Decoding the Word Problem

Following up from my previous blog, I draw my attention to the decoding of the word problem. This would include both step 4 and step 7. Some people, perhaps most, concede that this process is the most difficult part of the problem. I will take some time to focus on this task with several specific examples. I will start with the simplest problems working my way up to the more complex, including the notorious multivariable rate problem.

Let us start with a basic subtraction problem:

Jerry started the day with 16 marbles. After he played with his friends, he counted 11 marbles. How many marbles did Jerry lose?

After looking at the final question, we find that the answer we are looking for is a certain number of marbles. The given facts are marbles he had before and marbles he had after. The fact that he ended up with less marbles is inferred by the question. So, let’s say we assign marbles before and marbles after with B and A, respectively. The answer we are looking for, marbles lost, we assign with L. Logically we deduct marbles after from marbles before by setting up the equation B-A= L. Inserting the given information we get 16-11=L, which should give us the answer we are looking for.

Now let’s try a two-step problem with ages. Age problems used to be the problems that tripped me up in grade school.

Henry is four years younger than George. In three years, George will be 37. How old is Henry now?

OK, notice what the question is: How old is Henry now? Since that is what we are looking for, let’s set that answer as our variable, H = Henry’s age now, and we can relate everything else from our given information. We’ll call George’s age “G”. The difference between ages will not change over time at this time of year, but it is important to assume our variables are valued at the present time; that is when our answer will be. If Henry is 4 years younger than George, then H = G - 4. Note that G-4 does not make the variable “G” smaller; in fact, “G” must be larger to balance the equation. This can be a source of confusion, so if you look at this equation and you at first think “G” is the younger man, plug in an arbitrary value for the ages and see if it makes sense. George’s age (remember “G” is present age) in three years is 37, so G + 3 = 37. Getting “G” by itself from that equation by subtracting 3 results in G = 34. Substituting this for “G” in the first equation gives us H = 34 - 4, and H = 30.

Now we will look at decoding a problem with multiplication involved. Age problems may not be very practical, but they make interesting trivia puzzles. Let’s consider this one:

Max is 10 years older than Henry. In two years, Max will be three times Henry’s age. What are their ages now?

We can start by assigning variables to the unknown ages we are asked for, Max and Henry’s current ages. We will call them M and H, respectively. Their ages in two years will be M+2 and H+2. At these ages, Max is 3 times Henry’s age, so we will need to multiply Henry’s future age by 3 to make it equal Max’s age. The relationship should look like this: M+2 = 3(H+2). Distributing the multiplication and subtracting 2 from both sides of the equation will give us M = 3H+4. We are also given the age difference. Max is older, so we will need to add 10 to the younger one, Henry, to make them equal. Now if H+10 is equal to M, then we can substitute H+10 for M in the previous equation, giving us H+10 = 3H+4. Subtracting 4 and H from both sides reveals 6 = 2H, where we easily determine Henry’s age is 3, and Max’s age is 13.

Next, we should look at ratios of different units called rates. Examples of rates are miles per hour, parts per minute, and gallons per minute. The “per” is equivalent to saying “for each”, and describes which unit goes in the denominator of the ratio. It is a good idea to think of rates as a division of the different units instead of the rate being a unit itself. For example, don’t think of 60 mph, instead, think of it as 60 miles divided by 1 hour. This way it is easy to see that multiplying the rate by 2 hours results in 120 miles. If the gas mileage was 20 miles per gallon, we could divide, or multiply by the reciprocal (.05 gallons per mile), the speed rate to give us 3 gallons per hour. When we average a rate, we add up all the numerator units, and divide it by the total denominator units.

Problem: A car is driven 50 miles at 60 miles per hour and then drives another 50 miles at 90 miles per hour (assuming no state trooper pulls him over). What was the average speed?

If you averaged 60 and 90 to get 75 mph, you are incorrect. Think about it, if the car 50 miles at 60 mph, and 100 miles at 90 mph, would the average still be 75?

So, we know the miles traveled in the original problem was 100 miles. Next, we need to know how long it took. We can take the reciprocal of the speed, 0.167 hours per mile, and multiply by 50 miles. That would be .833 hours. The hours per mile for the second half would be 1/90, or 0.011 hours per mile. Multiplying by 50 gives us .555 hours for a total trip time of about 1.39 hours (0.833 + 0.555). The average speed would then be 100/1.39, or about 72 mph.

Here we will wrap up with some combined rate problems involving three rates:

A classic word problem is having a task that is done by individuals in a certain amount of time. The question often asked is: if we know how long each individual takes, how long will it take for them to do it together (assuming logistically they don’t interfere with each other). If we want to combine their work rates and we know how long it takes them by themselves, we need to add up the portions of the job they can do for each unit of time. For example, if Mark, Jack, and Cindy can paint a shed in “T” hours, and we use M, J, and C, respectively, for the hours it took for each of them to paint a similar shed, 1/M would be the part of a shed Mark could paint in an hour (shed per hour, if you will). Shed per hour is the reciprocal of hours per shed. Now, if we know M, J, and C, and want to find T, the total time it takes for all three to paint a shed, we should multiply the unknown total time, T, by each worker’s shed per hour (their portion of the one shed), and the sum of these portions should be 1, that is, one completed shed. So, T/M +T/J + T/C = 1. So, if Mark paints a shed in 12 hours, Jack paints one in 8 hours, and Cindy paints one in 6 hours, T/12 + T/8 + T/6 = 1. To find the time it takes for all of them to finish a shed, we find the lowest common denominator, in this case 24, and multiply both sides by that amount. That will give us 2T + 3T + 4T = 24, or 9T = 24. The time it takes for all three workers to paint the shed is 2.67 hours, or 2 hours and 40 minutes. Check to see if that makes sense.

Finally, we’ll look at another problem like the last one, but with more unknown values to determine, so it will be a little bit more complicated. Matt, John, and Cathy are the painters. Matt takes twice as long to paint a shed by himself than it takes Cathy to paint one by herself. Also, Cathy’s solo painting time is three fourths of John’s time. When all three paint together, it takes them 4 hours. How long does it take them each to paint a shed by themselves? We can set up the summing equation, but this time we know the total time while painting together. Assigning appropriate variable symbols, it will look like this: 4/M + 4/J + 4/C = 1. Now we should be concerned; we have three variables and only one equation. We need three equations to solve these unknowns, but we do know something about the relationship between these variables. We need to pick one variable to compare the others to; it doesn’t really matter which one, so I pick J, John’s shed painting time. Cathy’s painting time is 3/4 of John’s time, so C = .75J. Matt takes twice as long as Cathy, so M = 2C = (2)0.75J = 1.5J. Now we can set up our sum: 4/1.5J + 4/J + 4/.75J = 1. We can clean this up by multiplying both sides by 1.5J. That gives us 4 + (1.5)4 + (2)4 = 1.5J or 4 + 6 + 8 = 1.5J. 18/1.5 = J = 12. Going back to the other relationships we find that C = (.75) 12 = 8, and M = (2) 8 = 16. So, Matt takes 16 hours to paint a shed, John takes 12, and Cathy takes 8. Check to see if that makes sense.

Well, that’s all I have for now. Maybe in another exposition we can explore quadratic word problems and related rates. I hope this has helped your understanding of the decoding of mathematic word problems.