I'm doing a little reading about freight rates (the number of miles 1 metric ton can be transported using 1 gallon of diesel fuel). I came across this equation which calculates the freight rate of a trip given its miles traveled and gallons of fuel used:
$Freight\;Rate = \frac{(cargo \; weight \; in \; tons \quad * \quad miles \; traveled)}{gallons}$
The example they provide is:
$Freight\;Rate = \frac{(19 \; tons \quad * \quad 500 \; miles)}{71 \; gallons\;of\;diesel}$
$Freight\;Rate = 1\;ton\;can\;be\;transported\;134\;miles\;using\;1\;gallon\;of\;diesel\;fuel\;$
If final measure states "1 ton can be transported...", then why doesn't the equation involve dividing by the total tonnage of the cargo?
Let's just suppose that we can transport $1$ ton of cargo for $134$ miles using $1$ gallon of fuel.
If we spend $2$ gallons of fuel, we could move $2$ tons of cargo for $134$ miles. Or we could take the original $1$ ton and move it $268$ miles.
So it's the same effort to move $2$ tons of cargo for $134$ miles or $1$ ton and move it $268$ miles. Notice that $2 \times 134 = 1 \times 268.$ In other words, the effort required to move cargo for a distance can be measured by
$$ (\text{weight of cargo}) \times (\text{distance moved}). $$
Now back to the original supposition: transporting $1$ ton of cargo for $134$ miles uses $1$ gallon of fuel.
Suppose we want to move $1$ ton of cargo for $500$ miles instead of $134$ miles. Since $500 \approx 134 \times 3.7313,$ we are moving the cargo $3.7314$ times as far so it will use $3.7313$ times as much fuel. That is, $1$ ton for $134$ miles uses $1$ gallon, so $1$ ton for $500$ miles uses $3.7313$ gallons.
But now suppose instead of just $1$ ton moved $500$ miles, we want to move $19$ tons for the same $500$ miles. Now that's $19$ times as much fuel, so we use $19 \times 3.7313 \approx 70.89$ gallons. That rounds to $71$ gallons.
So moving $19$ tons for $500$ miles uses $71$ gallons, which is the same as each $1$ ton moved for $134$ miles used $1$ gallon.
The key thing here is that we suppose the amount of fuel used is always proportional to the to effort done (weight $\times$ distance), that is, there is a some constant ratio between fuel and effort. Now there are two ways to express a ratio between two quantities $A$ and $B.$ You can write $\frac AB$ or $\frac BA.$ In this case the source you're quoting from has chosen to write the ratio with the effort (weight $\times$ distance) in the numerator and the fuel in the denominator.
One could write it the other way around: fuel in the numerator, effort in the denominator. In that case the original data would look like this:
$$ \frac{71}{19 \times 500} $$
and after simplifying it one might have this:
$$ \frac{1}{1 \times 134}. $$
Now that certainly looks like something got divided by $19.$ But that's just because of the way we chose to write the ratio.
A more familiar real-life example might be the fuel efficiency of automobiles. In the U.S. this is expressed in miles per gallon. In Europe it's liters per $100$ kilometers. One is distance/fuel, the other is fuel/distance. Both are perfectly legitimate ratios that express fuel efficiency. They just express it in two different ways.