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Median problem

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by psm12se » Tue Aug 13, 2013 12:35 pm
4, 8, 12, y, 5, 10, 15

If y is a positive integer such that 3 < y < 16, but y does not equal 9, the median of the list above must be:

a. either 5 or 8
b. either 10 or 12
c. either 8 or 10
d. y
e. 9

Source: Grockit
OA: C
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by Brent@GMATPrepNow » Tue Aug 13, 2013 1:15 pm
psm12se wrote:4, 8, 12, y, 5, 10, 15

If y is a positive integer such that 3 < y < 16, but y does not equal 9, the median of the list above must be:

a. either 5 or 8
b. either 10 or 12
c. either 8 or 10
d. y
e. 9

Source: Grockit
OA: C
When posting questions, please use the spoiler function to hide the correct answer. This will allow others to attempt the question without seeing the final answer.

First list the numbers in ascending order {4, 5, 8, 10, 12, 15}
We need to add y to this set.

When we add y to the set, the median will be the middlemost value (since there is an odd number of values in the set)

Since y is an integer, and since y cannot equal 9, we need only consider two cases.
case 1: y < 8
case 2: y > 10

case 1: If y < 8, then the median = 8
case 2: If y > 10, then the median = 10

So, the median must be either 8 or 10

Answer: C

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Brent
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by psm12se » Sat Aug 24, 2013 7:32 am
Thank you Brent for the solution.

Also I will take care to use the spoiler next time.
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by ganeshrkamath » Sat Aug 24, 2013 7:50 am
psm12se wrote:4, 8, 12, y, 5, 10, 15

If y is a positive integer such that 3 < y < 16, but y does not equal 9, the median of the list above must be:

a. either 5 or 8
b. either 10 or 12
c. either 8 or 10
d. y
e. 9

Source: Grockit
OA: C
Arrange the other numbers in ascending order:
4,5,8,10,12,15

Now, for y to be the median, it has to be 9.
But since y cannot be 9, the median is either 8 or 10.

Choose C

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by Brent@GMATPrepNow » Sat Aug 24, 2013 8:15 am
psm12se wrote:Thank you Brent for the solution.

Also I will take care to use the spoiler next time.
No problem.
FYI, you can still go back and add the spoiler to your original post. Just go to that post and click EDIT

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by vinay1983 » Sat Aug 24, 2013 9:58 am
I observed that by plugging the answer choices, the median value for option A & B is 1 single number, while for C 2 values could be arrived at, hence i chose C.(Am I ok with this?)

But now, the methods listed above seem much more logical!
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
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by suk123 » Sat Aug 24, 2013 6:00 pm
Arrange the Data in ascending (or descending)order: 4,5,8,10,12,15
For this data the Median is: (8+10)/2 = 9
But the condition stops that option.
Now there are two Scenarios
First, Y<9, which makes 8 the Median (try plugging in any value for Y from 4 to 8 in to the data)
Second, 9<Y, which makes 10 the Median (again try plugging any value from 10 to 15 in to the data)

So the Median is either 8 or 10
and your answer is C
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