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

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If the average(arithmetic mean) of four different numbers is [#permalink] New post 22 Jun 2006, 17:13
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56% (02:13) correct 43% (00:34) wrong based on 66 sessions
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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Mar 2014, 07:39, edited 1 time in total.
Edited the question.
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 [#permalink] New post 22 Jun 2006, 17:21
Let x1, x2, x3, x4 be the #s

Given average = 30 i.e. sum = 120

S1: All #s < 60
Can come up with two sets {10,10, 50, 50} or {35, 35, 35, 15} which has 2 different answers (2 & 3).
Not sufficient.

S2: x1 = 9, x2 = 10
x3+x4 = 120-19 = 101
Possible Values :
x3 = 100, x4 = 1 OR
x3 = 50, x4 = 51
Two different answers. Not sufficient.

S1 & S2:
x3+x4 = 101
Max. possible value for {x3, x4} = 59
As sum has to be 101, other value has to be 41
Therefore enough to answer question: 2

Sufficient.

Answer : C
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Re: DS - mean [#permalink] New post 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.
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Re: DS - mean [#permalink] New post 22 Jun 2006, 18:24
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.


we know each st alone is not suff.
togather 1 and 2, we know that there are 2 numbers greater than 30.

total = 120.
sum of rest two integers = 120 - 10 - 9 = 111.
lets say one is 59.99, the last one must be 111-59.99 = 50.01.
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Re: GMATPrep questions - Arithmetic Mean [#permalink] New post 02 Sep 2007, 08:14
C.

1: it could be anything such as:

i: 35 + 35 + 35 + 15 = 120
ii. 59 + 58 + 2 + 1 = 120
111. 60 + 22+ 21 + 20 = 120. nsf


2: it could be 2 or 3:
i. 9 + 10 + 100 + 1 = 120
ii. 9 + 10 + 40 + 61 = 120. so insufficient.

togather: 9 + 10 + 60 + 41 = 120. suff.
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Re: If the average(arithmetic mean) of four different numbers is [#permalink] New post 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?

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

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