DS Practice test question #11

Question

DS Practice test question

California4jx · Answer

IMO: E

lunarpower · Answer

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. -- 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 · Answer

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 · Answer

lunarpower wrote: 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. -- 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 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 · Answer

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 · Answer

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 · Answer

cramya, how would you substitute y>4 in 2y - x > 6? Can you show how the calculations will work out? Thx

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DS Practice test question #11

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how to know which inequality to subsitute into

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