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What is the value of

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Senior Manager
Joined: 22 Feb 2018
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What is the value of  [#permalink]

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13 Apr 2018, 02:18
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95% (hard)

Question Stats:

39% (01:52) correct 61% (02:30) wrong based on 67 sessions

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What is the value of $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$, Given x,y are positive integer.

Statement 1: $$x^3 + y^ 3 + \frac{512}{x^3.y^3}= 24$$
Statement 2: $$x=y$$

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Re: What is the value of  [#permalink]

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13 Apr 2018, 05:56
4
Quite intersting question
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What is the value of  [#permalink]

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13 Apr 2018, 08:11
1
Princ wrote:
What is the value of $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$, Given x,y are positive integer.

Statement 1: $$x^3 + y^ 3 + \frac{512}{x^3.y^3}= 24$$
Statement 2: $$x=y$$

gmatbusters Nice Approach

This is another approach.
We have to find value of $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$, Given x,y are positive integer.
So we have to get individual value of x and y to get the definite value of expression .

Statement 1 : $$x^3 + y^ 3 + \frac{512}{x^3.y^3}= 24$$
Taking Arithmatic mean and Geometric mean of $$x^3$$, $$y^ 3$$, $$\frac{512}{x^3.y^3}$$, we get
A.M=$$\frac{x^3 + y^3 +\frac{512}{x^3.y^3}}{3}$$ ; G.M = $$\sqrt[3]{x^3*y^ 3*\frac{512}{x^3.y^3}}$$ further simplifying, we get G.M =8.
As $$A.M\geq{G.M}$$

$$\frac{x^3 + y^3 +\frac{512}{x^3.y^3}}{3}$$≥8

$$x^3 + y^3 +\frac{512}{x^3.y^3}$$≥24

So Statement 1 is implying , that A.M is equal to G.M.This occurs when all terms are equal i.e

$$x^3$$=$$y^ 3$$=$$\frac{512}{x^3.y^3}$$

From $$y^ 3$$=$$\frac{512}{x^3.y^3}$$ and substituting $$x^3$$ with $$y^ 3$$, we get $$y^9-512=0$$

$$y^9-2^9$$= $$(y^3)^3-(2^3)^3$$$$=0$$
$$(y^3)^3-(2^3)^3$$=$$(y^3-2^3)(y^6+y^3.2^3+2^6)$$$$=0$$,further expanding we get
$$(y-2)(y^2+2.y+2^2)(y^6+y^3.2^3+2^6)=0$$
So $$y =2$$, as $$(y^2+2.y+2^2)(y^6+y^3.2^3+2^6)$$ is +ve qty, cannot be equal to 0.

Similarly or using $$x^3=y^3$$, we can get $$x=2$$
As we know both $$x$$ and $$y$$ , we can calculate expression $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$
Statement 1 is alone sufficient

Statement 2: $$x=y$$
As $$x$$ and $$y$$ are positive integer, They can take infinite different values even if $$x=y$$.
Expression $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$ will give infinite different value.
Statement 2 is not sufficient.

So OA should be A.
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Joined: 27 Oct 2017
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Re: What is the value of  [#permalink]

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13 Apr 2018, 10:44
That's pretty impressive explanation mate...

Nice concept, keep posting conceptual problems like this.

Princ wrote:
Princ wrote:
What is the value of $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$, Given x,y are positive integer.

Statement 1: $$x^3 + y^ 3 + \frac{512}{x^3.y^3}= 24$$
Statement 2: $$x=y$$

gmatbusters Nice Approach

This is another approach.
We have to find value of $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$, Given x,y are positive integer.
So we have to get individual value of x and y to get the definite value of expression .

Statement 1 : $$x^3 + y^ 3 + \frac{512}{x^3.y^3}= 24$$
Taking Arithmatic mean and Geometric mean of $$x^3$$, $$y^ 3$$, $$\frac{512}{x^3.y^3}$$, we get
A.M=$$\frac{x^3 + y^3 +\frac{512}{x^3.y^3}}{3}$$ ; G.M = $$\sqrt[3]{x^3*y^ 3*\frac{512}{x^3.y^3}}$$ further simplifying, we get G.M =8.
As $$A.M\geq{G.M}$$

$$\frac{x^3 + y^3 +\frac{512}{x^3.y^3}}{3}$$≥8

$$x^3 + y^3 +\frac{512}{x^3.y^3}$$≥24

So Statement 1 is implying , that A.M is equal to G.M.This occurs when all terms are equal i.e

$$x^3$$=$$y^ 3$$=$$\frac{512}{x^3.y^3}$$

From $$y^ 3$$=$$\frac{512}{x^3.y^3}$$ and substituting $$x^3$$ with $$y^ 3$$, we get $$y^9-512=0$$

$$y^9-2^9$$= $$(y^3)^3-(2^3)^3$$$$=0$$
$$(y^3)^3-(2^3)^3$$=$$(y^3-2^3)(y^6+y^3.2^3+2^6)$$$$=0$$,further expanding we get
$$(y-2)(y^2+2.y+2^2)(y^6+y^3.2^3+2^6)=0$$
So $$y =2$$, as $$(y^2+2.y+2^2)(y^6+y^3.2^3+2^6)$$ is +ve qty, cannot be equal to 0.

