# Problem Solving

## Teaching Word Problems

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## The Scheme

Polya’s list is meant to be general, applicable to both proofs and numerical problems. My list is directed more to the standard algebra problems. Some comments are in order. My students often claim to be at a loss as to what to do, or how to start. Many act as if they should be able to read the problem once and immediately write the solution. They must be reassured that rereading is behavior. Also those students who skip numbers in their reading, see the math anxiety references, may be prepared to deal with them on a second reading. If the student expects to read the problem more than once, perhaps, the anxieties can be broken into manageable issues.

I. Understand the Problem. II. Devise a Plan. III. Carry out the Plan. IV. Examine the Solution. Look Back. - My experience is that one must be more explicit in giving the steps

I. Understand the Problem. 1. Read the problem. 2. What are we looking for? 3. Use the answer to step 2 to introduce variables. II. Devise a Plan. 1. What relationships exist between the variables and the givers? 2. Use the answer to step 1. to write an equation. III. Carry out the Plan. 1. Solve the equation. IV. Examine the Solution. Look Back. 1. Read the problem again to see how the solution of the equation relates to the question. Sometimes the answer is Just yes of no, not a number. 2. Check your answer with the words of the problem, not your equation.

If students are still stumped for what to do next, they could write their reactions to the problem and make comments about it. The comments would then be starting points from which the instructor may ask leading questions. Also if students are attempting to deal with anxiety by avoiding word problems they may come to realize this by reading their own words.

We learn best from examples. Here are some with discussion of the scheme.

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Problem
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Two trains are 500 miles apart. They start towards each other at the same time, one’s speed is 20 miles per hour faster than the other. They meet in 4 hours. How fast is each going?

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Understand the Problem
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. One aid is to draw a picture or a map.

(figure available in print form)

To understand the problem we must answer some questions. At the start the teacher may have to do all the questioning, with time it is expected that the students will do all the work on their own. The following dialogue is conceivable.

I: What does the picture show? S: Together the trains traveled 500 miles in 4 hours. I: What are we looking for? S: How fast was each train going? I: Do we have a relationship? S: I don’t see any. I: You said the picture showed the distance the trains went and that we were looking for the speed of the trains. So, is there a relationship between speed and distance? S: Yes, distance equals speed times the time traveled. I: Time, did you mention that before? Do you know anything about the time? S: Yes, both trains went for 4 hours. I: Seems like you have a plan. S: You are rushing me. We have not introduced any variables.

Hopefully the reader gets the point. So let us introduce some variables.

Let x = speed in miles per hour of first train, 4x its distance in miles.

x +20 = speed of second train, 4(x +20) its distance.

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Equation
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Together the trains went 500 miles.

4x + 4(x+20) = | 500 | |

4x + 4x +80 = | 500 | |

8x +80 = | 500 | |

8x = | 500 80 | |

8x = | 420 | |

x = | (420)/8 | |

x = | 52 1/2 mph | |

x+20 = | 72 1/2 mph. |

*. This is the part that needs more emphasis in teaching. Is it clear? Can we answer more questions than those called for? Looking Back seems hard when reading Polya, but once started it can be an endless task. Well, the first obvious thing to do is to check the answer.*

Check: First train 52 1/2 mph for four hours goes | 210 miles Second train 72 1/2 mph for 4 hours goes | 290 miles |

Total | 500 miles. |

We asked the question, “Is it clear?” Many students and teachers find it convenient to set the problem up in tabular form, using the key idea D = RT, distance equals rate times time as the organizing principle.

We still have the same variables: x is the rate of the first train, x+20 the rate of the second. So here is the table and its explanation.

(figure available in print form)

We have two trains so a row for each. We have literal expressions for the speeds so fill in the rate column, x and x+20. Both trains traveled for 4 hours so 4 goes into the T boxes. The D boxes are filled in by using the formula, the justification for the table. The first column equals the product of the second and third. The map still tells us to add the distances to get 500, this is our equation same as before.

Tables are intended to clarify our thinking to give us an organized way to attack the problem. If the student finds the idea more of a burden than a help, the student need not use it. The math anxiety people sounded as if they had been forced to use such techniques, only to become more confused. Perhaps, the formula was not stressed enough by the instructor. Even good ideas can be abused.

Consideration must be paid to the students’ styles of learning. This is where rereading Polya comes in. One must hesitate to offer a solution. At the same time the student who is floundering needs help.

The train problem leads into a class of problems: rate problems. The standard algebra problems of Mixture, Coin, Work, Solution, and Interest can all be thought of as rate problems, Just like Distance problems. In each case one quantity equals the product of two others. One of the factors is usually time. This relationship sets up a table for each of these problems. Let us see this in the following example of a work problem.

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Example
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If one card sorter will sort 27,000 cards in an hour and another will sort 39,000 in an hour, how long will it take them to sort 55,000 working together?

