Блог Анкара

In This Chapter

^ Going step-by-step through a problem using suggested techniques

^ Using both sketch and table to illustrate the problem

^ Solving a problem using Pythagoras’s theorem and a quadratic equation

Solving a math word problem may seem daunting at first, but it doesn’t have to be if you have a plan and the proper mindset. You don’t have to use all the steps and procedures in this chapter for every math problem, but here you see how different techniques are useful when attempting to solve a problem.

In this chapter, I tell you what to look for in terms of the question and information needed. I fill you in on all the steps used to solve an equation — and the proper order to do them in. And I show you the need for checking your answer at the end.

Laying Out the Steps to a Solution

A math word problem presents challenges in understanding, organization, and launching the mathematical problem to be solved. To illustrate all these steps (and more), consider a problem involving two friends and their walking adventure. They both leave the same place at the same time; one walks north and the other walks east. One walks faster than the other. And, for some reason known only to them, they can determine how far apart they are after a period of time.

The Problem: Shelly and Shirley leave their dorm at 8 a. m. and start walking in different directions. Shelly starts walking due north, and Shirley walks due east. Shirley takes her time to smell the flowers and Shelly walks at a pretty steady pace. So, at noon, when they stop walking, Shelly has gone 1 mile less than twice as far as Shirley. At noon, they are 17 miles apart. How far did Shelly walk?

How old was the wife?

At the time a man and woman married, the man wife’s age was now % of his age. How old was determined that his wife’s age was % of his age. his wife when they were married? After 12 years of marriage, he found that his

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.VLAiy Words, words, words. Somewhere in the description of the problem, you find the question that needs to be answered, information that needs to be organized, other information that can be ignored, an appropriate drawing that can be made, and an algebra problem begging to be written. In the following sections, I walk you through the various steps — steps you can apply to any math word problem.

Step 1: Determine the question

The main thing that empowers you when you’re attempting a math word problem is determining the question. You wade through all that information and wonder what in the world you’re going to do with everything. So, find the question, and you’ll have more direction. The question is usually at the end of the problem description.

In this problem, the question is, indeed, at the end. How far did Shelly walk? The only unit mentioned is Miles — because they’re 17 miles apart — so your answer should be something like: "Shelly walked_miles."

Step 2: Organize the information

After you’ve determined what the question is (see the preceding section), you can go back to the problem and decide what information is needed and in what order you want to do the various steps. If you have an answer guess or goal in mind, it’ll help you determine how the steps get arranged.

Eliminating the unneeded

The problem mentions the distance that Shelly and Shirley walked and the distance they’re apart. Also, in the problem description you see times listed. You don’t need the time they started and finished, because there’s no mention of rates — requiring you to use D = Rt. So just eliminate or ignore the mention of the times. It’s just fluff. It also doesn’t matter why Shirley walked slower; it could have been that her legs are shorter.

Shelly and Shirley leave their dorm at 8 a. m. and start walking in different directions. Shelly starts walking due north, and Shirley walks due east. Shirley takes her time to smell the flowers, and Shelly walks at a pretty steady pace. So, at noon, when they stop walking, Shelly has gone 1 mile less than twice as far as Shirley. At noon, they are 17 miles apart. How far did Shelly walk?

Making an educated guess

The problem involves two people walking in different directions and ending up 17 miles apart. If they walked in opposite directions — one north and one south — you’d expect the sum of the distances they walked to be 17. Neither could have walked more than 17 miles. In this case, they walked at right angles from one another, so the sum of the distances that they walked has to be less than 17. Keeping in mind a Reasonable Answer, you’re more likely to set up the process (equations and operations) properly. So some possible answers are that Shelly walked 10 miles and Shirley walked 5 miles. The numbers don’t really need to fit or be the right answer. You just have a relationship in mind — that Shelly walked farther and the sum of their distances is less than 17.

Getting organized

The question asks for the distance that Shelly walked. So you want some algebraic expression involving that distance. You could let the distance be represented by X. But look back at the problem to see what other distances are involved. You have the number 17 representing the distance between them, so you don’t need a symbol for that distance. The only other distance is the distance that Shirley walked. You could let the distance that Shirley walked be represented by Y, But then you’d have two different variables — X And Y — to worry about. You want to keep to one variable, if possible.

The last bit of information that hasn’t been used is that Shelly has gone 1 mile less than twice as far as Shirley. The two distances are compared to one another. Letting the distance that Shirley traveled be represented by Y, You can let Shelly’s distance be written in terms of Y. Shelly went 2y – 1 miles: twice Shirley’s less 1. Now you have Y For Shirley’s distance and 2y – 1 for Shelly’s distance.

Step 3: Draw a picture or make a chart

This problem just begs to have a picture drawn. But a chart or table may be helpful, too. In the following sections, I show you both options and let you decide which works best.

Providing some artwork

Shelly walks north and Shirley walks east. You have to assume that it’s possible for them to stay exactly on track and walk due north and due east, respectively, and not have to veer off. Math problems are usually about optimal situations, not the reality of how roads are laid out.

Figure 4-1 shows you a sketch of the paths taken by Shelly and Shirley. The representations of their distances — Y For Shirley and 2Y - 1 for Shelley — are shown, as is their distance apart.

Shelley’s path 2y – 1

Figure 4-1:

The paths are at right angles.

\ 17

Y

Shirley’s path

The picture of their paths seems to suggest a right triangle. And a right triangle comes with that most famous formula, the Pythagorean theorem. If you hadn’t thought of that theorem before you saw the picture, you probably did after seeing it.

