B. 12 --> i.e. x=2, y=3.

Both fractional terms are equal. The first term would constitute 1/3 of k, the second term would constitute 2/3 of k.

B: k=12, therefore (x/(x+y))*10 = 4, (x/(x+y))*20 = 8.
Therefore, x=4, x+y=10, y=6

A: can't work because 10 is not divisible by 3.
C: can't work because both fractional terms have to equal 1/2 (which only works when x=y)
D: can't work because 1/2 of k would be 6, 2/3 of k would be 12.
(x/(x+y))*10=6, (x/(x+y))*20=12
Therefore, x=6, x+y=10, y=4, and x>y (which violates the rule)
E: can't work because both fractional terms have to equal 1 (which is impossible since x<y, so y can't=0; if x=0, the entire fraction equals 0.)

x = 1, y =4, k = 18. Answer is D.

givemeanid wrote:x = 1, y =4, k = 18. Answer is D.

Umm.. x=1, y=4, k=6....

Answer is B.

Thanks for the response! the answer is D)

How is the answer D? What's the OE?

(x/x+y)*10 + (y/x+y)*20 = k
10(x+2y) = k(x+y)
Since x and y are +ve and x < y, k > 10.

Lets start with Choice C.
10(x+2y) = 15(x+y)
5y = 5x
Since x < y, this value of k is not big enough.

Try D.
10(x+2y) = 18(x+y)
2y = 8x
y = 4x
CORRECT.


You can also verify other choices and see that they do not hold up.

Oh... it's a y in the second numerator.. my b, kept reading both fractions with x in num.

Candidates from the US apply to an average of 20 B-schools each, whereas candiates from outside the US apply to an average of 10 B- schools each. Which of the following could be the average number of B- schools that candiates (American and otherwise) apply to if US candiates outnumber foreign candidates?

This would have been an easier way to rephrase the same question!