if xyz > 0, is x > 0?
xy > 0
xz > 0
from gmat software
can you please give detailed explanation as this problem has been very challenging. Is there more than one way to solve?
thanks
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- Guest
I get C.
xyz > 0 in two cases:
1. all the numbers are positive.
2. two of the numbers are negative and the other is positve.
st 1:
xy > 0
This statement is insufficient because x could be positive or negative to make it work.
It does give a useful piece of information though:
if x and y are both postive, xy > 0 and therefore z would have to also be positve to make xyz > 0.
or
if x and y are both negative, xy > 0 and therefore z would have to be positve to make xyz > 0.
Basically, the statement is saying that z must be positive.
st 2:
xz > 0
This statement is insufficient because x could be positive or negative to make it work.
It does give a useful piece of information though:
if x and z are both postive, xz > 0 and therefore y would have to also be positve to make xyz > 0.
or
if x and z are both negative, xz > 0 and therefore y would have to be positve to make xyz > 0.
Basically, the statement is saying that y must be positive.
Combined
From the two statements we know that z must be positive (st 1) and y must be postive (st 2).
If y and z are both positive, the only way that xyz will be > 0 is if x is positive or x > 0.
Therefore C.
xyz > 0 in two cases:
1. all the numbers are positive.
2. two of the numbers are negative and the other is positve.
st 1:
xy > 0
This statement is insufficient because x could be positive or negative to make it work.
It does give a useful piece of information though:
if x and y are both postive, xy > 0 and therefore z would have to also be positve to make xyz > 0.
or
if x and y are both negative, xy > 0 and therefore z would have to be positve to make xyz > 0.
Basically, the statement is saying that z must be positive.
st 2:
xz > 0
This statement is insufficient because x could be positive or negative to make it work.
It does give a useful piece of information though:
if x and z are both postive, xz > 0 and therefore y would have to also be positve to make xyz > 0.
or
if x and z are both negative, xz > 0 and therefore y would have to be positve to make xyz > 0.
Basically, the statement is saying that y must be positive.
Combined
From the two statements we know that z must be positive (st 1) and y must be postive (st 2).
If y and z are both positive, the only way that xyz will be > 0 is if x is positive or x > 0.
Therefore C.
- RonPurewal
- Students
- Posts: 19744
- Joined: Tue Aug 14, 2007 8:23 am
the easiest way to approach many problems involving the signs of numbers, like this one, is simply to enumerate all possibilities. in this problem, the question prompt requires xyz > 0; there are only four combinations of x, y, and z that satisfy that requirement:
(listed in order x y z)
+ + +
+ - -
- + -
- - +
since the prompt condition must be satisfied, you can examine each given statement by choosing from this list, rather than slogging through lots of theory.
(1) of the options on the above list, (+ + +) and (- - +) work here. x is positive in the former and negative in the latter, so this is insufficient.
(2) of the options on the above list, (+ + +) and (- + -) work here. x is positive in the former and negative in the latter, so this is insufficient.
(together) only (+ + +) works, so, sufficient.
answer = c
(listed in order x y z)
+ + +
+ - -
- + -
- - +
since the prompt condition must be satisfied, you can examine each given statement by choosing from this list, rather than slogging through lots of theory.
(1) of the options on the above list, (+ + +) and (- - +) work here. x is positive in the former and negative in the latter, so this is insufficient.
(2) of the options on the above list, (+ + +) and (- + -) work here. x is positive in the former and negative in the latter, so this is insufficient.
(together) only (+ + +) works, so, sufficient.
answer = c
- brynat
Great
Perfect explanation. I am recommending Manhattan GMAT to all of my friends.
- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9360
- Joined: Wed Oct 19, 2005 9:05 am
- Location: Montreal
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