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

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Manager
Joined: 09 Jul 2008
Posts: 111
Location: Dallas, TX
Schools: McCombs 2011
If a < y < z < b, is |y - a| < |y - b| ? (1) |z [#permalink]

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12 Feb 2009, 19:11
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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...
Director
Joined: 14 Aug 2007
Posts: 726

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12 Feb 2009, 21:44
1
KUDOS
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
[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.
Manager
Joined: 27 May 2008
Posts: 200

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12 Feb 2009, 22:07
1
KUDOS
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
Manager
Joined: 09 Jul 2008
Posts: 111
Location: Dallas, TX
Schools: McCombs 2011

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13 Feb 2009, 07:16
Thanks alpha and salvea. The number line method makes it lot easier... +1
Re: DS: From GMATPREP   [#permalink] 13 Feb 2009, 07:16
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