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Is r > s ?

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Is r > s ?

by November Rain » Sun Apr 04, 2010 2:06 pm
Hi guys. This sounds like a simple question but i actually can't understand it:

Is r > s ?

(1) -r + s < 0

(2) r < | s |


The OA is A but to me it should be D

The first statement is easy to see why.

The second however, is the one which is giving me a hard time to understand why it isn't enough:

If r < | s | then r < s or r > - s

So, for this condition to be true then -s < r < s. So this is sufficient to determine that r < s

The source of this question is the DS 1000 file.
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by student22 » Sun Apr 04, 2010 2:22 pm
Let's pick number for statement 2. Let r = -2, and let s = -1.

-2 < |-1| or -2 < 1

However, -2 is not > -1. Not Sufficient. Remember for weird problems like this to pick negative numbers.
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by Stuart@KaplanGMAT » Sun Apr 04, 2010 3:15 pm
student22 wrote:Let's pick number for statement 2. Let r = -2, and let s = -1.

-2 < |-1| or -2 < 1

However, -2 is not > -1. Not Sufficient. Remember for weird problems like this to pick negative numbers.
Picking numbers is a great idea - unfortunately the numbers you picked only show that you can get a "yes" answer; to disprove a statement we need both a "yes" AND a "no".

You chose r=-2 and s=-1; is -2 < -1? YES.

We can also choose r=2 and s=-5, since:

2 < |-5|

is 2 < -5? NO.

Now that we have both a yes and a no answer, we see that (2) is insufficient.
If r < | s | then r < s or r > - s
The math here is off - let's go through it.

r < |s| can be broken down into two cases:

1st: r < +s

and

2nd: r < -s

You reversed the inequality in the second solution, but there's no reason to do so. We only reverse the inequality if we're multiplying or dividing both sides by a negative, which there's no reason to do in this particular example.

If were were to multiply both sides of the second equation by -1, you'd get:

2nd: -r > s

which doesn't allow us to join the two inequalities together, since one centres around "r" and the other around "-r".

The best way to handle statement (2) is via either logic or picking numbers.

Using logic, r < |s| means that r is closer to 0 on the number line than is s. However, we have no idea on which side of 0 they lie, so we don't know which one is bigger. All of the following are possible:

-- s ---- r--- 0

------r---0----------s

--s--------0--r--------

-----0----r--------s

As you can see, in two of the cases we have r>s, in two we have s>r, so we can't definitively answer the question based on statement (2).
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by November Rain » Sun Apr 04, 2010 3:17 pm
Hi, thanks for your answer.

I also noticed that my equation is wrong: it should be r < s or r < -s.

But anyway, the best way to solve this is indeed to pick values.
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by ssuarezo » Sun Apr 04, 2010 7:39 pm
Stuart Kovinsky wrote:
You chose r=-2 and s=-1; is -2 < -1? YES.

We can also choose r=2 and s=-5, since:

2 < |-5|, is 2 < -5? NO.

Now that we have both a yes and a no answer, we see that (2) is insufficient.
If r < | s | then r < s or r > - s
The math here is off - let's go through it.

r < |s| can be broken down into two cases:

1st: r < +s

and

2nd: r < -s

You reversed the inequality in the second solution, but there's no reason to do so. We only reverse the inequality if we're multiplying or dividing both sides by a negative, which there's no reason to do in this particular example.
Hi Stuart:

Now, I'm confused. Please, check this video: https://www.ehow.com/video_4756710_solvi ... ities.html
The guy inverses the inequality sign for his absolute value problem, so |x+2| > 5 becomes x>3 and x< -7.
Big question: In which cases should I reverse the inequality?? Is the guy on the video wrong??

Thanks
Silvia
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by Stuart@KaplanGMAT » Mon Apr 05, 2010 11:32 am
ssuarezo wrote: Hi Stuart:

Now, I'm confused. Please, check this video: https://www.ehow.com/video_4756710_solvi ... ities.html
The guy inverses the inequality sign for his absolute value problem, so |x+2| > 5 becomes x>3 and x< -7.
Big question: In which cases should I reverse the inequality?? Is the guy on the video wrong??

Thanks
Silvia
I haven't watched the video, but when solving absolute values, especially those involving inequalites, you should always break the absolute value into the positive and negative case. Then, depending on exactly what you're solving for, you may end up swapping the inequality.

Using the example you cited:

|x+2|>5

we look at two cases:

+(x+2) > 5 and -(x+2) > 5

Solving the first:

x + 2 > 5
(subtract 2 from both sides)
x > 3

Solving the second:

-x - 2 > 5
(add 2 to both sides)
-x > 7
(multiply both sides by -1, which swaps the inequality)
x < -7

Automatically swapping the inequality can lead to a lot of confusion, so it's just safer all-around to go step by step.
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by ssuarezo » Mon Apr 05, 2010 3:10 pm
Stuart Kovinsky wrote: I haven't watched the video, but when solving absolute values, especially those involving inequalites, you should always break the absolute value into the positive and negative case. Then, depending on exactly what you're solving for, you may end up swapping the inequality.

Using the example you cited:

|x+2|>5

we look at two cases:

+(x+2) > 5 and -(x+2) > 5

Solving the first:

x + 2 > 5
(subtract 2 from both sides)
x > 3

Solving the second:

-x - 2 > 5
(add 2 to both sides)
-x > 7
(multiply both sides by -1, which swaps the inequality)
x < -7

Automatically swapping the inequality can lead to a lot of confusion, so it's just safer all-around to go step by step.
The clearest explanation. Thank you so much Stuart.
Silvia
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