Math 300

Fraction unit


Dr. Wilson


Carmen gets a dollar and a half per hour for doing chores. If she works for three and one third hours, how much money does she earn?


This is a classical example of multiplication viewed as repeated addition. The first hour Carmen works, she gets a dollar and a half. The next hour she gets another dollar and a half. She gets as many dollar and a halves as the number of hours she works. So you would need to multiply the number of hours times a dollar and a half to get the total amount of money she would earn.


= 5


The dollar and a half per hour is called the rate. The rate is how much you get each hour. Rates can be recognized by the fact that their units can be stated using the word "per" as in dollars per hour or miles per gallon or grams per cubic centimeter.

In our example where the rate is measured in dollars per hour, the number of hours is called the base. In general the base is the number of things which is measured by the word which comes after the "per". One multiplies the rate times the base to get the total amount.





23 miles per gallon x 15 gallons = 345 miles


3/4 g/cc x 12 cc = 8 g




Many applications of multiplication can be recognized as rate times base problems. The rate is the amount in each piece and the base is the number of equal pieces. It is helpful to have this terminology when discussing applications of division.

There are two types of applications of division. In one we are given the total amount and the number of pieces we are dividing it into. In this case we find the rate by dividing the total by the base. In the other type of application, we are again given the total amount but this time we are given the amount in each piece i.e. the rate. We divide the total by the rate to find the base. Thus each multiplication or division problem is a member of a family of three problems. For example:


23 miles per gallon x 15 gallons = 345 miles


345 miles / 15 gallons = 23 miles per gallon


345 miles / 23 miles per gallon = 15 gallons.



If we denote the rate by R and the base by B, then RB will be the rate times the base which we will call the amount. The three problems in each family can be described as


A = RB


R = A/B


B = A/R.


Since the base times the rate is the amount, if you know either the rate or the base, you would divide that into the amount to find the other one.