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Source: — Data Sufficiency |
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by neelgandham » Thu Jan 12, 2012 10:45 am
If x and y are nonzero, is (x^3)(y^5)/(x^2)(y^3)>0
The question can be rephrased to
Is x*(y^2)>0 ?
1). x>0
y^2 > 0
x > 0
x*(y^2)>0
Sufficient
2). y>0
y^2 > 0
x ??
Insufficient

IMO A
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by chieftang » Thu Jan 12, 2012 4:20 pm
sud21 wrote:If x and y are nonzero, is (x^3)(y^5)/(x^2)(y^3)>0
1). x>0
2). y>0
1. Let y<0, then

(positive)(negative) / (positive)(negative) = positive

Let y>0, then

(positive)(positive) / (positive)(positive) = positive

SUFFICIENT

2. Let x < 0, then

(negative)(positive) / (positive)(positive) = negative

Let x > 0, then

(positive)(positive) / (positive)(positive) = positive

INSUFFICIENT



ANSWER A
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