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DS - Inequality with exp.

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DS - Inequality with exp.

by karthikpandian19 » Fri Jul 06, 2012 4:24 pm
If x and y are non-zero integers, is y2 - x3 > 0?

x > y > -4
|y| > |x|
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by eagleeye » Fri Jul 06, 2012 4:48 pm
karthikpandian19 wrote:If x and y are non-zero integers, is y2 - x3 > 0?

x > y > -4
|y| > |x|
We are asked if y^2>x^3. x and y are non-zero integers. I did this by testing numbers.

Let's look at statement 1:

x> y> -4.

For y=-3, x=-2 we have y^2 > x^3 (9 > -8)
For y=1, x=5 we have y^2 < x^3 (1 < 125)
Insufficient.

For statement 2:
|y| > |x|
For y = 3, x=-2 we have y^2>x^3 ( 9>-8)
For y = 4, x=3 we have y^2 < x^3 ( 16 < 27)
Insufficient.

Together is where the problem becomes interesting.
We are told that x>y>-4
and that |y|>|x|
This works only when:
y=-3 and x= 2, -2, -1 or 1
y=-2 and x=-1, 1
For all other x, y combinations where x>y>-4, |x|>|y| does not work.
For cases when x is negative (-2, -1) y^2 is always positive here and x^3 is negative so y^2 > x^3.
For cases where x is positive, we need to check only for two cases:
For y=-3, x=2 here y^2 = 9, x^3 = 8. so y^2 > x^3
For y=-2, x=1 here y^2 =4, x^3 =1. so y^2>x^3.

So we get that in all cases, y^2 > x^3 when the two are taken together. Sufficient.

Hence C is the answer.

Let me know if this helps :)
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by karthikpandian19 » Fri Jul 06, 2012 6:23 pm
Thts a lot to test for a single sum...hmmmm
eagleeye wrote:
karthikpandian19 wrote:If x and y are non-zero integers, is y2 - x3 > 0?

x > y > -4
|y| > |x|
We are asked if y^2>x^3. x and y are non-zero integers. I did this by testing numbers.

Let's look at statement 1:

x> y> -4.

For y=-3, x=-2 we have y^2 > x^3 (9 > -8)
For y=1, x=5 we have y^2 < x^3 (1 < 125)
Insufficient.

For statement 2:
|y| > |x|
For y = 3, x=-2 we have y^2>x^3 ( 9>-8)
For y = 4, x=3 we have y^2 < x^3 ( 16 < 27)
Insufficient.

Together is where the problem becomes interesting.
We are told that x>y>-4
and that |y|>|x|
This works only when:
y=-3 and x= 2, -2, -1 or 1
y=-2 and x=-1, 1
For all other x, y combinations where x>y>-4, |x|>|y| does not work.
For cases when x is negative (-2, -1) y^2 is always positive here and x^3 is negative so y^2 > x^3.
For cases where x is positive, we need to check only for two cases:
For y=-3, x=2 here y^2 = 9, x^3 = 8. so y^2 > x^3
For y=-2, x=1 here y^2 =4, x^3 =1. so y^2>x^3.

So we get that in all cases, y^2 > x^3 when the two are taken together. Sufficient.

Hence C is the answer.

Let me know if this helps :)
Regards,
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON

---If you find my post useful, click "Thank" :) :)---
---Never stop until cracking GMAT---
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