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# Is y < (x + z)/2 ? (1) y - x < z - y (2) z - y > (z - x)/2

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Re: Is y < (x + z)/2 ? (1) y - x < z - y (2) z - y > (z - x)/2 [#permalink]
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You can divide both side of an inequality with a constant or a variable expression

If what you are dividing by is positive, nothing changes. If it is negative, the sign of the inequality flips.

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Re: Is y < (x + z)/2 ? (1) y - x < z - y (2) z - y > (z - x)/2 [#permalink]
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The key is to simplify the inequality before going on to the answer choices.

y<(x+z)/2 simplifies to 2y<x+z

Now that we have our simplified inequality 2y<x+z, let's look at the answers.

1) y-x<z-y
Add y and x to both sides gives you 2y<z+x Sufficient

2) z-y>(z-x)/2
Multiply 2 to the right to get 2z-2y>z-x
Move (z-x) over to the right and -2y to the left to get 2z-z+x>2y
Simplified: z+x>2y Sufficient
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Re: Is y < (x + z)/2 ? (1) y - x < z - y (2) z - y > (z - x)/2 [#permalink]
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Is y < (x + z)/2 ?

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

Attachment:
DS GMATPrep.JPG

Here again, rearranging the inequalities helped cut down the time to 48sec.

y < (x+z)/2

is 2y < x+z OR x+z > 2y?

stmt-1:
y-x < z-y

2y < z+x suff

stmt-2:
z-y > (z-x)/2

2z- 2y < z-x

z+x > 2y suff
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Re: Is y < (x + z)/2 ? (1) y - x < z - y (2) z - y > (z - x)/2 [#permalink]
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Re: Is y < (x + z)/2 ? (1) y - x < z - y (2) z - y > (z - x)/2 [#permalink]
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