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If the average (arithmetic mean) of four different numbers is 30, how

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If the average (arithmetic mean) of four different numbers is 30, how [#permalink]

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New post 23 Jul 2010, 20:11
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If the average (arithmetic mean) 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.

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[Reveal] Spoiler: OA

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Re: If the average (arithmetic mean) of four different numbers is 30, how [#permalink]

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New post 23 Jul 2010, 22:58
Hi,

A is not sufficient.
Consider 20, 20, 40, 40 the average is 30 and non is greater than 60 (2 #s are greater than 30)
Consider 0, 40, 40, 40 the average is 30 and non is greater than 60 (3 #s are greater than 30)
Thus B alone is not sufficient.

B. Two of the four numbers are 9 and 10 then the sum of the other two numbers are 4*30 - (9+10) = 101
if one number is zero the other is 101 (1 # is greater than 30)
if one number is 60 the other other is 51 (2 #s are greater than 30)
Thus B alone is not sufficient.

Taking A and B, the answer will be sufficient and we will have two numbers greater than 30.


Regards,
Jack

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Re: If the average (arithmetic mean) of four different numbers is 30, how [#permalink]

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New post 10 Oct 2010, 16:39
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1) set can be {30 30 30 30} or {10 10 50 50}.... multiple options

2) set can be {9 10 10 91} or {9 10 50 51}..... multiple options

together: { 9 10 x y}

x+y should be 101 and both need to be less than 60. if x is 59 then y will be 101-59 = 42.

if x is 31 then y is 80 - not good. increase x which in turn will reduce y but, both will be greater than 30 for sure.
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Re: If the average (arithmetic mean) of four different numbers is 30, how [#permalink]

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New post 16 Aug 2017, 02:27
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.

Answer: C.

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Re: If the average (arithmetic mean) of four different numbers is 30, how   [#permalink] 16 Aug 2017, 02:27
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