The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard A. Medeiros next Lesson 23

Application to Word Problems Application to Word Problems © 2007 Herbert I. Gross next

A Non-Algebraic Approach It is not uncommon for a student to say… © 2007 Herbert I. Gross next “I can do everything except the word problems.” Yet the importance of being able to solve “word problems” cannot be overemphasized. In particular, in the “Textbook of Life”, there are only “word problems”!

The remarkable thing, however, is that we do not have to know algebra in order to be able to solve “word problems”. © 2007 Herbert I. Gross next In many cases, if we have basic arithmetic skills and good reading comprehension, we can solve “word problems” by a combination of arithmetic and trial-and-error. In fact, especially if we are not dealing with large numbers, trial-and-error is a rather effective way of solving problems.

Erin and Bill together have 30 marbles. Erin has 20 marbles more than Bill. Answer: Erin has 25, and Bill has 5 © 2007 Herbert I. Gross next How many marbles does each one have? Example 1 Example 1 next

Don’t read too quickly! Many people tend to read this problem too quickly and they subtract 20 from 30 to get 10. Notice, however, that if Erin had 20 marbles and Bill had 10, Erin would only have 10 marbles more than Bill -- not 20 more. © 2007 Herbert I. Gross next Caution Therefore, once you get an answer, check to make sure that it meets the conditions stated in the problem. next

For example, suppose you obtained as your “answer” that Erin had 27 marbles and Bill had 7 marbles. This would certainly make the difference 20 (as the problem requires), but the total number of marbles would be 34, rather than the required 30 that the problem states. © 2007 Herbert I. Gross next

If we read the problem correctly, it gives us two “clues” for getting started. © 2007 Herbert I. Gross next One clue is that the total number of marbles the two children have is 30. The other is that one (Erin) has 20 more marbles than the other (Bill). We may start with whichever clue we wish. next

Suppose we start with the clue that the two children have a total of 30 marbles. We must then have enough reading comprehension to understand that this means… © 2007 Herbert I. Gross next The number of marbles Bill has The number of marbles Erin has + = 30 Approach 1

© 2007 Herbert I. Gross next The least number of marbles Bill could have is 0. The formula tells us (by simple arithmetic) that if Bill has no marbles, Erin must have 30. But if we suppose that Erin had 30 marbles and Bill had none, the difference between their amounts would be 30. Since the problem tells us that the difference has to be 20, we know that we have made an incorrect guess.

© 2007 Herbert I. Gross next Now we keep guessing, realizing that to keep the sum at 30; every time we add 1 to Bill’s amount we must subtract 1 from Erin’s amount. We’ll know we’ve guessed the correct answer when the difference between their two amounts is 20.

The chart includes the “Total Marbles” column to emphasize that the total number of marbles must always be 30 in this example. © 2007 Herbert I. Gross next In the highlighted row of the chart, the difference is 20. That is, the chart tells us that the correct answer to this example is that Erin has 25 marbles, and Bill has 5. Marbles Bill hasMarbles Erin hasTotal MarblesDifference In terms of a chart… next

This time we start with the fact that Erin has 20 marbles more than Bill. © 2007 Herbert I. Gross next The number of marbles Bill has The number of marbles Erin has + 20 = Approach 2 Be sure you interpret the problem correctly. The fact that Erin has 20 more marbles than Bill means that we have to add 20 to Bill’s amount in order to obtain Erin’s amount. That is… next

We could then have written the problem’s information as the formula below: © 2007 Herbert I. Gross next The number of marbles Bill has The number of marbles Erin has – 20 = Approach 3 Or we may start with the information that Erin has 20 less marbles than Bill.

