Box 2.0: Rate Problem Remediation: Learn It Well! KCT logo

© 1996 by Karl Hahn

Like it or not, you have to be able to do rate problems without a squirm or a whimper in order to do calculus. If you're weak on these, you'd better practice now and get strong, because otherwise you will drown later on. If you decide to skip this remediation, be very sure that you know this stuff cold.

A rate problem always relates two things by a rate. Take the example of a car going down the road. At different times it has gone different distances down the road. The time and distance are related by a rate, which, in this case is the speed of the car.

Rates will always have units that look like "something per something else." In our example, the rate might be "miles per hour". The word "per" always means division. So when the rate of the care is, say, 60 miles per hour, that means miles traveled divided by the hours it took to travel those miles is going to be 60.

It is very important that you always pay attention to the units something is given in when doing real-world problems. If a problem says, "A car travels for 3 hours," don't write down "time = 3." Write down, "time = 3 hours." Always carry the units to each new calculation that you do. Whenever you add or subtract two quantities, they'd better have the same units. If they don't, you made a mistake somewhere. If you multiply

                miles
   3 hours * 60 -----
                hour
you'd better be expecting to get an answer in miles. Why? Remember that "per" is division. "Miles per hour" is miles divided by hours, just as I have shown above. If you multiply that by hours, the hours cancel just the way any other factor in both the numerator and denominator does. You are left with miles. Likewise, if you divide miles by miles per hour, you expect the answer to be in hours. Why?

             180 miles
           -------------
               miles
            60 -----
               hour
is the same as:
                 1 hours
    180 miles * -- -----
                60 mile
(remember that dividing by a fraction is the same as turning it on its head and multiplying) and the miles cancel, leaving you with hours.

If your units don't come out right, then you screwed something up. Go back and check your work.

Not every rate problem deals with speed. If you are paid by the hour, the rate is in dollars per hour. Your car burns fuel at some number of miles per gallon. A rate of currency exchange is some number of yen per Deutschmark. But the operative word is "per".

You remember in algebra, you discussed the equation of a straight line. Remember the equation

y = mx + b
That is the mother of all rate problems. In pure algebra, we don't carry units around. But they are still there in an abstract sort of way. I have whatever abstract units y is in, call them "y-ers", and whatever abstract units x is in, call them "x-ers". In this case, m is the rate, given in the abstract units of y-ers per x-er. b has units of y-ers because it has to match the units of y.

So if I have the problem, A car starts 3 miles past the tollbooth at noon. It travels at 60 miles per hour. How far past the tollbooth is it at half-past noon?

Well, in this case, the x-ers are hours and the y-ers are miles. m is 60 miles per hour, b is 3 miles, and x is half an hour. So just substitute those expressions into y = mx + b to get

             miles
      y = 60 ----- * 0.5 hours  +  3 miles
             hour
So what is y? If you didn't get 33 miles, go back and check your work.

Well, so far these have been pretty easy rate problems. Remember in algebra class when the teacher asked, "Stan can shovel a ton of manure in an hour. Ollie can shovel 2 tons of manure in an hour. How long does it take them working together to shovel a ton of manure?" Some clown always pops up with "three hours." Bzzzt! That answer is only right if they took a 2 hour and 40 minute beer break.

Rates add.

      ton       tons       tons
   1 ----  +  2 ----  =  3 ----
     hour       hour       hour
That is Stan and Ollie's combined rate. So to turn 1 ton into hours, you have to divide by the rate. The units tell you to do it that way.
          1 ton                1 hour     1
       -----------  =  1 ton * - ----  =  - hours
            tons               3 ton      3
          3 ----
            hour
Note that the tons cancelled to give you hours, and a time is what the problem asked for. But suppose it asked for time in minutes. Now we get into units conversion rates.

There are 60 minutes in an hour, right? So

60 minutes = 1 hour
Now divide both sides by 1 hour. On the right, you have something divided by itself, which always equals 1, provided you're not dividing zero by zero. So that leaves you with:
   60 minutes
   ----------  =  1
    1 hour
In algebra, you are allowed to multiply or divide anything you like by 1 and leave it unchanged. So you have:
    1           1               1         60 minutes
    - hours  =  - hours * 1  =  - hours * ----------  =  20 minutes
    3           3               3          1 hour
with hours cancelling. If you do units conversion using this method of multiplying by something that is identical to the number 1, but is a fraction with the new units on top and the equivalent value of the old units on the bottom, the old units will always cancel, and you'll always get the right answer.

