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# If x and y are positive, is x^3 > y?

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Manager
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If x and y are positive, is x^3 > y? [#permalink]  23 Jun 2010, 11:30
00:00

Difficulty:

55% (hard)

Question Stats:

41% (02:57) correct 59% (00:26) wrong based on 37 sessions
If x and y are positive, is x^3 > y?

(1) $$\sqrt{x} > y$$
(2) x > y

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-is-x-3-y-100086.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Jun 2014, 01:22, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Manager
Joined: 03 May 2010
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WE 1: 2 yrs - Oilfield Service
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Re: If x and y are positive, is x^3 > y? [#permalink]  23 Jun 2010, 23:29
Statement 1 is actually "root(x) > y"

x>0 and y>0 Is x^3 > y?

From statement 1:
root(x) > y
Test cases:
x = 4 and y = 1 => root(x) = 2 > 1. x^3 = 64 > 1 ... Answer : Yes
x = 0.01 and y = 0.005 => root(x) = 0.1 > 0.05. x^3 = 0.001 < 0.005 ...Answer: No
INSUFFICIENT

From statement 2:
x > y
Test cases:
x = 2 and y = 3 => x^3 = 8 > 3. Answer -Yes
x = 0.1 and y = 0.01 => x^3 = 0.001 < 0.01. Answer - No
INSUFFICIENT

Both 1 and 2 together:
Test cases:
x = 4 and y = 1 satisfies both statements, and x^3 > y is true.
x = 0.01 and y = 0.005 satisfies both statements and x^3>y is false.
INSUFFICIENT.

Pick E.
Manager
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Re: If x and y are positive, is x^3 > y? [#permalink]  26 Jun 2014, 23:36
Hi Bunuel,

Can we do this ques based on graphs .. I am working on a technique that could help me to hit all in equality questions.

Thanks
Math Expert
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Re: If x and y are positive, is x^3 > y? [#permalink]  27 Jun 2014, 01:27
Expert's post
GmatDestroyer2013 wrote:
Hi Bunuel,

Can we do this ques based on graphs .. I am working on a technique that could help me to hit all in equality questions.

Thanks

I wouldn't suggest to use graph approach for this question because x^3 and $$\sqrt{x}$$ functions are not easy to plot and compare. Below are two approaches which are good for it.

If x and y are positive, is x^3>y?

NUMBER PLUGGING:

(1) $$\sqrt{x}>y$$ --> if $$x=1$$ and $$y=\frac{1}{2}$$ then the answer will be YES but if $$x=\frac{1}{4}$$ and $$y=\frac{1}{5}$$ then the answer will be NO. Two different answers, hence not sufficient.

(2) $$x>y$$ --> if $$x=1$$ and $$y=\frac{1}{2}$$ then the answer will be YES but if $$x=\frac{1}{4}$$ and $$y=\frac{1}{5}$$ then the answer will be NO. Two different answers, hence not sufficient.

(1)+(2) Both examples are valid for combined statements, so we still have two answers. Not sufficient.

ALGEBRAIC APPROACH:

For $$1\leq{x}$$: ------$$\sqrt{x}$$----$$x$$----$$x^3$$, so $$1\leq{\sqrt{x}}\leq{x}\leq{x^3}$$ (the case $$\sqrt{x}=x=x^3$$ is when $$x=1$$). $$y$$ is somewhere in green zone (as $$y<\sqrt{x}$$ and $$y<x$$), so if we have this case answer is always YES: $$y<x^3$$.

But:

For $$0<x<1$$: $$0$$----$$x^3$$----$$x$$----$$\sqrt{x}$$----$$1$$, so $$0<x^3<x<\sqrt{x}$$. $$y$$ is somewhere in green or red zone (as $$y<\sqrt{x}$$ and $$y<x$$), so if we have this case answer is sometimes YES: $$y<x^3$$ (if $$y$$ is in green zone), and sometimes NO: $$x^3<y$$ (if $$y$$ is in red zone). In fact in this case $$y=x^3$$ is also possible, for example when $$x=\frac{1}{2}$$ and $$y=\frac{1}{8}$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-is-x-3-y-100086.html
_________________
Re: If x and y are positive, is x^3 > y?   [#permalink] 27 Jun 2014, 01:27
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