Filling large and small jugs of water from 50L This is Question 21 from the AMC Junior Division 2016:
Angelo has a $50$ litre barrel of water and two sizes of jug to fill, large and small. Each jug, when full, holds a certain amount of litres. 
Angelo fills three large jugs, but does not have enough to fill a fourth one. With the remaining water he fills three small jugs, but does not have enough left to fill a fourth. What is the capacity of the small jug in litres?

$L$ = Large jug
$S$ = Small jug
$R$= Remainder left after third small jug was filled
So
$50$ L = $3L$ + $3S$ + $R$
What do I do now? I have a feeling that this question will involve simultaneous equations, but I'm not sure. How do I go about this?
P.S. Please explain in the math skill of the participants which are taking this competition e.g. Year 7&8 (although Year 9 is fine too.) I've also got no idea about which tag to use, other than contest-question.
 A: Let me try at explanation that would hopefully work for kids.

Angelo fills three large jugs, but does not have enough to fill a
  fourth one.

This sentence can be written as $3L < 50 < 4L$. 
You could explain it by expanding on the sentence in a fashion: "Angelo fills three jugs, i.e. there is enough water to fill three large jugs, or 50 litres is more than volume of three large jugs.", or something similar. Also, $4L>50$ because if it would not be, Angelo would be able to fill another large jug.
(Though, not quite, it is actually $3L\leq 50 < 4L$, but that first inequality is strict is implicit in the later text, and also, equality is impossible if integral values are assumed)

With the remaining water he fills three small jugs, but does not have
  enough left to fill a fourth.

This gives us $3S \leq 50-3L <4S$.
The explanation is the same as before, with addendum that we now only have the remainder of water after filling large jugs, which is $50-3L$.
(Note that now we cannot exclude equality case in our inequality.)

Now, we have well defined problem in mathematical language. So, all that we have to do now is see how to deal with the system of inequalities
$$3L < 50 < 4L\\3S\leq 50-3L < 4S$$
and it depends on how familiar your students are with these kind of things.
Low level method of solving would be guessing that $13\leq L \leq 16$ since if $L \leq 12$, then $4L\leq 48$, and if $L\geq 17$, then $3L\geq 51$.
But, if your students are familiar with linear inequalities, just solve them to get $$\frac{50}4< L < \frac{50}3.$$
Now, it is basically casework: $L\in\{13,14,15,16\}$. Again, if your students are familiar with solving linear inequalities, first write
$$\frac{50-3L}{4}< S \leq \frac{50-3L}{3}$$ and then the casework becomes:
$1.$ $\frac{11}{4} < S \leq \frac{11}{3}$
$2.$ $\frac{8}{4} < S \leq \frac{8}{3}$
$3.$ $\frac{5}{4} < S \leq \frac{5}{3}$
$4.$ $\frac{2}{4} < S \leq \frac{2}{3}$
with only the first case giving us solution $L = 13$, $S = 3$.
If your students don't know how to solve linear inequalities, then the casework becomes
$1.$ $3S \leq 11 < 4S$
$2.$ $3S \leq 8 < 4S$
$3.$ $3S \leq 5 < 4S$
$4.$ $3S \leq 2 < 4S$
and still it is not too hard to guess the solution.

If your students are familiar with graphing linear functions, you can show them this graph:

A: Interesting question! However, I would suggest using "whole amount of litres" instead of "certain amount of litres". Also for the remainder: "not enough whole amount of litres left to fill the fourth".
You started well:
$$3L+3S+R=50 (*)$$
If $R$ is dropped, LHS becomes smaller:
$$3L+3S<50 \Rightarrow L+S<\frac{50}{3}\approx 16.67  \ \ \ (1)$$
The water after the three large jugs was not enough for the fourth:
$$3S+R<L \ \ \ (2)$$
If in $(*)$, the value $3S+R$ is changed to the bigger $L$, LHS becomes bigger:
$$3L+L>50 \Rightarrow L>12.5 \ \ \ (3)$$
Also from $(1)$:
$$L\le 15 \ \ \ (4)$$
so that $S\ge 1$.
The water after the three small jugs was not enough for the fourth:
$$R<S \ \ \ (5)$$
If in $(2)$, the value $S$ is changed with smaller, LHS stays smaller:
$$3R+R<L \Rightarrow R<\frac{L}{4}\le \frac{15}{4}=3.75.$$
Thus, possible values of $R$ are $1, 2, 3$. However, from $(*)$ it is obvious $R$ can be only $2$ for $50-R$ to be divisible by $3$. Hence from $(*)$, $L+S=16$. Finally from $(2)$ and $(5)$, we can easily find $S=3$ and $L=13$.
A: We are given that $3L < 50 < 4L$ and $3S < 50 - 3L < 4S$. Since $S < L$, we get $50 < 3L + 4S < 7L$ and thus $L \geq 8$. Also, $3L < 50$ gives $L \leq 16$. We now try putting values for $L$ starting with $16, 15, \ldots$ and check we can get a value for $S$. Clearly, if $L = 16, 15, 14$ we can not find a value for $S$. When $L = 13$ and $S=3$, we get a possible solution.   
A: You can restrict the values of $L$ and $S$ based on the information in the question:
You know that $3L < 50 < 4L$, which can be rewritten as $\frac{50}{4} < L < \frac{50}{3}$, and that $50 - 3L = R_{L}$, with $R_{L}$ being the remainder from filling the large jugs, from which the small jugs are filled.
Similarly, you then have $3S < R_{L} < 4S$. For each possible value of $L$, $R_{L}$ has a corresponding value, and you should be able to find out for each value of $R_{L}$ which values of $S$ are permitted, if any.
