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PS :- GMAT Prep

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Re: PS :- GMAT Prep

by gabriel » Tue Jul 10, 2007 6:27 am
sahaldar wrote:1. xy+z=x(y+z) which of the following must be true.
a) x=0 and z=0
b) x=1 and y=1
c) y=1 and z=0
d) x=1 or y=0
e) x=1 or z=0

All the given anwsers satisfies the equation. why the anwser is (d)
we have xy+z=x(y+z) .. which boils down to z=xz .. to satisfy this condition there are 2 possiblities

first x = 1 .. in which case z = 1*z ..that is z=z ...

second z= 0 .. in which case 0=x*0 .. that is 0=0 ..

so the the answer should be e
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Re: PS :- GMAT Prep

by Shadow » Tue Jul 10, 2007 11:05 pm
gabriel wrote:
sahaldar wrote:1. xy+z=x(y+z) which of the following must be true.
a) x=0 and z=0
b) x=1 and y=1
c) y=1 and z=0
d) x=1 or y=0
e) x=1 or z=0

All the given anwsers satisfies the equation. why the anwser is (d)
we have xy+z=x(y+z) .. which boils down to z=xz .. to satisfy this condition there are 2 possiblities

first x = 1 .. in which case z = 1*z ..that is z=z ...

second z= 0 .. in which case 0=x*0 .. that is 0=0 ..

so the the answer should be e

It should be D

The equation boils down to z=xz
Cancelling z on both sides we get 1=x

so the value of x=1

so we can eliminate choice A and C

Again having a look into the equation,
xy+z=xy+xz

Put y=0, we get 1=x

So the choice should be D.
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Re: PS :- GMAT Prep

by gabriel » Wed Jul 11, 2007 9:49 am
Shadow wrote:
gabriel wrote:
sahaldar wrote:1. xy+z=x(y+z) which of the following must be true.
a) x=0 and z=0
b) x=1 and y=1
c) y=1 and z=0
d) x=1 or y=0
e) x=1 or z=0

All the given anwsers satisfies the equation. why the anwser is (d)
we have xy+z=x(y+z) .. which boils down to z=xz .. to satisfy this condition there are 2 possiblities

first x = 1 .. in which case z = 1*z ..that is z=z ...

second z= 0 .. in which case 0=x*0 .. that is 0=0 ..

so the the answer should be e

It should be D

The equation boils down to z=xz
Cancelling z on both sides we get 1=x

so the value of x=1

so we can eliminate choice A and C

Again having a look into the equation,
xy+z=xy+xz

Put y=0, we get 1=x

So the choice should be D.

our aim over here is not to prove that x= 1 .. our aim is to find the condition under which the equation xy+z=x(y+z) is true ..

...by putting x= 1 we get y+z=y+z .. that is the equation is true if x = 1

... by putting z=0 we get xy = xy that is we again prove that the equation is true .. if z= 0

remeber we are not trying to find the value of any variable over here .. but are trying to find the conditions under which a seemingly unequal equation is equal ..
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Re: PS :- GMAT Prep

by Sadowski » Mon Jul 23, 2007 7:01 am
gabriel wrote:
Shadow wrote:
gabriel wrote:
sahaldar wrote:1. xy+z=x(y+z) which of the following must be true.
a) x=0 and z=0
b) x=1 and y=1
c) y=1 and z=0
d) x=1 or y=0
e) x=1 or z=0

All the given anwsers satisfies the equation. why the anwser is (d)
we have xy+z=x(y+z) .. which boils down to z=xz .. to satisfy this condition there are 2 possiblities

first x = 1 .. in which case z = 1*z ..that is z=z ...

second z= 0 .. in which case 0=x*0 .. that is 0=0 ..

so the the answer should be e

It should be D

The equation boils down to z=xz
Cancelling z on both sides we get 1=x

so the value of x=1

so we can eliminate choice A and C

Again having a look into the equation,
xy+z=xy+xz

Put y=0, we get 1=x

So the choice should be D.

our aim over here is not to prove that x= 1 .. our aim is to find the condition under which the equation xy+z=x(y+z) is true ..