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Understanding the problem
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. What are we asked for? The time it takes the machines working together. Can we get an upper limit on the time? An estimate? Well if they both worked for an hour 27,000 + 39,000 = 66,000 cards would be sorted. So, we know the time has to be less than an hour. This example was introduced as showing work problems are rate problems, so let us calculate the number of cards sorted per minute by each machine.

Some discussion of rates is in order; Some students will find an appeal to arithmetic satisfying. Per means divide, so cards per minute means the number of card in a certain time interval divided by the number of minutes in the time interval. Other students will look to the equation C = RT, Cards equals rate of sorting times Time sorting, and solve for the rate.

Returning to the example, 27,000 cards in an hour means 27,000 cards in 60 minutes. 27,000/60 = 450 cpm; 39,000/60= 650 cpm. Now fill in the table. We are looking for the time the machines work together it is unknown so call it t.

Cards | = | Rate | Time | |

Sorter I | 450t | 450 cpm | t min. | |

Sorter II | 650t | 650 cpm | t min. | |

Together | 55,000 |

450t + 650t = | 55,000 | |

1100t = | 55,000 | |

t = | 55,000/1100 | |

t = | 50 minutes. |

*. Check it. 450 cpm for 50 min. gives 22,500 cards. 650 cpm for 50 min. gives 32,500 cards. These amounts total to 55,000 cards as was required, we have the solution.*

Solution problems are a different type of rate problem. They do not involve time, but still we have one variable is the product of two others. The amount of solute equals the product of the percent concentration times the amount of solution. The percent concentration serves in the role of the rate.

When looking for a relationship between the variables, givens and unknowns, think of common sense general principles. Solution problems are examples where this type of thinking can be used to make an intimidating problem manageable. If I have two pails of water one contains 5 gallons and the other contains 4, how many gallons would I have in a third pail if I poured the first two into it? If you said nine you can do solution problems. Here is an example.

Example
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. If 5 gallons of 20% alcohol is mixed with 4 gallons of 25% alcohol, what will be the percent concentration of the new mixture?
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Understanding the Problem
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. The previous discussion tells us with three quantities: amount of solute, alcohol; the percent concentration; and the amount of solution. Also the first quantity is the product of the other two. Recall your arithmetic, to multiply by a percent change it to a decimal. 20% = .20; 25% = .25.

Amount | Percent | Amount of | ||

Alcohol | concentration | Solution |

Solution | 1 gal. | 20% | 5 gal. |

Solution | 1 gal. | 25% | 4 gal. | |

Mixture | 2 gal. | P | 9 gal. |

2 = | P(9) | |

2/9 = | P | |

P = | 22.22% |

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Example:
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If a car travels 5 hours at a rate of 20 mph and another 4 hours at a rate of 25 mph, what is its average speed for the 9 hours? ‘

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Understanding the problem:
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Many students find the average between 25 and 20 getting

22.5 mph. They do not understand the concept of average speed. The average speed is the total distance divided by the total time. So we must know how far the car went. We need our relationship: D= RT.

D | = | R | . | T | ||

First part | 100 m | 20 mph | 5 hrs. | |||

Second part | 100 m | 25 mph | 4 hrs. | |||

Total trip | 200 m | R | 9 hrs. |

200 = | R(9) | |

200/9 = | R | |

R = | 22.22 mph |

*. The definition average speed is less than the common error “average”: 22.22 versus 22.50. While the problem may be considered arithmetic, we should still point it out to our students. Why should they miss easy problems?*

Let us conclude these examples with a more algebraic solution problem.

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Example:
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How many gallons of pure alcohol must be added to 5 gallons of 20% alcohol solution to make a 50% alcohol solution?

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Understanding the Problem:
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As in the previous solution problem, we know that the amount of alcohol equals the product—of the percent and the amount of solution. We are mixing two solutions to get a new solution so we are adding amounts. We want to know how much pure alcohol to add, the quantity, so let that be x. What is the percent alcohol of a pure quantity of alcohol? Pure means 100% alcohol. Now, fill in a table.

Amt. Alc. | = | % Alc. | . | Amt. Solution | ||

First Solution | 1 gal | 20% = 0.20 | 5 gal. | |||

Pure Alcohol | X | 100% = 1.0 | X | |||

The Mixture | 1 + X | 50% = 0.50 | 5 + X |

1 + X = | 0.50 ( 5 + X) | |

100 + 100X = | 250 + 50 X | |

50X = | 150 | |

X = | 3 gallons. |

*It 3 gallons of pure alcohol are added to 5 gallons of solution, the new mixture is 8 gallons. At the same time the 3 gallons of pure alcohol were added to the one gallon of alcohol that was already in the solution for a total of 4 gallons of alcohol. 4 gallons out of 8 is 50%. We have solved the problem.*