Problem Solving

If x, y and k are positive numbers such that 10*x/(x+y)+20*y/(x+y)=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

Putting the sum over a common denominator, we get:
(10x + 20y) / (x+y) = k.

Let x = the number of $10 shirts purchased at a certain store.
Let y = the number of $20 shirts purchased at a certain store.
Total cost of the $10 shirts = 10x.
Total cost of the $20 shirts = 20y.
Total number of shirts purchased = x+y.
Thus, the AVERAGE cost per shirt is equal to the following:

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

In the problem above, the value of k is equal to the AVERAGE cost per shirt.
Since each shirt costs either $10 or $20, the average cost per shirt must be BETWEEN 10 and 20.
Since y>x, the number of $20 shirts purchased is GREATER than the number of $10 shirts purchased, with the result that the average cost per shirt must be CLOSER TO 20 than to 10.
Of the answer choices, the only viable option is k=18.

The correct answer is D.

An alternate approach is to plug in the answers, which represent the value of k.
Let x=1.
When we plug in the correct answer choice for k, x<y.

Answer choice C: k=15
(10*1 + 20y)/(1+y) = 15
10 + 20y = 15 + 15y
5y = 5
y = 1.
Since x=y, eliminate C.

Answer choice D: k=18
(10*1 + 20y)/(1+y) = 18
10 + 20y = 18 + 18y
2y = 8
y = 4.
Since x<y, success!

The correct answer is D.

abhasjha wrote:

The Easiest Explanation goes as follows:

Given : 10*x/(x+y)+20*y/(x+y)=k and if x<y

Question : Which of the following COULD be the value of k?

The word 'COULD' suggests that there are many possible values however one of them is Hidden among options.

No information about x+y is given so it's safe to assume x+y=10 as it makes equation much easier

@ x+y=10 the equation, 10*x/(x+y)+20*y/(x+y)=k and if x<y becomes
x+2y = k

but x+y = 10 and x<y therefore
Case 1: x =0, y=10 then x+2y = 20 = k NOT AMONG OPTIONS
Case 2: x =1, y=9 then x+2y = 19 = k NOT AMONG OPTIONS
Case 3: x =2, y=8 then x+2y = 18 = k AVAILABLE AMONG OPTIONS

Answer: Option D