Is |S-R|>|T-U|? (1)|R-T|>|S-U| (2)|R-U|>|S-T|

I am sorry to say that the following logic is wrong.
[S-R] > [T-U]
opening the inequality gives:
A. S-R > T-U
B. S-R < U-T

for example take s = -16, r = -10, t = -15, u = -10
|s-r| = | -16+10 | = |-6| = 6
|t-u| = | -15+10 | = |-5| = 5
this satisfies the condition |s-r| > |t-u|
but
s-r = -16+10 = -6
t-u = -15+10 = -5
This does not satisfy ( s-r > t-u ) so we cannot use the above logic.

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Actually
|S-R| >= |S| - |R|
or
|S+R| <= |S| + |R|

Applying this logic to condition 1

(1)|R-T|>|S-U|

( |R-T| >= |R|-|T| ) > ( |S-U| >= |S|-|U| )

Let us assume that |R-T| > |R|-|T| and not greater than or equal to
then
|R|-|T| > |S|-|U| let us rearrange this equation
-|S|+|R| > |T|-|U|
so |S|-|R| < |T|-|U| this gives the answer to the question
but the problem is
|R-T| can be equal to |R|-|T| and |S-U| can be greater then |S|-|U| then
we cannot say for sure |R|-|T| > |S|-|U|
The same rule applies to condition 2, and for this reason alone we cannot combine the derivatives of conditions 1 and 2.
Thus we can conclude that neither of the conditions can be used to solve the inequality.

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Thanks Anand.
I tried to over simplify the problem.

Anand I see a problem with your objection.

putting the values that you chose

S - R > T- U => 6 > 5
S - R < U -T => -6 < 5 (sign of inequality reverses)

so the technique used by Racer seems to be correct.
Your explantion is elaborate and correct too. For GMAT we need short cut methods for problems.

Racer how did you solve the two (A. and B.) to get
S > R.
Could you please explain this.

Thanks.

I could not agree with you more. I am not sure I would have solved the question in two minutes. We definitely need shortcuts. I just feel that arriving at correct answer does not mean the method of solving the problem is correct. A slight twist in the question can make the solution invalid and can give you wrong answers.
I tried to put the solution because new comers can learn little bit about the absolute inequialities. I have learnt a lot from people like you by going through the post and I would like to contribute wherever I can.

Thanks,
Anand.

You are a good man Anand.

Hey, could any one verify IF this is correct.

say we have

a-b > c-d
a-b > d-c

so can we say

a-b> 0 => a>b.

Any takers for this....

I think this problem should be approached in the following way. I know that the method looks very long but I think if understood once, it should not take long to do similar problems. In a way, it is just an extension of Racers approach.

We know that |x| = x, if x > 0 but |x| = -x if x |y| can result in the following possibilities:

x is +ve and y is +ve => x > y
x is -ve and y is -ve => -x > -y
x is +ve and y is -ve => x > -y
x is -ve and y is +ve => -x > y

Applying this logic, what we need to find here is Is |S-R|>|T-U|? In other words,

Is S-R > T-U? or
Is R-S > U-T? OR
Is S-R > U-T? OR
Is R-S > T-U? ========================(Q)

Now applying the logic to statement 1,

|R-T|>|S-U|, is this sufficient to say that ANY ONE of the above 4 possibilities is correct?

Let us see, this statement will give following possibilities

R-T > S-U => R-S > T-U (This is one of the possibilities in Q above. So the answer is YES)
T-R > U-S => S-R > U-T (So the answer is YES again)
R-T > U-S => R+S > T+U (Can not say )
T-R > S-U => T+U > S+R (Can not say)

Statement 1 NOT SUFF

Similarly statement 2 will give 4 possibilities, 2 of which will answer the question in YES and remaining 2 will not answer. So statement 2 NOT suff.

TOGETHER

Both the statments taken togethet will also result in the 4 possibilities that are mentioned in BOLD above.

So together NOT SUFF.

This will give answer E.

Can you guys comment on this method?

Akmai, can you also jump in here? Thanks