If x, y, and k are positive numbers such that (10x/x+y) + (20y/x+y) and x<y, which of the following could be he value of k? (inequalities, ratios)

10 12 15 18 30

Could you explain the formal answer better? I don’t understand how the 10(x+y)/x+y + (10y/x+y) appeared.

Formal Answer:

(10x/x+y) + (20y/x+y) =
10(x+y)/x+y + (10y/x+y) =

10+(10y/+y)

Since x,y, x/2 + y/2 < y < x+y. Therefore 5<(10y/x+y)<10.
It follows that 15 < y < 20.

Alternatively, think that k is the average value of x+y bills, x of which are $10 bills and y of which is @40 bills. Will be between 15 and 20, i.e. between 10 and 20 and closer to 20 than to 10.

OK, I plugged in numbers, considering x<y, I got k = roughly just over 16 almost every time. I chose 15 for the answer. While studying the formal answer, I noticed that there is a step missing that I need. The bold portion—how did the problem get to that answer?

THANK YOU!!! for helping me with everything! A

[kevin]

20y=10y + 10y, so we can write (10x+10y)/(x+y)+10y((x+y)

Alternatively, pick a value for x (e.g 1)

k=10/(y+1)+20y/(y+1)

y=(10-k)/(k -20)

Try the values of k