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jasonmaehara
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Is |x-y| > |x|-|y|

by jasonmaehara Mon Apr 04, 2011 8:26 pm

See question in title.

1) Y < X
2) XY < 0

Answer is B, could use some help as to why!
tim
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Re: Is |x-y| > |x|-|y|

by tim Tue Apr 05, 2011 6:35 pm

this is a really interesting problem. tell us a little about what you tried and what worked and what didn't, and where you got stuck. if you got stuck right at the beginning and couldn't do anything at all, i'll recommend doing a review of absolute values and inequalities before trying the problem again. but if you made some progress and then got stuck we can totally help you..
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jasonmaehara
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Re: Is |x-y| > |x|-|y|

by jasonmaehara Tue Apr 05, 2011 9:46 pm

What I knew from the question was that from #2, XY is negative so obviously the signs are different for X and Y.

Unfortunately this was one of the questions that I skipped during my test because I didn't know a solution off the bat, didn't want to get stuck on a difficult question that I was likely to get wrong anyways (based on what I knew at that point in time).

In terms of what I know, I'm aware that with absolute values there is the need to test cases. I just wasn't sure how to proceed from there...
RonPurewal
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Re: Is |x-y| > |x|-|y|

by RonPurewal Thu Apr 07, 2011 4:54 am

jasonmaehara Wrote:What I knew from the question was that from #2, XY is negative so obviously the signs are different for X and Y.

Unfortunately this was one of the questions that I skipped during my test because I didn't know a solution off the bat, didn't want to get stuck on a difficult question that I was likely to get wrong anyways (based on what I knew at that point in time).


pretty much all across the math section, but especially on number properties, plugging in specific cases is the best way to discover patterns.

it's really hard to explain why statement (2) is sufficient in symbols, but, if you toss in a few examples, you will start to see exactly what is happening:
you need opposite signs.

first, let's take a few examples where x is positive and y is negative:

x = 1, y = -1 -->
left hand side = |1 + 1| = 2
right hand side = 1 - 1 = 0

x = 10, y = -1 -->
left hand side = |10 + 1| = 11
right hand side = 10 - 1 = 9

x = 8, y = -20 -->
left hand side = |8 + 20| = 28
right hand side = 8 - 20 = -12

again, it's very difficult to express what's happening here in symbols (and, frankly, even more difficult to comprehend such an explanation if it is tendered). but pattern recognition is pretty easy once you see a few examples -- on the left-hand side you are adding two positive numbers in each case, while on the right-hand side you are subtracting those numbers. therefore, the left-hand side is clearly going to be greater in all of these cases.

now, let's take a few examples where x is negative and y is positive:

x = -1, y = 1 -->
left hand side = |-1 - 1| = 2
right hand side = 1 - 1 = 0

x = -10, y = 1 -->
left hand side = |-10 - 1| = 11
right hand side = 10 - 1 = 9

x = -8, y = 20 -->
left hand side = |-8 - 20| = 28
right hand side = 8 - 20 = -12

again, you can see the pattern -- it's exactly the same as the pattern for the first case, except the left-hand side gets there by adding two negative numbers and then reversing the sign rather than just adding two positive numbers.


don't depend completely on algebra! be able to employ plug-in methods when algebra lets you down!