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by pradeepkaushal9518 » Mon Aug 23, 2010 5:30 pm
If the average (arithmetic mean) of six different numbers is 25, how many of the numbers are greater than 25?

None of the six numbers is greater than 50.
Three of the six numbers are 7, 8, and 9, respectively.
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by beatthegmatinsept » Tue Aug 24, 2010 7:51 am
pradeepkaushal9518 wrote:If the average (arithmetic mean) of six different numbers is 25, how many of the numbers are greater than 25?

None of the six numbers is greater than 50.
Three of the six numbers are 7, 8, and 9, respectively.
IMO C.
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by Fiver » Tue Aug 24, 2010 9:06 am
C Agree with the above poster.

St1] None of the six numbers is greater than 50.

We can have contradictory possibilities here.
The max value any of the 6 nos. is 50. We know that the sum of all 6 = 150
If 1 is 50, the avg. of the remaining 5 is 20. So we could have an AP of 5 nos. with 20 as the median.
In this case if the common difference is 1, then all these 5 nos. have to be less than 25.
Hence our answer is only 1 no. of the 6nos is greater than 25.

If the common difference is 3 then there are 2 nos greater than 25.

Insuff.

St2] Three of the six numbers are 7, 8, and 9, respectively.
This means that the sum of the remaining 3 is 126 and the avg is 42.
Here i could have 1 extremely large no. such as 100 & 2 nos. each smaller than 25 or one as 26 & the other as 1.

Insuff.

Together, We know that the max value of the sum of 2 of the other 3 nos. is 99, hence the 3rd has to be greater than 25.
So we must have 3 nos. greater than 25.
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