Simplifying algebraic expressions

[jeffedwards]

Is there a better way to solve it than what the book says? It makes sense, but I doubt I could duplicate that on the test. Any tips?

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

• 1
  10
  12
  15
  18
  30

OA: D
Source: OG 12th E; PS #148

[gmatv09]

Is there a mistake in the problem...
there eqn should be [x/(x+y)]10+[y/(x+y)]20=k

[palvarez]

(10x+20y)/(x+y) = k, which is an integer

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

k > 10

10y/(x+y) = 10/(1+(y/x))

y/x's max value 1.

10/(1+(y/x))'s min value = 5

therefore, k > 15

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

here, k < 20


We got one value in that range. 18

10y = 8x+8y
y = 4x, which is consistent with x < y.

[KICKGMATASS123]

(10x+20y)/(x+y) = k, which is an integer

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

k > 10

10y/(x+y) = 10/(1+(y/x))

y/x's max value 1.

10/(1+(y/x))'s min value = 5

therefore, k > 15

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

here, k < 20


We got one value in that range. 18

10y = 8x+8y
y = 4x, which is consistent with x < y.

dont get 10 + (10y/(x+y)) = k step!

[palvarez]

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

divide by (x+y), you are left with 10 + 10y/(x+y)

[okigbo]

This is just a weighted average problem. No need to do calculations - average of 10 and 20 is 15 - equation is weigthed on the 20 side so must be greater than 15 - pick 18 and move on.