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D
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Difficulty:
15%
(low)
Question Stats:
80% (01:16) correct
20%
(01:48)
wrong
based on 739
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Is \(y < \frac{x + z}{2}\) ?
(1) \(y - x < z - y\)
(2) \(z - y > \frac{z - x}{2}\)
DS GMATPrep.JPG [ 3.96 KiB | Viewed 16084 times ]
(1) \(y - x < z - y\)
(2) \(z - y > \frac{z - x}{2}\)
Attachment:
DS GMATPrep.JPG [ 3.96 KiB | Viewed 16084 times ]
17 Apr 2007, 03:34
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nick_sun
From Statement 1, y-x<z-y
Adding y on either sides gives 2y-x<z.
Adding x on either sides gives 2y<x+z
Dividing either sides by 2 gives y<x> (z-x)/2
Multiplying either sides by 2 we get 2z-2y>z-x.
Adding 2y on either sides we get 2z>z-x+2y.
Subtracting z and adding x on either sides we get z+x>2y.
Dividing either sides by 2 we get (z+x)/2>y. Therefore sufficient.
So answere is D.
Javed.
Cheers!
22 Oct 2010, 23:34
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While solving an inequality problem, is it acceptable to divide or multiply both sides of the inequality with a constant (say an integer or fraction)?
Specifically in this problem, for St. 1
1) (y - x) < (z - y)
=> 2y < (x+z)
=> So can we divide both sides by 2 (though we don't know if x,y,z are +ve, -ve. zero, integers or fractions)?
Or is there an alternate solution without dividing by 2 on both sides?
2) z - y > (z- x)/2
( I'm able to get this statement since this doesn't required multiplication or division.)
Specifically in this problem, for St. 1
1) (y - x) < (z - y)
=> 2y < (x+z)
=> So can we divide both sides by 2 (though we don't know if x,y,z are +ve, -ve. zero, integers or fractions)?
Or is there an alternate solution without dividing by 2 on both sides?
2) z - y > (z- x)/2
( I'm able to get this statement since this doesn't required multiplication or division.)
