If x and y are positive, is x^3 > y? : GMAT Data Sufficiency (DS)
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# If x and y are positive, is x^3 > y?

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

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30 Aug 2010, 08:38
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If x and y are positive, is x^3 > y?

(1) $$\sqrt{x} > y$$
(2) x > y
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Dec 2012, 02:58, edited 2 times in total.
Edited the question.
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30 Aug 2010, 09:31
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udaymathapati wrote:
If x and y are positive, is $$x^3$$ > y?
(1) \sqrt{x} > y
(2) x > y

My approach:
1.\sqrt{x}>y ....>x>$$y^2$$....NS
2. x>y ....NS

Combine...$$y^2$$>x>y....> $$x^3$$>y. This can be verified by taking no like 4>3>2..where y=2 and x=3

Any other apporach.

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 $$y=\frac{1}{8}$$

Hope it's clear.
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31 Aug 2010, 03:07
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When there are squaring or cubing is involved, always remember -ve values as well as the values between -1 and 1
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18 Mar 2011, 06:11
Bunuel wrote:
udaymathapati wrote:
If x and y are positive, is $$x^3$$ > y?
(1) \sqrt{x} > y
(2) x > y

My approach:
1.\sqrt{x}>y ....>x>$$y^2$$....NS
2. x>y ....NS

Combine...$$y^2$$>x>y....> $$x^3$$>y. This can be verified by taking no like 4>3>2..where y=2 and x=3

Any other apporach.

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 $$y=\frac{1}{8}$$

Hope it's clear.

+1 for u
Great Post Thanks Bunuel
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18 Mar 2011, 20:25
x = 1/4, y = 1/8

From (1)

1/2 > 1/8, but 1/8 = 1/8

but (9)^3 > 2

and sqrt(9) > 2

In (2) also, the same is true.

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Re: If x and y are positive, is x^3 > y? [#permalink]

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30 May 2014, 06:10
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udaymathapati wrote:
If x and y are positive, is x^3 > y?

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

My method without number plugging:

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

Square both sides. Since x & y are both positive, we know this doesn't change the sign at all (because $$f(x) = x^2$$ is a strictly increasing function for $$x \geq 0$$). Thus $$x > y^2$$. We're trying to get this in a form that resembles our original question, so we multiply each side by $$x^2$$(which we know is positive). Thus, $$x^3 > x^2y^2$$, or that$$x^3 > y (x^2 y).$$ Therefore, our question is TRUE if and only if $$x^2y\geq 1$$. Since we don't know anything about the behavior of$$x^2y$$, we mark 1) as INSUFFICIENT.

2) $$x>y.$$

Multiplying both sides by $$x^2$$, we get that $$x^3 > yx^2$$. Thus, if $$x^2 \geq 1,$$ we get TRUE, otherwise, we get FALSE. Since we have no information about $$x^2$$, we mark this as INSUFFICIENT.

Taking 1) and 2) together, we still have no information about $$x^2$$ or $$x^2y$$, so 1) & 2) are INSUFFICIENT.

Last edited by speedilly on 30 May 2014, 14:37, edited 1 time in total.
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30 May 2014, 13:46
Bunuel wrote:
udaymathapati wrote:
If x and y are positive, is $$x^3$$ > y?
(1) \sqrt{x} > y
(2) x > y

My approach:
1.\sqrt{x}>y ....>x>$$y^2$$....NS
2. x>y ....NS

Combine...$$y^2$$>x>y....> $$x^3$$>y. This can be verified by taking no like 4>3>2..where y=2 and x=3

Any other apporach.

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 $$y=\frac{1}{8}$$

Hope it's clear.

What if the question says x and y are positive INTEGERS ??
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30 May 2014, 14:40
ShajibAhmed wrote:

What if the question says x and y are positive INTEGERS ??

If they were both integers, then the answer would be D: either of them are sufficient, since we'd know that$$x^2y \geq 1$$ and $$x^2 \geq 1$$(see my solution above). The question doesn't state that, however, so we must consider all cases.
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