Figuring out possible values with a table

This problem has Shelly traveling one less than twice as far as Shirley. You could make a table of possible values for the distances they traveled. You’d probably be interested only in whole-number values, which won’t solve the problem if the answer is a fraction, but you may get fairly close to the answer. Table 4-1 has some possible numbers or distances, starting with Shirley going 1 mile and ending to keep the total distance from getting larger than 17.

Counting cars

On a warm, sunny day, Clark went to the car races to watch an automobile race. As the cars sped around the track and Clark tried to watch them, he got dizzy. So he decided to keep his eyes on one particular car — the bright green one.

Clark then decided to count how many cars were in the race. He noticed that the total number of cars was equal to one-third of the cars in front of the green car plus three-quarters of the cars behind the green car plus the green car. How many cars were in the race?

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This table doesn’t really get you any closer to the answer, but it shows you how the distances are related. You know that the total distance has to be less than 17, so several of the entries in the table give possible solutions to the problem.

Step 4: Align the units

Many problems come with two or more different units. You may have information on time in terms of minutes and seconds, information on distance in terms of feet and yards, or some combination of all these. You can’t really

Change minutes and feet into the same unit, but feet and yards can be changed to feet, and minutes and seconds can be changed into minutes (or seconds). (Turn to Chapter 3 for a full coverage of how to deal with units.)

In this problem, because the only unit mentioned (after the time was eliminated) is miles, you just leave the unit as is. You’re finished in terms of the units.

Step 5: Set Up the operations or tasks

This is where all the previous steps should be coming together. You have the question in mind and a picture of what’s happening. You have an estimate or guess of how the answer should come out. You have variable expressions representing the distances traveled. Now you need to set up a process or equation. Sometimes the process is no more than multiplying a number by 2 or 3. That’s the best-case scenario. But it’s much more fun when you set up an equation to be solved.

The previous work on this problem suggests an equation. You have two different distances represented by expressions involving a Y. Write an equation and solve for Y. What equation? Why Pythagoras’s, of course!

In a right triangle whose shorter sides are A And B In length and whose longest side is C Long, the following is always true: A2 + B2 = C2.

The distances that Shirley and Shelly walked are the A And B Of a right triangle. The distance that they’re apart, 17 miles, is the C Value. Substituting into the Pythagorean theorem, the equation A2 + B2 = C2 Becomes (y)2 + (2y – 1)2 =172.

Finding a concrete solution

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Solving the Problem

A math word problem is different from other arithmetic and algebra problems, because you first have to translate from the words to the symbols before you do the operations or solve the equation for the answer.

Step 6: Perform the operations or solving the equation

The problem involving Shelly and Shirley boils down to an equation to solve. The equation is quadratic — requiring either factoring (coupled with the multiplication property of zero) or the quadratic formula. Some operations are performed on the equation first — squaring the binomial and simplifying terms, and the result of the simplifying is a quadratic equation set equal to 0. The equation (y)2 + (2y – 1)2 =172 can actually be solved by factoring, but I’ll show it both that way and with the quadratic formula, because the factoring may not be entirely obvious to you.

Solving the equation using factoring

The equation (y)2 + (2y – 1)2 =172 is quadratic. You first square each of the terms, including the binomial, and then simplify the terms by combining what you can. Then move all the terms to the left to set it equal to 0.

(Y )2 + (2y – 1)2 = 172 Y2 – 4y2 – 4y + 1 = 289 5y2 – 4y – 288 = 0

The factorization of this quadratic is the product of two binomials. The first terms in the binomials have to be 5y and Y. There’s no other choice. It’s the second numbers that will be the challenge. You have to find two numbers whose product is 288 — that’s challenge enough. But then you have to figure out how to arrange the factors so that the difference between the outer and inner products is 4Y.

5y2 – 4y – 288 = 0 (5y )((y ) = 0

The two factors that work are 36 and 8. Their product is 288, and, when you multiply the 8 by 5 and the 36 by 1 you get 40 and 36, respectively. The difference between the two products is 4. The product 40Y Has to be negative for

The difference 4y to come out negative. So the factor 8 is negative and the factor 36 is positive in the factorization.

(5y + 36)(y – = 0

Setting the two factors equal to 0, you get Y =— Or Y = 8. The negative

Fraction doesn’t make any sense if this is supposed to represent distance, so you go with the solution Y = 8, only.

Solving the equation using the quadratic formula

Not all quadratic equations can be solved by factoring. And sometimes those that can be solved by factoring are more easily solved using the quadratic formula. On the other hand, All Quadratic equations can be solved using the formula. It’s just that factoring is usually quicker, easier, and more accurate (not as many opportunities for error).

The quadratic formula says that if a quadratic equation is written in the form

Ax2 + Bx + C = 0, then its solutions are found with X -

B ±Jb2 — 4ac

2a

Solving the equation 5y2 – 4y – 288 = 0 with the quadratic formula, you get

-(— 4) ±7(— 4)2 — 4(5)(—288)

Y =-oTF\-

2(5)

4 ± /16 + 5,760

= -

10

4 ± 5,776

= -

10

= 4 ± 76 = 80 or —72 10 10 10

This gives you the same two answers that you get with the factoring method: 8 and — -55-.

Step 7: Answer the question

After working hard at solving the quadratic equation, it’s tempting to just sit back, relax, and think that your work is done. Not so. The solution of the equation that you think will work is that Y = 8. What is Y? Does its value answer the question?

First, Y Represents the distance that Shirley walked. The question asks how far Shelly walked, so the number 8 is not the answer to the question. To determine how far Shelly walked, use 2y – 1 and replace the Y With 8. You get that 2(8) – 1 = 16 – 1 = 15. So Shelly walked 15 miles.

