In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 35.0mi/gal , how many liters of gasoline would you need to buy to complete a 142-km trip in Europe? Use the following conversions: 1km=0.6214mi and 1gal=3.78L.
My conversion pathway:
$$(35.0 \mathrm{mi}/1 \mathrm{gal}) \times (1 \mathrm{km} / 0.6214) \times (1 \mathrm{gal} / 3.78 \mathrm{l})= 15 \mathrm{km / l}$$
So: $$(15 \mathrm{km/l})\times (142 / \mathrm{km}) = 2130 \mathrm{km}$$
Am I right?
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$\begingroup$I noticed that you multiplied by $(1$ km$/0.6214)$ and $(1$ gal$/3.781)$. From a style point of view, I would recomment putting the units of measure in the denominators as well (e.g. expressing the km-mi ratio as $(1$ km$/0.6214$mi)).
The following chaining method is one that I've used in high school as well as in university for performing conversions. It is especially useful in Physics as well as in Chemistry class (Stoichiometric calculations in particular). Notice how each ratio cancels a unit from the ratios that precede it.
Anyways, here goes $$\frac{142\color{blue}{\text{ km}}}{1}\cdot\frac{0.6214\color{green}{\text{ mi}}}{1\color{blue}{\text{ km}}}\cdot\frac{1\color{red}{\text{ gal}}}{35.0\color{green}{\text{ mi}}}\cdot\frac{3.78\text{ L}}{1\color{red}{\text{ gal}}}$$
$\endgroup$ $\begingroup$The best way to know if your right is to realize wether or not you're multiplying by 1.
If you have:
$$a=b$$
Where $a$ and $b$ are two different values in a unit
Than with basic algebra:
$$\frac{a}{b}=\frac{b}{a}=1$$
And notice that:
$$(35mi/gal) • (1) • (1)=35mi/gal$$
So as long as you multiplied by 1 your good, don't be afraid to multiply by 1 more than once. Make sure the units reduce too or else you'll end up with a weird unit.
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