In a group of 120 persons there are 80 elements of B and 40 elements of G. Further 70 persons in the group are M and the remaining are H. Then the number of elements that are both in B as well as M.
I have made the following approach:
Let BM be the elements of B which are also in M; similarly for all the combinations.
Now can we write:
BM + BH = 80
GM + GH = 40
MB + MG = 70
HB + HG = 50
We need to find BM.
Which further brings us to the following system of equations:
a + b =50
c + d =70
c + a =80
d + b =40
This is where I am stuck. I find it fairly impossible to find any solution from the above equations. ( I tried calculating the row echelon form of the matrix, with disastrous results) This makes me doubt the correctness of my approach.
Maybe a kind soul could give me a heads up.
There has been some talk about incomplete information so I am going to cut the crap and attach the original question:
There isn’t enough information to answer the question. To see this, note that both of the following arrangements are possible (among many others), and one has $70$ people in $B\cap M$, while the other has only $30$.
$$\begin{array}{c|cc|c} &B&G&\text{Total}\\ \hline M&70&0&70\\ H&10&40&50\\ \hline \text{Total}&80&40&120 \end{array}$$
$$\begin{array}{c|cc|c} &B&G&\text{Total}\\ \hline M&30&40&70\\ H&50&0&50\\ \hline \text{Total}&80&40&120 \end{array}$$
Added: Now that we know that it’s a multiple choice question, the situation is a little different. The arrangements above show that the only choice that could possibly be correct is (A), and we can prove without much difficulty that (A) is correct: if $|B\cap M|<30$, then $|B\cap H|>50$, which is impossible, since there are only $50$ Hindus. Thus, we must certainly have $|B\cap M|\ge 30$.