Блог Анкара

In This Chapter

^ Deciphering between fact and fiction

^ Getting organized and planning an approach

^ Turning guessing into a science

Etting ready to solve a math word problem involves more than sharpening your pencil, aligning your sheets of paper, and taking a deep breath. Math word problems have the reputation of being obscure, difficult, confusing — you name it. The only way to overcome this bad rep is to be ready, able, and willing to take them on. Start with the right attitude and preparation so that you can approach math word problems in a confident, organized fashion.

In this chapter, I review the importance of isolating the question, determining just what information is needed, and ignoring the fluff. Math word problems often contain information that makes the wording of the problem more interesting but adds nothing to what’s needed for the solution. Also in this chapter, you see the importance of doing a reality check — does the answer really make sense? Is it what you expected? If not, why not?

Singling Out the Question

A math word problem is full of words — big surprise! Word problems really represent the real world. When you have a problem to solve at the office involving ordering new file cabinets, you don’t sit down to write out your times tables, and you aren’t handed a piece of paper with an algebra problem asking you to solve 2x + 3 = 27. To be successful with a word problem, you

Have to translate from words to symbols, formulate the equation or problem needed, and then perform the operations and arithmetic correctly to find the answer.

Wading through the swamp of information

One of my favorite sayings is: "When you’re knee deep in alligators, it’s hard to remember that your mission is to drain the swamp." This certainly applies to math word problems and sorting out what the question is from all the other stuff. Consider the following problem. See if you can find the question in all the verbiage.

The Problem: The 17 office workers and 4 managers at Super Mart all need new file cabinets before the end of this month, which has 31 days. The file cabinets cost $300 each, and the supplier of file cabinets will haul away the old cabinets for $5 each. How much will it cost to replace the file cabinets if each of the office workers takes his old file cabinet home and each of the managers elects to have them hauled away?

There’s a lot of information in this problem, but you need to first look for the question — what is it that you want to find? Look for What, how many, how much, when, find, Or other questioning or seeking words. Ignore the rest for now until you determine what you’re looking for. Is it a number of file cabinets? Is it an amount of money? Is it a number of people? Get your question nailed down, and then worry about how to put it all together. In this problem, the question is basically: "How much money?"

If you can’t stand to let a problem go unanswered and you want to solve this one, the answer is $6,320. You figure out how much money it costs by determining how much was spent on each office worker and then adding the total to how much the managers spent. Each of the 17 office workers spent just $300 (just the cost of the cabinets), so 17 x $300 = $5,100. The managers had their file cabinets hauled away, which added $5 on to the cost. The total for each manager is $305, so 4 x $305 = $1,220. Add the amounts together, and the cost of the file cabinets for the entire office is $5,100 + $1,220 = $6,320.

Going to the end

In 95 percent of all math word problems, the question is at the end of the description. No, I didn’t do a scientific survey. This is just my best guess based on years of experience and reading way too many problems. Just trust me. Most of the questions in word problems are at the end. That’s just the way word problems are most efficiently constructed.

Reading the last sentence first isn’t a bad idea. When you find the question that you need to answer, you can go back and wade through all the information and sort out what’s needed to find the answer. The following questions are examples of how word problems are constructed and how the question seems to come more naturally at the end.

The Problem: In a particular parking garage, you have spaces for regular-size cars, compact cars, and large vans. The regular-size cars can park in the spaces designed for a van, and the compact cars can fit in the spaces designed for either a regular-size car or a van. If there are 200 spaces for regular-size cars, 40 spaces for compacts, and 30 spaces for vans, How many Regular-sized cars can park in the garage if vans are not allowed on a particular day?

The Problem: A magazine salesman gets a 5 percent commission on all one-year subscriptions, a 10 percent commission on all two-year subscriptions, a 20 percent commission on all three-year subscriptions, and a flat fee of $5 for every subscription he sells in excess of 100 subscriptions in any one week. On Monday, he sold 14 one-year subscriptions, 23 two-year subscriptions, and 6 three-year subscriptions. On Tuesday, Wednesday, and Thursday, he sold 12 of each type of subscription. On Friday, he sold 60 two-year subscriptions. One-year subscriptions cost $20, two-year subscriptions cost $35, and three-year subscriptions cost $52. On Which day Did he earn the most?

The Problem: If Tom is twice as old as Dick was ten years ago, and if the sum of their ages five years ago was 90, then How old Is Tom now?

You want the answers to these questions? Okay, you’ll get them, but you’ll have to read on in this chapter where I cover eliminating the unwanted and doing the operations in order.

