Word or story problems give us a first glimpse into how mathematics is used in the real word. To be solved, a word problem must be translated into the language of mathematics, where we use symbols for numbers - known or unknown, and for mathematical operations. When all is said and done, a word problem, stripped from inessential details, translates into one or more mathematical equations of one kind or another. After the equations have been solved, the result can be translated back into the ordinary language.

Mathematical language is clear and succinct. Mathematical formulation contains only what is important in a problem. But how does one know what is important and what is not, what is essential and what can be dispensed with? One simple approach to find out is to try changing the problem a small piece at a time. Play by ear: pay attention to those small modifications that do not distort the meaning of the problem.

It was René Descartes (1596-1650) who first used letters at the end of the alphabet for unknowns. By now, Descartes' usage evolved into a well established tradition.

A symbol used to denote an unknown quantity (or the quantity itself) is called a variable. This does not sound right as there is nothing that actually varies. But this is another tradition, a part of mathematical culture.

Examples:

1. A total of $20,000 was invested in two bond mutual funds, a junk bond fund and a government bond fund. The junk bond fund is risky and yields 11% interest. The safer government bond fund yields only 5%. The total income for the year from the two investments was $1300. How much was invested in each fund?

Solution. Let x = the amount of money invested in the junk bond fund

20,000 – x = the amount of money invested in the government bond fund

To calculate interest, we use the simple interest formula: interest = principle · rate · time

Since the time is one year, in this problem, interest = principle · rate

	Amount invested (Principal)	Interest Rate	Interest Earned
Junk Bonds	x	0.11	0.11x
Government Bonds	20,000 – x	0.05	0.05(20,000 – x)
Total	20,000		1300

The total interest earned is the sum of the interest earned from each fund. We obtain the following equation from the "Interest Earned" column of the chart:

0.11x + 0.05(20,000 – x) = 1300
0.11x + 1000 – 0.05x = 1300	Distributive Property
0.06x + 1000 = 1300	Combine like terms
0.06x = 300	Subtract 1000 from each side
x = 5000	Divide both sides by 0.06

The amount invested in the junk bond fund, x, is $5000.

The amount invested in the government bond fund, 20000 – x, is $15,000.

2.  Two travelers left a restaurant in Oklahoma City and traveled in opposite directions on Interstate 40. If one driver averaged 65 mph and the other averaged 60 mph, how long was it before they were 400 miles apart?

Solution. This problem involves distance, rate, and time; these variables are related by the distance formula: distance = rate · time

Let t = the time the two travelers drive until they are 400 miles apart

	Rate	Time	Distance
One driver	65	t	65t
Other driver	60	t	60t
Total			400

The sum of the distances of each driver is the total, 400. From the "Distance" column in the chart, we obtain the equation:

65t + 60t = 400
125t = 400	Combine like terms
	Divide both sides by 125
t = 3.2	Simplify

The travelers will be 400 miles apart in 3.2 hours.

3.  Braums Dairy mixed two grades of milk, one containing 3% butterfat and the other containing 4.5% butterfat, to obtain 150 gallons of milk that contained 4% butterfat. How many gallons of each were used in the mixture?

Solution. Let x = number of gallons of the 3% butterfat milk

150 – x = number of gallons of the 4.5% butterfat milk

We multiply the number of gallons of solution by the percent of butterfat to obtain the number of gallons of butterfat in each solution.

	Gallons of Solution	Percent of Butterfat	Gallons of Butterfat
3% Milk	x	0.03	0.03x
4.5% Milk	150 – x	0.045	0.045(150 – x)
Mixture	150	0.04	0.04(150)

The total amount of butterfat in the mixture is the sum of the amounts of butterfat in each of the two original milks. Thus, from the "Gallons of Butterfat" column in the chart, we obtain the equation:

0.03x + 0.045(150 – x) = 0.04(150)
0.03x + 6.75 – 0.045x = 6	Distributive Property
-0.015x + 6.75 = 6	Combine like terms
-0.015x = -0.75	Subtract 6.75 from both sides
x = 50	Divide both sides by –0.015

The number of gallons of 3% butterfat milk in the mixture, x, is 50 gallons. The number of gallons of 4.5% butterfat milk in the mixture, 150 – x, is 100 gallons.

4. A winery has a vat to hold Chardonnay. An inlet pipe can fill the vat in 9 hours, while an outlet pipe can empty the vat in 12 hours. How long will it take to fill the vat if both the outlet and the inlet pipes are open?

Solution. In solving a work problem, begin by using the following fact to express each rate of work: If a job can be done in t units of time, then the rate of work is job per unit of time.

Let x = number of hours it will take to fill the vat with both pipes open

= rate of the inlet pipe

- = rate of the outlet pipe (This rate is negative because the outlet pipe is working

against the inlet pipe rather than with it)

	Rate	Time	Part of Job Done
Inlet Pipe		x
Outlet Pipe		x

Part of the job done Part of the job done

by one pipe + by other pipe = 1 (Whole job done)

	Multiply both sides by the LCD, 36
	Distributive Property
4x – 3x = 36	Simplify
x = 36	Combine like terms

It would take 36 hours to fill the vat with both pipes open.

Summary

1. When translating a word problem into the mathematical language seek the essential. Inessential can be modified without affecting the meaning of the problem.
2. A variable in an equation is just an unknown quantity. Its name is quite arbitrary.
3. Think of a problem class to which the given problem belongs.

An equation may be less restrictive than the original problem. Check the answer against the problem's background.