Inequality and Absolute Value

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Inequality and Absolute Value

by sourabh33 » Fri May 20, 2011 7:59 pm
If

Both A & B are either positive or negative
Both X & Y are either positive or negative


Is A+B = X+Y ?

(1) -(A+B)<= (X+Y) <=(A+B)
(2) |A+B| >= (X+Y)
Last edited by sourabh33 on Fri May 20, 2011 8:42 pm, edited 2 times in total.
Source: — Data Sufficiency |

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by uslalas22 » Fri May 20, 2011 8:09 pm
I know as usual I am probably missing something, or something is missing from the question, but doesn't it ask is
(A+B) >= (X+Y) ... but then statement 1) explicitly notes....(X+Y) <= (A+B)....

Too good/obvious to be true, it probably isn't, so I'm curious as to the correct answer and explanation. [/quote]

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by pemdas » Fri May 20, 2011 8:31 pm
translating question into algebra
|a+b| >= |x+y|?
(1) -|a+b|<= |x+y| <=|a+b| is Sufficient to answer Yes, as it is obvious from mods
(2) |a+b| >= |x+y| is Sufficient to answer Yes

IOM d

sourabh33 wrote:If

Both A & B are either positive or negative
Both X & Y are either positive or negative


Is (A+B) >= (X+Y) ?

(1) -(A+B)<= (X+Y) <=(A+B)
(2) |A+B| >= (X+Y)
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by sourabh33 » Fri May 20, 2011 8:36 pm
There was a typo in the question. My apologies for the confusion.

Instead of a+b >= x+y the question is a+b = x+y

Regards

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by pemdas » Fri May 20, 2011 8:48 pm
with the question modified, how it's possible to solve equation with changing signs within inequalities; then answer turns e
sourabh33 wrote:There was a typo in the question. My apologies for the confusion.

Instead of a+b >= x+y the question is a+b = x+y

Regards
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