In a class of 65, there are twice as many maths students as biology students. I have a task:
In a class of $65$, there are twice as many maths students as biology students. If $12$ biology students do not take maths and $15$ students take neither of these subjects, how many students take maths but not biology?  
A. 7
B. 19
C. 26
D. 31
E. This is impossible 
I found out that the result is $31$ (Answer D) but only by using the answer as help.  Is there another mathematical way? I tried out a lot but I couldn't find out.
Thanks!
 A: Effectively there are only $50$ students to worry about.
There are $50-12=38$ students who take math, and therefore $19$ who take biology. Thus $19-12$, that is, $7$ take both. That leaves $38-7$ students who take math but not biology.
Remark: A picture (Venn diagram) may be useful in keeping track of things.
A: Let the number of students who take strictly Biology be $b$, the number who take strictly maths be $m$, the number who take both be $d$, and the number who take neither be $n$.  Then
$$
b + m +  d + n = 65
$$ We are also given that
$$
m + d = 2(b + d)
$$
and that $b = 12$ and $n = 15$.  Then 
$$
m + d = 38
$$ and 
$$
m  - d = 2b = 24
$$ So
$$
2m = 24 + 38 = 62
$$ So $m = 31$.
A: Let $B$ be the set of biology students and $M$ be the set of Math students and $U$ be the set of all the students. Then we are given that 
$$|U|=65, \quad |M|=2|B|, \quad |B-M|=12, \quad |U-(M \cup B)|=15.$$
What you want is $|M-B|$. Let $y=|M-B|$ and $x=|M \cap B|$, then $|M|=x+y$, so we have
$$(x+y)=2(12+x).$$
and
$$y+x+12+15=65.$$
Now solve for $y$ to get your answer. If you are familiar with Venn diagrams then you will be able to follow the argument I have stated fairly easily. 
A: Let t be total no. of students in the class, m be no. of students having maths only, b be students having biology only, n be students having neither and x be students having both Maths and Biology.
We are given that

t= 65
b=12
n=15
(m+x)=2*(b+x)

Also

(m+x) + b + n = t

Replacing it with with values.

2*(b+x) + 12 + 15 = 65
(b+x) = 38/2 = 19, No. of biology students

There for
No. of Maths students = t  - n  - (b+x)
                      = 65 - 15 - 19
                      = 31
A: Total number of students, T = 65
Number of biology students, B
Number of maths students, M = 2B
Number of other students, O = 15
Number of students studying either maths or biology = T - O = 65 - 15 = 50
Number of students studying just biology = 12 (given)
Number of maths students( just Maths + both maths and biology), M = 50 - 12 = 38
As, M = 2B <=> B = 38/2 = 19
Number of students who do both maths and biology = 19 - 12 = 7
Therefore, Number of maths students, M = 38 - 7 = 31 (Option D)
A: Making a picture often helps. For sets such pictures are called Venn diagrams.
Consider for instance the Venn diagram below:

We know that there are twice as many math students as there are biology students, so it must hold that $|A_1| + |A_3| = 2 (|A_1| + |A_2)$ ($|S|$ denotes the number of elements in the set $|S|$), or equivalently that $|A_3| \overset{|A_2|=12}{=} |A_1| + 2 |A_2| = 24 + |A_1|$. Also, we know that $|A_1| + |A_2| + |A_3| = 50$, or equivalently, that $|A_3| \overset{|A_2|=12}{=} 38 - |A_1| $.
So now we have two equations:


*

*$|A_3| = 24 + |A_1|$

*$|A_3| = 38 - |A_1|$


From 2 it follows that $|A_1| = 38 - |A_3|$,
so that from 1 it follows that $|A_3| = 24 + |A_1| = 24 + 38 - |A_3|$, or equivalently
$2|A_3| = 62 \rightarrow |A_3| = 31$.