So here are some rate problems:

1) Water pours out of a hose at 200 gallons per hour. A 20,000 gallon swimming pool starts out with 5,000 gallons. How many days does it take to fill it to capacity? Use your old friend, y = mx + b What units are your x-ers? Hours, right? What units are your y-ers? Gallons, right? Gallons are given. The rate, m, is given. What are its units? y-ers per x-er or in other words, gallons per hour. b is also given. It's units are y-ers too, or in other words gallons -- 5000 gallons, in fact. The only thing not given in the problem is x. What are its units? x-ers, or in other words, hours. Didn't the problem ask for a time? Yes but it wanted days. That's ok, solve it in hours then convert to days. Substitute all the numbers and units into the equation.

                             gallons
        20,000 gallons = 200 ------- * x  +  5000 gallons
                              hour
So subtract 5000 gallons from both sides (note that on the left you are you are subtracting gallons from gallons -- the units match, which they must always to when you add or subtract). Now divide out the rate from both sides. On the right, x stands alone. On the left, gallons cancel, leaving you with 75 hours. One day is equal to 24 hours. The problem wants the answer in days. So old units are hours, new units are days. Multiply 75 hours by:
      1 day
     --------
     24 hours
to get 3.125 days.

2) A line has slope of m = 13 and intercept of b = -3. At what x value does the y value of the line equal 62? No units are given in this problem, but use the abstract units. Just use them in your head to check your work. Your instructors may be too dense to understand if you actually write them on your paper. The slope is 13 x-ers per y-er. The intercept is -3 y-ers.

Or you can imagine this to be a real rate problem: "A car starts at 3 miles before the tollbooth, traveling at 13 miles per hour. How many hours does it take to get 62 miles beyond the tollbooth?" Or, "You get paid 13 dollars for each basket of flowers you sell. You owe your mother 3 dollars, and she takes her money as soon as you have it. You want to buy a pair of shoes that costs 62 dollars. How many baskets do you have to sell in order to afford the shoes?"

Did you get x = 5 or 5 hours or 5 baskets?

3) Ollie shovels manure one and a half times as fast as Stan. Together they shovel 5 tons in 3 hours. How fast does Stan shovel? This problem gives you the two related things and asks for a rate -- how many tons per hour does Stan shovel? Tons per hour -- that means divide tons by hours. Well we know only the combined tons, which is 5 tons. So together they shovel at how many tons per hour? The number you get is Ollie's rate plus Stan's rate. But Ollie's rate is one and a half times Stan's rate.


    combined rate  =  

        Ollie's rate  +  Stan's rate = 1.5 * Stan's rate  +  Stan's rate

          =  2.5 * Stan's rate
But you already know the combined rate. So divide that by 2.5 to get Stan's rate. Did you get two thirds of a ton per hour? Can you convert that to tons per day? And that number, 2.5 -- what were its units? That's the combined rate in tons per hour divided by Stan's rate in tons per hour. You divide tons per hour by tons per hour, and all the units cancel out. You have a value that has no units. Such a value is often referred to as dimensionless.

If you still feel you need more practice, make some problems up for yourself. Try hard to relate your rate problems back to your old friend, y = mx + b Find an old algebra book and do some of the rate problems in it. Also do some of the problems in the chapter on the equation of a straight line. These are the same problem. One is concrete, the other abstract, but otherwise they are exactly the same.

There are countless people in this world who are completely stumped by algebra problems until you rephrase them in terms of money. Then these same people can give you the correct answer right away. Money is what they are familiar with -- they use it in business every day. You cannot afford to suffer their malady. Make the connection from the concrete to the abstract. The meat of what you are doing is the same.

A large part of calculus has to do with a kind of rate called a derivative. You have been solving problems, both concrete and abstract, where the rate stays the same. In calculus we deal with rates that are constantly changing. Your instructors will want you to deal with them mostly as abstractions -- that is until they hit you with a word problem on the exam. You must be comfortable on either bank of this river, and be able to swim across it at any time. If you can't deal with fixed rates in both ways, changing rates will blow you away.

But you are too smart for that to happen. Believe it. You really are.


Return to main text