In a certain class 1/3 of students are honors 1/4 of

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.

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

x/4.

the official answer is 48.

thanks
praetorian

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?

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.

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.

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

amarsesh wrote


But that is why I had written:


so we consider x/3 and not x/4

In a certain class 1/3 of students are honors 1/4 of students play varisity sports. If 12 students are honors and play varisty sports , what is the least number of students in class?

20

22

36

48

144


Please explain your solution..

Backsolve.
Using 36, 1/3(36)=12
1/4(36)=9
Impossible, since 12 are both.


Using 48, 1/3(48) = 16
1/4(48) = 12

Possible, since we have 12 playing sports now. We are asked for least possible number, so this has to be the one, since 144 will satisfy the condition given but have an excess of students who are only playing one role.

The least number comes when both the categories have the highest overlap. Thus, the maximum overlap is all the students that play varsity sports are honors. => x/4 = 12. i.e., x = 48 which is the minimum.

My answer is 144

Total=Group1+group2-both+neither
144=48+36-12+60
144=144.

My answer is 144

Total=Group1+group2-both+neither
144=48+36-12+60
144=144.

I got the math wrong, please ignore last post.

Mallelac, your approach is very cool!!

Thank you to both!!