DS - integers + absolute value

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DS - integers + absolute value

by joanjgonzalez » Tue Dec 16, 2008 7:03 am
Is x an integer?

1. |x+2| = |x-3|
2. x>0

answer is [spoiler]A

Thanks![/spoiler]

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by [email protected] » Tue Dec 16, 2008 7:21 am
Solving this question requires us to first recognize the 4 situations in which two absolute values can equal each other. I'll show this with 4 examples:
|4|=|4| both values inside are already positive
|4|=|4| both values inside are negative
|-4|=|4| one value is negative and the other is positive
|4|=|-4| one value is positive and the other is negative

So, (1) gives us 4 possible situations:

a. x+2 = x-3 (both sides already yield positive outcomes)
b. -(x+2) = -(x-3) (both sides yield negative outcomes, which we "undo" or make positive, by negating both sides, which is the same as finding the abs value of each side)
c. -(x+2) = x-3 (the left side yields a negative outcome, which we "undo" by negating)
d. x+2 = -(x-3) (the right side yields a negative outcome, which we "undo" by negating)

Note: c is actually impossible since it could never be the case that x+2 is negative while x-3 is positive, but it doesn't really matter here.

If we solve each of the 4 equations we find that a and b cannot be solved. c and d yield x=1/2

Since x=1/2, we can conclude that (1) is sufficient.

(2) is not sufficient.

Answer: A

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by joanjgonzalez » Tue Dec 16, 2008 7:24 am
Absolutely amazing!

Thanks!

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by vittalgmat » Tue Dec 16, 2008 5:49 pm
Brent Hanneson wrote:Solving this question requires us to first recognize the 4 situations in which two absolute values can equal each other. I'll show this with 4 examples:
|4|=|4| both values inside are already positive
|4|=|4| both values inside are negative
|-4|=|4| one value is negative and the other is positive
|4|=|-4| one value is positive and the other is negative


Answer: A
From what I learnt on BTG, I think we dont need to solve all the 4 conditions.
for example: if we are given
|x -y| = |a -b|
we need to solve only the foll cases
(x -y) = + (a-b) AND
(x-y) = -(a -b)

Let me know if my understanding is wrong.

thanks

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by [email protected] » Tue Dec 16, 2008 6:00 pm
You're absolutely right, vittalgmat.
I did more work than was necessary.
Good catch!
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Re: DS - integers + absolute value

by logitech » Tue Dec 16, 2008 7:22 pm
St1)

Squaring both sides will cancel the X^2 and give us ax+b=0 SUF

St2)

Insuf

Choose A
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by maihuna » Tue Apr 14, 2009 8:11 am
I will go it this way:

1: |x+2| = |x-3| has following two possibilities:

1. x+2 = x-3 impossible as 2 != -3
2. x+2 = -(x-3) => 2x = 1 or x = 1/2 so x is not an integer. A

2. Says nothing:

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Re: DS - integers + absolute value

by Ian Stewart » Tue Apr 14, 2009 2:20 pm
joanjgonzalez wrote:Is x an integer?

1. |x+2| = |x-3|
2. x>0

answer is [spoiler]A

Thanks![/spoiler]
Or, using a distance approach, from Statement 1 we have |x - (-2)| = |x - 3|, which says, in words, "the distance between x and -2 is equal to the distance between x and 3". So x must be halfway between -2 and 3, and x = 1/2.
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by gmat740 » Tue Apr 14, 2009 7:46 pm
Thanks Ian

I was also thinking something like that, but was not able to strike the problem in the same way

If we use the non-conventional methods like this, we save lot of time.

Karan

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Re: DS - integers + absolute value

by maihuna » Wed Apr 15, 2009 5:34 am
Ian Stewart wrote:
joanjgonzalez wrote:Is x an integer?

1. |x+2| = |x-3|
2. x>0

answer is [spoiler]A

Thanks![/spoiler]
Or, using a distance approach, from Statement 1 we have |x - (-2)| = |x - 3|, which says, in words, "the distance between x and -2 is equal to the distance between x and 3". So x must be halfway between -2 and 3, and x = 1/2.
Thats superb Ian.