Six drums are used to store water. Five drums are of equal c

Question

Six drums are used to store water. Five drums are of equal capacity, while the sixth drum has double the capacity of each of these five drums. On one morning, three drums are found half full, two are found two-thirds full and one is found completely full. It is attempted to transfer all the water to the smaller drums. How many smaller drums are adequate to store the water?

A. Four but not three
B. Three or four, depending on which drum had how much water initially
C. Five but not four
D. Five may be inadequate, depending on which drum had how much water initially
E. Three but not two

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IanStewart · Answer

That's a strange question. We have five small drums, and one large one that is the size of two small drums. We want to know how many small drums we need to contain the water currently in the drums. We can just work out the minimum possible amount of water we might have -- we'll get that when the huge drum is as empty as possible. Then it's half full, which is like having one full small drum. We then have two 1/2-full, two 2/3-full, and one full small drum, which gives us the equivalent of (2)(1/2) + (2)(2/3) + 1 = 2 + 4/3 small drums, and adding on the 1 additional drum (from the half-full big drum), we have 3 + 4/3 = 4 and 1/3 small drums of water. So we need at least 5 drums to contain all that water, and either C or D is correct.

We can then find the maximum amount of water we could have. We want the big drum to then be the full drum, so it contributes 2 small drums worth of water. We then have 3 small drums that are 1/2 full, and 2 that are 2/3-full, for a total of 3/2 + 4/3 = 17/6 small drums worth, and adding the additional 2 (from the big drum, which is full), we get a total of 4 and 5/6 small drums of water. That's the maximum, so no matter what, 5 drums will contain all the water, and D is wrong, so C is right.

chetan2u · Answer

Let us check for the minimum and maximum requirements

1)  Minimum : When the largest is filled up to the least possible volume. 
So largest is filled up half, but that is equal to full of one small. 
So  \frac{L}{2}+\frac{S}{2}+\frac{S}{2}+\frac{2S}{3}+\frac{2S}{3}+S 
 \frac{2S}{2}+2*\frac{S}{2}+2*\frac{2S}{3}+S=4S+\frac{S}{3} ….At least 5 required

2)  Maximum : When the largest is filled up to the maximum possible volume. 
So largest is filled up full, but that is equal to full of two small. 
So  \frac{S}{2}+\frac{S}{2}+\frac{S}{2}+\frac{2S}{3}+\frac{2S}{3}+L 
 3*\frac{S}{2}+2*\frac{2S}{3}+2S=4S+\frac{5S}{6} ….At least 5 required

In both cases, five of small are sufficient.

C

IanStewart · Answer

I think this should say "+ 5S/6" at the end, rather than "+2S/3".

That makes sense, because the min and max cases only differ in one way:

- in min case, a small drum is full and a big drum is 1/2 full
- in max case, a small drum is 1/2 full and a big drum is full

so going from the min to the max case, we lose 1/2 of a drum when we reduce the amount in the small drum, but gain a full drum when we fill up the big drum, for a net gain of 1/2 of a drum. So the min and max cases should differ by 1/2 a small drum, which I point out here because it's a slightly faster way than the one I posted above to arrive at the answer.

chetan2u · Answer

Ian,
You are correct and the logic that the bigger one full and smaller one full will make a final difference of half of the smaller one was exactly why I answered 4 2/3. Didn’t calculate and in hurry just made 1/3 as 2/3, but of course that is when we add 1/3 and not 1/2. 
max is 4 5/6.

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Six drums are used to store water. Five drums are of equal c

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