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median of 3 positive number - GMATPrep(TM) question

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median of 3 positive number - GMATPrep(TM) question

by Andrei » Sat Sep 05, 2009 12:01 am
"Is z equal to the median of the three positive integers, x, y and z?
(1) x < y + z;
(2) y = z."

The official answer is B.

My answer is E.
Why?

A - no clear answer, as it can be:
- x < y + z => 4< 2+3 => z is median
- x< y + x => 4< 2 + 7 => z is not median;

B - no clear answer, as is can be:
- x = 4, y=z=4 so median is: (4+4)/2 = 4 => z is median; (it is not stated that x MUST be different than y=z);
- x = 10, y=z=2, so median is: (2+10)/ = 6 => z is not median;

C - no clear answer, as it can be:
- x = 4, y=z=4 so median is: (4+4)/2 = 4 => z is median; (it is not stated that x must be different than y=z);
- x = 4, y=z=6 so median is: (4+6)/2 = 5 => z is not median;

So I think the only correct answer must be E.
Please tell me where my fault is, thank you in advance.
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by heshamelaziry » Sat Sep 05, 2009 1:55 am
X,Y,Z ----> 2,3,3 - WE CAN'T THAT THE MEDIAN IS (3+2)/2. MEDIAN IS 3, SO AS LONG AS Y=Z Z IS THE MEDIAN OF X,Y,Z
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by shadowsjc » Sat Sep 05, 2009 10:50 am
Your method for solving for the median is incorrect. If you have an odd number of terms, the median is simply the middle term.

Assuming you have x = 1, y = z = 2, then

1, 2, 2.

The median is 2 (which is equal to z)

If you have x = 3, y = z = 2, then you have

2, 2, 3

and again, your median is 2 (which is Z).

Therefore, the correct answer is B (you know that no matter what the numbers, z will always be the median).

Your analysis on the first part of the problem was correct, as there is no way to figure out from the problem whether z is the median.
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by Andrei » Sat Sep 05, 2009 2:15 pm
Thank you for the answers, I forgot that in an odd string, the median is the middle ... irrespective of the values
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Re: median of 3 positive number - GMATPrep(TM) question

by enniguy » Sat Sep 05, 2009 7:53 pm
Andrei wrote:"Is z equal to the median of the three positive integers, x, y and z?
(1) x < y + z;
(2) y = z."

The official answer is B.

My answer is E.
Why?

A - no clear answer, as it can be:
- x < y + z => 4< 2+3 => z is median
- x< y + x => 4< 2 + 7 => z is not median;

B - no clear answer, as is can be:
- x = 4, y=z=4 so median is: (4+4)/2 = 4 => z is median; (it is not stated that x MUST be different than y=z);
- x = 10, y=z=2, so median is: (2+10)/ = 6 => z is not median;

C - no clear answer, as it can be:
- x = 4, y=z=4 so median is: (4+4)/2 = 4 => z is median; (it is not stated that x must be different than y=z);
- x = 4, y=z=6 so median is: (4+6)/2 = 5 => z is not median;

So I think the only correct answer must be E.
Please tell me where my fault is, thank you in advance.
You are calculating Median wrongly. A median is just the middle number if there are odd number of elements in the the "ordered" set and a median is the average of 2 middle numbers if there are even number of elements in the "ordered" set.

So, let's consider the examples you have taken to prove B wrong.
1.x = 4, y=z=4. Then your set will become,
{4, 4, 4}. Median = 4. z = 4.

2. x = 10, y=z=2. Then your set will become,
{2, 2, 10}. Median = 2. z = 2. (Note the set here is {y,z,x} or {z,y,x}. Why? Because this set needs to be ordered)

3. Consider, x=1, y=z=73. Then the set will become,
{1, 73, 73} Again Median=73. z=73.

There you go. B is sufficient.
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