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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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

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

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

Have you read this: if-the-average-arithmetic-mean-of-four-different-numbers-is-100604.html#p1322958 ?

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.
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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)

You can't pick the same numbers because one of the conditions in this question is that the each number is different. The answer remains C in this question but keep an eye out for this detail for future questions.
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Hi,
I have a tendency to think of any weird avg problem as a weighted average problem, then I got stuck on this question trying to solve it using the W.A seesaw method.

Does anyone has tips how to identify if a problem is NOT for W.A?
Plus when i tried solving reguraly I totally forgot about the max min issue (which i know how to solve).

Sum of the 4 numbers = (quantity)(average) = 4*30 = 120.

Statement 1:
Case 1: The four numbers are 32, 31, 30, 27
In this case, two of the numbers are greater than 30.
Case 2: The four numbers are 33, 32, 31, 24
In this case, three of the numbers are greater than 30.
INSUFFICENT.

Statement 2:
Case 1: The four numbers are 9, 10, 31, 70
In this case, two of the numbers are greater than 30.
Case 2: The four numbers are 9, 10, 30, 71
In this case, one of the numbers is greater than 30.
INSUFFICENT.

Statements combined:
Since the sum of the four numbers = 120, and two of the numbers are 9 and 10, the sum of the other two numbers = 120-9-10 = 101.
Since neither of the two remaining numbers may exceed 60, a sum of 101 is possible only if each of the two remaining numbers is greater than 30.
Thus, two of the numbers must be greater than 30.
SUFFICIENT.


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Why the value of two numbers cannot be (1,100);(10,91) or any other number where one no. Is less than 30 and other obviously greater than 30 ??

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The question asks: how many of the numbers are greater than 30?

When considering the statements together, we got that 2 numbers will be more than 30. How many of the numbers are greater than 30 in the example you are giving?
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Do we have consider maximum and minimum values to solve such type of questions?? Then both numbers are definitely greater than 30.

Thanks in advance for clearing my doubt. Bunuel

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