Organizing the Facts, Ma’am, Just the Facts

Some writers of word problems seem to need to wax poetic — they go on and on with unnecessary facts just to make the problem seem more interesting. You know that it isn’t necessary to make these more interesting — they’re fascinating enough already, right? Okay, don’t answer that.

Eliminating the unneeded

After you’ve isolated the question, you go back to the problem to sort out the needed information from the extra fluff. Look at the three problems from the previous section again. I’ve drawn lines through the interesting-but-unnecessary.

The Problem: In a particular parking garage, you have spaces for regular size cars, compact cars, and large vans. The regular-size cars can park in the spaces designed for a van, and compact cars can fit in the spaces designed for either a regular-size car or a van. If there are 200 spaces for regular-size cars, 40 spaces for compacts and 30 spaces for vans, how many regular-size cars can park in the garage if vans are not allowed on a particular day?

The Problem: A magazine salesman gets a 5 percent commission on all one-year subscriptions, a 10 percent commission on all two-year subscriptions, a 20 percent commission on all three-year subscriptions, and a flat fee of $5 for every subscription he sells in excess of 100 subscriptions in any one week. On Monday, he sold 14 one-year subscriptions, 23 two-year subscriptions, and 6 three-year subscriptions. On Tuesday, Wednesday, and Thursday, he sold 12 of each type of subscription. On Friday, he sold 60 two year subscriptions. One-year subscriptions cost $20, two-year subscriptions cost $35, and three-year subscriptions cost $52. On which day did he earn the most?

The Problem: If Tom is twice as old as Dick was ten years ago, and if the sum of their ages five years ago was 90, then how old is Tom now?

You see that quite a bit is eliminated in the first problem, just a bit is eliminated in the second problem, and nothing is eliminated in the third problem. The information that’s been eliminated may be useful to answer some other question, but it isn’t needed for the question at hand.

Doing the chores in order

You’ve isolated the question and eliminated the riff-raff. Now it’s time to set up the arithmetic problems or equations needed to solve the problems. The order in which you perform the operations is pretty much dictated by what the question is. The last operation performed is what gives you the final answer. I’ll take the problems one at a time.

The Problem: In a particular parking garage, you have spaces for regular size cars, compact cars, and large vans. The regular-size cars can park in the spaces designed for a van, and compact cars can fit in the space designed for either a regular size car or a van. If there are 200 spaces for regular-size cars,

40 opacoo for compacto and 30 spaces for vans, How many Regular-sized cars can park in the garage if vans are not allowed on a particular day?

To solve this problem, you need to answer the question How many, Which is a total of two types of parking spaces. The total is the sum of 200 spaces plus 30 spaces, which is 200 + 30 = 230 spaces. For more problems of this type, turn to Chapter 5.

The Problem: A magazine salesman gets a 5 percent commission on all one-year subscriptions, 10 percent commission on all two-year subscriptions, a 20 percent commission on all three-year subscriptions, and a flat fee of $5 for every subscription he sells in excess of 100 subscriptions in any one week. On Monday he sold 14 one-year subscriptions, 23 two-year subscriptions, and 6 three-year subscriptions. On Tuesday, Wednesday, and Thursday, he sold 12 of each type of subscription. On Friday, he sold 60 two-year subscriptions. One-year subscriptions cost $20, two-year subscriptions cost $35, and three-year subscriptions cost $52. On which day did he earn the most?

Solving this problem requires comparing the commissions made on three different days. Only three days are necessary, because the commissions are the same for Tuesday, Wednesday, and Thursday. Comparing the commissions requires finding the total commission for each day by multiplying the number of each type times the rate for that type times the price; then the commissions from the different types are all added together. A table or chart is helpful in this case to keep everything in order (see Table 2-1). You can do the commission amount computations ahead of time. For one-year subscriptions, 5 percent of $20 is $1. I got that by multiplying 0.05 X $20. Two-year subscriptions earn the salesman 10 percent of $35, or $3.50; and three-year subscriptions are worth 20 percent of $52, which equals $10.40. Refer to Chapter 6 for more on percentage problems involving decimals and percents.

He had his highest commission on Friday. In Chapter 14, you see many problems involving Quality x quantity — more like this type of problem.

The Problem: If Tom is twice as old as Dick was ten years ago, and if the sum of their ages five years ago was 90, then how old is Tom now?

This problem is more like the word problems that most people remember from their high school years. The problem almost seems like double-speak when you first read it. I hate to give too much away right here, because I cover age problems in great detail in Chapter 15, but let this be a Tickler — Something to get you all enthused about doing more age problems.

