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by tanya161 » Wed Sep 19, 2012 12:31 am
Q1 Is 2X-3Y< X2?

(1) 2X - 3Y = -2
(2) X>2 and Y>0

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by everything's eventual » Wed Sep 19, 2012 12:39 am
Is 2X - 3Y < X^2. ( Am I right in assuming it is x^2?)

1) 2X - 3Y = -2

This statement says that 2x-3y is negative. x^2 will always be positive.

Therefore, 2x - 3y < x^2. Sufficient.

2) X >2 and Y>0

Let x = 3 and y = 1

2(3) - 3 (1) = 3
x^2 = 3*3 = 9

3 is less than 9

Let x = 3 and y = 10

2(3) - 3(10) = -24
x^2 = 9

-24 is less than 9

Sufficient.

IMO D)

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by GMATGuruNY » Wed Sep 19, 2012 1:43 am
tanya161 wrote:Q1 Is 2X-3Y< X2?

(1) 2X - 3Y = -2
(2) X>2 and Y>0
Statement 1: 2x-3y = -2
Substituting 2x-3y = -2 into the question stem, we get:
-2 < x²?
Since the least possible value of x² is 0, we know that x² > -2.
SUFFICIENT.

Statement 2: X>2 and Y>0.
Let's rephrase the question stem:
2x - 3y < x²
0 < x² - 2x + 3y
0 < x(x-2) + 3y.

Question stem rephrased: Is x(x-2) + 3y > 0?
Since x>2 and y>0, every value on the lefthand side (x, x-2 and 3y) is positive.
Thus, we know that x(x-2) + 3y > 0.
SUFFICIENT.

The correct answer is D.
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