Introduction - Basic Mathematics for Water and Wastewater Operators - page 7

and cancel common terms to obtain:

3

ft × gal

= gal

3

ft

By multiplying the two dimensions (ft 3 and gal/ft 3 ), the answer will be in gallons,
which is what we want. Thus, because the math setup is correct, we would then mul-
tiply the numbers to obtain the number of gallons:

(140 0 ft 3 ) × (7.48 gal l /f ft3) 3 ) = 10,472 gal

Now, let's try another problem with exponents. We wish to obtain an answer in
square feet. If we are given the two terms—70 ft 3 /sec and 4.5 ft/sec—is the following
math setup correct?

(70 ft 3 /sec) × (4.5 f t /sec)

First, only the dimensions are used to determine if the math setup is correct. By
multiplying the two dimensions, we get:

3

ft
sec

ft
sec

3

ft /sec × ft/sec = ×

Multiply the terms in the numerators and denominators of the fraction:

3 4

ft ft
sec ec

×
×

ft
sec

=

2

The math setup is incorrect because the dimensions of the answer are not square
feet; therefore, if we multiply the numbers as shown above, the answer will be wrong.
Let's try division of the two dimensions instead:

3

ft
sec
ft
sec

3

ft /sec =

Invert the denominator and multiply to get:

3

ft
sec

×= ×××
×

sec
ft

ft ft ft sec

2

= ft

sec t