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What type of question ?

Expert replies
Source: — Problem Solving |

by vr4indian » Mon Sep 22, 2008 1:48 am
I will go for E

Explanation.

XY+Z = X(Y+Z)

xy +z = xy + xz ------------eq1

To cover max. values for X, Y, Z we will go with OR option instead of And option

Take 4 as first option

for X =1 its perfectly okay but if we take Y =0 means z = xz , which can be false.

Take 5 as second option
If we consider X =1 then equation will be y + z = y +z , this will be valid for all the values of Y and Z

Now considerZ =0 , so eq1 will become xy = xy , which is again valid for any valiues of X and Z

..
PS.
I think B can be valid answer.. but just because of AND clause i rulled out it.

Vishal Shah
Pune
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by Xbond » Mon Sep 22, 2008 4:31 am
OK, Could you clarify how you select between AND & OR ?
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Re: What type of question ?

by lunarpower » Tue Sep 23, 2008 1:38 am
Xbond wrote:If XY+Z = X(Y+Z), which of the following must be

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

Can you help me to understand how to resolve this kind of PS ?
well, the first thing you want to do is simplify the original "given" statement.
distribute to get xy + z = xy + xz
then subtract xy from both sides (remember, subtracting a quantity from both sides is always "safe") to give z = xz.

DO NOT DIVIDE BY Z at this point. if you don't know this, then this should be a MAJOR MAJOR takeaway: NEVER divide by a variable, unless you are absolutely sure that the variable is nonzero.
if you have a variable that could possibly be zero (which is most variables), then FACTOR IT OUT instead of dividing by it.

z = xz
0 = xz - z
0 = z(x - 1)
0 = z or 0 = x - 1
0 = z or 1 = x

ans = (e)

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note that you could just plug in numbers to test these answers. for instance, if you wanted to test (e), the correct answer, you'd plug in z = 0 and x other than 1, and then x = 1 and z other than 0. (the fact that you're only obeying one of the conditions at a time cements the "or" part; if it were "and", then it would only work if you satisfied both conditions at once.)
Ron has been teaching various standardized tests for 20 years.

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Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

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by khudania » Tue Sep 23, 2008 4:33 am
E

Looking at the statement, its clear that Z has to be 0 for the statement to be true. So eliminate answer choices that do not have 0.

Plug in the other numbers in the remaining options & E works out.
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by Xbond » Tue Sep 23, 2008 12:07 pm
Thks a lot
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