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The most important and most difficult part of solving word problems is creating the equation that is to be solved. The following describes how to do this for for several different standard types of word problems as found in most algebra textbooks. The specific strategy being advocated here is to break the equation-writing process down into two stages:

(1)  Write a Rough Equation which is stated in a mixture of English and mathematical notation. For standard types of problems, the rough equation is found in a standard way.  Sometimes (for motion problems in particular), the rough equation can be arrived at only by a careful reading of the problem and/or by referring to a picture that you have drawn.

(2)  Refine the rough equation into a Finished Equation. The finished equation will be stated entirely algebraically, and you should be able to solve it using standard algebraic methods.

The following examples to be shown would be solved using a 5-step format.  Since we are focusing on the equation-writing process, only the first three steps are shown. The complete 5-step format is:

#1: definition of the unknown
#2: a picture, if applicable
#3: the rough and finished equations
#4: solution and check
#5: state the answer

Solution problems are "chemistry-like" problems that refer to the concentration (e.g. 20%) of a solute (e.g. salt) in a solvent (e.g. water) to create a solution (e.g. saltwater, or brine). For example, 10 ounces of a 20% solution of salt in water would contain

(.20)(10) = 2 ounces of the solute (salt)

These problems usually involve mixing two or more component solutions of various concentrations to create a combined solution of yet another concentration.

The rough equation will always be a solute equation of the following form (assuming we are mixing two component solutions to create a combined solution; there could be three or more components, but the idea is the same):

Example:

A chemist has two solutions of sulfuric acid, a 20% solution and an 80% solution. How much of each should be used to obtain 100 liters of a 62% solution?

These are the familiar problems involving mixtures of coffee, nuts, etc., and refer to the unit price of an item (e.g. $2.00 per pound for coffee), a certain number of items (e.g. 10 lbs.) and the total cost (e.g. $20.00) of that many items. For example, if we have 12 lbs. of cashew nuts at $1.50 a pound, the total cost is (12)(1.50) or $18.00.

The equation to be written will always be a cost equation, of the following form:

Example:

Jim sells almonds for $6.00 per pound, and walnuts for $5.20 per pound. He has 12 pounds of almonds on hand. How many pounds of walnuts should he add to his almonds to make a mixture that sells for $5.80 per pound?

These are problems involving "mixtures" of coins, stamps, or tickets, and refer to the denomination of these objects (e.g. $.10 for a dime), the total number of each denomination (e.g. 12 dimes), and the total value (e.g. $1.20) of all items of that denomination. For example, if there are 1300 tickets sold for seats that cost $5.50 each, the total value of those tickets is (1300)(5.50) = $7150.00.

The equation will always be a value equation, of the following form:

Example:

Victoria has a total of 38 dimes and quarters. The total value of the collection is is $5.60. How many coins of each denomination does she have?

These problems have to do with placing money in bank accounts at certain rates of interest, or buying stocks that will increase in value at given rates, etc. The terms used are principal (the amount invested, e.g. $100), rate (the interest rate, e.g. 6%), time (the amount of time over which interest is accumulated, e.g. 2 years), and interest (the amount of money that an investment earns). The formula that is used to compute simple interest is

For the above example, the interest earned is (100)(.06)(2) = $12.00.

A common source of confusion is the use of the similar terms interest and interest rate. "Interest rate" is always a percent, e.g. "8%", whereas just plain "interest" is money (the amount an investment earns), e.g. $12.00. Problem solvers need to be careful about recognizing the distinction between the two.

In so-called "blended interest" problems, two principal amounts are placed in two separate accounts at different rates of interest, which together earn a total amount of interest.

The equation to be written for such problems is an interest equation, as follows:

Example:

Elmer Fudd invested $8000 for 1 year, part at 7% and part at 5 1/4% simple interest. If the accounts earn $458.50 total interest, how much was invested at each rate?

Uniform motion problems are those that deal with objects (trains, bicycles, etc.) that are moving at a constant rate of speed. The equation to be written cannot be characterized in the same way that those for solution, mixture, or other types of problems can be. These problems usually require a careful analysis of the problem (including drawing a picture) to "discover" the equation to be written. For this reason, these problems are the most difficult to solve.

Motion problems refer to distance, rate, and time. For example, if a train travels for 3 hours (time) at a rate of 80 mph, it will cover a distance of 240 mi. The basic formula is

Sometimes, other forms of the same equation will need to be used: t = d/r or r = d/t.

For example, if an inchworm travels at .12 centimeters per second, and crawls three minutes, the distance travelled will be (.12)(180) = 21.6 cm. Note that since the rate was expressed using seconds as a unit, we had to covert the time to seconds (3 min = (3)(60) = 180 seconds) to be consistent in our use of units.

Example:

Hilary and Chelsea jog down Pennsylvania Ave. every day. Hilary jogs at 5 mph, and Chelsea jogs at 4 mph. Chelsea starts at noon, and Hilary starts at 12:30. At what time will Hilary overtake Chelsea? How far will they have jogged?

Work problems are the familiar problems of the kind:

It takes Sam 8 hours to mow the lawn and Sue 6 hours. How long does it take if they do it together?

The basic item that needs to be computed is the fraction of work done by each of the parties. This is rather easily expressed:  for example, if Sam takes 8 hours to do the whole job, then in 4 hours he can obviously do 1/2 the job. But what if he works only 3 hours? Then he completes 3/8 of the job. In general, the fraction that he completes would be:

And in fact, if he spends 8 hours, he will get 8/8 = 1, (or 100%) of the job done! So 1 is the "fraction" that stands for the whole job.

Carrying this one step further, if we don't know how long Sam works, and just call it "x", then "x/8" is the best we can do at describing the fraction of work done by Sam.

Many work problems are of the type for which there are two agents working together to complete a job. Under those conditions, the two fractions of work done by each of them must add up to 1 (the whole job). For example, if Sam does 1/4 of the job, then Sue must do the remaining 3/4 of the job (1/4 + 3/4 = 1 = the fraction representing the whole job).

Based on these ideas, you can see that the equation to be written for many work problems must be a cooperative work equation:

Example (the one about Sam and Sue stated above):

It takes Sam 8 hours to mow the lawn and Sue 6 hours. How long does it take if they do it together?

Some work problems have "agents" that are working not cooperatively, but against each other, or uncooperatively.

Example:

A faucet can fill the bathtub in 30 minutes, but the drain can empty it in 20 minutes. How long does it take to fill the tub if faucet and drain are both open?

Here, the "work" is to fill the tub, but the faucet is trying to fill it while the drain is trying to empty it. The drain is undoing the faucet's work; it is doing "negative work".