If x and y are positive, is x^3>y? (1) sqrt x > y (2) : DS Archive
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# If x and y are positive, is x^3>y? (1) sqrt x > y (2)

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VP
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If x and y are positive, is x^3>y? (1) sqrt x > y (2) [#permalink]

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22 May 2006, 22:07
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If x and y are positive, is x^3>y?

(1) sqrt x > y
(2) x > y
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SVP
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22 May 2006, 22:45

1 is not sufficient.

The first equation can be written as x> y^2
lets say x=1/2 and y=1/2. this satisfies the equation above. But x^3 = 1/8 and y=1/2. Hence x^3 is not greater than y
Say x =2 and y =2, Satisifies (1) but x^3 = 8 and y= 2, hence x^3> y
We cannot give the answer with certainty

2 is not sufficent.

Say X =1/2 and Y =1/4, satisfies (2), but x^3 < y
Say X=2 and Y=1 , satisfies (2) but X^3 > y

Both together also are not sufficient hence E
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22 May 2006, 23:06
St1:
If x = 16, y = 2,then x^3 > y

If x = 1/4, y = 1/3, then x^3 < y.

Insufficient.

St2:
x > y
Insufficient. If x = 2, y = 1, x^3 > y. If x = 1/2, y = 1/3, x^3 < y.

Using St1 and St2:
If x = 2, y = 1, sqrt(x) > y and x > y, x^3 > y

If x = 1/2, y = 1/3, sqrt(1/2) > y and x > y, x^3 < y

Insufficient.

Ans E
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23 May 2006, 07:18
Agree with the explanations. The key is to substitute fractions into the equations.
VP
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23 May 2006, 18:17
Great job guys, OA is E
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23 May 2006, 18:17
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