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M26-36

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M26-36  [#permalink]

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New post 16 Sep 2014, 01:26
1
10
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

32% (01:08) correct 68% (01:23) wrong based on 202 sessions

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Re M26-36  [#permalink]

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New post 16 Sep 2014, 01:26
3
2
Official Solution:


(1) \(xyz=-6\). Infinitely many combinations of \(x\), \(y\) and \(z\) are possible which will give different values of the expression in the stem: try \(x=y=1\) and \(y=-6\) or \(x=1\), \(y=2\), \(z=-3\). Not sufficient.

(2) \(x+y+z=0\). Rearrange: \(x=-(y+z)\) and substitute this value of \(x\) into the expression in the stem to get \(\frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}\), as \(x=-(y+z)\) then:

\(\frac{-3yz(y+z)}{xyz}=\frac{-3yz*(-x)}{xyz}=\frac{3xyz}{xyz}=3\). Sufficient.

Must know for the GMAT:\((x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3\) and \((x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3\).


Answer: B
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Re: M26-36  [#permalink]

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New post 04 Nov 2014, 18:21
1
Once it gets to this stage:

−3yz(y+z)
xyz

yz can be canceled out so we're only left with:

−3(y+z)
x

And since x = -(y+z)

3x
x

= 3
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Re: M26-36  [#permalink]

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New post 30 Aug 2017, 01:43
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Can I expect such type of questions in GMAT
this is more or less like CAT(Entrance Exam for Indian Institute of Management) Question
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Re: M26-36  [#permalink]

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New post 17 Dec 2017, 13:29
3
1
THERE IS A SIMPLE PROPERTY
IF X+Y+Z = 0
THEN X^3+Y^3+Z^3 = 3XYZ
HOPE IT HELPS
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M26-36  [#permalink]

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New post 05 May 2018, 18:46
THERE IS A SIMPLE PROPERTY
IF X+Y+Z = 0
THEN X^3+Y^3+Z^3 = 3XYZ
HOPE IT HELPS

I assume we only expect the property above when NONE of X,Y and Z is equal to zero.
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Re: M26-36  [#permalink]

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New post 24 Jul 2018, 06:19
Use the property -> x^3+y^3+z^3 - 3xyz=(x+y+z)*(x^2+y^2+z^2-xy-yz-zx) . If x+y+z=0 , The required expression comes to 3. I doubt though if such a formulaic question will ever come on the GMAT !!!
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Re: M26-36  [#permalink]

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New post 05 Sep 2018, 09:23
statement 1 not sufficient as many combinations of x,y and z are possible which can result into different values.
statement 2 implies:
\(x+y+z=0\)
\(x=-(y+z)\)
substitute for x
\(\frac{(-y^3 - z^3 - 3y^2z -3yz^z +y^3 + z^3)}{xyz}\)
\(\frac{-3yz(y+z)}{xyz}\)
\(\frac{-3yz(y+z)}{-yz(y+z)}\)
\(3\)
Sufficient

Answer is B
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Re: M26-36  [#permalink]

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New post 06 Sep 2018, 21:05
Any OG/GMAT question which employs (x+y)^3 formula in its solution?
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Re: M26-36 &nbs [#permalink] 06 Sep 2018, 21:05
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