Is 2x-3y<x^2 ?
1. 2x-3y<-2
=> 2x-3y<0
x^2>0
=>2x-y<x^2 sufficiennt
2. x>2, y>0
for x>2, 2x<x^2
=>2x-3y<x^2 (we are subtracting a positive number from something that is already less than x^2, therby making it still lesser
sufficent
hence D
OG 11th, DS #119
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Source: Beat The GMAT — Data Sufficiency |
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scoobydooby
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Statement 1:
A square of any number will always be non-negative. So, x^2 must be a non-negative number. If we already know that 2x-3y is less than -2, then we are sure that 2x-3y is less than x^2. Sufficient.
Statement 2:
Is 2x - 3y < x^2 ?
Is x^2 - 2x > -3y ?
Is x(x-2) > -3y ?
Since y>0, -3y can only be a negative number.
Since x>2, x(x-2) can only be positive.
Therefore, we can prove that x(x-2) is always larger than -3y. Sufficient.
Choose D.
-BM-
A square of any number will always be non-negative. So, x^2 must be a non-negative number. If we already know that 2x-3y is less than -2, then we are sure that 2x-3y is less than x^2. Sufficient.
Statement 2:
Is 2x - 3y < x^2 ?
Is x^2 - 2x > -3y ?
Is x(x-2) > -3y ?
Since y>0, -3y can only be a negative number.
Since x>2, x(x-2) can only be positive.
Therefore, we can prove that x(x-2) is always larger than -3y. Sufficient.
Choose D.
-BM-
