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# Is x > 3?

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Math Expert
Joined: 02 Sep 2009
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10 Sep 2017, 20:41
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25% (medium)

Question Stats:

70% (01:14) correct 30% (01:11) wrong based on 47 sessions

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Is x > 3?

(1) $$\frac{(x^2y)^3}{(xy^3)^2} = 3$$

(2) $$x^3y^{(–3)} = 1$$
[Reveal] Spoiler: OA

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10 Sep 2017, 21:54
(1) $$\frac{(x^2y)^3}{(xy^3)^2} = 3$$
Simplifying,
$$\frac{(x^6*y^3)}{(x^2*y^6)} = 3$$ => $$\frac{x^{(6-2)}}{y^{(6-3)}} = 3$$ => $$\frac{(x^4)}{(y^3)} = 3$$
Because y can take any value, we cannot say anything clearly about the value of x(Insufficient)

(2) $$x^3y^{(–3)} = 1$$
This can also be written as $$\frac{(x^3)}{(y^3)} = 1$$
Because y can take any value, we cannot say anything clearly about the value of x(Insufficient)

On combining the information from both the statements,
If we divide equation in statement 2 from statement 1
$$\frac{{(x^4)}}{{(y^3)}}*\frac{{(y^3)}}{{(x^3)}} = \frac{3}{1}$$
This simplifies to give x = 3.

Therefore, the equation x>3 will always be false(Sufficient)(Option C)
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Is x > 3?   [#permalink] 10 Sep 2017, 21:54
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