Step S: Check for accuracy and common sense

Accuracy and common sense go hand in hand. You can’t have one without the other. The accuracy part goes to the arithmetic or algebra involved in solving the problem and whether the relationships hold. The common sense refers to whether the answer — even though it’s the solution of an equation — really fits the problem and the real world.

In the case of Shirley and Shelly who walked away from each other in directions that are right angles from one another, you find that Shirley walked 8 miles and Shelly walked 15 miles. They’re only 17 miles apart — even though they walked a total of 23 miles — because their journeys and the distance between them formed a right triangle. They didn’t walk in opposite directions. As far as the accuracy of the solution, check out the Pythagorean theorem with the distances:

82 + 152 = 172 64 + 225 = 289

289 = 289

The equation checks, the answer makes sense, and the problem is solved with the question answered.

Part II

In This Chapter

^ Adjusting units for ease in computations

^ Changing from English to metric units and back again

^ Squaring off with square and cubic units

Any mathematical problems involve units of length, weight, volume, or money. You incorporate the units into your computations and

Then report them in the answers so the solution makes sense and is useful. Sometimes you’re confronted with problems that have two or more different units — such as feet and inches or pounds and ounces — and you have to make a decision as to which unit to use.

In this chapter, I offer suggestions on how to choose the unit or units and then how to work with the unit or units you’ve chosen. This chapter also covers the tricky conversions of square feet to square inches or cubic yards to cubic feet. And, of course, no discussion of units is complete without introducing meters and kilograms, so you get conversions involving metric and English measures.

Choosing the Best Measure

When a problem involves two different measures, you choose one or the other measure to work with and convert the unchosen measure to the unit you want so that they’re all the same. You may even decide to change measures when they’re already all the same — just because you think that another measure may work better.

Using miles instead of inches

A mile is much longer than an inch. In fact, there are 12 X 5,280 = 63,360 inches in 1 mile. If you’re measuring how far it is from one side of the desk to another, you’ll use inches. If you’re measuring how far it is from your home to your workplace, you’ll usually measure in miles.

The Problem: Train tracks are made of metal, and metal expands when it gets hot and shrinks when the weather is cold. When the tracks are put in place, a gap should be left between the adjacent tracks to allow for expansion. A metal track that’s 1 mile long expands with the heat and increases in length by 12 inches. There was no gap between the tracks, however, so it buckled in the middle and formed a V shape. (See Figure 3-1 for a picture of what the track looks like.) How high up did the track rise?

Figure 3-1

The train track expanded with the heat.

1 mile

^VLA/J! Two different measures are given: 1 mile and 12 inches. You can work with inches, and change the 1 mile to 63,360 inches, or you can work with miles

And write the 12 inches (1 foot) as 5 mile. The choice here is between

Using really large numbers and using fractions or decimals.

I choose to go with the fractions — to work with pieces of a mile. To find out how high the rise in the track is, I use the Pythagorean theorem (Chapter 18 is completely devoted to that theorem of Pythagoras) and one right triangle going halfway down the track. The bottom segment of the triangle is >2 Mile

Long, and the Hypotenuse (longest side) is >2 Mile plus 6 inches or 10 650 mile

Long. To solve for the rise, which I’ll represent with X, I solve the equation for X.

/1\2 /-« 1 \

‘ T ) + x2 = D 2 + Tn^ 2 2 10,560

This looks pretty nasty, but a scientific calculator makes short work of the problem, and you get

0.25 + X2. 0.2500947 X2 = 0.0000947 X = 0.009732

The height or rise of 0.009732 doesn’t seem like much, but, remember, this is in miles. Multiply by 5,280 feet and you get over 51 feet. Whoa! That’s quite a rise!

Working with square feet instead of square yards

When you buy carpeting, you usually buy it in square yards — 3 feet by 3 feet. But you probably bought your last tile floor in terms of a number of square feet.

The Problem: Using your yardstick, you measure the length of a room to be 6 yards and its width to be 5 yards. You plan on putting in 1-foot-square tiles. How many tiles will you need?

Before determining the area of the room, first change the yards to feet using 1 yard = 3 feet. So 6 yards is 6 x 3 = 18 feet. Five yards is 5 x 3 = 15 feet. A room that’s 18 feet by 15 feet is 18 x 15 = 270 square feet.

But what if you preferred finding the area in square yards, first, and then changing the area to square feet? The area of the room is 6 x 5 yards or 30 square yards. A square yard is equal to 9 square feet (3 feet x 3 feet). So multiply 30 x 9 to get 270 square feet.

Converting from One Measure to Another

When a problem contains more than one measure, you change everything to the same measure before doing the computing on the problem or solving the equation. You can’t add 6 inches to 4 feet and get 10 — you have to change the inches to feet or feet to inches. Knowing when to multiply and when to divide sometimes gets confusing, so your best bet is to write down the equivalence or change of units and then work from the equation.

Changing linear measures

First, here’s a list of some common equivalences used when working with lengths. I cover the English and metric equivalences later, in "Mixing It Up with Measures."

1 foot = 12 inches 1 yard = 3 feet 1 mile = 5,280 feet

The measure equivalences are used to convert from one measure to another. You may need to do more than one computation if there isn’t a direct equivalence between units — such as changing inches to yards or yards to miles.

The Problem: Cheryl has 48 rolls of satin ribbon, each containing 15 yards of ribbon. She plans to wrap packages to send overseas as gifts, and each package requires 30 inches of ribbon. How many packages can she wrap?

First, determine how many yards of ribbon are in those 48 rolls. Then change the yards to feet using 1 yard = 3 feet and the feet to inches using 1 foot = 12 inches. After you have the total number of inches, you can divide by 30 to get the number of packages that can be wrapped.

