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Absolute value and inequality

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Absolute value and inequality

by sakshi7591 » Tue Sep 13, 2016 6:34 am
Hi,

I came across this question. Can you please answer it?

Suppose that |x|<|y+2|<|z|. Suppose further that y>0 and xz>0. Which of the following could be true?

Indicate all statements that apply.

a) 0<y<x<z

b) 0<x<y<z

c) x<z<0<y

d) 0<y+1.5<x<z

e) z<x<0<y
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by [email protected] » Tue Sep 13, 2016 9:20 am
Hi sakshi7591,

What is the source of this question? I ask because it appears to be a GRE prompt (so it really shouldn't be posted on this site - as this type of prompt won't appear on the GMAT). That having been said, you can solve it by TESTing VALUES, but you'll have to consider a number of different possibilities.

Here's an example to get you started...

We're told that |X| < |Y+2| < |Z| AND that Y>0 and XZ> 0. We're clearly meant to consider NEGATIVE values for X and Z at some point, but I'm going to start with an obvious example first.

IF....
X=1, Y=2, Z=5
then Answer B is true.

Staying with this example, and looking at the other answer choices, does Y have to be greater than X? Do they have to be integers? Can you come up with another example using all positive numbers that would confirm one (or more) of the remaining answers? Now how about if X and Z are NEGATIVE?

With a little bit of 'playing around' and some basic arithmetic, you should be able to find all of the answers that COULD be true.

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by sakshi7591 » Wed Sep 14, 2016 12:16 am
Thanks for your response Rich.C

Source of the question is Manhattan GRE

I understood that Answer "B" must be true. But the answer says "A,B,D and E".
I can't seem to figure out how "E" can be the answer.
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by [email protected] » Wed Sep 14, 2016 9:38 am
Hi sakshi7591,

To start, you have to recognize that XZ > 0 means that X and Z could both be NEGATIVE. Once you realize that, and you prove that Answer B could be true, then you can just turn the X and Z into negatives...

IF....
X= -1, Y=2, Z= -5
then Answer E is true.

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by Matt@VeritasPrep » Thu May 11, 2017 7:58 pm
For E, try a few values:

x = -1
z = -10
y = 3

The trick is that |x| < |z| means that z is further from 0 than x is from 0. So in the positive world, this would give us something like z = 10, x = 1, and in the negative world, this would give us something like z = -10, x = -1. In both cases, |x| < |z|, so we can try either one where appropriate.

The problem does seem really time consuming, but one shortcut I'd use is the relationship between x and z. A, B, C, and D all have x < z, so those are nice to start: with A, B, and D we make both numbers positive, and with C we make them both negative. From there, finding y values to try isn't TOO bad ... but the problem is still a time drain!
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