...by putting x= 1 we get y+z=y+z .. that is the equation is true if x = 1

... by putting z=0 we get xy = xy that is we again prove that the equation is true .. if z= 0

remeber we are not trying to find the value of any variable over here .. but are trying to find the conditions under which a seemingly unequal equation is equal ..
I must be missing something. Your logic makes sense, but it doesn't prove why other answers are incorrect. For instance, choosing option C (y=1, z=0) we get x=x. This proves that the equation is true just as well as choice D proves it.
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by Sadowski » Mon Jul 23, 2007 7:52 am
Anonymous wrote:In fact according to D it's either X=1 or Y=0....

Now if we put only Y=0..equation comes to Z=XZ....

this can't be true since X can hold any value....it means D can't be answer...let me know if I am doing mistake.....
Actually, y=0 cancels out the x on both sides, so Z=Z. However, I'm still not sure why this is the best scenario compared to all other answers.
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by UmanG » Mon Jul 23, 2007 8:57 am
Ohhh man,

XY+Z = X(Y+Z)
XY+Z = XY + XZ
putting Y = 0
Z = XZ

How Y eliminates XZ....Can you please explain buddy???
Thanks,
UmanG - restless mind..
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by Sadowski » Mon Jul 23, 2007 9:01 am
UmanG wrote:Ohhh man,

XY+Z = X(Y+Z)
XY+Z = XY + XZ
putting Y = 0
Z = XZ

How Y eliminates XZ....Can you please explain buddy???
You're right, my bad.
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by gabriel » Mon Jul 23, 2007 12:04 pm
Sadowski wrote:
gabriel wrote:
Shadow wrote:
gabriel wrote:
sahaldar wrote:1. xy+z=x(y+z) which of the following must be true.
a) x=0 and z=0
b) x=1 and y=1
c) y=1 and z=0
d) x=1 or y=0
e) x=1 or z=0

All the given anwsers satisfies the equation. why the anwser is (d)
we have xy+z=x(y+z) .. which boils down to z=xz .. to satisfy this condition there are 2 possiblities

first x = 1 .. in which case z = 1*z ..that is z=z ...

second z= 0 .. in which case 0=x*0 .. that is 0=0 ..

so the the answer should be e

It should be D

The equation boils down to z=xz
Cancelling z on both sides we get 1=x

so the value of x=1

so we can eliminate choice A and C

Again having a look into the equation,
xy+z=xy+xz

Put y=0, we get 1=x

So the choice should be D.

our aim over here is not to prove that x= 1 .. our aim is to find the condition under which the equation xy+z=x(y+z) is true ..

...by putting x= 1 we get y+z=y+z .. that is the equation is true if x = 1

... by putting z=0 we get xy = xy that is we again prove that the equation is true .. if z= 0

remeber we are not trying to find the value of any variable over here .. but are trying to find the conditions under which a seemingly unequal equation is equal ..
I must be missing something. Your logic makes sense, but it doesn't prove why other answers are incorrect. For instance, choosing option C (y=1, z=0) we get x=x. This proves that the equation is true just as well as choice D proves it.
... the question over here is not what can be true but it is what must be true ...

... take a close look at the given equation .. xy+z=x(y+z) .. that is xy + z = xy +xz ...

.. as we can see xy is eliminated from both the sides and the eqn boils down to z = xz .. that means y is out of the picture .. that is it doesnt matter what value y takes bcoz eventually the terms with y are eliminated from the eqn .. y can take a value of 10 million ..still doesnt matter ..

now look at the equation that is left that is z = xz ... suppose z = 200 ... we have 200 = x 200 ... for this to be true x must be 1 ... what if the value of x = 2 ..we have z = 2*z .. which can never be true .. so for the eqn xy+z = x(y+z) to be true x has to be 1 ...

.. the other possiblity for the eqn z = xz to be true is that both sides are equal to 0 ... for that to be true z must be 0 ..

.. these are the only 2 ways the above eqn is true .. so the answer is z=0 or x = 1 .. remember it is about what must be true not what can be true ...
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by beny » Mon Jul 23, 2007 11:40 pm
gabriel is right. The answer is E. This question is easy. I don't know why there's so much debate and confustion.
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