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Bunuel
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shouldnt the question be more clear in saying that the max 4 and min 4 are part OF the 7 list?

it could be that there is a master list of 15 nos, and we have 4 min, 4 max and 7 unrelated nos in between
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C. 48
a+b+c+d+e+f+g=12(7)=84
a+b+c+d=8(4)=32
d+e+f+g=20(4)=80
so we can find d
d = (32+80)-84
d = 28
To get the sum of the 3 greatest and smallest numbers, we have,
a+b+c = 80-d
a+b+c = 80-28 = 52
e+f+g = 32-28 = 4
Thus,
52 - 4 = 48 is the difference

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rahulkashyap
shouldnt the question be more clear in saying that the max 4 and min 4 are part OF the 7 list?

it could be that there is a master list of 15 nos, and we have 4 min, 4 max and 7 unrelated nos in between

Hi rahulkashyap,
IMO,
The average (arithmetic mean) of 7 numbers in a certain list is 12. The average of the 4 smallest numbers in this list is 8, while the average of the 4 greatest numbers in this list is 20.

"In this list"---> refers to "list of 7 numbers"
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Bunuel
The average (arithmetic mean) of 7 numbers in a certain list is 12. The average of the 4 smallest numbers in this list is 8, while the average of the 4 greatest numbers in this list is 20. How much greater is the sum of the 3 greatest numbers in the list than the sum of the 3 smallest numbers in the list?

(A) 4
(B) 14
(C) 28
(D) 48
(E) 52

The sum of all the numbers is 7 x 12 = 84, the sum of the 4 smallest numbers is 32, and the sum of the 4 greatest numbers is 80.

We note that the 4 smallest numbers and the 4 greatest numbers both contain the middle number; if we were to add the sum of the 4 smallest numbers and the sum of the 4 greatest numbers, we see that we will obtain the sum of all the numbers plus the repeated middle number, since it was added twice. So if we let n = the repeated, middle number, then we can create the equation:

84 = 32 - n + 80

n = 112 - 84

n = 28

So the sum of the 3 smallest numbers is 32 - 28 = 4 and the sum of the largest 3 numbers is 80 - 28 = 52, so the difference is 52 - 4 = 48.

Answer: D
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Bunuel
The average (arithmetic mean) of 7 numbers in a certain list is 12. The average of the 4 smallest numbers in this list is 8, while the average of the 4 greatest numbers in this list is 20. How much greater is the sum of the 3 greatest numbers in the list than the sum of the 3 smallest numbers in the list?

(A) 4
(B) 14
(C) 28
(D) 48
(E) 52

\(a + b + c + d + e + f + g = 84\) { Let the numbers be in ascending order }

\(a + b + c + d = 32\)------------------------( I )

\(d + e + f + g = 80\)------------------------( II )


Now, \(( a + b + c + d ) + ( d + e + f + g ) = 112\)

So, \((a + b + c + d + e + f + g ) + d = 112\)

Or, \(84 + d = 112\)

Or, \(d = 28\)

Now substitute \(d = 28\) in equation (II) we get \(e + f + g = 52\) --------> Sum of 3 greatest integers

Now substitute \(d = 28\) in equation (I) we get \(a + b + c = 4\) --------> Sum of 3 smallest integers

So, the required difference is \(52 - 4 = 48\) , Answer must be (D)
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