Is y < (x+z)/2?

[andrei_mart]

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.

[nitin_prakash_khanna]

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.


To answer your above question, Not neccessarily x,y& z have to be positive.

Because while simplifying your ineuqality in both the statements you havent multiplied your inequality by a variable without knowing its sign. And adding or subtracting variables (or constants) on both sides doesnt impact the ineuality sign as long as you do the same on both the sides.

lets look at St.2 for example.
z - y > (z -x)/2
=> 2z - 2y > z -x (Step 1 you multiplied both sides by 2, a positive number so no need to change inequality sign)
=> z + x > 2y
your above step can be broken down in two sub steps.

After you were at 2z-2y>z-x
subtract z from both sides 2z-2y-z>z-x-z => z-2y>-x (since we subtracted a z from both sides we dont need to worry about flipping the sign, if we would have multiplied then off course things will be different)

and from z-2y>-x can be simplied as z+x>2y
which can be further simplified to 2y<x+z or y <(x+z)/2.

HTH.

[Ben Ku]

In your work simplifying the statements, you're only adding and subtracting terms. With addition and subtraction, we don't need to worry about whether the variable is positive or negative.

We cannot multiply or divide by a variable in an inequality unless we know that they are positive.

[Radix]

I have 2 questions related to your statement above :
1> I guess multiply and divide with a constant value( like 2, -2, 0.5 etc) is acceptable? of course, changing the sign when it is a negative constant value.
2> if I have a/b > c/d is it ok to do a+x/b+x> c+y/d+y assuming x and y to be positive values? Is there any other trap?

Please let me know.

[RonPurewal]

I have 2 questions related to your statement above :
1> I guess multiply and divide with a constant value( like 2, -2, 0.5 etc) is acceptable? of course, changing the sign when it is a negative constant value.

yes

2> if I have a/b > c/d is it ok to do a+x/b+x> c+y/d+y assuming x and y to be positive values? Is there any other trap?

not sure what you mean by "any other trap", but, no, this definitely doesn't work.
for instance, 2/3 is greater than 1/2. but, if you let x = 1 and y = 2 above, then you get 3/4 > 3/4, which is obviously not true.