Is xy > 0 ?

[Guest]

Is xy > 0 ?

1) x - y > -2

2) x - 2y < -6

Answer is C but I have no clue how anyone gets to this answer.

[RonPurewal]

ok, well, first establish that (a) and (b) <i>individually</i> are insufficient by testing cases:
(1)
x = y = 1 --> yes
x = y = 0 --> no
insufficient

(2)
x = y = 100 --> yes
x = -10, y = 10 --> no
insufficient

so think about the two statements together:
SWITCH THE SIGNS of the latter inequality, so that the signs face the same way:
x - y > -2
-x + 2y > 6

add them (note that you can add inequalities whose signs are facing the same way - a useful fact):
y > 4
this means y is positive.
back to the first inequality
x > -2 + y
since y is more than 4, this means that x is more than 2
so x is positive

so x and y are both positive.

[rschunti]

Hi Ron,

If we substitue value of "y" as 4 in equation#--> 2) x - 2y < -6 we get a wide condition as x < 2. This means that value of variable "x" can be any thing below "2" when y is 4. So answer should be "E" not "C". Pls can you help explain this more?

Thanks

[answer]

Hi,

You need to consider that both cases need to satisfy in C

If you add both inequalities, you will get
y>4

This means Y needs to be greater than 4, and NOT equal to 4. You got your analysis by making y=4

So now lets say y=5

(1) x-5>-2
x>3

(2) x-10<-6
x<4

Therefore 3<x<4

So X is positive. and XY is positive. Hope this helps...

And yes.. I agree that this question sux..

[rfernandez]

Nice work.

[cracker]

Great explanation RON!!!!!!!!!!!!!

[Guest]

ok, well, first establish that (a) and (b) <i>individually</i> are insufficient by testing cases:
(1)
x = y = 1 --> yes
x = y = 0 --> no
insufficient

(2)
x = y = 100 --> yes
x = -10, y = 10 --> no
insufficient

so think about the two statements together:
SWITCH THE SIGNS of the latter inequality, so that the signs face the same way:
x - y > -2
-x + 2y > 6

add them (note that you can add inequalities whose signs are facing the same way - a useful fact):
y > 4
this means y is positive.
back to the first inequality
x > -2 + y
since y is more than 4, this means that x is more than 2
so x is positive

so x and y are both positive.

Ron,

I'm having serious troubles with these types of questions. My major problem is testing numbers. For #1, you randomly chose to test:

X=Y=0
&
X=Y=1

Then, for statement #2, you chose a completely different set of numbers (how you knew to choose these is beyond me):
x = y = 100
x = -10, y = 10

When I do these problems, I get lost because I plug in numbers and then think, "oh wait, what about fractions?"... "What about negative fractions..." "What if X isn't equal to Y? How would that affect the outcome (especially in statement 2). It takes me 3 or 4 minutes just to try out all the numbers. Even if I'm able to prove both statements insufficient, I easily get lost in the muck of the different scenarios that I may or may not have forgotten. That means I can't decide between C & E. Can you PLEASE offer some advice on these problems and how to choose smart numbers?

[esledge]

Some number testing principles for DS:

The question may provide a clue to what the question is "about," and therefore what numbers to try.

Translate the question to see what I mean:
Is xy > 0?
Is the product of x and y positive?
Do x and y have the same sign?
Insight: This is about the sign of x and y, NOT so much about their values.
Number Picking Strategy: Only 4 basic cases (positive x and y, negative x and y, pos x and neg y, neg x and pos y)

You may not (probably won't) need to try every possible scenario.

It's generally easier to prove insufficiency than to prove sufficiency! For insufficient, you just need to find one Yes case and one No case (y/n question), or find two cases that produce different values (value question). Thus, to save time, remember that you have something to prove...If you have a "yes" case, your next goal is to find a "no" case.

Take statement (1) for example:

A "yes" case will be one where x and y have the same sign. Can x-y>-2 if x and y have the same sign? Sure, take x = big positive and y = smaller positive. That would make x-y=positive, which is >-2.

Now we need a "no" case to prove insufficiency. Can x-y>-2 if x and y have different signs? Sure, take x = positive and y = negative. That would make x-y=pos-neg=pos, which is >-2.

As you can see, you can "pick numbers" without picking specific numbers. Thinking about types of numbers is a time-saver.

Similar approach to (2):
"yes" case: Can x-2y <-6 if x and y have the same sign? Sure, if x = small pos (e.g. 1) and y = big positive (e.g. 10). Again, don't get hung up on the specific values, just check for the existence of some numbers that would work.

"no" case: Can x-2y<-6 if x and y have different signs? Sure, think x = negative and y = positive. That would make x-2y=neg-pos=neg. As long as x and -2y are negative enough, there are plenty of x-2y examples that are <-6.

There are two ways to combine statements when picking numbers:
1. Sometimes, the numbers or types of numbers we check on (1) are the same as those we check on (2). When we combine, the same cases apply. If that's not the situation, then you must...
2. Combine the statements algebraically as much as possible, then try new numbers from scratch.

There is no way around the algebra to combine (1) and (2) (see Ron's explanation of combining the inequalities). Trying numbers is simply too inefficient, perhaps because you have to meet both constraints and because they end up being sufficient together.

Thus, the final principle is that picking numbers is not the optimal approach, and sometimes it won't help much at all. Use the technique when it works, solve algebraically when it doesn't, and accept it for what it is.

[ells1986]

To Ron: Ron, if you get a chance, could you look at the other GMAT prep ineq question I posted: "Is x less than 20?" As you'll note in my question there, I have always used the rule you refer to in your response above when solving ineqs: namely, if the ineq signs are pointed in the same direction, you may add two or more inequalities. I thought I had cracked the "Is x less than 20?" question in rapid fashion and was feeling good b/c I did it with algebra. Then, I looked at the OA and I got it wrong. On that problem, I can see in retrospect that picking numbers makes the answer of E an easy one, but I'm baffled why the "add two inequals if their sign is in the same direction" approach doesn't work. Any thoughts?

[ranjeet_1975]

Thank you Emily Sledge

[Ben Ku]

Glad it helped.