Counting / System of Equations A barrel can be filled to the top by pouring 20 small, 5 medium and 8 large buckets of water, or 2 small, 2 medium and 11 large buckets of water. How many large buckets does it take to fill the barrel? I have been able to show the 1 L = 6 s + m  but I cannot seem to proceed to the next step.
I have been told the answer should be 12. But, I cannot work out a linear combination of the given constraints with 12 as the correct answer. I think the problem boils down to showing that
$$ 2m + 2 s = L $$ where m is medium, s is small and L is large. But, it seems impossible. So I am wondering if perhaps the answer key is wrong.
 A: You're correct. It's impossible to solve this problem to a specific result with this information.
We don't know the exact barrel volume, so the proportions are the only relevant information at this point. We can make the equations simpler by setting the smallest bucket as our unit.
Then we can establish that the barrel must contain more than 81 units total, because otherwise the medium bucket becomes smaller than the small bucket!
However, we still have three unknown values in two equations, leaving one of them free. This cannot be solved for a unique solution. Some values are give below, showing that a range of values are possible, and therefore $L=12$ might be a correct solution, but it is definitely not the correct answer.
Medium bucket volume; Large bucket volume; Barrel total volume; Ratio of medium buckets to one barrel; Ratio of large buckets to one barrel.




M volume
L volume
B volume
M ratio
L ratio




1
7
81
81
11.57


2.46
8.46
100
40.65
11.82


4
10
120
30
12


7
13
159
22.71
12.23


10
16
192
19.2
12.375



A: We are given these two equations for the volume $V$:
$V = 20S + 5M+8L$
$V = 2S + 2M + 11L$
Combining these two expressions leads to the relation $L = 6S + M$. Now let us divide the volume of the barrel V by the volume of the large bucket L. This gives us the following expression:
$$\frac{V}{L} = \frac{68 + 13(M/S)}{6+(M/S)}$$
Now $M$ must be larger than or equal to S, hence $V/L \ge 81/7 > 11$. On the other hand from the fact that $M$ can be very large compared to $S$ (but of course it can not be not infinitely larger) we get $V/L < 13$. It follows that $V/L = 12$ is the only integer solution. This solution corresponds to the values $M = 4S$ and $L = 10S$.
A: The question lacks sufficient information, probably what was intended was to specify that the size of the buckets, and the number of buckets of each size needed to fill the barrel were all integers.
Under the above presumption, we can, without loss of generality,
take the size of the smallest bucket to be $1$ unit,
and that of medium/large to be respectively M/L units
Then
$20+5M+8L = 2+2M+11L \Longrightarrow \;M = L-6$
Substituting in one of the original equations,
$20 +5(L-6) +8L = 13L - 10$
We need $\frac{13L-10}{L}\;\,$ to be an integer,
and $M$ to be a positive integer thus  $L=10$,
and finally, number of large buckets to fill the barrel $=\frac{120}{10} = 12$