© 2007 Herbert I. Gross next The number of marbles Bill has The number of marbles Erin has + 20= next The number of marbles Bill has The number of marbles Erin has – 20 = In terms of algebra, the formula above is obtained by subtracting 20 from both sides of the formula… There is usually more than one way to paraphrase a word problem into a verbal equation. Note For example, next

© 2007 Herbert I. Gross next The number of marbles Bill has The number of marbles Erin has – 20 = The number of marbles Bill has The number of marbles Erin has + 20= or 20 The fact that Erin has 20 more marbles than Bill tells us that Erin must have at least 20 marbles. So, let’s guess that Erin had 20 marbles. 0 But the formulas tell us that Bill would have no marbles; and if Bill had none, they would only have a total of 20 marbles. Since we know that the total number of marbles must be 30, we know that our guess is incorrect. next

So we keep guessing, realizing that to keep the difference at 20, every time we add 1 to Bill’s amount we must also add 1 to Erin’s amount. © 2007 Herbert I. Gross next Marbles Bill hasMarbles Erin hasTotal MarblesDifference next Thus, this time, our chart might look like…

The “Difference” column is included just to emphasize that, in this example, the difference always has to be 20. © 2007 Herbert I. Gross next Marbles Bill hasMarbles Erin hasTotal MarblesDifference next The highlighted row tells us that the total number of marbles will be 30 and the difference will be 20 when Erin has 25 marbles and Bill has

While these trial-and-error methods will always work, they become more and more tedious as the numbers become greater. Note © 2007 Herbert I. Gross next For instance, suppose we tried the same method in the following illustration…

Erin and Bill, together, have 200 marbles. Erin has 40 marbles more than Bill. © 2007 Herbert I. Gross next How many marbles does each have? Answer: Erin has 120, and Bill has 80 Illustration

If we wanted to use the same approach that we used in the previous example, our chart might now look like… © 2007 Herbert I. Gross next Marbles Bill hasMarbles Erin hasTotal MarblesDifference next While the theory hasn’t changed, we can see that at the rate we’re going it will take a “a lot of rows” before we get to the row in which the difference is 40.

However, there are more efficient ways of guessing. For example, instead of guessing next that Bill has 6 marbles, we might take a big leap forward and guess that he has, for example, 70 marbles. © 2007 Herbert I. Gross next Marbles Bill hasMarbles Erin hasTotal MarblesDifference (130 – 70) next Since the total number of marbles still has to be 200, this would mean that Erin had 130. In this case, one row of the chart would look like… Since the difference is still more than 40, we know that Bill must have more than 70 marbles.

So let’s take another guess, for example, that Bill has 90 marbles. Because the total number of marbles still has to be 200, it means that Erin must have 110 marbles. © 2007 Herbert I. Gross next Marbles Bill hasMarbles Erin hasTotal MarblesDifference (110 – 90) In this case, another row of the chart would look like… next Since the difference is now less than 40, we know that Bill must have less than 90 marbles.

We could continue in this way until we ultimately guessed the correct answer. © 2007 Herbert I. Gross next Marbles Bill hasMarbles Erin hasTotal MarblesDifference (120 – 80) However, the fact that the relationships in this example are linear allows us to use more “clever” forms of trial-and-error. next The point is that with just these two guesses (Bill had either 70 or 90 marbles), we have found that Bill must have more than 70 but fewer than 90 marbles.

If we examine the chart below, we see that the relationship between the “Difference” and the “Marbles Bill has” is linear. A Note on Linear Relationships © 2007 Herbert I. Gross next Marbles Bill hasMarbles Erin hasTotal MarblesDifference That is, we can see from the chart that as we move down from one row to the next, the “Difference” column decreases by 2 each time.

