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Algebra - DS

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Source: — Data Sufficiency |
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by WhiteWNNoise » Sun Feb 05, 2012 10:57 am
The correct answer is A.

Option 1:
=> x=2x+2y => x+2y=0 => x=-2y
Putting the value of x in (2x+y)/(x-2y), we get
=> (-3y)/(-4y) = 3/4 .......... Option 1 sufficient.

Option 2:
=> x=2y+4 => (2x+y)/(x-2y) = (5y+8)/(4) ............... Option 2 insufficient.
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by Anurag@Gurome » Sun Feb 05, 2012 7:23 pm
vinni.k wrote:Answer is A

Thanks & Regards
Vinni
(1) x/(x + y) = 2
x = 2x + 2y
x + 2y = 0 or x = -2y
So, (2x + y)/(x - 2y) = [2(-2y) + y]/[-2y - 2y] = -3y/-4y = 3/4; SUFFICIENT.

(2) x - 2y = 4
x = 4 + 2y
So, (2x + y)/(x - 2y) = [2(4 + 2y) + y]/[(4 + 2y) - 2y] = (5y + 8)/4, which is in terms of y; NOT sufficient.

The correct answer is A.
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by GMATGuruNY » Mon Feb 06, 2012 4:57 am
vinni.k wrote:Answer is A

Thanks & Regards
Vinni
What is (2x+y)/(x-2y)?

An alternate approach:
-- simplify the statements
-- plug different combinations of values into the question stem

Statement 1: x/(x+y) = 2.
First, a bit of algebra:
x = 2x + 2y
x = -2y.

Case 1: y=1, x=-2
(2x+y)/(x-2y) = (2(-2) + 1)/(-2 - 2*1) = -3/-4 = 3/4.
Case 2: y=2, x=-4
(2x+y)/(x-2y) = (2(-4) + 2)/(-4 - 2*2) = -6/-8 = 3/4.

Since the result in each case is 3/4, SUFFICIENT.

Statement 2: x-2y=4.
First a bit of algebra:
x = 2y+4

Case 1: y=1, x=6
(2x+y)/(x-2y) = (2*6 + 1)/(6 - 2*1) = 13/4.
Case 2: y=2, x=8
(2x+y)/(x-2y) = (2*8 + 2)/(8 - 2*2) = 18/4 = 9/2.

Since the result in the first case is 13/4 and the result in the second case is 9/2, INSUFFICIENT.

The correct answer is A.
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by vinni.k » Tue Feb 07, 2012 12:51 am
Thank you so much Mitch and Anurag. Appreciate your replies. :D . Now I can understand my mistake.

Regards
Vinni
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