Multiplying 48 rolls x 15 yards you get 720 yards. Start with the equivalence involving yards and feet. To change 720 yards to feet, you multiply each side of the equation 1 yard = 3 feet by 720.

1 yard x 720

3 feet

X720

720 yards = 2,160 feet

You have 2,160 feet of ribbon. Change this to inches by using the equivalence involving feet and inches, 1 foot = 12 inches.

1 foot = 12 inches X 2,160 x 2,160 2,160 feet = 25,920 inches

That’s 25,920 inches of ribbon. Divide 25,920 by 30 to get 864 packages that Cheryl can wrap.

Adjusting area and volume

Area is a two-dimensional measure. You’re counting up how many squares — all the same size — fit into some flat region. You use area measures for floors in buildings and spaces in parking lots, as well as when you want to find out how much room there is in a backyard.

Volume is a three-dimensional measure and tells you how many cubes of a particular size fit into an object. Volume measures tell you about the inside of a refrigerator or the size of a cardboard carton.

IBE# 1 square foot = 144 square inches (12 inches X 12 inches)

1 square yard = 9 square feet (3 feet X 3 feet) 1 square mile = 640 acres

1 cubic foot = 1,728 cubic inches (12 inches X 12 inches X 12 inches) 1 cubic yard = 27 cubic feet (3 feet X 3 feet X 3 feet)

The Problem: Jimmy is going to play a prank on his dad and fill the refrigerator with ice cubes that are 1 inch on a side. The refrigerator can hold 6 cubic feet. How many ice cubes will Jimmy need?

Determine the number of cubic inches in 6 cubic feet by using the equivalence 1 cubic foot = 1,728 cubic inches and multiplying each side of the equation by 6.

1 cubic foot = 1,728 cubic inches

X 6 X 6

6 cubic feet = 10,368 cubic inches

That’s over 10,000 ice cubes. Jimmy had better rethink his plan. He’ll never get the ice cubes all stacked inside the refrigerator before they start melting — and he gets frostbite.

The Problem: Timothy bought 3,200 acres of land and intends to plant seedling trees on it. If each seedling requires an area of 9 square yards to grow properly, how many seedlings can he plan on his new acreage?

YUUV First, change the acres to square miles using 1 square mile = 640 acres and then the square miles to square feet. Determine how many square feet are in 9 square yards using 1 square yard = 9 square feet and dividing the result into the number of square feet in the acreage.

Changing the acres to square miles:

1 square mile = 640 acres X Square miles = 3,200 acres

Make a proportion of the equivalences, lining up the numbers and the X Exactly as they appear — opposite one another. Solve for X.

1 640

- = -

X 3,200 3,200 = 640X 3,200 640

Ran = 77777 X

640 640 5 = X

So 3,200 acres is equivalent to 5 square miles. Change the square miles to square feet by multiplying each side of the equation 1 square mile = 5,280 X 5,280 square feet by 5. You get that 5 square miles = 5,280 X 5,280 X 5 = 139,392,000 square feet.

Now find the number of square feet that each tree needs. If each seedling needs 9 square yards, use the equivalence that 1 square yard = 9 square feet and multiply each side of the equation by 9 to get 9 square yards = 81 square feet.

Now divide 139,392,000 square feet by 81 square feet to get the number of trees that will fit on the acreage. 139,392,000 divided by 81 = 1,720,888.89 trees. That’s a lot of seedlings.

Keeping It All in English Units

Many countries, including the United States, use primarily the English units of measurement. Pressure to change to metric hasn’t been strong enough, even though advocates have proposed changing to metric for over 40 years. The awkwardness of the English units is that they have all sorts of different numbers in their equivalences — as compared to the metric system where all the numbers are multiples of 10.

Comparing measures with unlikely equivalences

As disjointed as the English measurement system seems to be, it has a long tradition and some interesting and charming equivalences. Here are some more uncommon but historic measures, plus, to finish it off, a rate.

1 rod = 16>2 Feet 1 fathom = 6 feet 1 furlong = 220 yards 1 hand = 4 inches 1 league = 3 miles 1 pica = 12 points

1 mile per hour = 88 feet per minute = 1 -JF Feet per second

The Problem: The Preakness, one of the horse races in the Triple Crown, has a distance of 9.5 furlongs. How many miles is that?

Change the furlongs to yards using 1 furlong = 220 yards and the yards to feet using 1 yard = 3 feet. Then change the feet to miles using 1 mile = 5,280 feet. The race is 9.5 furlongs, so multiply 9.5 x 220 to get 2,090 yards. Multiply the number of yards by 3 to get the number of feet: 2,090 x 3 = 6,270. Now

A – a ,u Co7n( ,u Coon. , ,u u t I 6,270 .,

Divide the 6,270 feet by 5,280 to get the number of miles: r OOA = 1.1875 miles.

3 3 5,280

The decimal 0.1875 is equal to Tf, so the race is 1Tf.

16 16

To change a terminating decimal to its fractional equivalent, create a fraction that has all the digits to the right of the decimal point in the numerator and, in the denominator, a power of 10 that has as many zeros as there are digits in the numerator. Then reduce the fraction. For the decimal 0.1875, you write 1875 in the numerator and a 1 followed by four zeroes in the denominator.

0.1875 =

1875 10,000

Now reduce the fraction. You can first divide both numerator and denominator by 25 and then divide the resulting numerator and denominator by 25.

