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

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nick_sun
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Is y < (x + z)/2 ? (1) y - x < z - y (2) z - y > (z - x)/2 [#permalink] New post   Updated on: 20 Oct 2019, 03:21
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79% (01:17) correct 21% (01:50) wrong based on 665 sessions
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Is \(y < \frac{x + z}{2}\) ?


(1) \(y - x < z - y\)

(2) \(z - y > \frac{z - x}{2}\)

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D

Originally posted by nick_sun on 16 Apr 2007, 01:09.
Last edited by Bunuel on 20 Oct 2019, 03:21, edited 6 times in total.
Renamed the topic, edited the question and added the OA.
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javed
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Re: Is y < (x + z)/2 ? (1) y - x < z - y (2) z - y > (z - x)/2 [#permalink] New post   17 Apr 2007, 03:34
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nick_sun wrote:
Is y < (x+z)/2 ?


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.

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Re: Is y < (x + z)/2 ? (1) y - x < z - y (2) z - y > (z - x)/2 [#permalink] New post   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.)
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Re: Is y < (x + z)/2 ? (1) y - x < z - y (2) z - y > (z - x)/2 [#permalink] New post   23 Oct 2010, 00:21
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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] New post   01 Oct 2015, 19:04
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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] New post   09 Apr 2016, 08:10
nick_sun wrote:
Is y < (x + z)/2 ?

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

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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] New post   13 May 2023, 02:44
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Re: Is y < (x + z)/2 ? (1) y - x < z - y (2) z - y > (z - x)/2 [#permalink]
13 May 2023, 02:44
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