Do you see the linear relationship between the “difference” (D) and the “marbles Bill has” (B)? © 2007 Herbert I. Gross Marbles Bill hasMarbles Erin hasTotal MarblesDifference What the chart shows is that the rate of change of D with respect to B is 2 (that is, the difference decreases by 2 when the number of marbles Bill has increases by 1) next

On the top line of the chart, the “Difference” is 200. We want to get to the line for which the “Difference” is 40. That is… © 2007 Herbert I. Gross next Bill’s AmountErin’s AmountDifference 0200 We know that the difference between 200 and 40 is 160, next ??40 and we also know that as we go down from one row to the next, the difference decreases by 2. next

Since 160 ÷ 2 = 80, we see that we have to move 80 rows down from the top before the number in the “Difference” column will be 40. © 2007 Herbert I. Gross next Since each row down from the top adds 1 more marble to Bill’s amount; by the time we move 80 rows down, we will have added 80 to the number of marbles Bill has. Since the total number of marbles has to stay constant at 200, it means that we must subtract the 80 we gave Bill from the amount that Erin had.

In summary… © 2007 Herbert I. Gross next Bill’s AmountErin’s AmountDifference – next The highlighted row tells us that the total number of marbles Erin has is 120 marbles, and Bill has

The fact that the relationship is linear gives us yet another way to guess the answer quickly. In one of our previous charts we saw that… © 2007 Herbert I. Gross next Marbles Bill hasMarbles Erin hasTotal MarblesDifference (130 – 70) ??? (?? – ?) (110 – 90) Because the relationship is linear, the fact that 40 is halfway between 60 and 20, means that the number of marbles Bill has must be halfway between 70 and 90. Hence, Bill must have 80 marbles. next

Note that this technique requires that we have a linear relationship. For example, the relationship y = x 3 is not linear; and while 2 is half way between 1 and 3, 2 3 is not halfway between 1 3 and 3 3. In terms of a chart… © 2007 Herbert I. Gross next yx3x is halfway between 1 and 3, but 8 isn’t halfway between 1 and next

The fact that every time we take a marble away from Bill and give it to Erin, we increase the difference by 2. This gives us yet another way to solve the Illustration. © 2007 Herbert I. Gross next Namely, let’s now pretend that they have equal amounts of marbles. Let’s suppose Erin and Bill each have 100 marbles (in which case the difference is 0). To create a difference of 40, we then take 20 away from Bill and give them to Erin.

© 2007 Herbert I. Gross next Bill’s AmountErin’s AmountDifference – Taking 20 marbles from Bill’s amount and adding them to Erin’s amount; we see that…

In Lessons 1 and 2, we introduced the idea of letting a rectangle (which we called a “corn bread”) represent the unit we were measuring. We can use this model to solve the above examples more visually. © 2007 Herbert I. Gross A Geometric Solution (“Corn Bread” Model) For example, let’s revisit Problem 1 using the corn bread model. Recall that the problem was… Erin and Bill, together, have 30 marbles, and Erin has 20 more marbles than Bill. How many marbles does each one have? next

So, we let the corn bread represent the total number of marbles (30). We then cut the cornbread into pieces. © 2007 Herbert I. Gross = number of marbles Bill has. next B = number of marbles Erin has. B Thus… B 20 B 30 we see that, of the 30 marble total, the two pieces of size “B” represent 10 marbles; and thus that B itself equals 5 marbles. next 55

The beauty of the corn bread model is that it remains virtually the same no matter how many marbles the two of them have or what the difference in their numbers is. Also as we shall see in our next lesson, the corn bread model is a good way to introduce the transition from arithmetic to algebra. © 2007 Herbert I. Gross For example, let’s use the same model to solve the Illustration. Recall that… Erin and Bill, together, have 200 marbles, and Erin has 40 more marbles than Bill. How many marbles does each one have? next

To use the corn bread model in this problem we can do verbatim what we did in the case of Problem 1; except that this time we replace 20 by 40 and 30 by 200. To indicate this more explicitly, we have reproduced the previous corn bread solution with the appropriate changes… © 2007 Herbert I. Gross next

That is, we let the corn bread represent the total number of marbles (200). We then cut this cornbread into pieces. © 2007 Herbert I. Gross = number of marbles Bill has. next B = number of marbles Erin has. B Thus… …from which we see that, of the 200-marble total the two pieces of size “B” represent 160 marbles; or that B itself equals 80 marbles. next 40BB80

In summary, there are many ways of solving word problems by trial-and-error. © 2007 Herbert I. Gross next The idea is to find ways that are comfortable for you to use and to proceed from there. In fact, you may find other correct ways that work for you but which have not been described in this Lesson. In essence, thinking is a personal thing and what works best for one person might not work at all for another person.