1875 75 10,000 400

_3_ 16

The Problem: A popular method for determining how far away a bolt of lightning has struck is to count the number of seconds between the lightning flash and the sound of the thunder. If sound travels at about 1,100 feet per second, and if it’s 6 seconds between the flash of lightning and the roar of the thunder, then how far away was the lightning strike in miles? And what is the speed of the sound in miles per hour?

First, determine how many feet the sound traveled by multiplying 1,100 feet by 6 to get 6,600 feet. Determine the number of miles using 1 mile = 5,280 feet. You divide 6,600 feet by 5,280 and you get 1.25 miles. According to what I was told, the number of seconds you count is the number of miles away. I don’t think I was told right.

Now, to the speed of sound in miles per hour, use the equivalence that 1 mile

Per hour is equal to 1 feet per second. The speed of sound is about 1,100

Feet per second. Write a proportion using these figures, letting the speed of sound in miles per hour be represented by X. Then solve for X.

1 mile per hour

1feet per second

X Miles per hour 1,100 feet per second

1 – _J5_ _ 22 _

X

1,100 15 (1H)0100) 1,500 750 11

- — ———

X 750

The speed of sound comes out to be about 750 miles per hour. This is a bit over the speed you usually see quoted. The textbooks say that the speed of sound is actually 1,088 feet per second at 32° F. I rounded the number up to 1,100 feet for ease in computation and assumed that the temperature during a thunderstorm would be a bit warmer than 32°F.

Working for a bar of gold

Television’s favorite billionaire is interviewing yet another group of potential employees. To keep from being told, "You’re fired!" the finalists have the following problem posed to them, and the first person to come up with the solution will not hear the dreaded words. The problem: You’ve hired someone to work for you for the

Next seven days. You must pay him VJ Of a bar of gold per day, but he requires a daily payment of VJ Of that bar of gold — no credit. It’s expensive to cut through a bar of gold, so what are the fewest number of cuts necessary to meet his requirements?

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Loving you a bushel and a peck

Volumes and weights take on some historically interesting values when working with equivalences. Do you buy your apples by the bushel, or is a peck all you need?

1 quart = 2 pints = 4 cups 1 gallon = 4 quarts = 32 gills 1 bushel = 4 pecks 1 pound = 16 ounces

1 ton = 2,000 pounds = 20 hundredweights

The Problem: According to a recent census, the typical American ate 13.8 pounds of turkey in a recent year. If this represents the total amount of turkey eaten at 12 different meals, then how many ounces of turkey were consumed at each meal?

VLAiV Change the number of pounds to ounces using 1 pound = 16 ounces by multiplying 16 X 13.8. Then divide the total number of ounces by 12. The product of 16 X 13.8 = 220.8 ounces. Divide 220.8 12 and you get 18.4 ounces per meal. That’s over a pound of turkey at a sitting!

The next problem involves a phenomenon of agriculture and requires that fruit growers adopt a good balance in their orchards. Consider an orchard where a certain number of apples are produced by each tree in an average year. If the number of trees in the orchard is increased, there’ll be more trees producing apples, but the crowding causes a reduction in the number of apples per tree. The balancing act for growers amounts to adding enough trees to increase production but not too many to decrease the production per tree by too much.

The Problem: An orchard contains 240 apple trees, which produce, on average, 2 bushels of apples per year. If the orchard manager increases the number of trees in the orchard by 60, she calculates that the amount of apples will be reduced by 1 peck per tree. If she goes ahead and plants those 60 additional trees, will the total crop be greater or smaller than the crop obtained from the original 240 trees?

^VLA* First, determine how many bushels of apples are produced by 240 apple trees by multiplying 240 x 2 = 480 bushels. Increasing the number of trees by 60 results in 300 apple trees. The production of 300 trees will be 2 bushels less 1 peck. Change pecks to bushels with 1 bushel = 4 pecks, giving you that 1 bushel is >4 Peck. Subtract >4 Bushel from 2 bushels, and the yield per tree

Will be 13 bushels. Multiply the number of trees by the new yield, and you 43

Get a total of 300 x 1-4 = 300 x 1.75 = 525 bushels. There’s an increase of 525 – 480 = 45 bushels of apples.

Mixing It Up with Measures

Most of the problems in this book use the English measures of length, volume, and weight. But metric measures are very important to know, because of the great incidence of foreign travel and trade with other countries that use metrics.

Matching metric with metric

The metric measurement system is extremely easy to use, because all the units and equivalents are powers of 10. A kilogram is 1,000 times as big as a gram, and a centimeter is 0.01 as big as a meter. The multiplication and division problems using metric measures are really a piece of cake. When you learn what the different prefixes stand for, you can navigate your way through the metric measurement system.

In the metric system, Kilo Means 1,000 times as much, Hecto Means 100 times as much, Deca Means 10 times as much, Deci Means K0 As much, Centi Means M™ As much, and Milli Means K000 As much.

The Problem: Stephen is the track manager at a race-car competition that’s 16 kilometers long. If he wants to put a spotter every 25 meters for the length of the race, then how many spotters will he need?

Change kilometers to meters using 1 kilometer = 1,000 meters and then divide the total number of meters by 25. The 16-kilometer race is 16 x 1,000 = 16,000 meters long. Divide 16,000 ■ 25 to get 640 spotters. That’s a lot of people!

The Problem: Stephanie works for a candy company and got permission to produce a piece of licorice that’s 12 meters long. She’s going to take the licorice to a party (just for the effect) and then divide it into individual pieces that are each 3 centimeters long. How many pieces of licorice will she have?

A centimeter is M™ Of a meter, so there are 100 centimeters in a meter. Multiply the 12 meters by 100, and you get 1,200 centimeters of licorice. Divide that total by 3, and Stephanie will have 400 pieces of candy.

