Can this be done under 2 mins?

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Topic: Can this be done under 2 mins?

Fri Oct 30, 2009 4:34 pm

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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

truplayer256
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Fri Oct 30, 2009 4:58 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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Mon Nov 02, 2009 6:44 am

xy + z = xy + xz z = xz => either x=1 or z =0

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Mon Nov 02, 2009 9:24 pm

Answer should be 'E'

heshamelaziry
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Mon Nov 02, 2009 10:35 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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Mon Nov 02, 2009 11:39 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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Tue Nov 03, 2009 12:36 am

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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Tue Nov 03, 2009 9:14 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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Tue Nov 03, 2009 10:44 am

So, the answer can be A or E !!!!! I think the question has a typo.

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Tue Nov 03, 2009 3:56 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". _________________ Stuart Kovinsky, B.A. LL.B. Academic Manager Admissions Consultant Kaplan Test Prep & Admissions Toronto Office 1-800-KAP-TEST GMAT Blogs Learn more about me

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Tue Nov 03, 2009 4:17 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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Tue Nov 03, 2009 4:23 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. _________________ Stuart Kovinsky, B.A. LL.B. Academic Manager Admissions Consultant Kaplan Test Prep & Admissions Toronto Office 1-800-KAP-TEST GMAT Blogs Learn more about me

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Tue Nov 03, 2009 9:29 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.