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Is y < (x+z)/2 ?

Expert replies
by Andrei » Fri Oct 30, 2009 9:05 am
Hi,

I found the following problem in the Gmatprep test:

Is y < (x+z)/2?

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

OA is D.


I tried the following way:
1. y - x < z - y => 2y < Z +x => y < (z+x)/2 - SUFFICIENT

2. z - y > (z -x)/2 => 2z - 2y > z -x => z + x > 2y
=> (z+x)/2 > y - SUFFICIENT

So, the answer should be D.

BUT ... the above rationale is correct only if x,y,z are positive, so the signs of the inequalities do not change.
Please help, thanks.
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Source: — Data Sufficiency |

by mp2437 » Fri Oct 30, 2009 9:46 am
Isn't the answer B?
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Re: Is y < (x+z)/2 ?

by ayashlaha » Fri Oct 30, 2009 11:00 am
Andrei wrote:Hi,

I found the following problem in the Gmatprep test:

Is y < (x+z)/2?

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

OA is D.


I tried the following way:
1. y - x < z - y => 2y < Z +x => y < (z+x)/2 - SUFFICIENT

2. z - y > (z -x)/2 => 2z - 2y > z -x => z + x > 2y
=> (z+x)/2 > y - SUFFICIENT

So, the answer should be D.

BUT ... the above rationale is correct only if x,y,z are positive, so the signs of the inequalities do not change.
Please help, thanks.
Hi Andrei,

Your logic is correct. In either case you are dividing by 2 ( a positive number)

Additions,subtractions and (dividion or multiplication by positive numbers) do not require you to change directions of the inequality.

Hope this helps.
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