MGMAT Advanced Quant - Ch 4, In Action Q 13 (Pg 124)

[leighton.d.powell]

Is (a-k)/(b-k)>(a+k)/(b+k)?

(1) a>b>k

(2) k>0

The correct answer is C, but I'm not sure how they arrived there. I know that with inequalities with variable expressions, if you don't know the sign of the variables, you must consider both a positive and negative case.

The answer explanation considers the following cases:

1) If (b-k)/(b+k)>0
2) If (b-k)/(b+k)<0

I actually don't understand why they've done this or their explanation in general. Can someone please help?

Thanks!

[tim]

How much were you able to do on your own with this one? The reason I ask is that the explanation I could offer depends heavily on whether you were able to eliminate 1 and 2 alone. In other words, do you just need an explanation of why they work together, or do you need help understanding why each statement is insufficient on its own?

[leighton.d.powell]

Hi Tim,

Since posting this question, I've revisited it and figured out the answer. What was tripping me up was the fact that I'm used to seeing a single variable on one side in the the denominator of such inequalities:

Eg. Is x<y?

(1) x/y<1

For this statement, if y>0, we have x<y, but if y<0, we have x>y.

I wasn't used seeing variable expressions in both denominators on both sides of the inequality. Now I know if that's the case, I must consider whether the product of the two variable expressions in the denominators are greater than zero or less than zero. Is that correct?

Thanks!

[tim]

Not only that, but you have to consider any information you know about the two variables separately, specifically as it relates to positive/negative issues..

[leighton.d.powell]

What do you mean by that? Do you mean, for example, if one variable is greater than the other and the possible effects that may have on the positive/negative issues?

[tim]

that, among other things. i'm not going to list them, because the point i'm trying to make is that rather than memorizing a list of scenarios and exceptions you really just need to be careful and think your way through problem without assuming something that might be a trap. you'll figure out with experience which situations call for extra caution; until then, assume they all do..

[jonvindjohnsen]

I have a question regarding this question. I tried plugging in numbers so they satisfy both statements and then PI the question stem.

And for this PI I found a case where C is insufficient:

3>2>1/2 , so, we get --> 5/3>9/5

Is this approach wrong? If not, where did I go wrong?

Thanks.

[tim]

your math is incorrect. when it became clear that you had the wrong answer, that should have been the first thing you checked..

[jonvindjohnsen]

your math is incorrect. when it became clear that you had the wrong answer, that should have been the first thing you checked..

I knew my math was wrong, but couldn't figure out where until I had another look today. --> 5/3 > 7/5 .
As for my other question, plugging in is a valid approach for these kind of problems? And I'm wondering if we actually have to check pos. non-integers as well because of the constraint: a>b>c>0 ?

[jnelson0612]

your math is incorrect. when it became clear that you had the wrong answer, that should have been the first thing you checked..

I knew my math was wrong, but couldn't figure out where until I had another look today. --> 5/3 > 7/5 .
As for my other question, plugging in is a valid approach for these kind of problems? And I'm wondering if we actually have to check pos. non-integers as well because of the constraint: a>b>c>0 ?

Glad that you figured things out!

Yes, plugging works for this type of problem. Given the constraints, it's smart to try fractions for your variables, just to get a look at all the possibilities.