Ian Stewart wrote:

iikarthik wrote:Q17:
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

The trick is to recognize that the equation just gives a weighted average of x and y. The equation above is the same as you'd use if you were asked "If x pounds of peanuts and y pounds of cashews are mixed together, and peanuts cost $10/pound and cashews cost $20/pound, what is the price per pound of the resulting mixture?" The answer must be between 10 and 20, and if y > x, the answer must be closer to 20 than to 10. So 18 is the only possible answer.

Ian Stewart wrote:

iikarthik wrote:Q17:
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

The trick is to recognize that the equation just gives a weighted average of x and y. The equation above is the same as you'd use if you were asked "If x pounds of peanuts and y pounds of cashews are mixed together, and peanuts cost $10/pound and cashews cost $20/pound, what is the price per pound of the resulting mixture?" The answer must be between 10 and 20, and if y > x, the answer must be closer to 20 than to 10. So 18 is the only possible answer.

iikarthik wrote:Q17:
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

GMATGuruNY wrote:

iikarthik wrote:Q17:
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

For those who worry that they wouldn't be able to recognize that this is a weighted average problem, an easy and efficient approach would be to plug in the answer choices, which represent the value of k:

Answer choice C:
(10x + 20y)/(x+y) = 15
10x + 20y = 15x + 15y
5y = 5x
y = x
Doesn't work because the problem states that x<y.

We need y to be larger, so let's try 18:
10x + 20y = 18x + 18y
2y = 8x
y/x = 8/2
Success! x<y.

The correct answer is D.

The process would be very quick if you recognized right away that since k = one of the answer choices, you could simply plug them in for k.

Ian Stewart wrote:

rahul.s wrote: Ian,

In such a problem, where we need to experiment with numbers, how would we know which are the right numbers to choose?

Well, you don't need to experiment with numbers in the question above; I didn't in my post above. That said, if you don't see a conceptual or algebraic solution fairly quickly, plugging in numbers is a good fallback option. It would be a bit lucky to find numbers that give the exact answer to this question, but if you plug in a few very simple sets of numbers (you don't want to waste any time on complicated numbers), making sure that x is less than y, you'll always find that k is between 15 and 20, which may lead you to the correct answer here.

There are also algebraic solutions to the question:

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

10x + 20y = kx + ky

20y - ky = kx - 10x

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

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

and since 0 < x < y, then 0 < x/y < 1, and it must be that 0 < (20 - k) / (k - 10) < 1. From the answer choices we can be sure k - 10 isn't negative, so we can multiply through this inequality by k-10 to find that 0 < 20 - k < k - 10, or that 15 < k < 20.