If wxyz does not equal 0, is w/x > y/z? (1) wz > xy (2) xz > 0

If wxyz does not equal 0, is w/x > y/z?

(1) wz > xy

(2) xz > 0

St 1 .
(1) wz > xy

Don't know the sign of variable , Hence cant cross multiple ...hence not possible .

St 2
(2) xz > 0

Both are either positive or negative .

hence alone not sufficient .

Together ...
not possible because if get two different ans .
When xz are positive .
then
w/x > y/z yes

But XZ are both negative then

w/x < y/z no ...

hence E ans .

KAPLAN OFFICIAL SOLUTION:

First, you should note that none of our numbers can equal zero, because wxyz does not equal 0. Whenever you see a data sufficiency problem with inequality signs, you should immediately start thinking about positives and negatives, as multiplying or dividing an inequality by a negative number will cause the sign to flip.

Statement 1 tells us that wz > xy. In order to answer the question, most students divide both sides of the inequality by x and z, which produces the inequality w/x > y/z. However, this operation assumes that both x and z are positive numbers. If one of the numbers is positive and the other is negative, the inequality sign will flip once, when we divide by the negative number. This produces w/x < y/z. Therefore, statement 1 is not sufficient, as our answer is ‘yes’ if both x and z are positive, but ‘no’ if one of x and z is positive and the other is negative.

Statement 2 tells us that xz > 0. This means that x and z are either both positive or both negative. However, it tells us nothing about w and y and which of x and z is larger. Statement 2 is, therefore, also insufficient.

Looking at both statements together, we know that x and z can both be positive or both be negative. When we divide by x and z in the inequality in statement 1, we know we will end up with w/x > y/z if both are positive. Likewise, if both x and z are negative, the inequality sign will flip when we divide by the first negative number, but then flip back to its original position when we divide by the second negative number. Therefore, we always end up with w/x > y/z, which answers our question ‘always yes.’ Thus, the statements are sufficient together, or answer choice (C) or (3) in Data Sufficiency terms.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.

If wxyz does not equal 0, is w/x > y/z?

(1) wz > xy

(2) xz > 0

Multiplying by negative figures change the direction of the inequality equation. Therefore, multiplying both sides by square values will maintain the direction of the inequality sign.
Transforming the original condition and the question, w/x>y/z? and multiplying both sides by (xz)^2 we have x(z^2)w>y(x^2)z?, x(z^2)w-y(x^2)z>0?,
xz(zw-xy)>0?. Therefore the answer is C가 답이 된다. Using both 1) & 2) together we have xz>0, wz-xy>0 and the answer is yes. Transforming the original condition and the question for these types solves 30% of DS questions

My 2 cents:
wxyz != 0 => w != 0, x != 0, y != 0 & z != 0.
Also question asks -> w/x > y/z
=> w/x - y/z > 0
=> (wz - xy)/xz > 0
Now, if xz & (wz - xy) must have same sign
Case-1: xz > 0 then wz - xy > 0 => wz > xy
&
Case-2: xz < 0 then wz - xy < 0 => wz < xy

1) wz > xy but we don't know anything about xz. NS
2) xz > 0 but we don't know anything about (wz - xy). NS
1+2) We have case-1. Thus C

(2) Insufficient info; x and z have same sign.

(1) wz > xy can't be deduced from stem as we don't know signs of x and z.

1+2

x,z can be both +ve. Stem holds true.
x,z can be both -ve. Stem holds true. Let me show you how

w/x > y/z, multiplying both sides by z(which is -ve)
wz/x < y, multiplying both sides by x(which is -ve)
wz > xy proved. C

GMATPrepNow that concept we were discussing yesterday.

\(\frac{w}{x}>\frac{y}{z}…\frac{w}{x}-\frac{y}{z}>0…\frac{wz-xy}{xz}>0?\)

(1) wz > xy: insufic.
\(\frac{wz-xy}{xz}>0…\frac{positive}{xz}>0?\)
\(xz>0:\frac{positive}{positive}>0…answer=yes\)
\(xz<0:\frac{positive}{negative}<0…answer=no\)

(2) xz > 0: insufic.

(1&2): sufic.
\(xz>0:\frac{positive}{positive}>0…answer=yes\)

Answer (C)

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.