DS Practice test question #11

[missionmba]

DS Practice test question

[California4jx]

IMO: E

[lunarpower]

<annoying administrative comment>

just fyi, you could easily type this problem into the forum directly, in something like 10-20% of the time it would take to take a screen shot and upload it to the forum.

and that would be easier for everyone.

--

</annoying administrative comment>

step 1
REPHRASE THE QUESTION

xy > 0 means that x and y have the same sign.

so, the question can be rephrased as
do x and y have the same sign?

--

step 2
INDIVIDUAL STATEMENTS

these are 2 linear inequalities; ALL linear inequalities include at least two neighboring quadrants of the xy-plane.
if you have two neighboring quadrants, then the signs of x and y are going to be the same in one of those quadrants, and opposite in the other one. (this is actually a very useful general takeaway; remember it)

therefore, each of the individual statements is insufficient.

--

step 3
TWO STATEMENTS TOGETHER

here's the best, and simplest, rule to follow as far as combining inequalities:
you can add two inequalities if the inequality signs face the same way.
don't worry about subtracting inequalities - the rules are confusing and annoying. if you're in a situation where the inequalities where one is "<" and the other one is ">", and/or you want to subtract them in order to cancel something, just multiply by -1 and add them.

in these two inequalities, both contain 'x', and the inequality signs are opposite. either one of these things alone is a good enough reason to multiply by -1 and add, but, together, the case is overwhelmingly persuasive.

multiply inequality (2) by -1, giving
x - y > -2
2y - x > 6

add them:
y > 4
so, y is positive.

plug back into the first one:
x - 4 > -2
x > 2
so, x is also positive.

so, x and y have the same sign.

sufficient.

answer = c

[lunarpower]

by the way, you can also just GRAPH these inequalities, if you're comfortable with that sort of thing. if you graph the 2 inequalities together, you'll note that their intersection lies entirely within the first quadrant (it's a narrow wedge between two upward-sloping lines).
if anyone wants to see this, post back and i'll draw it and upload it.

don't forget that the official laminated yellow pad contains graph-paper grid lines, so graphing is much easier than it would be if the pad were unlined.

--

even if you're the graphing type, make sure that you still learn the takeaway above about adding inequalities, because it's MUCH more generally applicable and useful. for example, if they throw inequalities at you with 3 or more variables, then graphing is futile, but you can still add the inequalities in exactly the same way that you can in the post above.

[missionmba]

<annoying administrative comment>

just fyi, you could easily type this problem into the forum directly, in something like 10-20% of the time it would take to take a screen shot and upload it to the forum.

and that would be easier for everyone.

--

</annoying administrative comment>

step 1
REPHRASE THE QUESTION

xy > 0 means that x and y have the same sign.

so, the question can be rephrased as
do x and y have the same sign?

--

step 2
INDIVIDUAL STATEMENTS

these are 2 linear inequalities; ALL linear inequalities include at least two neighboring quadrants of the xy-plane.
if you have two neighboring quadrants, then the signs of x and y are going to be the same in one of those quadrants, and opposite in the other one. (this is actually a very useful general takeaway; remember it)

therefore, each of the individual statements is insufficient.

--

step 3
TWO STATEMENTS TOGETHER

here's the best, and simplest, rule to follow as far as combining inequalities:
you can add two inequalities if the inequality signs face the same way.
don't worry about subtracting inequalities - the rules are confusing and annoying. if you're in a situation where the inequalities where one is "<" and the other one is ">", and/or you want to subtract them in order to cancel something, just multiply by -1 and add them.

in these two inequalities, both contain 'x', and the inequality signs are opposite. either one of these things alone is a good enough reason to multiply by -1 and add, but, together, the case is overwhelmingly persuasive.

multiply inequality (2) by -1, giving
x - y > -2
2y - x > 6

add them:
y > 4
so, y is positive.

plug back into the first one:
x - 4 > -2
x > 2
so, x is also positive.

so, x and y have the same sign.

sufficient.

answer = c

@lunarpower

You said it right <annoying administrative comment> </annoying administrative comment>

lol

jokes apart

I had to save the intereseting questions before re-installing GMAT-Prep.
n typing the questions again for posting was not a good option

Hope this weakens your argument

and thanks for the reply

[xyz21]

Ron,

If we substitute y = 4 in second inquality (2y - x > 6 ) then we'll get x < 2. So how do we know which inequality we should choose to substitute?

With graphical method there isn't any such confusion.

Thanks a lot.

[cramya]

If we substitute y = 4 in second inquality (2y - x > 6 ) then we'll get x < 2. So how do we know which inequality we should choose to substitute?

With graphical method there isn't any such confusion.

Go with what u r most comfortable with. The reason why u ran in to the contradiction is because u took y=4 and it should actually be y>4

[xyz21]

cramya,

how would you substitute y>4 in 2y - x > 6?

Can you show how the calculations will work out?

Thx