Problem Solving

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

This question if from OG question no-148
Is there a quicker way to solve this problem?

(x/x+y)*10 + (y/x+y)*20, Can be simplified as shown below.
(x/x+y)*10 + (y/x+y)*10 + (y/x+y)*10
((x+y)/(x+y))*10 + (y/x+y)*10
10 + (y/x+y)*10

0 < x < y and x,y are positive numbers. The the boundaries are x=0 and x=y

10 + (y/y+y)*10 < 10 + (y/x+y)*10 < 10 + (y/0+y)*10
15 < 10 + (y/x+y)*10 < 20

The only number which is between 15 and 20 is 18, IMO Option D

You can do this problem very quickly without doing any math. If you look closely at the problem, you'll see that the equation is simply a weighted average of 10 and 20.

x/(x+y) + y/(x+y) will equal 1 --> x/(x+y) + y/(x+y) = (x+y)/(x+y) = 1

So we can simplify this equation by saying x/(x+y) = X and y/(x+y) = Y where X + Y = 1. This now writes 10X + 20Y = k. This is the weighted average equation above. Since X + Y = 1 and X and Y are both positive, they will both be percentages and k will be between 10 and 20.

Since we know that x < y, we know that Y has to be greater than 50%. This also means that k has to be between 15 and 20. The only answer that fits that description is D.

nidhis.1408 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

This question if from OG question no-148
Is there a quicker way to solve this problem?

I posted a solution here:

https://www.beatthegmat.com/tough-algebr ... tml#292229