You want to answer the question "How old is Tom?" So let Tom’s age be represented by T. You’re comparing Tom’s age to Dick’s age. So let Dick’s age be represented by D. But Tom’s age is compared to Dick’s age Ten years ago. How old was Dick ten years ago? You get that by subtracting 10, so Dick was D - 10 ten years ago, and Tom is twice that or 2(d – 10). So you can say that T = 2(d – 10). Well, there are lots of numbers that satisfy the equation T = 2(d – 10), so you need a bit more information. You’re told that Five years ago. . . . Their ages five years ago were T - 5 and D - 5. The sum of their ages at that time was 90, so T - 5 + D - 5 = 90, which simplifies down to T + D = 100. So their ages today must add up to 100. The algebraic way of solving the system of equations T = 2(d – 10) and T + D = 100 is found in Chapter 17. For now, I’ll let you off the hook and tell you that the solution is that Tom is 60 and Dick is 40.

Estimating an Answer to Check for Sense

No matter how easy or complicated a math word problem, you should always have some guess or inkling or idea as to what the answer may be before you even get started on solving it. Even if you’re way off with your guess, this exercise is very useful. You’re more apt to check the work if you think that the answer is Way out there. Many times, you just find that you didn’t do a very good job of guessing. Other times, you find that you made a mistake in the arithmetic and you can go back and correct your error.

Guessing an answer

One of a student’s favorite instructions on a test is: Must show work. And here I am, encouraging you to guess the answer. I’m a firm advocate of showing work, too. But guessing is a great skill to have, and it’s useful when doing these word problems. Showing the work and all the steps involved is good practice, too, so you can apply the same steps and techniques on problems that aren’t quite as easy to guess for the answers.

The Problem: You have $10,000 to invest over the next year and will put some of the money in an account earning 12 percent (risky) and three times as much as that in an account earning 4 percent (safer). How much interest will you earn?

You’re going to multiply some money by 12 percent and the rest by 4 percent. (Refer to Chapter 10 on using formulas and Chapters 6 and 14 for more problems like this.) You make an estimate of the total amount of interest you’ll earn by picking a percentage somewhere between 12 percent and 4 percent and multiplying the $10,000 by that percent. Because more money is going to be invested at the lower rate, you’ll probably use either 5 percent or 6 percent. Let your guess be the 6 percent, making your estimate on the interest earned be 6 percent of $10,000, or $600. Now, when you actually do the problem, if you get some out-of-the-world answer like $8,000, you’ll know that you’ve put a decimal point in a wrong place or done some other silly computation. What’s the actual answer, you ask? It really is $600. Sometimes I even surprise myself. I multiplied $2,500 by 12 percent and $7,500 (three times as much as $2,500) by 4 percent and got a total of $600.

Doing a reality check

Some math word problems have answers that just don’t seem right. These answers are the ones that definitely need checking. Other answers are just obviously wrong, meaning that the problem needs reworking.

Being struck by the obvious

Math problems sometimes require solving algebraic equations. The algebra is wonderful, but you run the risk of creating some Extraneous roots. Extraneous roots or solutions are answers that satisfy the algebraic equation but don’t really mean a thing in the situation.

The Problem: If Folabo leaves home and drives 400 miles north while her sister Fatima leaves home at the same time and drives 300 miles east, how far apart are they?

Y\Mi A line segment drawn 400 miles upward connected to a line segment drawn 300 miles eastward forms two sides of a right triangle. You find out how far apart they are by finding the hypotenuse of a right triangle. (In Chapter 18, Pythagoras and his famous triangle are discussed at length. Pardon the pun.) Anyway, to solve this you solve 4002 + 3002 = C2 For C And get C = ±500. Obviously, the distance can’t be negative, so you just use the positive value and determine that they’re 500 miles apart.

Getting a firm grip on reality

Sometimes a weird answer looks good at first, and it takes careful checking to show that it doesn’t work.

The Problem: I’m thinking of a number. If you subtract 1 from the number and find the square root of the difference, you get the same answer as when you subtract the number from 7. What is my number?

This problem requires a nice algebraic equation. (Chapter 11 has lots of problems like this, if these are up your alley.) Letting the number be represented by X, The equation to use is Jx — 1 = 7 — X. Squaring both sides of the equation and then solving the quadratic equation by factoring, you get

(yX—T j=(7—X )2

X — 1 = 49 — 14x + X2 0 = 50 — 15x + X2 0 = (10 — X)(5 — X)

The two answers to this equation are X = 10 and X = 5. They both look perfectly wonderful, don’t they? Not! If you put 10 back in for X In the equation, you get that 3 = -3. Doesn’t work. But putting the 5 in for X, You get that 2 = 2. The 5 is a solution, and the 10 is Extraneous — it satisfies the quadratic equation, but it makes no sense in the answer. (If you like this type of number problem, you’ll find lots of them in Chapter 11.)

Chapter 3