How can we solve this elementary school math problem without using equation or simultaneous equations?

Question

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 $C$?

Answer 1

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 $C$.

Answer 2

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.

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

2. 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.

3. 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.

4. 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.

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

6. 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?"

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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.

Answer 3

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.