Changing from metric to English

You’ve decided to go to Europe and you want to be sure that you order the right size beverage, know how far you’ll be traveling by car, and dress appropriately for the weather on any particular day. All these functions relate to changing from English units of measure to metric measure. Here are some of the more useful conversion equivalences you’ll need for your travels. For help with the temperatures, refer to Chapter 10 for conversions from Celsius to Fahrenheit and back again.

1 meter = 39.37 inches 1 kilometer = 0.621 mile 1 liter = 1.057 quarts 1 kilogram = 2.205 pounds

The Problem: You’re in Europe and about to take a day trip with a rented car and trusty map. You’re going to drive from your hotel to a famous cathedral. According to the map, the distance is 500 kilometers. How far is that in miles?

Use the equivalence 1 kilometer = 0.621 mile and multiply each side of the equation by 500. You get that 500 kilometers = 310.5 miles. That’s a pretty long trip, depending on what kinds of roads you’re going to find. You may want to check on an overnight stop.

The Problem: You’re driving along and notice that you’ll be needing fuel very soon. You spot a service station and pull over to buy fuel. The price on the sign is $2.25. You gulp after you realize that the price is for 1 liter of fuel. What is the price per gallon?

First, use the equivalence 1 liter = 1.057 quarts in a proportion with the price of 1 liter = $2.25 to determine how much the fuel costs per quart. Then you multiply that price by 4 because 1 gallon = 4 quarts.

1 liter 1.057 quarts

$2.25 1

2.25

X Dollars

1.057

X

2.25 (1.057) = 2.37825

The fuel is about $2.38 per quart. Multiply by four, and 2.37825 x 4 = 9.513 or about $9.51 per gallon. And you thought gas prices were bad in the United States!

Zz

X

Changing from English to metric

You’re on your European tour and you’ve brought some fabric samples to make curtains for your hostess and a recipe so you can cook up a thank-you dinner. Now you have the challenge of converting some of your measures into the measures of the country you’re visiting.

1 yard = 0.9144 meter 1 pound = 0.454 kilogram 1 cup = 0.2365 liter

Part of the challenge of cooking in another country is trying to find the ingredients that you’re used to working with at home. The other challenge comes when you need to measure those ingredients.

The Problem: Your recipe for lasagna calls for a 16-ounce jar (2 cups) of tomato sauce. You find a can of tomato puree (which you’ll have to spice up a bit), and the can contains % liter of puree. How many cans of the puree will you need to buy?

Create a proportion using 1 cup = 0.2365 liter and 2 cups = X Liters. Solve for X, Which will be the amount of tomato sauce you need in terms of liters. Then compare that amount to % or 0.75 liter.

1 cup 0.2365 liter

2 cups X Liters 1 0.2365

—— — -—-

— :

2

2 (0.2365)=0.473

The can contains 0.473 liter of tomato sauce. You need 0.75 liter. You’ll have to buy 2 cans, giving you 0.946 liter, and just save the extra for the next project. Good luck with the measuring part!

You’ve had way too good of a time on your trip to Europe. You’ve been avoiding getting on the scale to see if you’ve gained any weight, but you finally decide, the day before leaving for home, to get on the scale to see what the damage is. Omigosh! You’ve lost weight! You’ve lost a Lot Of weight! Then you realize that the scale is in kilograms.

The Problem: You weigh yourself on a metric scale and it says 68. How many pounds do you really weigh?

Use the equivalence 1 kilogram = 2.205 pounds and multiply each side of the equation by 68. You get that 68 kilograms = 149.9 or 150 pounds.

X

Chapter 4

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

In This Chapter

^ Introducing terminology and mathematical conventions

^ Comparing sentence and equation structure for more clarity

^ Using pictures for understanding

^ Looking to tables and charts for organization of information

Athematicians decided long ago to conserve on words and explanations and replace them with symbols and single letters. The only problem is that a completely different language was created, and you need to know how to translate from the cryptic language of symbols into the language of words. The operations have designations such as +, -, X, And Algebraic equations use letters and arrangements of those letters and numbers to express relationships between different symbols.

In this chapter, you get a refresher of the math speak you’ve seen in the past. I review the vocabulary of algebra and geometry and give examples using the appropriate symbols and operations.

Latching onto the Lingo

Words used in mathematics are very precise. The words have the same meaning no matter who’s doing the reading of a problem or when it’s being done. These precise designations may seem restrictive, but being strict is necessary — you want to be able to count on a mathematical equation or expression meaning the same thing each time you use it.

For example, in mathematics, the word Rational Refers to a type of number or function. A person is Rational If he acts in a controlled, logical way. A number is Rational If it acts in a controlled, structured way. If you use the word Rational To describe a number, and if the person you’re talking to also knows what a rational number is, then you don’t have to go into a long, drawn-out explanation about what you mean. You’re both talking in the same language, so to speak.

Defining types of numbers

Numbers are classified by their characteristics. One number can have more than one classification. For example, the number 2 is a Whole Number, an Even Number, and a Prime Number. Knowing which numbers belong in which classification will help you when you’re trying to solve problems in which the answer has to be of a certain type of number.

Naming numbers

Numbers have names that you speak. For example, when you write down a phone number that someone is reciting, you hear Two, one, six, nine, three, two, seven, And you write down 216-9327. Some other names associated with numbers refer to how the numbers are classified.

Natural (counting): The numbers starting with 1 and going up by ones forever: 1, 2, 3, 4, 5, . . .

Whole: The numbers starting with 0 and going up by ones forever. Whole numbers are different from the natural numbers by just the number 0: 0, 1, 2, 3, 4, . . .

