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For how many integer values of x, is |x – 8| + |5 – x| &

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by Jay@ManhattanReview » Sat Apr 29, 2017 4:21 am
ziyuenlau wrote:For how many integer values of x, is |x - 8| + |5 - x| > |x + 7|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

OA=E
It can be dealt swiftly with non-algebraic way.

We have |x - 8| + |5 - x| > |x + 7|.

One of the possibilities is: (x-8) + (5-x) > (x+7)

=> -3 > x+7

=> -10 > x

We see that at least any integer less than -10 is qualified for the value of x, thus x can take infinitely as many integer values. There is no need to test for the more values.

You may test a large random value for x in case there is any restriction with x.

Say = -100. At x = -100, |x - 8| + |5 - x| > |x + 7|.

The correct answer: E

Hope this helps!

Relevant book: Manhattan Review GMAT Number Properties Guide

-Jay
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by [email protected] » Sat Apr 29, 2017 9:30 am
Hi ziyuenlau,

This is a great 'concept' question, meaning that if you recognize the concepts involved, you don't have to do much math to get to the correct answer.

To start, notice that we're adding two absolute values that must sum to a total that is greater than a third individual absolute value. Consider what happens when X becomes really large (for example, X = 1,000). The differences in the values of the individual absolute values becomes negligible, meaning that we're essentially ending up with 1000 + 1000 > 1000. This will occur for every integer value of X at higher and higher values - thus, there's really no reason to try to count them all up - there's an infinite set of solutions.

Final Answer: E

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