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Seems so Easy???

Expert replies
by vladmire » Wed Dec 03, 2008 7:24 pm
This looks like a basic math concept of Laws of Operations but I'm missing something still.

If 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


Shouldn't I just be able to plug in the numbers for the variables and pick any number for the unknown variable and see if the equal on both sides of the equation?
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Source: — Problem Solving |

Re: Seems so Easy???

by hwiya320 » Wed Dec 03, 2008 7:34 pm
vladmire wrote:This looks like a basic math concept of Laws of Operations but I'm missing something still.

If 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


Shouldn't I just be able to plug in the numbers for the variables and pick any number for the unknown variable and see if the equal on both sides of the equation?
I would break it down like this,

xy+z = x(y+z)
then
xy+z = xy+xz, subtract xy
z = xz
so the question asks what MUST be true.
let's go one by one
a. x=0 and z=0 Not true, if z=0, then x does NOT have to be 0 to be true. (notice the "and" in the choices?)
b. x=1 and y=1 Not true, we don't even have to know what y is, we need to focus on Z
c. y=1 and z=0 Not true, again, we really want to focus on x or z
d. x= 1 or y= 0 Not true, same as above
e. x=1 or z = 0 I would pick this, if x was 1, then z=1z, true or if z=0, then 0=x0, true again.

Let's even try E in original equation.
if x=1, then 1y+z=1(y+z) , correct
if z=0, then xy+0=x(y+0), xy=xy

so answer is E.

You have to see the key words, and vs. or
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by raajan_p » Thu Dec 04, 2008 8:47 am
Just a tip :

For "which of the following" questions, try to solve from the bottom...

as in solve E first, then D....

You would probably find the answer between E and D...

but may not work always...
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by vittalgmat » Thu Dec 04, 2008 2:14 pm
from the question we have
xy +z = x(y+z)
-> xy +z = xy + xz
the xy cancells out so
z = xz
-> z -xz = 0
-> z(1 -x) = 0
So either z = 0 OR 1 -x = 0
ie z = 0 or x = 1

E
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