
Matrix my buddy allow me to explain if u dont mind


the question says ( ratio is GREATER) than 3:80 ( a fraction to increase is by one of two things

nominator constant and incr deniminator or



deniminator constant while decrease nominator

in our problem

3:80 is equivelant to 5:133.3 to make it greater as mentioned in the stem

decrease 133.3 to 133

I hope this helps
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 [#permalink] New post 17 Oct 2006, 11:42
We need to find the max students such t:s ratio is always greater than 3:80
if we chose D the ratio 5:133 which is 0.03759
if we chose E the ratio will be 0.0373 thus s>134 can not be an option..
thus POE says D but have a doubt s>133 would actually mean 134 which decreases ratio below 3:80..
Thus guess the answer choices should be less..

Please correct me if I am wrong..
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 [#permalink] New post 17 Oct 2006, 11:49
yezz wrote:
Matrix my buddy allow me to explain if u dont mind


the question says ( ratio is GREATER) than 3:80 ( a fraction to increase is by one of two things

nominator constant and incr deniminator or



deniminator constant while decrease nominator

in our problem

3:80 is equivelant to 5:133.3 to make it greater as mentioned in the stem

decrease 133.3 to 133

I hope this helps



Thanks Yezz. Very good explanation :-D
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 [#permalink] New post 17 Oct 2006, 12:16
At certain university ,the ratio of teachers to students for evry course is always greater than 3:80. At this universty ,what is te maximum number of students possible in a course that has 5 teachers?
a.>130
b.>131
c.>132
d.>133
e.>134

3 teachers means 80 students
1 teacher means 80/3 students
5 teachers means 80*5/3 = 400/3 =133.33

Since number of teachers can not be in decimal. Rounding off means D.
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 [#permalink] New post 17 Oct 2006, 15:07
Should be E.

For 3 teachers it is always more than 80 students.

For 5 teachers, it is more than at least 80*5/3 = 133.3.
Since it has to be more than, it is 134.
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 [#permalink] New post 17 Oct 2006, 15:11
yezz wrote:
if we assumed we have 3 teachers / 80 students the ratio can increase but never decrease ( THE TEACHERS NUMBER WONT BE ENOUGH TO TEACH TO A BIGGER NUMBER OF STUDENTS

so if 5 teachers can teach to 133.3 students

then they can teach to less for sure but not for more

my answer is D 133

You're right Yezz, it should be D.
  [#permalink] 17 Oct 2006, 15:11
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