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If a < y < z < b, is |y - a| < |y - b| ? (1) |z

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Manager
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If a < y < z < b, is |y - a| < |y - b| ? (1) |z [#permalink]

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New post 12 Feb 2009, 20:11
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If a < y < z < b, is \(|y - a| < |y - b| ?\)

(1) \(|z - a| < |z - b|\)

(2) \(|y - a| < |z - b|\)

No clue how to attack this...

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Re: DS: From GMATPREP [#permalink]

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New post 12 Feb 2009, 22:44
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D.

Look at the question in terms of distance between points

irrespective of +ve/-ve signs the a < y < z < b will appear on number line in the same order.

1) tells us that distance between z&a is less than distance between z&b
y lies somewhere in between z and a. i.e y is closer to a than to b
which answers the question
[quote="kman"] is \(|y - a| < |y - b| ?\) quote]
suff

2) similarly, this states that distance between y and a is smaller than distance between z & b.
since y is lower than z, the distance between y and b will be more than that between z&b => |y-a| < |y-b|
suff.

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Re: DS: From GMATPREP [#permalink]

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New post 12 Feb 2009, 23:07
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kman wrote:
If a < y < z < b, is \(|y - a| < |y - b| ?\)

(1) \(|z - a| < |z - b|\)

(2) \(|y - a| < |z - b|\)

No clue how to attack this...


to prove \(|y - a| < |y - b| ?\), we need to know the distance between y and a is less than distance between y and b.

So Stmt1: dist. b/w z and a is less than dist. b/w z and b.
Well we can conclude by saying \(|y - a| < |y - b| ?\), consider a,y,z,b in a number line.
dist(a,z) < dist(z,b) which also means dist(a,y) < dist(y,b). y can very anywhere between a and z. and since y<z it will always obey the rule.

Stmt2:
\(|y - a| < |z - b|\) also means dist(a,y) < dist(z,b)
and since y<z it will always obey the rule.

So
D

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Manager
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Re: DS: From GMATPREP [#permalink]

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New post 13 Feb 2009, 08:16
Thanks alpha and salvea. The number line method makes it lot easier... +1

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Re: DS: From GMATPREP   [#permalink] 13 Feb 2009, 08:16
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