Question

If x, y, and k are positive numbers such that ((x)/(x+y))(10) + ((y)/(x+y))(20) = k and if x < y, which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30

Answer 1

$$10\cdot\frac{x}{x+y}+20\cdot\frac{y}{x+y}=k$$Observe we can pull out \(10\) to get:$$10\left(\frac{x}{x+y}+\frac{2y}{x+y}\right)=k\\10\left(\frac{x+2y}{x+y}\right)=k\\10\left(1+\frac{y}{x+y}\right)=k$$... so we know \(k>10\). Knowing that \(y>x\) tells us that \(y/(x+y)>1/2\) so we can infer further than \(k>10(1+1/2)=15\). However, since obviously \(y<x+y\) we know \(y/(x+y)<1\) so \(k<10(1+1)=20\). Thus we've realized \(15<k<20\) and find there is only one possible solution listed, \(18\).