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In a certain class, 1/3 of the students are honors students,

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In a certain class, 1/3 of the students are honors students, [#permalink] New post 05 Dec 2003, 02:51
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In a certain class, 1/3 of the students are honors students, and 1/4 of the students play varsity sports. If 12 students play varsity sports and are honors students, what is the least possible number of students in the class?


20
22
36
48
144


please explain

Last edited by Praetorian on 05 Dec 2003, 05:06, edited 1 time in total.
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 [#permalink] New post 05 Dec 2003, 04:50
Please, clarify the wording. Hardly can people understand it.
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Students [#permalink] New post 05 Dec 2003, 18:17
My logic:

The question stem doesn't specify that only 12 students play varsity sports or are honor students; it just says that 12 are both.

If we want the least possible number of total students we have to assume the smaller of the percentages to be only what is specified.

We assume there are only 12 students who play varsity sports and thus these students account for 25% of the entire class. And of course if there are 12 students who play varsity sports, there are a total of 16 students who are honor students (12 just happen to also play varsity sports).

The least number of students is 48.

Is this correct?
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 [#permalink] New post 05 Dec 2003, 19:08
I say 48 students.

I reckon with csperber on the logical approach.

Here's a theoretical approach:
For now let's assume all students must be involved in at least one of the categories.

x = total students

(x/3 - 12) + (x/4 - 12) = x
4x - 144 + 3x - 144 = x
6x = 288
x = 48
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 [#permalink] New post 05 Dec 2003, 22:02
wonder_gmat wrote:
I say 48 students.

I reckon with csperber on the logical approach.

Here's a theoretical approach:
For now let's assume all students must be involved in at least one of the categories.

x = total students

(x/3 - 12) + (x/4 - 12) = x
4x - 144 + 3x - 144 = x
6x = 288
x = 48


why have you excluded 12 (both) from the individual numbers?
it should be like:
x/3 + (x/4 - 12) = x ..... this would give -ve value :(

try drawing this on the venn diagram.
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 [#permalink] New post 05 Dec 2003, 22:19
wonder_gmat wrote:
I say 48 students.

I reckon with csperber on the logical approach.

Here's a theoretical approach:
For now let's assume all students must be involved in at least one of the categories.

x = total students

(x/3 - 12) + (x/4 - 12) = x
4x - 144 + 3x - 144 = x
6x = 288
x = 48



i had trouble understanding why you subtracted 12 from both x/3 and

x/4.

the official answer is 48.

thanks
praetorian
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 [#permalink] New post 13 Dec 2003, 07:00
48

we know from the problem stem that the number of students playing varsity sports and the number of students who are honor students taken together has (should it be have or has here ?? explain ... :) ) to be greater than or equal to the number of students from each category taken alone.

Also, 1/3 of students represents a greater number than 1/4 of students.

So, x/3 + x/4 - 12 >= x/3
x/4 - 12 >= 0
x/4 >= 12
x >= 48

what say?
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 [#permalink] New post 14 Dec 2003, 09:42
calculation done by wonder_gmat is wrong.
-------------------------------------------------
x/3 - 12) + (x/4 - 12) = x
4x - 144 + 3x - 144 = x
-------------------------------------------------
Actually it is 4x - 144 + 3x - 144 = 12x
You will get -ve result here.

The explaination given by pitts20042006 makes more sense.
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 [#permalink] New post 15 Dec 2003, 06:46
The fact that x>=48, does not mean the minimum of X is indeed 48.

It's better to go the other way: we need at least 12 students form both categories to satisfy the condition.
However, sport students is smaller group, meaning we have to take this group as reference. So, we need at least 12 sport students, and all of them are honors students. => We need at least 48 students (48/4=12). Consequently, we have 48/3=16 honors students.
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Re: PS : Students [#permalink] New post 15 Dec 2003, 10:33
praetorian123 wrote:
In a certain class, 1/3 of the students are honors students, and 1/4 of the students play varsity sports. If 12 students play varsity sports and are honors students, what is the least possible number of students in the class?


20
22
36
48
144


please explain


My approach was like this. First of all, the answer should be divisible by both 3 and 4. So, A and B are out, leaving us with C, D and E

Starting with 36,
Honors students = 12
Students that play sports = 9

The problem mentions that 12 students do both. If so, -3 students should be playing sports alone, which is not feasible, while no students are honor students alone, which is okay.

Looking at 48,
Honors students = 16
Students that play sports = 12

In this case, there could be no students that play varsity sports alone and is still valid and

So, I went with 48.



Also, pitts ....

if we consider x/3 + x/4 - 12 >= x/4 then we get the answer as x >= 36
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 [#permalink] New post 15 Dec 2003, 13:02
amarsesh wrote
Quote:
Also, pitts ....
if we consider x/3 + x/4 - 12 >= x/4 then we get the answer as x >= 36


But that is why I had written:
Quote:
Also, 1/3 of students represents a greater number than 1/4 of students.


so we consider x/3 and not x/4
  [#permalink] 15 Dec 2003, 13:02
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