OG PS #148

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OG PS #148

by warraich » Sun May 29, 2011 10:08 am
The official guide explanation goes through a very lengthy algebraic acrobatics to get to the answer.
During the exam, I dont think this would be the best approach given the time constraints. I am wondering if there is an alternate faster way of solving this problem.

Thanks,
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by cans » Sun May 29, 2011 1:33 pm
First half is same i.e. k = (10(x+y) + 10y)/(x+y) = 10 + 10*y/(x+y)
but then you can remove option A and E (clearly k>10 and also k<20)
as x>y, y/(x+y)>1/2 and thus 10*y/(x+y)>5 and thus k>15
Only option available is D (18)
Thus OA D.

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by GMATGuruNY » Sun May 29, 2011 1:41 pm
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
An easy and efficient approach would be to plug in the answer choices, which represent the value of k:

Answer choice C: k=15
(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 answer choice D.
Answer choice D: k=18
10x + 20y = 18x + 18y
2y = 8x
y/x = 8/2
Success! x<y.

The correct answer is D.
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by cans » Sun May 29, 2011 10:15 pm
But suppose the options were
A)12
B)13
C)14
D)15
E)16
if we start with C) and then go to D and then to E, it can take sufficient time.

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by GMATGuruNY » Mon May 30, 2011 1:04 am
cans wrote:But suppose the options were
A)12
B)13
C)14
D)15
E)16
if we start with C) and then go to D and then to E, it can take sufficient time.
Given your set of answer choices:

Answer choice C (k=14) makes x>y:
10x + 20y = 14x + 14y
6y = 4x
y/x = 4/6.

Answer choice D (k=15) makes x=y:
10x + 20y = 15x + 15y
5y = 5x
y/x = 5/5.

Since k=14 makes x>y and k=15 makes x=y, the process is complete: we know that the next largest answer choice (k=16) makes x<y.

Quick and efficient. And no algebraic insight required.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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