Integer: The positive and negative whole numbers and 0: . . . ,-3, -2, -1, 0, 1, 2, 3, 4, . . .

Rational: Numbers that can be written as p where both P And Q Are

Integers, but Q Is never 0: t,1^9, —-J5-,24, and so on

Even: Numbers evenly divisible by 2: . . . ,-4, -2, 0, 2, 4, 6, . . .

Odd: Numbers not evenly divisible by 2: . . . ,-3, -1, 1, 3, 5, 7, . . .

Prime Numbers divisible evenly only by 1 and themselves: 2, 3, 5, 7, 11,

13, 17, 19, 23, 29, . . .

Composite: Numbers that are not prime; numbers that are evenly divisible by some number other than just 1 and themselves: 4, 6, 8, 9, 10, 12,

14, 15, . . .

Relating numbers

Numbers of the same or even different classifications are often related in another way that makes them usable in problems. For example, if you want only Multiples of five, You draw from evens, odds, and integers — several types to create a new relationship.

Consecutive: A listing of numbers, in order, from smallest to largest, that have the same difference between them: 22, 33, 44, 55, . . . are consecutive multiples of 11 starting with 22.

Multiples: Numbers that all have a common multiplier: 21, 28, and 63 are multiples of 7.

Geometric figures appear frequently in mathematical applications — and in life. Geometric figures have names, classifications, and characteristics. The figures are also measured in two or more ways. Flat figures have the lengths of their sides, their whole perimeter, or their area measured. Solid figures have their surface area and volume measured. You can find all the formulas you need on the Cheat Sheet and in Chapters 18, 19, and 20. What you find here is a description of what the measures mean.

Plying perimeter

The Perimeter Is a linear measure: inches, feet, centimeters, miles, kilometers, and so on. Perimeter is a measure of distance — the distance around the outside of a flat figure. The perimeter of a figure made up of line segments is equal to the sum of the length of all the segments. The perimeter of a circle is also called its Circumference And is always slightly more than three times the circle’s diameter. In Figure 1-1, you see several sketches and their respective perimeters.

Gauging the geometric

Figure 1-1: (

Add up \ the lengths \ of the 3" segments to get the perimeter.

P = 3 + 5 + 5 + 3 = 16"

P = 8 + 9 + 4 + 2 + 3 = 26 mi.

C = 30n ~ 94.2′

Assembling the area

The Area Of a figure is a two-dimensional measure. The area is a measure of how many squares you can fit into the figure. If the figure doesn’t have 90-degree or squared-off angles, then you have to count up pieces of squares — break them up and put them back together — to get the whole area. Think about putting square tiles in a room — you have to cut some of them to go around cabinets or fit along a wall. The formulas that you use to compute areas help you with the piecing-together of squares.

In Figure 1-2, you see a triangle with an area of exactly 12 square units. See if you can figure out how the pieces go together to form a total of 12 squares. If that doesn’t work, you can compute the area by just looking up the formula for the area of a triangle.

Figure 1-2:

How many squares are in the triangle?

Coming to the surface with surface area

The Surface area Of a solid figure is the sum of the areas of all the sides. A four-sided figure has a triangle on each side, so you add up the areas of each of the triangles to get the total surface area. How do you get the area of each triangle? You go back to the formula for finding the area of triangles of that particular size — or just count how many squares! Figure 1-3 shows three of the six sides of a right rectangular prism and how each side has its area determined by all the squares it can fit on that side.

Figure 1-3:

How much paper will you need to wrap the package?

The prism in Figure 1-3 has a surface area of 112 square units. That’s how many squares cover the six surfaces of this solid figure. Formulas are much easier to use than actually trying to count squares.

Vanquishing volume

The Volume Of a solid figure is a three-dimensional type of unit. When you compute the volume of something, you’re determining how many cubes (like sugar cubes or dice) will fit inside the figure. When the sides slant, of course, you have to slice, trim, and fit to make all the cubes go inside — or you can use a handy-dandy formula. Figure 1-4 shows how you can set cubes next to one another and then stack them to determine the volume of a solid.

Figure 1-4:

Cubes all in a row.

Formulating financials

Most people are interested in money, in one way or another. Money is the way people keep count of whether they can trade for what they want or need. Financial formulas aid with the computation of money-type situations.

The financial formulas here are divided into two different types: interest formulas and revenue formulas. The interest formulas both involve a percentage that needs to be changed into a decimal before being inserted into the formula. To change a percent into a decimal, you move the decimal point two places to the left. So 3.4 percent becomes 0.034 and 67 percent becomes 0.67.

The interest formulas are of two types: simple interest and compound interest. The simple-interest formula is I = Prt. The I Indicates how much interest your money has earned — or how much interest you owe. The P Is the principal — how much money you invested or are borrowing. The R Represents the interest rate — the percentage that gets changed to a decimal. And the T Stands for time, which is usually a number of years.

"Tomorrow, tomorrow, I love ya, tomorrow. . ."

If yesterday had been Wednesday’s tomorrow, and if tomorrow is Sunday’s yesterday, then what day is it today?

■Aepu-i S, i\:jaMsu\f

Compounding interest means that you split up the rate of interest into a designated number of subintervals (every three months, twice a year, daily, and so on), figure the interest earned during that subinterval, add the interest to the principal, and then figure the next interval’s interest on the sum of the original principal plus the interest you’ve added. As you may expect, you’ll have more money in the end if you deposit it where you can earn compound interest rather thntan just a flat amount. The formula for compound interest

Is A = P C1 + N J . The A Represents the total amount of money — all the

Principal plus the interest earned. The R And T Are the same as in simple interest. The N Represents the number of times each year that the interest is compounded. Most banks compound quarterly, so the value of N Is 4 in those cases.

