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 (15961650) 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
 When
translating a word problem into the mathematical language seek the essential. Inessential can be modified without affecting
the meaning of the problem.
 A
variable in an equation is just an unknown quantity. Its name is quite arbitrary.
 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.