Seems so Easy???

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vladmire: Master | Next Rank: 500 Posts; Posts: 132; Joined: Tue Oct 07, 2008 4:59 pm; Thanked: 4 times

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: Beat The GMAT — Problem Solving | Wed Dec 03, 2008 7:24 pm

hwiya320: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Tue Sep 02, 2008 5:34 pm; Thanked: 7 times

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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raajan_p: Senior | Next Rank: 100 Posts; Posts: 75; Joined: Wed Aug 27, 2008 7:19 am; Location: Orlando; Thanked: 6 times

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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vittalgmat: Legendary Member; Posts: 621; Joined: Wed Apr 09, 2008 7:13 pmFollowed by: 4 members; Thanked: 33 times

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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