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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.
(1) None of the four numbers is greater than 60.
(2) Two of the four numbers are 9 and 10, respectively.
28 Jan 2014, 00:41
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artugoca
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.
22 Jun 2006, 18:21
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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
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
