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:
