How can we solve this elementary school math problem without using equation or simultaneous equations? I got an elementary math problem this morning. The students have not learnt linear equation as well as simultaneous equations yet.
I have solved it with simultaneous equations and I have no idea whether or not it is possible to solve it without using algebra. I means that I am not sure that only by using some words or simple arithmetic can we solve it.
The problem can be rephrased as follows. 

There are 3 identical grass fields $A$, $B$, and $C$. Three elephants
  and 10 horses are released simultaneously on the field $A$ and its
  grass is completely eaten in 3 hours. Two elephants and 5 horses are
  released simultaneously on the field $B$ and its grass is completely
  eaten in 5 hours. How many hour do one elephant and one horse, that are released simultaneously, take to
  completely eat all grass on the field $C$?

 A: Here's a way for you to explain it to them. As mentioned, the question at heart is about rate and proportion. With that perspective in mind, you can see how I set it up to be explained.
(I'm assuming that grass doesn't grow in the time spent, which would be too complicated otherwise.)
Ask them these questions. You can release them 1 at a time, or all at a go, depending on your aims.


*

*How long does it take 9 elephants and 30 horses to eat up a field?  (Ans: 1 hour)   

*How long does it take 10 elephants and 25 horses to eat up a field? (Ans: 1 hour)
The idea of the first 2 questions was to get an equal time comparison (in this case I chose 1 hour). This tells us the relative rates of the elephants and horses.

*Hence, 1 elephant is 'equal' to how many horses? (5 horses)
In simultaneous equations, this is the step of expressing one variable in terms of the other to eliminate it.

*Hence, how many horses take it take to eat up a field in 1 hours? (75 horses)
Now, we actually find the rate of a horse.

*Hence, how long does it take 6 horses to eat up a field? ($\frac{75}{6}$, I hope you taught them fractions)   

*Hence, how long does it take 1 elephant and 1 horse to eat up a field? ($\frac{75}{6}$)
Depending on the ability of the kids, I would leave out this question, or perhaps replace it with "Why do I care about 6 horses?"

Note that algebra is just a fancy way for shorthand. Back in the day, they worked with equations as
" The quantity when cubed, and added to itself thrice, is equal to four."
Of course, now we just say $x^3 + 3x = 4$. If they could do math then, it's not impossible to leave out algebra when talking with young children.
A: I look at it and notice that $5$ horses is half of $10$ horses. That suggests considering what would have happened had we doubled the zoo on field $B$ to $4$ elephants and $10$ horses; on the standard assumptions for such problems the field would have been cleared in half the time, or $2.5$ hours. Since $3$ elephants and $10$ horses required $3$ hours to clear field $A$, our fourth elephant must eat as much in $2.5$ hours as $3$ elephants and $10$ horses ate in their extra half hour, which is one-sixth of the field. In one hour, then, an elephant must denude $\frac1{2.5}\cdot\frac16=\frac1{15}$ of the field.
Now go back to the $2$ elephants and $5$ horses that were actually released on field $B$. In $5$ hours the two elephants denuded $2\cdot5\cdot\frac1{15}=\frac23$ of the field, and the $5$ horses must have accounted for the remaining $\frac13$ of the field. Each one therefore accounted for $\frac1{15}$ of the field in $5$ hours, or $\frac1{75}$ in one hour.

As a check, note that this implies that $3$ elephants and $10$ horses should denude $3\cdot\frac1{15}+10\cdot\frac1{75}=\frac15+\frac2{15}=\frac13$ of a field in an hour, so it should indeed take them $3$ hours to clear a field.

Now we can finish it off: one elephant and one horse will denude $\frac1{15}+\frac1{75}=\frac6{75}=\frac2{25}$ of the field in an hour and will therefore require $\frac{25}2=12.5$ hours to clear field $C$.
A: This is based on the other answers. So I make it as a CW.
Step 1: Using a table to make it easier to analyze.
\begin{align*}
\text{Elephants} && \text{Horses} &&\text{Time} \\
3 && 10 && 3 \\
2 && 5 && 5 \\
\end{align*}
Step 2: Finding the equivalence of elephant and horse to know each rate separately.
\begin{align*}
\text{Elephants} && \text{Horses} &&\text{Time} && \text{Description}\\
9 && 30 && 1 && \text{multiplying animals by 3, reducing time by $1/3$ } \\
10 && 25 && 1 && \text{multiplying animals by 5, reducing time by $1/5$ } \\
\end{align*}
Reducing 5 horses while adding 1 elephant or reducing 1 elephant while adding 5 horses does not change the time taken. It implies one elephant is equal to 5 horses.
Step 3: Using the equivalence, recreate the table as follows. 
\begin{align*}
\text{Elephants} && \text{Horses} &&\text{Time} && \text{Description}\\
15 && 0 && 1 && \text{30 horses are converted to 6 elephants} \\
0 && 75 && 1 && \text{10 elephants are converted to 50 horses} \\
\end{align*}
An elephant consumes $1/15$ portion of the field per hour while a horse consumes $1/75$ portion of the field  per hour. They consume $1/15+1/75=2/25$ portion of the field per hour if they works together. To consume the whole field, they take $25/2=12.5$ hours.