Similarly or using $$x^3=y^3$$, we can get $$x=2$$
As we know both $$x$$ and $$y$$ , we can calculate expression $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$
Statement 1 is alone sufficient

Statement 2: $$x=y$$
As $$x$$ and $$y$$ are positive integer, They can take infinite different values even if $$x=y$$.
Expression $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$ will give infinite different value.
Statement 2 is not sufficient.

So OA should be A.

_________________
Senior DS Moderator
Joined: 27 Oct 2017
Posts: 904
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
Re: What is the value of  [#permalink]

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13 Apr 2018, 10:49
I want to say that after this step "So Statement 1 is implying , that A.M is equal to G.M.This occurs when all terms are equal "
It can be done like this...
hence $$x^3 = y^3 = 512/(x^3*y^3) = 24/3$$
or $$x^3 = y^3 = 8$$
or x= y=2.
hence we can find the value of the expression asked in the question stem.

Sufficient.

Princ wrote:
Princ wrote:
What is the value of $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$, Given x,y are positive integer.

Statement 1: $$x^3 + y^ 3 + \frac{512}{x^3.y^3}= 24$$
Statement 2: $$x=y$$

gmatbusters Nice Approach

This is another approach.
We have to find value of $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$, Given x,y are positive integer.
So we have to get individual value of x and y to get the definite value of expression .

Statement 1 : $$x^3 + y^ 3 + \frac{512}{x^3.y^3}= 24$$
Taking Arithmatic mean and Geometric mean of $$x^3$$, $$y^ 3$$, $$\frac{512}{x^3.y^3}$$, we get
A.M=$$\frac{x^3 + y^3 +\frac{512}{x^3.y^3}}{3}$$ ; G.M = $$\sqrt[3]{x^3*y^ 3*\frac{512}{x^3.y^3}}$$ further simplifying, we get G.M =8.
As $$A.M\geq{G.M}$$

$$\frac{x^3 + y^3 +\frac{512}{x^3.y^3}}{3}$$≥8

$$x^3 + y^3 +\frac{512}{x^3.y^3}$$≥24

So Statement 1 is implying , that A.M is equal to G.M.This occurs when all terms are equal i.e

$$x^3$$=$$y^ 3$$=$$\frac{512}{x^3.y^3}$$

From $$y^ 3$$=$$\frac{512}{x^3.y^3}$$ and substituting $$x^3$$ with $$y^ 3$$, we get $$y^9-512=0$$

$$y^9-2^9$$= $$(y^3)^3-(2^3)^3$$$$=0$$
$$(y^3)^3-(2^3)^3$$=$$(y^3-2^3)(y^6+y^3.2^3+2^6)$$$$=0$$,further expanding we get
$$(y-2)(y^2+2.y+2^2)(y^6+y^3.2^3+2^6)=0$$
So $$y =2$$, as $$(y^2+2.y+2^2)(y^6+y^3.2^3+2^6)$$ is +ve qty, cannot be equal to 0.

Similarly or using $$x^3=y^3$$, we can get $$x=2$$
As we know both $$x$$ and $$y$$ , we can calculate expression $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$
Statement 1 is alone sufficient

Statement 2: $$x=y$$
As $$x$$ and $$y$$ are positive integer, They can take infinite different values even if $$x=y$$.
Expression $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$ will give infinite different value.
Statement 2 is not sufficient.

So OA should be A.
[/quote]
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Joined: 26 Mar 2013
Posts: 1839
Re: What is the value of  [#permalink]

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15 Apr 2018, 14:10
Princ wrote:
What is the value of $$x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y$$, Given x,y are positive integer.

Statement 1: $$x^3 + y^ 3 + \frac{512}{x^3.y^3}= 24$$
Statement 2: $$x=y$$

Nice question.

what is the source of the question?
Director
Joined: 02 Oct 2017
Posts: 651
Re: What is the value of  [#permalink]

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28 Apr 2018, 09:43
I) solving equation we get

x^3*y^3(x^3+y^3-24)=-512
So
On further analysis 512 can be made via
(xy)^3=64 x^3+y^3-24=-8
So x and y=2
Put value in eqn and find and
Sufficient

II)terms are still there only replace y with x .no possible solution via this.
Insufficient

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Re: What is the value of &nbs [#permalink] 28 Apr 2018, 09:43
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