## Word Problems

Here's an easy word problem:

Suzy is ten years older than Billy, and next year she will be twice as old as Billy. How old are they now?

If you don't use algebra, you probably have to solve it by trial and error. This sometimes works fine. Here it is a little slow. And, in other problems, it is next to impossible.

1. Translate into equations

With algebra, the solution is easy. The only problem is to convert the above sentence into equations, because equations are what we need to use algebra. How is this?

S=10+B

S+1=2(B+1)

That is a direct translation of the word problem. I used S to represent Suzy's age (this year), and B for Billy's age. I could use other letters, but these are easier to remember. Do you see how this translation is done? "Suzy is ten years older than Billy" is an equation (S=10+B), but is in words instead of symbols. "Next year she will be twice as old as Billy" is a little more complicated, but is just another equation (S+1=2(B+1)).

The translation process seldom gets much more difficult than the above. But, you may have to weed out extra information. If I had started the word problem with, "Suzy is six inches taller than Billy, and...," you are getting extra info which has nothing to do with the problem. You would have an extra equation (Z=6+L, where Z is Suzy's height, and L is Billy's height). You would find that this equation does not affect the other two equations, at all. You could write down this equation, but you would end up solving the other two equations.

2. Solve the equations

This is just regular algebra (two equations and two unknowns). There are several ways to continue:

1. Solve for one variable (in one equation) and substitute in the other equation.
2. Subtract one equation from another (after changing their form, perhaps), to solve for one variable.
3. Use Matrices.
4. Use Determinants.
5. Graph the equations (not as easy).

If we solved for one variable, then we go back and solve for the other variable. Let's use the first method. S=10+B. Substitute that into the second equation:

10+B+1=2(B+1)

B=9

Going back into the first equation, we find that S=19. Billy is 9, Suzy is 19.

Checking your work may seem like a waste of time, especially when you get good at algebra. But, remember that you went through a translation process above. That is not an exact science. Check your work, see that you didn't get something backward.

We can put the numbers (the answers) back into the word problem to see if they are correct. Above, we notice that next year Billy will be 10 and Suzy will be 20. And she will indeed be twice as old as Billy.

The mystery seems to be in step #1, "Translate into equations". It is sometimes easier said than done. And you really should practice doing word problems. But there are (fairly obvious) clues (key words) in the statement of the problem:

• "and" often means "plus"
• "difference" often means "larger-smaller"
• "five more than" means "+5"
• "next year" means "(this year)+1"
• "twice" means "times 2"
• ages are normally in whole numbers

We see statements like "Billy has two and a half dollars less than Sally." So, B=S-2.50. How about this:

In seven years, Billy's age will be one and a half times Sally's age.

What will Billy's age be in seven years? B+7. What will Sally's age be? S+7. What do we get from the above sentence? B+7=(1.5)(S+7). Right?

Note: I will write a separate article on units (like miles or degrees), and keeping them straight.

In geometry, we have formulas for areas and volumes; see Areas and Perimeters and Some Triangle Formulas. Also see The Pythagorean Theorem. In science, we have d=vt or distance is velocity (speed) times time, and other well-known formulas. Velocity is a "rate;" we also have rates of interest involving money, which work the same way. We have formulas for converting units; see Fahrenheit to Celsius. If we don't know these formulas (or if we cannot figure them out from simpler ideas), then we cannot solve certain algebra problems.

And we can use these formulas backward. In other words, we can solve for any of the variables, if we know the value of the other variables.

Example #1: The first stage of a rocket burns 28 seconds longer than the second stage. If the total burning time for both stages is 152 seconds, how long does each stage burn?

This is not really about rockets (did the word "rockets" scare you, making the problem more difficult?). It is about numbers. I restate the problem: We have two numbers, one is 28 more than the other, and the sum is 152. No matter how we state the problem in words, algebraically it is x+x+28=152, and we can solve for x (or t, as many people would do).

Example #2: In a student election, 584 students voted for one or the other of two candidates for president. If the winner received 122 more votes than the loser, how many votes were cast for each candidate?

This is the same problem as #1, with different numbers. x+x+122=584.

Example #3: How many liters of a 10% solution of acid should be added to 20 liters of a 60% solution of acid to obtain a 50% solution?

We might state this formula for the use of percent solutions: A=RM, where A is the amount of the substance (acid), R is the ratio (a percentage is a ratio times 100, so we have to divide by 100) of the substance in the mixture, and M is the amount of the mixture (solution). This is just the definition of "ratio" or "percentage." Well, using A=RM, we have a 10% solution and a 60% solution. A1=0.10M1 and A2=0.60(20), and we want to mix them: A1+A2=0.50(M1+20). We can calculate A2, and that gives us two equations with two unknowns (A1 and M1), so we can probably solve for both.

Example #4: Jim can row his boat upstream at four kilometers per hour, and downstream at six kilometers per hour (Jim is very strong). He rowed upstream a certain distance and then rowed back downstream to his starting point in one hour. How far did he row upstream?

We might ask how fast the river is flowing. The answer is one km/hr, and Jim would row five km/hr on an unmoving lake. His speed upstream is 5-1 km/hr, and his speed downstream is 5+1 km/hr. This is important to all problems involving movement with or against some other motion (like walking inside a train car). Well here, rowing against a current produces my rowing speed minus the speed of the current. The above problem is a d=st, or distance equals speed (or rate or velocity) times time problem. The distance is the same both ways, so 4t=6(1-t). We can solve for t. Then we can use d=st to solve for d.

Example #5: A rectangular swimming pool measures 20 meters by 10 meters. How wide must a concrete walk around the pool be if the walk is to cover an area of 124 square meters?

We have to assume that the walk must have the same width on each of the four sides of the pool. We often have unstated assumptions like this in a math problem. With a width of w, the area of the entire walk is 20w+20w+10w+10w+4w^2 (where w^2 is w squared), or 60w+w^2. This is 124 square meters, so we can solve for w using the Quadratic Formula.

Example #6: A man is nine times as old as his son. In nine years, he will be only three times as old as his son. How old are each now?

If the son's age is s, then the father's age is 9s. In 9 years, the son's age will be s+9, and the father's age will be 9s+9. The father's age will also be 3(s+9). So 3(s+9)=9s+9. We can solve for s, and then solve for s+9 (don't forget that part of the problem).

Example #7: I have \$11.60, all dimes and quarters, in my pocket. I have 32 more dimes than quarters. How many dimes, and how many quarters, do I have?

For non-Americans (not un-Americans of course), a dime is \$0.10 and a quarter is \$0.25. We can deduce 0.25q+0.10(q+32)=11.60. And we can then solve for q, and then for q+32 (don't forget that part of the problem).

Example #8: I have balanced a weightless beam (there are a lot of them in physics), ten meters long, on a fulcrum by putting a 6 kg mass on one end, and an unknown mass on the other end. To balance this, I had to place the fulcrum 1.5 meters from the 6kg mass. What is the unknown mass?

Our lever principle states that F1d1=F2d2, where the F's are forces. Well our 6 kg mass is not a force but a mass, but force is proportional to mass. So, 6(1.5)=x(8.5), and we can solve for x.

Example #9: Bill can mow his mother's lawn in 6 minutes. His brother Jim can mow it in 9 minutes. How long will it take them to do it together, if each has his own lawnmower.

This is called a work problem. In one minute, Bill mows 1/6 of the lawn, and Jim mows 1/9. T/6+T/9=1 (the whole lawn). We then solve for T. We can accept fractions of a minute, or minutes and seconds (with maybe fractions of a second?).

Also see Sally's Age, for another example.