Interpreting the Operations

What would mathematics be without its operations? The basic operations are addition, subtraction, multiplication, and division. You then add raising to powers and finding roots. Many more operations exist, but these six basic operations are the ones you’ll find in this book. Also listed here are some of the special names for multiplying by two or three.

Naming the results

Each operation has a result, and just naming that result is sometimes more convenient than going into a big explanation as to what you want done. You can economize with words, space, time, and ink. The following are results of operations most commonly used.

IU Sum: The result of adding IU Difference: The result of subtracting

I Product: The result of multiplying I Twice or double: The result of multiplying by two IU Thrice or triple: The result of multiplying by three I Quotient: The result of dividing I Half: The result of dividing by two

Assigning the variables

A variable is something that changes. In mathematics, a variable is represented by a letter — usually one from the end of the alphabet — and it Always Represents a number (usually an unknown number). For example, if you’re doing a problem involving Jake and Jim and their ages, you can let X Represent Jake’s Age, But you can’t let X Represent Jake.

As you work on a problem, it’s a good idea to make a notation as to what you’re letting the variable or variables represent, so you don’t forget or get confused when constructing an equation to solve the problem.

Aligning symbols and word forms

One of the things that people see as a challenge in word problems is that they’re full of Words! After you’ve changed the words to symbols and equations, it’s smooth sailing. But you have to get from there to here. Table 1-1 lists some typical translations of words into symbols and an example of their use.

(continued)

Drawing a Picture

One of the most powerful tools you can use when working on word problems is drawing a picture. Most people are very visual — they understand relationships between things when they write something down and/or draw a picture illustrating the situation.

Visualizing relationships

The words in a math problem suggest how different parts of the situation are connected — or not connected. Drawing a picture helps to make the connections and, often, suggests how to proceed with a solution.

For example, consider a word problem starting out with: "A plane is flying east at 600 mph while another plane is flying north at 500 mph. . . ." You need more information than this to determine what the question and answer are, but a picture suggests what process to use. Look at Figure 1-5, where two possible scenarios for the statement are illustrated.

600

Figure 1-5:

The planes are leaving or

Approaching one another.

500

500

The precise relationship between the planes has to be given, but both sketches suggest that a triangle can be formed by connecting the ends of the arrows. Right triangles suggest the Pythagorean theorem, and other triangles come with their respective perimeter and area formulas. In any case, the picture solidifies the situation and makes interpretations possible.

Another example where a picture is helpful involves a situation where you’re cutting a piece of paper. The word problem starts out with: "A rectangular piece of paper has equal squares cut out of its corners. . . ." You draw a rectangle, and you show what it looks like to remove squares that are all the same size. Figure 1-6 illustrates one interpretation.

Figure 1-6:

Cut squares from the corners.

With the figure in view, you see that the lengths of the outer edges are reduced by two times some unknown amount. The picture helps you write expressions about the relationships between the original piece of paper and the cut-up one.

Labeling accurately

Pictures are great for clarifying the words in a problem, but equally important are the labels that you put on the picture. By labeling the different parts — especially with their units in feet, miles per hour, and so on — you improve your chances of writing an expression or equation that represents the situation.

You’re told "A trapezoidal piece of land has 300 feet between the two parallel sides, and the other two sides are 400 feet and 500 feet in length, while the two parallel bases are 600 feet and 1,200 feet." This statement has five different numbers in it, and you need to sort them out. Figure 1-7 shows how the different measures sort out from the statement.

600 ft.

Figure 1-7:

The trapezoid has a height of 300 feet.

1200 ft.

Watch out for the black cat

A completely black cat was ambling down the street during a total blackout — electricity was down throughout the entire town. Not a single streetlight had been on for hours. Just as the cat was crossing the middle of the street, a Jeep with two broken headlights came racing toward

Where the cat was. Just as the Jeep got to where the cat was crossing, it swerved out of the way to avoid hitting the cat. How could the driver of the Jeep have seen the cat in time to swerve and avoid hitting it?

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Constructing a Table or Chart

A really nice way to determine what’s going on with a word problem is to make a list of different possibilities and see what fits in the list or what pattern forms. Patterns often suggest a formula or equation; the values in the listing sometimes even provide the exact answer. Just as with pictures, making a chart is a way of visualizing what’s going on.

Finding the values

When creating a table or chart, designate a variable to represent a part of the problem, and see what the results are as you systematically change that variable. For example, if you’re trying to find two numbers the product of which is 60 and the sum of which is as small as possible, let the first number be X.

Then the other number is 6×0. Add the two numbers together to see what you get. Table 1-2 shows the different values for the two numbers and the sum — if you stick to whole numbers.

3

20

3 + 20 = 23

I stop here, because I’m just going to get the same pairs of numbers in the opposite order. If the rule is that the numbers can’t be fractions, then the two numbers with a product of 60 and with the smallest possible sum are 6 and 10.

Increasing in steps

When making a table or chart, you want to be as systematic as possible so you don’t miss anything – especially if that Anything Is the correct answer. After you’ve determined a variable to represent a quantity in the problem, you need to go up in logical steps — by ones or twos or halves or whatever is appropriate. In Table 1-2, in the preceding section, you can see that I went up in steps of 1 until I got to the 6. One more than 6 is 7, but 7 doesn’t divide into 60 evenly, so I skipped it. Even though the work isn’t shown here, I mentally tried 7, 8, and 9 and discarded them, because they didn’t work in the problem. When you’re working with more complicated situations, you don’t want to skip any steps — show them all.

Chapter 2