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5 different numbers is 12

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by ska7945 » Thu Sep 11, 2008 4:59 am
If the average (arithmetic mean) of 5 different numbers is 12, what is the median of the 5 numbers?
(1) The median of the 5 numbers is equal to 1/3 of the sum of the 4 numbers other than the median.
(2) The sum of the 4 numbers other than the median is equal to 45.
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Source: — Data Sufficiency |
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Re: 5 different numbers is 12

by parallel_chase » Thu Sep 11, 2008 5:51 am
ska7945 wrote:If the average (arithmetic mean) of 5 different numbers is 12, what is the median of the 5 numbers?
(1) The median of the 5 numbers is equal to 1/3 of the sum of the 4 numbers other than the median.
(2) The sum of the 4 numbers other than the median is equal to 45.
I think the answer should be B.

Mean = 12
Median = ?

Statement I

The median of the 5 numbers is equal to 1/3 of the sum of the 4 numbers other than the median.

Median = 1/3 * (60-x)

x could be any number less than 60, hence insufficient.

Statement II

The sum of the 4 numbers other than the median is equal to 45.

sum of 5 numbers = 60
sum of 4 numbers except for median = 45

median = 15
Sufficient.

Hence B is the answer.

whats the OA?
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by ska7945 » Thu Sep 11, 2008 6:33 am
i thought b, too
but oa is d.
the problem is from gwd #23.
maybe oa is wrong.
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by bluementor » Thu Sep 11, 2008 7:29 am
IMO D.

Assume the sum of the other 4 numbers other than the median is x. And lets say the median is y.

(x+y)/5=12
x+y=60 (eqn A)

and statement 2 says that:

y = x/3 (eqn B)

Using equations A and B, you can solve for the median, y. Therefore statement 2 is sufficient on its own.

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by bluementor » Thu Sep 11, 2008 7:31 am
Assume the sum of the other 4 numbers other than the median is x. And lets say the median is y.

(x+y)/5=12
x+y=60 (eqn A)

and statement 2 says that:

y = x/3 (eqn B)

Using equations A and B, you can solve for the median, y. Therefore statement 2 is sufficient on its own.
Sorry I made a typo here... it should be read statement 1, not statement 2.
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by pseudononymous » Thu Sep 11, 2008 8:40 am
(1) tells you that the median is 1/3 the sum of the other 4 numbers.
other #s + median = 12 * 5
3 medians + median = 60
4 medians = 60
median = 15
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Re: 5 different numbers is 12

by sudhir3127 » Thu Sep 11, 2008 9:02 am
ska7945 wrote:If the average (arithmetic mean) of 5 different numbers is 12, what is the median of the 5 numbers?
(1) The median of the 5 numbers is equal to 1/3 of the sum of the 4 numbers other than the median.
(2) The sum of the 4 numbers other than the median is equal to 45.
D for me

assume the numbers to X1, X2, Y ( median) , X3 , X4

statement 1.

(X1+ X2+ X3+ X4)= 1/ 3*Y.....................................a

Question stem says

X1+ X2+ Y+ X3+ X4 = 60..........................................b

a-b

60- 4Y = 0
y = 15

hence sufficient

statemnt 2.

The sum of the 4 numbers other than the median is equal to 45.

sum of 5 numbers = 60
sum of 4 numbers except for median = 45

median = 15
Sufficient.

hence D

do let me know if u have any doubts...
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Re: 5 different numbers is 12

by canuckclint » Thu Sep 11, 2008 9:27 pm
sudhir3127 wrote:
ska7945 wrote:If the average (arithmetic mean) of 5 different numbers is 12, what is the median of the 5 numbers?
(1) The median of the 5 numbers is equal to 1/3 of the sum of the 4 numbers other than the median.
(2) The sum of the 4 numbers other than the median is equal to 45.
D for me

assume the numbers to X1, X2, Y ( median) , X3 , X4

statement 1.

(X1+ X2+ X3+ X4)= 1/ 3*Y.....................................a

Question stem says

X1+ X2+ Y+ X3+ X4 = 60..........................................b

a-b

60- 4Y = 0
y = 15

hence sufficient

statemnt 2.

The sum of the 4 numbers other than the median is equal to 45.

sum of 5 numbers = 60
sum of 4 numbers except for median = 45

median = 15
Sufficient.

hence D

do let me know if u have any doubts...
D is right my man
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