Question 1131586
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THE ABOVE SOLUTION IS WRONG!  It's 5/12ths of the students

Let J = the set of students wearing jeans
Let T = the set of students wearing T-shirts

Let x = the number wearing jeans but not T-shirts = N(J&#8745;T')
Let y = the number wearing both jeans and a T-shirt = N(J&#8745;T)
Let z = the number wearing T-shirts but not jeans = N(J'&#8745;T)

N(J) = N(J&#8745;T') + N(J&#8745;T) = x+y
N(T) = N(J&#8745;T) + N(J'&#8745;T) = y+z

This is the Venn diagram:

{{{drawing(300,200,-4,4,-2,4.8, locate(-2,1.8,x),locate(1.5,1.7,z), locate(-3.6,2.5,J), locate(-.1,1.8,y),blue(circle(-sqrt(2),sqrt(2),2)),blue(circle(-sqrt(2),sqrt(2),1.95)),blue(circle(-sqrt(2),sqrt(2),1.975)),
red(circle(sqrt(2),sqrt(2),2),circle(sqrt(2),sqrt(2),1.95),circle(sqrt(2),sqrt(2),1.975)),
locate(3.4,2.5,T)
 )}}}


So {{{system(x+y=expr(2/3)n,y+z=expr(3/4)n)}}}

After clearing of fractions:

{{{system(3(x+y)=2n,4(y+z)=4n)}}}

This says that n must be divisible by both 3 and 4, so n must be
divisible by 12, so n = 12k

Since two-third of the students are wearing blue jeans,
2/3 of 12k = 8k are wearing blue jeans.

Since three-fourths of the students are wearing T-shirts,
3/4 of 12k = 9k are wearing T-shirts.

N(J or T) = N(J) + N(T) - N(J and T)

       12k = 8k + 9k - N(J and T)

       12k = 17k - N(J and T)

N(X and Y) = 5k

So 5k students are wearing both jeans and T-shirts.

So there are many solutions depending upon the choice of k:

Since there are 8k students in set J, the blue circle, and 5k students
in the overlapping part of the two circles, there must be 8k-5k or 3k
in the left part of the blue circle.

Similarly, since there are 9k students in set T, the red circle, and 5k
students in the overlapping part of the two circles, there must be 9k-5k 
or 4k in the right part of the blue circle.

{{{drawing(300,200,-4,4,-2,4.8, locate(-2,1.8,3k),locate(1.5,1.7,4k), locate(-3.6,2.5,J), locate(-.1,1.8,5k),blue(circle(-sqrt(2),sqrt(2),2)),blue(circle(-sqrt(2),sqrt(2),1.95)),blue(circle(-sqrt(2),sqrt(2),1.975)),
red(circle(sqrt(2),sqrt(2),2),circle(sqrt(2),sqrt(2),1.95),circle(sqrt(2),sqrt(2),1.975)),
locate(3.4,2.5,T)
 )}}}

Since there are 5k students wearing both, out of 3k+5k+4k=12k students,
the answer is always 5/12 of the students. 

The smallest possible solution is when k=0, no students at all in the
classroom, 0 wearing blue jeans, and 0 wearing T-shirts. That is a
possibility because 0 is both 2/3rds as well as 3/4ths of 0.  Also 5/12ths
of zero is zero. It's probably not the solution your teacher wanted, but
mathematically it fits.

The next smallest possible solution is when k=1, 

{{{drawing(300,200,-4,4,-2,4.8, locate(-2,1.8,3),locate(1.5,1.7,4), locate(-3.6,2.5,J), locate(-.1,1.8,5),blue(circle(-sqrt(2),sqrt(2),2)),blue(circle(-sqrt(2),sqrt(2),1.95)),blue(circle(-sqrt(2),sqrt(2),1.975)),
red(circle(sqrt(2),sqrt(2),2),circle(sqrt(2),sqrt(2),1.95),circle(sqrt(2),sqrt(2),1.975)),
locate(3.4,2.5,T)
 )}}}

In that case, there are 12 students in the classroom, 8 wearing blue jeans
and 9 wearing T-shirts. That is a possibility because 8 is 2/3rds of 12 and
9 is 3/4ths of 12. The answer is 5 wearing both, which is 5/12ths of 12. 

The next smallest possible solution is when k=2

{{{drawing(300,200,-4,4,-2,4.8, locate(-2,1.8,6),locate(1.5,1.7,8), locate(-3.6,2.5,J), locate(-.1,1.8,10),blue(circle(-sqrt(2),sqrt(2),2)),blue(circle(-sqrt(2),sqrt(2),1.95)),blue(circle(-sqrt(2),sqrt(2),1.975)),
red(circle(sqrt(2),sqrt(2),2),circle(sqrt(2),sqrt(2),1.95),circle(sqrt(2),sqrt(2),1.975)),
locate(3.4,2.5,T)
 )}}}

In that case, there are 24 students in the classroom, 16 wearing blue jeans
and 18 wearing T-shirts. That is a possibility because 16 is 2/3rds of 24 
and 18 is 3/4ths of 24. The answer is 10 wearing both, which is 5/12ths of 24. 

The next smallest possible solution is when k=3

{{{drawing(300,200,-4,4,-2,4.8, locate(-2,1.8,9),locate(1.5,1.7,12), locate(-3.6,2.5,J), locate(-.1,1.8,15),blue(circle(-sqrt(2),sqrt(2),2)),blue(circle(-sqrt(2),sqrt(2),1.95)),blue(circle(-sqrt(2),sqrt(2),1.975)),
red(circle(sqrt(2),sqrt(2),2),circle(sqrt(2),sqrt(2),1.95),circle(sqrt(2),sqrt(2),1.975)),
locate(3.4,2.5,T)
 )}}}

In that case, there are 36 students in the classroom, 24 wearing blue jeans
and 27 wearing T-shirts. That is a possibility because 24 is 2/3rds of 36 
and 27 is 3/4ths of 36. The answer is 15 wearing both, which is 5/12ths of 36. 

We can let k be any integer and get a different solution.  The solution
in every case will be 5k, a multiple of 5.

Edwin</pre>