Math questions from any Manhattan Prep GMAT Computer Adaptive Test.
Capthan
 
 

x, y, and z are non zero-DS

by Capthan Sat Aug 30, 2008 8:11 am

If x, y, and z are nonzero numbers, is (x)(y+z)>0?

(1) |x+y|=|x|+|y|

(2) |z+y|=|y|+|z|

Please explain why the answer is C
Thanks
Blue_Lotus
 
 

by Blue_Lotus Sat Aug 30, 2008 11:29 am

Let us reprase the question

=> IS xy + xz >0

On first glance at the answer choices we can see that
choice 1 has x and y, and choice 2 has some information about y and z

BUT for solving this problem we need to know about x y and z.

At this point if there is no sufficient time to solve we can guess C ( both statement together)
or E( neither statement provide the solution)

============================
Now we have enough time so let us solve this:

choice 1: |x + y | = |x| + |y|

This means both x and y have the same sign. ( good to remember this as a rule)

let us prove it by taking some values.
let X = 2 and y = 3 ( both positive values)
L.H.S (left hand side) of equation becomes | 2+3| = 5
R.H.S = |2| + |3| = 2 + 3 = 5
Means with both x and y as positive , the equation stands true.


let X = -2 and y = -3 ( both negative values)
L.H.S (left hand side) of equation becomes | -2-3| = |-5| = 5
R.H.S = |-2| + |-3| = 2 + 3 = 5
Means with both x and y as negative, the equation stands true.

let X = 2 and y = -3 ( both different signs)
L.H.S (left hand side) of equation becomes | 2-3| = 1
R.H.S = |2| + |-3| = 2 + 3 = 5
Means with both x and y having different signs , the equation does not stand true.

This choice therefore actually implies that both x and y have same signs.
As it does not say anything about z , this statement alone cannot solve our problem

Choice 2 : |z + y| = |y| + |z|
using out thumb rule above we can now safely say that
y and z have the same sign.

Still choice 2 by itself cannot solve our problem.


combine choice 1 and 2, we get that x,y and y, z have same sign
=> x,y,z all have same sign.
=> x,y,z all are positive or all are negative.

Does this solve our problem ?
There are 2 cases when x,y,z all are +ve
and when x,y,z are all -ve

case 1: when all are +ve , the L.H.S of equation xy + xz becomes
= positive + positive = positive , this is greater than 0

case 2: when all values ae -ve, the L.H.S of the equation becomes
= (-ve)(-ve) + (-ve)(-ve) = positive + positive = positive , this is greater than 0


Answer therefore is C, BOTH satement together solve this D.S
jwinawer
ManhattanGMAT Staff
 
Posts: 76
Joined: Mon Aug 16, 2004 1:15 pm
 

by jwinawer Sat Sep 20, 2008 3:23 am

nice explanation!
kannan_m_80
Course Students
 
Posts: 9
Joined: Mon Mar 02, 2009 4:24 am
 

Re: x, y, and z are non zero-DS

by kannan_m_80 Mon Jun 06, 2011 4:23 pm

I went with the direction given in class which was to always simplyfy the equation
So i got
xy +xz >0
xy>-xz
y>-z
or
y<z?

So I felt that neither A or B helped answer this and hence chose E

What am I doing wrong here? Please direct me. I do know the solution provided above is simple but I was wondering why we should not simplify this equation

thanks
jnelson0612
ManhattanGMAT Staff
 
Posts: 2664
Joined: Fri Feb 05, 2010 10:57 am
 

Re: x, y, and z are non zero-DS

by jnelson0612 Thu Jun 23, 2011 8:27 pm

kannan, the problem with your approach is that you cannot divide an inequality by a variable unless you know that variable is positive. If the variable is negative you can divide but must flip the inequality sign.
Jamie Nelson
ManhattanGMAT Instructor