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MartyMurray KarishmaB Would you like to explain this question ?
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Bunuel
List A consists of four numbers. The smallest possible average (arithmetic mean) of three of these four numbers is 30 and the largest possible average (arithmetic mean) of three of these four numbers is 40. What is the range of possible averages of list A?

A. 2
B. 3
C. 5
D. 6
E. 10


Are You Up For the Challenge: 700 Level Questions
­sayan640, refer your PM

We have numbers, a, b, c and d.
Now, sum of the lower three or upper three is fixed. We can manipulate the values to increase or decrease the overall mean.

Let us work on the upper three => b+c+d = 3*40 = 120
a would depend on what we take b and c as.

Maximize the mean/sum: a should be the maximum, and that will be possible only when a=b=c=30 as 30 is the average of these three numbers.
so, Sum = a+b+c+d = 30+120 = 150..............a,b,c,d = 30,30,30,60

Minimize the mean/sum: a should be the minimum, and that will be possible only when b and c are maximized, that is b=c=d=40 as 40 is the average of these three numbers. Thus, a+40+40 = 3*30 = 90 or a=10
so, Sum = a+b+c+d = 10+120 = 130..............a,b,c,d = 10,30,30,60

Range of mean = \(\frac{150}{4}-\frac{130}{4}=\frac{20}{4}=5\)
 
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I think there is a typo.It would be 10 , 40 , 40 40.Kindly check the highlighted part.
chetan2u
Bunuel
List A consists of four numbers. The smallest possible average (arithmetic mean) of three of these four numbers is 30 and the largest possible average (arithmetic mean) of three of these four numbers is 40. What is the range of possible averages of list A?

A. 2
B. 3
C. 5
D. 6
E. 10


Are You Up For the Challenge: 700 Level Questions
­sayan640, refer your PM

We have numbers, a, b, c and d.
Now, sum of the lower three or upper three is fixed. We can manipulate the values to increase or decrease the overall mean.

Let us work on the upper three => b+c+d = 3*40 = 120
a would depend on what we take b and c as.

Maximize the mean/sum: a should be the maximum, and that will be possible only when a=b=c=30 as 30 is the average of these three numbers.
so, Sum = a+b+c+d = 30+120 = 150..............a,b,c,d = 30,30,30,60

Minimize the mean/sum: a should be the minimum, and that will be possible only when b and c are maximized, that is b=c=d=40 as 40 is the average of these three numbers. Thus, a+40+40 = 3*30 = 90 or a=10
so, Sum = a+b+c+d = 10+120 = 130..............a,b,c,d = 10,30,30,60

Range of mean = \(\frac{150}{4}-\frac{130}{4}=\frac{20}{4}=5\)
 

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Bunuel
List A consists of four numbers. The smallest possible average (arithmetic mean) of three of these four numbers is 30 and the largest possible average (arithmetic mean) of three of these four numbers is 40. What is the range of possible averages of list A?

A. 2
B. 3
C. 5
D. 6
E. 10


Are You Up For the Challenge: 700 Level Questions
­
Let's put 4 numbers in ascending order to get: Least, B, C, Greatest

We get the smallest possible average of 3 numbers when we pick the 3 smallest numbers. So
\((Least + B + C) = 3*30 = 90\)

We get the greatest possible average of 3 numbers when we pick the 3 greatest numbers. So
\((B + C + Greatest) = 3*40 = 120\)

Average of all 4 numbers = \(\frac{90 + 120 - (B+C)}{4}\)

The smallest value of this average is when B+C is greatest. The greatest values  they can take is 40 each in which case the Greatest number is 40 too. If B and C are more than 40, then the average of greatest 3 numbers will be more than 40 too. 
Least Average of all 4 numbers = \(\frac{90 + 120 - 80}{4} = \frac{130}{4}\)

The greatest value of this average is when B+C is smallest. The smallest values  they can take is 30 each in which case the Least number is 30 too. If B and C are less than 30, then the average of smallest 3 numbers will be less than 30 too. 
Greatest Average of all 4 numbers = \(\frac{90 + 120 - 60}{4} = \frac{150}{4}\)

\(Range = \frac{150}{4} - \frac{130}{4} = 5\)

 ­Answer (C)­

Mean has been discussed here: https://youtu.be/W-qhIZ29UIs
 
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