GMATPrep xy + z = x(y+z)

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GMATPrep xy + z = x(y+z)

by shilotilo » Mon Sep 08, 2008 9:55 pm
Hi,
Firstly, I have done a search and I couldn't find this question. (I searched for both "xy+z" and "x(y+z)" )
Secondly, I apologise if it is a silly question BUT ...
I ended up getting each of the options other than d as a “must be true”, so obviously I have missed out something basic in my Quant studies. Can someone please tell me what’s wrong with my calculations?

The question is:

If xy + z = x(y+z), which must be true?---

1) x=0 and z=0

substituting the values of x and z in the stem’s equation,
0*y + 0 = 0( y + 0)
and so 0=0 ….. True

2) x=1 and y = 1

substituting the values of x and y in the equation, we get
1*1 +z =1(1+z) and from there,
1+z =1+z …. True

3)y=1 and z =0
substituting the values of y and z in the equation, we get
x*1 +0 =x(1+0)
and so x = x….. True

4)x=0 or y=0
substituting the value of x, we get
0*y +z= 0(y+z)
and so, z=0, which may or may not be true…… so discarded.

5)x=1 or z=0 (official answer)


Thanks.
Shilo

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Re: GMATPrep xy + z = x(y+z)

by Stuart@KaplanGMAT » Mon Sep 08, 2008 10:44 pm
shilotilo wrote:Hi,
Firstly, I have done a search and I couldn't find this question. (I searched for both "xy+z" and "x(y+z)" )
Secondly, I apologise if it is a silly question BUT ...
I ended up getting each of the options other than d as a “must be true”, so obviously I have missed out something basic in my Quant studies. Can someone please tell me what’s wrong with my calculations?

The question is:

If xy + z = x(y+z), which must be true?---
The problem is with your interpretation of the question.

When a question asks "which of the following MUST" be true, we're looking for the option that's true in every possible case.

You interpreted the question as "which of the following COULD be true." Let's look at (a) as an example:
1) x=0 and z=0

substituting the values of x and z in the stem’s equation,
0*y + 0 = 0( y + 0)
and so 0=0 ….. True
You've shown that x and z COULD each equal 0. However, if we let x, y and z all = 1, we get:

1*1 + 1 = 1(1+1)
2 = 2

which also works.

As an aside, this question is a lot easier if you simplify the first equation:

xy + z = x(y+z)
xy + z = xy + xz
z = xz
z - xz = 0
z(1 - x) = 0

and we now know that either:

z = 0 or 1 - x = 0
z = 0 or x = 1

which is (e).
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by shilotilo » Mon Sep 08, 2008 11:32 pm
Thanks Stuart !

:)