Let the 4 barrels be A,B,C,D ---> A+B+C+D = 32 and B+C = 20---> A+D = 12 gallons.

Now lets test the values.

Start with C. Can we have 3 for A ---> 3, B,C,D ---> B+C+D = 29 ---> the maximum we can get is 3,29/3,29/3,29/3 but the median in this case is NOT equal to 10. Thus this choice is not correct and so eliminate choices D and E as well.

Now try choice B, if A =2, we can get B+C+D = 30 and the maximum is 10,10,10 each. This goes well with median = 10 and thus is the correct answer. B is the correct answer with the set as {2,10,10,10}
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4 barrels with average capacity of 8 gallons. Hence Total Capacity is 32 G.

Since Median is 10 i.e, average of 2nd and 3rd barrel is 10 => Sum of 2nd and 3rd barrel is 20, with 10 Gallons each (To keep 1st barrel maximum)

Hence 1st and 4th barrel = 32-20= 12
If the 1st Barrel has to be maximum, the fourth barrel should be as low as possible, but it should be minimum of 10.
Hence => 12-10 = 2

Option (B)

Avg capacity of four barrels= 4*8=32 gallons
Let the four barrels be a,b,c,d
Now, b+c/2=10(Since median is 10)
b+c=20
Now to maximize a, we need to minimize b and c which in turn will minimize d
Therefore, if a=5,d=7 and c and d can't be 8,12 or 10,10 or 7,13
Using the given options, if a=2, then d=10,c=10 and b=10. This works.
Answer B

An average of 8 gallons over 4 barrels leads to a total of 32 gallons.

If the Median of the set is 10
- we can have a set of 11, 11, 9, x ==> x would be 1 in this case
- we can have a set of 11, 10, 10, x ==> x would be 1 again
- we can have a set of 10, 10, 10, x ==> x would be 2 in this case

There is no other way to maximize x, having a median of 10 for the set.

Answer is (B) 2
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The sum is 32 (4*8). For the median to be 10, the second and third barrel must average to 10. Let's assume both of these are ten. That requires the capacities of the two unknown barrels to be 12. For the median to be maintained, the highest capacity barrel cannot be lower than 10. Knowing this, the barrels can be 2, 10, 10, 10. B is the answer.

You want to make the other 3 barrels as small as you can if you are going to make the other barrel as big as you can. The sum is limited to 4 x 8 = 32, and the smallest the other 3 barrels can be is 10 each in order to keep the median at 10. The final barrel will be 2 gallons
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Suppose 4 barrels are B1, B2, B3, B4 in ascending capacity.

(B1+B2+B3+B4)/4 = 8
B1+B2+B3+B4=32

median is 10 ....... as the no. of terms are even i.e 4, median = (B2+B3)/2
==> B2+B3 = 20
now we have B1+B4 = 32-20 =12

Only the following possible cases can be there :
1,9,11,11
1,10,10,11
2,10,10,10

Max value of smallest barrel = 2

800score Official Solution:

Let us denote the capacities of the barrels by v, x, y, z naming from the smallest one to the largest one. The median is 10 gallons. Therefore (x + y) / 2 = 10.
x + y = 20

The average is 8. Therefore (v + x + y + z) / 4 = 8.
v + (x + y) + z = 32
v + 20 + z = 32
v + z = 12

The capacity z must be no less than 10, since 10 is the median. Therefore the maximum possible capacity of the smallest barrel is 12 – 10 = 2 gallons (all the other barrels would have a capacity of 10 gallons in this case).

The correct answer is (B).
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the sum of the first and the last must be 20, which means that the minimum for the first can be 1 and the maximum for the last one 11. to maximize the first one, we need to minimize the last one. the minimum value of the last one is 10, and the maximum minimum of the first one is 2.

Great Question.
Here is my solution to this one ->


Let the Values in increasing order be -->
w1
w2
w3
w4



Now using ==> Mean = Sum/#


Sum(4)=8*4=32

Median =10

Hence w2+w3=20

Now we are asked to maximise w1 --> We must minimise all others .
w4=w3=Median = 10
Hence ww=10 too .

Now w1+30=32
So,w1=2

Hence B
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Let, 4 barrels' capacities (in increasing order) are a b c d = 8*4 = 32 Gallons

Median = b+c = 10*2 = 20
i.e. a+d = 32-20 = 12

For maximizing the capacity of smallest (i.e. a) we MUST minimize the capacity of Largest (i.e. d)

d can't be smaller than median i.e. 10 hence a maximum = 12-10 = 2

Answer: option B
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