BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Median problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 109
Joined: Sun Oct 21, 2012 5:10 am
Followed by:1 members

Median problem

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
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Master | Next Rank: 500 Posts
Posts: 109
Joined: Sun Oct 21, 2012 5:10 am
Followed by:1 members

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.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 283
Joined: Sun Jun 23, 2013 11:56 pm
Location: Bangalore, India
Thanked: 97 times
Followed by:26 members
GMAT Score:750

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

Cheers
Every job is a self-portrait of the person who did it. Autograph your work with excellence.

Kelley School of Business (Class of 2016)
GMAT Score: 750 V40 Q51 AWA 5 IR 8
https://www.beatthegmat.com/first-attemp ... tml#688494
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
User avatar
Legendary Member
Posts: 643
Joined: Wed Aug 14, 2013 4:27 am
Thanked: 48 times
Followed by:7 members

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!
Top
User avatar
Newbie | Next Rank: 10 Posts
Posts: 8
Joined: Sun Jul 31, 2011 11:41 am
Thanked: 2 times

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
Top
Post new topic Post Reply
• Page 1 of 1