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Re: If the average (arithmetic mean) of four different numbers is [#permalink]
22 Jun 2006, 18:10

kook44 wrote:

If the average of four different numbers is 30, how many of the numbers are greater than 30? (1) None of the four numbers is greater than 60 (2) Two of the four numbers are 9 and 10, respectively.

1) (29, 31, and 28, 32) or (0, 35, 40, 45) so not suff
2) (9, 10, and 0, 101) or (9, 10, and 50, 51)

combining both we know that 2 # should be greater then 30, so C.

Re: If the average (arithmetic mean) of four different numbers is [#permalink]
27 Jan 2014, 13:51

I have seen a lot of post in this question that are considering same numbers to proof the statement wrong. May be I am wrong but aren't we supposed to consider different numbers?

Re: If the average (arithmetic mean) of four different numbers is [#permalink]
27 Jan 2014, 23:41

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artugoca wrote:

I have seen a lot of post in this question that are considering same numbers to proof the statement wrong. May be I am wrong but aren't we supposed to consider different numbers?

Thanks

Yes, the numbers should be distinct though the answer still remains C.

If the average (arithmetic mean) of four different numbers is 30, how many of the numbers are greater than 30?

a+b+c+d=4*30=120

(1) None of the four numbers is greater than 60. Many combinations are possible. For example, numbers can be: 20-25-30-45 (only one number is greater than 30) OR 15-20-40-45 (two number are greater than 30). Not sufficient.

(2) Two of the four numbers are 9 and 10 respectively. Also not sufficient, consider: 0-9-10-101 OR 9-10-35-66. Not sufficient.

(1)+(2) As two of the four numbers are 9 and 10 then the sum of other two must be 120-(9+10)=101. Now, as the greatest number can be at most 60, then the least value of the other one is 41, so in any case two numbers will be more than 30. Sufficient.