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Can this be done under 2 mins?

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by okigbo » Fri Oct 30, 2009 3:34 pm
If xy + z = x(y + z), which must be true?
a. x = 0; z = 0
b. x = 1; y = 1
c. y = 1; z = 0
d. x = 1 or y = 0
e. x = 1 or z = 0
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by truplayer256 » Fri Oct 30, 2009 3:58 pm
Yes, it can be done under two minutes. This problem should really only take you 45-55 secs to solve.

Let's take a look at it:

xy+z=x(y+z)

xy+z=xy+xz

z=xz

xz-z=0

z(x-1)=0

z=0, x=1

The correct answer is E.
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by vikram4689 » Mon Nov 02, 2009 5:44 am
xy + z = xy + xz
z = xz

=> either x=1 or z =0
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by mehravikas » Mon Nov 02, 2009 8:24 pm
Answer should be 'E'
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by heshamelaziry » Mon Nov 02, 2009 9:35 pm
truplayer256 wrote:Yes, it can be done under two minutes. This problem should really only take you 45-55 secs to solve.

Let's take a look at it:

xy+z=x(y+z)

xy+z=xy+xz

z=xz

xz-z=0

z(x-1)=0

z=0, x=1

The correct answer is E.

Why not a ? it yields 0 ?
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by life is a test » Mon Nov 02, 2009 10:39 pm
heshamelaziry wrote:
truplayer256 wrote:Yes, it can be done under two minutes. This problem should really only take you 45-55 secs to solve.

Let's take a look at it:

xy+z=x(y+z)

xy+z=xy+xz

z=xz

xz-z=0

z(x-1)=0

z=0, x=1

The correct answer is E.

Why not a ? it yields 0 ?
In A if x=0 then it doesnt matter what value z has the outcome will always be 0=0 so z needn't be 0; z can be any value.
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by heshamelaziry » Mon Nov 02, 2009 11:36 pm
life is a test wrote:
heshamelaziry wrote:
truplayer256 wrote:Yes, it can be done under two minutes. This problem should really only take you 45-55 secs to solve.

Let's take a look at it:

xy+z=x(y+z)

xy+z=xy+xz

z=xz

xz-z=0

z(x-1)=0

z=0, x=1

The correct answer is E.

Why not a ? it yields 0 ?
In A if x=0 then it doesnt matter what value z has the outcome will always be 0=0 so z needn't be 0; z can be any value.
How can z be any number if x = 0, in answer A ? If x = 0 then the between brackets will be -1 and if z = 2, then the answer to our re-written answer is -2 not zero ?
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by life is a test » Tue Nov 03, 2009 8:14 am
heshamelaziry wrote:
life is a test wrote:
heshamelaziry wrote:
truplayer256 wrote:Yes, it can be done under two minutes. This problem should really only take you 45-55 secs to solve.

Let's take a look at it:

xy+z=x(y+z)

xy+z=xy+xz

z=xz

xz-z=0

z(x-1)=0

z=0, x=1

The correct answer is E.

Why not a ? it yields 0 ?
In A if x=0 then it doesnt matter what value z has the outcome will always be 0=0 so z needn't be 0; z can be any value.
How can z be any number if x = 0, in answer A ? If x = 0 then the between brackets will be -1 and if z = 2, then the answer to our re-written answer is -2 not zero ?
sorry I got the explanation the wrong way around, if z=0 then x can be any value since z=xz will always be 0 =0 in that case.

hope that helps.
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by heshamelaziry » Tue Nov 03, 2009 9:44 am
So, the answer can be A or E !!!!! I think the question has a typo.
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by Stuart@KaplanGMAT » Tue Nov 03, 2009 2:56 pm
heshamelaziry wrote:So, the answer can be A or E !!!!! I think the question has a typo.
No, the question is fine.

The question is "which of the following MUST be true", not "which of the following COULD be true".

It MUST be true that either x=1 or z=0.

While it COULD be true that x=0 and z=0, it could also be false (e.g. we could pick x=100 and z=0), so A is not a "MUST be true".
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by heshamelaziry » Tue Nov 03, 2009 3:17 pm
If x = 100 and z = 0 still gives 0 ! I do not think that this is a good study question.
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by Stuart@KaplanGMAT » Tue Nov 03, 2009 3:23 pm
heshamelaziry wrote:If x = 100 and z = 0 still gives 0 ! I do not think that this is a good study question.
You're missing the point.

x=100 and z=0 still give 0, which is exactly why (A) doesn't have to be true; we can pick values that violate (A) and still satisfy the original equation.
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by life is a test » Tue Nov 03, 2009 8:29 pm
heshamelaziry wrote:So, the answer can be A or E !!!!! I think the question has a typo.
the ans is E. you can think of it another way:

z=xz

xz-z=0

z(x-1)=0 hence x=1 or z=0 must be true.
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