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If a<y<z<b, is abs(y-a)<abs(y-b)?

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brianmcma
 Post subject: If a<y<z<b, is abs(y-a)<abs(y-b)?
 Post Posted: Tue Aug 11, 2009 12:57 pm 
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Course Students


Posts: 1
Apologies if this has already been posted, but I was not able to find it. This appeared for me as question 31 in GMAT Prep 2:

If a < y < z < b, is abs(y-a) < abs(y-b)?
1. abs(z-a) < abs(z-b)
2. abs(y-a) < abs(z-b)

Answer is D. Was wondering if there is a quicker method to solve this problem, other than simply plugging in numbers? Some sort of rules-based, or common-sense approach. Thanks!
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Ben Ku
 Post subject: Re: If a<y<z<b, is abs(y-a)<abs(y-b)?
 Post Posted: Tue Aug 18, 2009 11:16 pm 
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ManhattanGMAT Staff


Posts: 823
First, let's try to clear the absolute values.

Because we know that a < y < z < b, we know
abs(y - a) = y - a
abs(y - b) = b - y (since y - b is negative)

We can rephrase the question:
Is y - a < b - y?
or
Is 2y < a + b?

Statement (1) can be rephrased: z - a < b - z, so 2z < a + b. We also know that since y < z, then 2y < 2z. So 2y < 2z < a + b. (1) is sufficient.

Statement (2) can be rephrased: y - a < b - z, so y + z < a + b. Since y < z, we can add y to both sides: 2y < y + z. So 2y < y + z < a + b, so (2) is sufficient as well.

Hope this helps.

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Ben Ku
Instructor
ManhattanGMAT
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