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Re: If the average (arithmetic mean) of four different numbers is [#permalink]

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22 Jun 2006, 19:10

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]

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27 Jan 2014, 14:51

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?

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.

Re: If the average (arithmetic mean) of four different numbers is [#permalink]

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09 Feb 2015, 08:22

Hi Guys,

could one of you possibly explain to me why the from the two statements we are certain of the number of values greater than 30. From what I understand we have 101 left to a and d so couldnt a=1 and d=101?

could one of you possibly explain to me why the from the two statements we are certain of the number of values greater than 30. From what I understand we have 101 left to a and d so couldnt a=1 and d=101?

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 (from 1), then the least value of the other one is 41, so in any case two numbers will be more than 30. _________________

Re: If the average (arithmetic mean) of four different numbers is [#permalink]

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04 Mar 2015, 06:56

Freddy123 wrote:

Hi Guys,

could one of you possibly explain to me why the from the two statements we are certain of the number of values greater than 30. From what I understand we have 101 left to a and d so couldnt a=1 and d=101?

Really appreciate the help - thanks

Another solution is to use the data points of the average and their particular differentials:

Test (1)

Not sufficient because it is obvious: When you choose 30, 30, 30, 30 four numbers equal 30 and when you choose 10,10,50,50 two numbers are over 30

Test (2)

Data Point 9 leads to differential "-21"

Data Point 10 leads to differential "-20"

To balance these underweight you can choose

51 (30+21) and 50 (30+20) as additional data points or

71 (30+41) and 30 (30+0; no difference to the average) as additional data points

that is, not sufficient

Test (1) & (2)

Tells you that you cannot choose 71 and 0 because (1) says every number is less than 60. Hence you can choose only the following data points

51 and 50 (as described above)

60 (30 + 30) and 41 (30 + 11). This solution shows the maximum of one data point (60)

Re: If the average (arithmetic mean) of four different numbers is [#permalink]

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28 Jul 2016, 07:31

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