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Bunuel
Four barrels of water have an average capacity of 8 gallons. What is the maximum possible capacity of the smallest barrel if the median is 10 gallons?

A. 1
B. 2
C. 3
D. 4
E. 5

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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
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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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Bunuel
Four barrels of water have an average capacity of 8 gallons. What is the maximum possible capacity of the smallest barrel if the median is 10 gallons?

A. 1
B. 2
C. 3
D. 4
E. 5

Kudos for a correct solution.

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.
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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
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Bunuel
Four barrels of water have an average capacity of 8 gallons. What is the maximum possible capacity of the smallest barrel if the median is 10 gallons?

A. 1
B. 2
C. 3
D. 4
E. 5

Kudos for a correct solution.

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.
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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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Bunuel
Four barrels of water have an average capacity of 8 gallons. What is the maximum possible capacity of the smallest barrel if the median is 10 gallons?

A. 1
B. 2
C. 3
D. 4
E. 5

Kudos for a correct solution.

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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Bunuel
Four barrels of water have an average capacity of 8 gallons. What is the maximum possible capacity of the smallest barrel if the median is 10 gallons?

A. 1
B. 2
C. 3
D. 4
E. 5

Kudos for a correct solution.

Total sum = 32

Median is 10..

Which means avg of terms 2,3 is 10
It could be 10,10,9,11 or whatever.
Now we have 12 for 1st and 4th term.
Maximum possible value for 1st term is 2.( 2+10 = 12)

Posted from my mobile device
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Out of 4 barrels, capacity of 2nd and 3rd barrels taken together has to be 20 as the median is 10( so 2nd and 3rd could be 9and 11 or 10 and 10) so we are left with 1st and 4th barrels and combined capacity of both these barrels has to be 12(32-20). To make smallest barrels's capacity to maximum we need to check for lowest capacity of 4th barrels and that can be 10 and hence maximum capacity of smallest barrels is 2

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