Erin has three times as many marbles as Bill. Together they have 120 marbles. Answer: 90 © 2007 Herbert I. Gross next How many marbles does Erin have? Example 2 Example 2 next Let’s now do a few more examples to make sure that you understand the trial-and-error methods for solving word problems.

Solution © 2007 Herbert I. Gross next Don’t confuse “three times as many” with “three more than”. The fact that Erin has three times as many marbles as Bill means that we multiply the number of marbles Bill has by 3 to find the number of marbles Erin has. For example, if Bill has 20, Erin has 60; and if Bill has 40, Erin has 120. These two estimates already give us a hint as to how the trial-and-error process should proceed.

Because the number of marbles (120) is between 80 and 160, Bill’s amount must be between 20 and 40. Moreover, the fact that the relationship is linear tells us that since 120 is halfway between 80 and 160, Bill's amount must be halfway between 20 and 40. Hence, Bill must have 30 marbles. As a check, we see that… © 2007 Herbert I. Gross next Bill’s AmountErin’s AmountTotal Amount (less than 120) (more than 120) Namely… Bill’s AmountErin’s AmountTotal Amount

© 2007 Herbert I. Gross next Do you see why the relationship in this example is linear? The point is that since Erin has 3 times as many marbles as Bill, every time we give Bill 1 more marble, we have to give Erin 3 more marbles. In other words, whenever the number of marbles Bill has increases by 1, the total number of marbles increases by 4. Bill Erin Total

© 2007 Herbert I. Gross next If your arithmetic skills are good there is no need to even use a chart. Namely, the fact that Erin has three times as many marbles as Bill means that Erin has 3 / 4 of the marbles and Bill has 1 / 4 of the marbles. Since 3 / 4 of 120 is 90, we see that Erin must have 90 marbles. Bill Erin Total 1/41/4 3/43/4 4/44/4

© 2007 Herbert I. Gross next The corn bread model is a visual version of the arithmetic approach. Namely, if we let the corn bread represent the 120 marbles, 120EEEB we then divide it into 4 equally sized pieces, one of which represents the number of marbles Bill has and the other three of which represent the number of marbles Erin has. That is… next 4 pieces = 120 marbles 1 piece = (120 ÷ 4) = 30 marbles = Bill’s amount 30 3 pieces = 90 marbles = Erin’s amount 90 next

© 2007 Herbert I. Gross next Reading comprehension is still very important. Be Careful! Notice that the problem asks for the number of marbles Erin has. If you write “30” as the answer, it is incorrect. “30” would have been the correct answer if the problem had asked for the number of marbles Bill had.

The only money Mary has with her is 30 coins consisting of dimes and quarters, and she has twice as many quarters as dimes. Answer: $6.00 © 2007 Herbert I. Gross next What is the total value of the coins she has? Example 3 next Sometimes, the solution of one problem is a step toward the solution of another problem. For example…

Solution © 2007 Herbert I. Gross next Using any of our previous methods we can determine that Mary has 20 quarters and 10 dimes. For example, since the ratio of quarters to dimes is 2:1, 2 / 3 of the 30 coins must be quarters and the remaining 1 / 3 of the 30 coins must be dimes. However, the problem didn’t ask us to find the number of dimes and quarters but rather to find the value of the 30 coins.

