BANON wrote:

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

\(10*\frac{x}{x+y}+20*\frac{y}{x+y}=k\)

\(10*\frac{x+2y}{x+y}=k\)

\(10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k\)

Finally we get: \(10*(1+\frac{y}{x+y})=k\)

We know that \(x<y\)

Hence \(\frac{y}{x+y}\) is more than \(0.5\) and less than \(1\)

\(0.5<\frac{y}{x+y}<1\)

So, \(

15<10*(1+\frac{y}{x+y})<20\)

Only answer between \(15\) and \(20\) is \(18\).

Answer: D (18)There can be another approach:

We have: \(\frac{10x+20y}{x+y}=k\), if you look at this equation you'll notice that it's a weighted average.

There are \(x\) red boxes and \(y\) blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes?

k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition.

Answer: D (18)Hi, Bunuel! I have a silly question. Can you please explain how you arrived at the following?