 # How Do You Calculate Imperial Units of Measurement?

This entry uses the imperial system of measurement. For the metric system, click here.

Here, you are given some examples on what happens when you multiply numbers by different units of measurements.

Example 1

A room is $\text{}3\text{}\phantom{\rule{0.17em}{0ex}}\text{ft}$ wide and $\text{}4\text{}\phantom{\rule{0.17em}{0ex}}\text{ft}$ long. How many square feet is the room?

 $3\phantom{\rule{0.17em}{0ex}}\text{ft}\cdot 4\phantom{\rule{0.17em}{0ex}}\text{ft}=3\cdot 4\cdot \text{ft}\cdot \text{ft}=12\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{2}$ Example 2

You are celebrating your birthday and you want to serve Kool-Aid from a small plastic pool in your garden. You fill the pool with $\text{}80.5\text{}\phantom{\rule{0.17em}{0ex}}\text{pt}$ Kool-Aid. How many cups of Kool-Aid is that?

$\begin{array}{llll}\hfill 80.5\phantom{\rule{0.17em}{0ex}}\text{pt}& =\left(80.5\cdot 2\right)\phantom{\rule{0.17em}{0ex}}\text{cups}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =161\phantom{\rule{0.17em}{0ex}}\text{cups}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Thus, you have 161 cups of Kool-Aid for your guests.

Example 3

What is the volume of a cube-shaped container with sides of $\text{}6\text{}\phantom{\rule{0.17em}{0ex}}\text{ft}$? Give your answer in gallons (gal).

First, you have to find the volume of the container. That is, $V=l\cdot b\cdot h$. Hence, the calculation is like this: $\begin{array}{llll}\hfill V& =6\phantom{\rule{0.17em}{0ex}}\text{ft}\cdot 6\phantom{\rule{0.17em}{0ex}}\text{ft}\cdot 6\phantom{\rule{0.17em}{0ex}}\text{ft}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =6\cdot 6\cdot 6\cdot \text{ft}\cdot \text{ft}\cdot \text{ft}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =216\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Now, you have to calculate into gallon. Note that $1$ ft3 approximately equals $7.48$ gal. $\begin{array}{llll}\hfill 216\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}& =125\cdot 7.48\phantom{\rule{0.17em}{0ex}}\text{gal}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1615.68\phantom{\rule{0.17em}{0ex}}\text{gal}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The volume of the container is thus $1615.68$ gallons.

You are also going to experience situations with compound units which consists of several simple units. For example, the unit for speed is a compound unit, consisting of miles and hours (mi/h or mph).

Example 4

A car driver is driving 70 miles in $\text{}1.5\text{}$ hours. How fast has he driven on average per hour?

In this exercise, you need to know the formula for distance, velocity and time:

 $s=v\cdot t.$

Since you are going to find the speed, you must first solve the formula for $v$: $\begin{array}{llll}\hfill s& =v\cdot t\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{s}{t}& =\frac{v\cdot t}{t}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill v& =\frac{s}{t}=\frac{70\phantom{\rule{0.17em}{0ex}}\text{mi}}{1.5\phantom{\rule{0.17em}{0ex}}\text{h}}\approx 46.7\phantom{\rule{0.17em}{0ex}}\text{mi/h}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Thus, the car driver drove $46.7$ mi/h on average.