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Source: — Data Sufficiency |
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by akshatsingh » Tue May 13, 2008 10:06 pm
Easily, both are suficent
Aks
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by vinviper1 » Thu May 15, 2008 11:23 am
care to explain? Thanks
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by codesnooker » Thu May 15, 2008 12:11 pm
akshatsingh wrote:Easily, both are suficent
LOL anybody can just say both are sufficient as answer is given. Please help others.

Anyway, I will try to explain the answer:-

Our data set is {7, 2, 16, 11, x}

Now rearrange the data set in the ascending order. Forget the placement of the x as we don't know its value.

data set = {2, 7, 11, 16 and x (without placement for the moment)}

Arithmetic Mean = AM = (2 + 7 + 11 + 16 + x)/5
AM = (36 + x) / 5
Now according to the question, its given that AM = Median (M).
i.e. AM = M

Now median is the value from the dataset which come in the middle of the dataset.

Now According to first statement:

7 < x < 11 (it means x lies between 7 and 11 in the data set.)

Therefore Median = M = x

as M = AM = (36 + x)/5

therefore, x = (36 + x)/5
i.e. x = 9

SUFFICIENT

Now lets take the second statement:
x is the median of the numbers:
i.e. same as we have proved in statement 1. Hence again x = 9

SUFFICIENT

Therefore (D) is the answer.

PS: I will suggest you to study Mean, Mode, Median and Standard Deviation first and solve some problems.
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by vinviper1 » Mon May 19, 2008 8:56 am
Thank you! Tough to see but when you line up the numbers in ascending order it comes a bit clearer.

Thank you for your time in writing it out. Much appreciated.
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