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# Is y<(x+z)/2

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Joined: 17 Jul 2006
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Is y<(x+z)/2 [#permalink]  29 Jul 2006, 06:29
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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
1. y-x < z-y
2 z-y > (z-x)/2

Can anyone explain the easiest way to come work this.

Thanks.
VP
Joined: 25 Nov 2004
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Re: DS: Is y<(x+z)/2 [#permalink]  29 Jul 2006, 07:40
lacey.griffith wrote:
Is y<(x+z)/2?

1. y-x < z-y
2 z-y > (z-x)/2

just solve it. its easy one.

from 1: y-x < z-y
y+y < x+z
2y < x+z
y < (x+z)/2. suff

from 2: z-y > (z-x)/2
2z - 2y > z-x
2z - z +x > 2y
z +x > 2y
(z +x)/2 > y. also suff. so its D.
Re: DS: Is y<(x+z)/2   [#permalink] 29 Jul 2006, 07:40
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