© 2007 Herbert I. Gross next Since each dime is worth 10 cents, 10 dimes are worth $1, and since each quarter is worth 25 cents, 20 quarters are worth $5. Altogether, then, she has $6. That is… Number of Dimes Number of Quarters Value of Dimes Value of Quarters Total Value 1020$1.00$5.00$6.00

© 2007 Herbert I. Gross next Notice that this is a multi-step problem in which we first have to find the number of dimes and the number of quarters; after which we compute the amount of money the coins represent. Be Careful! Thus, although she has 20 quarters and 10 dimes, this is not the correct answer to this problem. Namely, the problem asks for the total value of the coins; not the number of each denomination.

Answer: 94 © 2007 Herbert I. Gross next How many marbles does Tom have? Example 4 Example 4 next The trial-and-error method works just as well even if there are more than two variables. This is illustrated in the following problem… Erin had three times as many marbles as Bill and Tom has four more marbles than Erin. Altogether they have 214 marbles.

© 2007 Herbert I. Gross next Since Bill has the fewest marbles, let’s work in terms of the number he has. Solution Obviously, the least number he can have is 0. Since 3 × 0 = 0, Erin would also have 0, and since Tom has 4 more than Erin, Tom would have 4. More generally, starting with the number Bill has, we multiply that number by 3 to find the amount Erin has. Then adding 4 to Erin’s amount gives us Tom’s amount.

© 2007 Herbert I. Gross next So our chart might look like… The chart reminds us that the relationship is again linear. Namely, every time Bill’s amount increases by 1 marble, Erin’s and Tom’s amount each increases by 3 marbles. Hence, the total number of marbles increases by 7. Bill’s Amount Erin’s Amount Tom’s Amount Total Amount

© 2007 Herbert I. Gross next We want to get to the row in which the total amount is 214. The total amount in the top row of our chart is 4. Hence, we want the “total amount” column to increase by 214 – 4 or 210. Since the total amount increases by 7 from row to row and since 210 ÷ 7 = 30, we see that we must move down 30 rows. That is, Bill must have 30 marbles. We then have… The problem asks that we find the number of marbles Tom has. We see from the chart that Tom has 94 marbles. Bill’s Amount Erin’s Amount Tom’s Amount Total Amount

© 2007 Herbert I. Gross next Remember again that reading comprehension is all-important. Be Careful! For example, the problem could have asked us to find the number of marbles Bill has or the number of marbles Erin has. The above chart tells us that Bill has 30 marbles and that Erin has 90. Bill’s Amount Erin’s Amount Tom’s Amount Total Amount

© 2007 Herbert I. Gross next We didn’t have to begin our trial-and-error method with Bill’s amount being 0. Since the total number of marbles (214) is between 144 and 284, we know that the number of marbles Bill has must be between 20 and 40. Bill’s Amount Erin’s Amount Tom’s Amount Total Amount For example, we might have tried the guess that Bill had 20 marbles or perhaps 40. In this case the chart would look like… next

© 2007 Herbert I. Gross next More specifically, since the relationship is linear and since 214 is halfway between 144 and 284, we know that Bill’s amount must be halfway between 20 and 40. Hence, Bill must have 30 marbles. Erin must have 90 (three times as many as Bill), and Tom must have 94 (three times as many as Bill plus four more).

© 2007 Herbert I. Gross next The corn bread model also works quite well in this problem. To simplify the diagram, let’s pretend that Tom had 4 fewer marbles. This would mean Tom and Erin would have equal amounts of marbles, and the total number of marbles would be 210. If we think of Bill’s number of marbles as being one piece of the corn bread, Tom’s share and Erin’s share would each be represented by three pieces of the corn bread.

