I do not know how to get to the answer for this question

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If the average (arithmetic mean) of 5 positive temperatures is x degress F, then the sum of the 3 greatest of these temperatures, in degress F, could be
A) 6x
B) 4x
C) 5x/3
D) 3x/2
E) 3x/5


Please explain to me. Thanks

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by vineeshp » Wed Jul 06, 2011 4:56 pm
This is an awesome question and I go with B.

My take:
Since arithmetic mean of 5 positive numbers is x, sum is 5x.

Now we need to go by the process of elimination. Logic applied here is that the sum of the 3 greatest numbers should definitely be greater than the sum of the 2 smaller numbers.

A) 6x - not possible since the sum of 5 numbers itself is 5x.
B) 4x - looks possible. Sum of smaller = x < Sum of 3 larger. yes this is our answer.
C) 5x/3 - Sum of 3 numbers is 5x/3. So sum of the remaining 2 numbers has to be 5x - 5x/3 = 10x/3 : Again not possible since sum of the 3 greater numbers should be greater than the sum of the 2 smaller numbers.
D) 3x/2 Similar logic applied. Sum of 2 smaller = 7x/2 > 3x/2.
E) 3x/5 - Similarly> Sum of 2 smaller = 22x/5 > sum of 3 larger.
Vineesh,
Just telling you what I know and think. I am not the expert. :)

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by tvtt2010 » Thu Jul 07, 2011 8:26 pm
Thank you

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by MBA.Aspirant » Fri Jul 08, 2011 3:41 am
tvtt2010 wrote:If the average (arithmetic mean) of 5 positive temperatures is x degress F, then the sum of the 3 greatest of these temperatures, in degress F, could be
A) 6x
B) 4x
C) 5x/3
D) 3x/2
E) 3x/5


Please explain to me. Thanks
say x = 10

sum = 5*10 = 50

50 could = 20+10+10+5+5

sum of greatest 3 = 20+10+10 = 40

plugin the answers with x =10

only B works

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by Tani » Fri Jul 08, 2011 1:17 pm
1. The sum of all five would be 5x, and since the temperatures are all positive, the sum of the three largest would have to be less than the sum of all five. (0 is not positive. Eliminate A

If all five temperatures were the same, then each one would be x degrees, and the sum of the three largest would be 3x. That establishes our minimum, eliminating C, D, and E.

The answer is B.
Tani Wolff