K = [10x/(x + y)] + [20y/(x + y)]satyavegi wrote:If X,Y and K positive numbers such that (10x/(x+y))+(20y/(x+y))=K and if x<y, then which of the following could be value of K?
Now, K is nothing but the weighted average of 10 ad 20 the weights being x and y respectively. Hence, K will be between 10 and 20. As x < y, the weighted average will be closer to 20 than 10.
Only such value among the options is 18.
The correct answer is D.

















