12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in

Let P1 be a and P2 be b for simplicity.....
Given that x kg is taken out and mixed with other...
So two different alloys will have iron content as...
1) ((12-x)*a+x*b)
% iron content \(\frac{(12-x)*a+x*b}{12}\)
2) ((6-x)*b+x*a)
%iron content=\(\frac{(6-x)*b+x*a}{6}\)

Now the above two must be equal..
SO
\(\frac{(12-x)*a+x*b}{12}=\frac{(6-x)*b+x*a}{6}\)
\(\frac{12*a-x*a+x*b}{12}=\frac{6*b-x*b+x*a}{6}\)
72a-6xa+6xb=72b-12xb+12xa
72(a-b)=18x(a-b).....
72=18x.....x=72/18=4..
D


Another way slightly easier is to substitute choices And find ratio Alloy1:Alloy2....
A) 1..
A1:A2...(12-1):1and 1:(6-1)..... ratios are not same..
Similarly for B and C..
D) 4..
(12-4):4 and 4:(6-4), which is 8:4 and 4:2...
Now both are in ratio 2:1..
So this is what we are looking for..

Ans D