Mixtures

Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a
result, 40% solution of acid was obtained, what part of the original solution was replaced?

Let us complete forget that a part of a 50% solution of acid is replaced and assume that X ml of a 50% solution of acid and Y amount of 30% solution of acid are mixed to form a 40% solution of acid.

Using allegations:
Quantity of the 50% acid solution/Quantity of the 30% acid solution = X/Y = (%strength of resultant - % strength of the 30% solution)/(% strength of the 50% solution - strength of resultant)
= (40 - 30)/(50 - 40) = 1:1

So, One part of 50% acid solution is mixed with one part of 30% acid solution. i.e. In two parts of 50% acid solution, one part is replaced with one part of 30% acid solution. So, the answer to this question is 1/2.

Nice Solution.Any other approaches.

Regards

We can see that 40% is in middle of of 50% and 30% , in weighted avg. when both the coefficients are same (a*50% + a*30%)2 mean would fall exactly in between the two numbers i.e. a*40%.
Therefore equal quantities of 30% and 50% solution must have been mixed, and to make it equal 1/2 of the qty. of original mixture must have been removed.