© 2007 Herbert I. Gross next Thus, the corn bread, which now represents 210 marbles, would be cut into seven equally sized pieces. Pictorially… 210 marbles ÷ 7 pieces = 30 marbles per piece TTTEEE B30

© 2007 Herbert I. Gross next 1 piece = 30 marbles = number Bill has pieces = 90 marbles = number Erin has. 3 pieces = 90 marbles = number Tom would have had. However, the actual number of marbles Tom has is 4 more than that. Hence, Tom has 94 marbles next

Answer: 360 miles © 2007 Herbert I. Gross next Find the distance between A and B if the round trip takes 30 hours. Let’s apply these ideas to one more problem before we move on to the next Lesson. An antique automobile travels from A to B at a constant speed of 30mph. It makes the return trip from B back to A at a constant speed of 20mph. Example 5 Example 5

© 2007 Herbert I. Gross Let’s take a guess at the distance. To make our arithmetic easier, we’ll pick one that is divisible by both 20 and 30. Solution If we assume that the one-way distance is 60 miles (the least common multiple of 20 and 30); the 60 mile trip from A to B would (at 30 mph) take 2 hours, and the return from B to A (at 20 mph) would take 3 hours. That is, the drive would have taken 5 hours. However, we were told the round trip actually takes 30 hours, so the guess of 60 miles in 5 hours was not a good guess. next

© 2007 Herbert I. Gross next Now let’s use the fact that the relationship between the distance and the time is linear (That is, every time the distance between A and B increases by 60 miles, the time for the round trip increases by 5 hours). Since 30 hours is the 6 th multiple of 5 hours, the distance between A and B must be the 6 th multiple of 60 miles. Since 60 × 6 = 360 we see that the distance between A and B must be 360 miles.

© 2007 Herbert I. Gross next Don’t be discouraged if you didn’t take advantage of the fact that the relationship was linear. You could still have constructed a chart similar to the one shown below… Notes Distance between A and B Round Trip Distance Time Spent at 30mph Time Spent at 20mph Total Time Spent 60 miles120 miles2 hours3 hours5 hours120 miles240 miles4 hours6 hours10 hours180 miles360 miles6 hours9 hours15 hours240 miles480 miles8 hours12 hours20 hours300 miles600 miles10 hours15 hours25 hours360 miles720 miles12 hours18 hours30 hours

© 2007 Herbert I. Gross next In the chart we see that the auto spent 12 hours at the greater speed and 18 hours at the lesser speed. Because more time was spent at the slower speed, the average speed for the round trip is closer to 20 mph than to 30 mph. In fact if we divide the total distance (720 miles) by the total time it took (30 hours) we see that the average speed for the round trip is 24 mph, not 25 mph. Distance between A and B Round Trip Distance Time Spent at 30mph Time Spent at 20mph Total Time Spent 360 miles720 miles12 hours18 hours30 hours Finding the average of two numbers by adding them and then dividing by 2 is valid only if the two numbers are equally weighted.

© 2007 Herbert I. Gross next In doing Example 5, people often fail to recall the correct way to calculate average speed. Instead, they “split the difference" between the two different speeds (20 and 30 mph), and assume that the average speed for the round trip is 25 mph. Be Careful! Then, they multiply this 25 mph by 30 hours to get 750 as the total trip distance. Next, dividing this 750 by 2 for the one-way distance, they incorrectly conclude that the distance between A and B is 375 miles rather than the correct distance (360 miles). next

© 2007 Herbert I. Gross next The trial-and-error methods described in this section are good ways to solve problems. Trial-and-Error In fact, in very complicated situations, there is no other way to solve a problem. Moreover, since the advent of modern computers, trial-and-error methods have been greatly improved.

© 2007 Herbert I. Gross next Trial-and-Error At the graduate school level, entire courses called “numerical analysis” are devoted to showing how this can be done. Even so, there are many times when trial-and-error techniques are too tedious, and there are other times when we need more exactness than we can get by using trial-and-error.

© 2007 Herbert I. Gross next Trial-and-Error For this reason, it is often helpful to translate problems into formulas and algebraic equations, and then solve the resulting equation. Because the equations we have discussed in this course are linear, the algebraic solution is not difficult. How we translate word problems into algebraic equations will be the subject of Lesson 24.