1. Key Stone Problem… Key Stone Problem… next Set 23 © 2007 Herbert I. Gross

2. You will soon be assigned problems to test whether you have internalized the material in Lesson 23 of our algebra course. The Keystone Illustration below is a prototype of the problems you’ll be doing. Work out the problems on your own. Afterwards, study the detailed solutions we’ve provided. In particular, notice that several different ways are presented that could be used to solve each problem. Instructions for the Keystone Problem next © 2007 Herbert I. Gross

3. As a teacher/trainer, it is important for you to understand and be able to respond in different ways to the different ways individual students learn. The more ways you are ready to explain a problem, the better the chances are that the students will come to understand. next © 2007 Herbert I. Gross

4. next © 2007 Herbert I. Gross Preface next The keystone problem is the same for both this lesson and Lesson 24. The difference is that in this lesson the problem will be solved by non-algebraic methods while in Lesson 24 it will be solved algebraically.

5. © 2007 Herbert I. Gross Preface next Which method(s) is preferable varies from person to person. Even within the same person there are times when one method seems preferable, and other times when a different method seems preferable. The key is to feel comfortable with many different methods. The more methods you know, the more likely it is that you will be able to solve a given problem.

6. © 2007 Herbert I. Gross Preface next While it may prove to be tedious in some cases, a method that can always be used is trial and error. That is, we make a guess at what the answer is and then see if our guess satisfies the given conditions. Moreover, in doing the arithmetic that is involved with trial and error, students have an opportunity to improve on both their computational skills and their ability to discover patterns.

7. © 2007 Herbert I. Gross Preface So let’s look at a specific problem and see how we can start with a trial-and-error solution and gradually refine it into a more efficient way.

8. next © 2007 Herbert I. Gross Three girls shared a sum of money. Cathy received $180 more than Betty. The total amount Betty and Cathy received was 3 times the amount Alice received. The amount Alice and Cathy received was 5 times the amount Betty received. Keystone Problem for Lesson 23 next What was the sum of money shared by the three girls?

9. © 2007 Herbert I. Gross In its purist form, trial and error has no prerequisites. That is, we are free to make whatever guesses we wish. However, while there are no prerequisites for starting a trial-and-error solution, it is often helpful to do some preliminary “research”. For example, we may make use of reading comprehension. next Note

10. next © 2007 Herbert I. Gross A Trial-and-Error Solution To this end, as an abbreviation for our table we’ll let next A denote the amount of money Alice has, B denote the amount of money Betty has, C denote the amount of money Cathy has, and T denote the amount of money they have altogether. And for convenience we will assume that A, B, C, and T are whole numbers.

11. © 2007 Herbert I. Gross A Trial-and-Error Solution For example, just by reading the problem we should see that… next A + B + C = T and.. B + C = 3A Hence, we may replace B + C in A + B + C = T by its value in B + C = 3A. next

12. © 2007 Herbert I. Gross A Trial-and-Error Solution A + B + C = T A + (B + C) = T We see that… A + 3A = T 4A = T A = 1 / 4 T From the equation, A = 1 / 4 T, we see that T has to be divisible by 4. next

13. A Trial-and-Error Solution A + B + C = T Similarly, since A + C = 5B, we may replace A + C in A + B + C = T by its value in A + C = 5B to obtain… 5B + B = T 6B = T From the equation, B = 1 / 6 T, we see that T has to be divisible by 6. next (A + C) + B = T B = 1 / 6 T © 2007 Herbert I. Gross

14. next © 2007 Herbert I. Gross We also know that Cathy has $180 more than Betty has, or equivalently that C – B = 180. Since T is divisible by both 4 and 6, it must also be divisible by 12. Hence, we may limit our guesses for T to the multiples of 12. next A Trial-and-Error Solution Moreover, since Cathy has $180 more than Betty has, we might guess that altogether they must have at least $200, but since 200 is not a multiple of 12, we might choose as our first guess that T = 240. next

15. © 2007 Herbert I. Gross In this case we see that… next A Trial-and-Error Solution The chart tells us that our guess is too small. That is, it led to C – B being only 100, and we know that C – B must be 180. next T A = ( 1 / 4 T) A + B C = (T – [A + B]) B = ( 1 / 6 T) C – B 2406010014040100

16. next © 2007 Herbert I. Gross Thus, our next guess should be a number that is greater than 240. One such number is 480. We could choose any number, but since 480 is twice 240, all we have to do is double each entry in the previous table to obtain… next A Trial-and-Error Solution The chart now tells us that our guess of 480 is too large. That is, it leads to C – B = 200, which is greater than 180. next T A = ( 1 / 4 T) A + B C = (T – [A + B]) B = ( 1 / 6 T) C – B 2406010014040100--- --------(180)48012020028080200 next

17. © 2007 Herbert I. Gross At this point, we already know that the exact value of T must be greater than 240 but less than 480. Our next guess might be based on the fact that since 200 is closer in value to 180 than 100 is, the correct value for T should be closer to 480 than to 240. next A Trial-and-Error Solution Halfway between 240 and 480 is 360. Thus, our next guess for T should be greater than 360 (but still be divisible by 12). So T can’t be less than 360 + 12, or 372.

18. next © 2007 Herbert I. Gross So if we choose T to be 372, we see that… next A Trial-and-Error Solution Eventually we would get to the correct answer. In fact without any further analysis, we could just fill out the chart in $12 increments. next T A = ( 1 / 4 T) A + B C = (T – [A + B]) B = ( 1 / 6 T) C – B 2406010014040100--- --------(180)372931552176215548012020028080200

19. next © 2007 Herbert I. Gross For example, starting with T = 372 our chart would look like… A Trial-and-Error Solution T A = ( 1 / 4 T) A + B C = (T – [A + B]) B = ( 1 / 6 T) C – B 372931552176215539699165231661653849616022464160408102170238681704321081802527218042010517524570175

20. next © 2007 Herbert I. Gross We can simplify our guessing by noticing that the above chart exhibits quite a few patterns. For example, every time T increases by 12… next Some Possible Refinements A increases by 3 T A = ( 1 / 4 T) A + B C = (T – [A + B]) B = ( 1 / 6 T) C – B 3729315521762155 3969916523166165 3849616022464160 40810217023868170 43210818025272180 42010517524570175 B increases by 2 C increases by 7B – C increases by 5

21. next © 2007 Herbert I. Gross So let’s see what happens if we restrict our guesses for T to be the multiples of 12. Our chart would then appear to be… next Row # 1 3 2 4 n --- T 12 36 24 48 n × 12 --- A = ( 1 / 4 T) 3 9 6 12 n × 3 --- C 7 21 14 28 n × 7 --- B = ( 1 / 6 T) 2 6 4 8 n × 2 --- C – B 5 15 10 20 n × 5 ---

22. next © 2007 Herbert I. Gross We want the row in which C – B is 180. We see from our chart that in the nth row the value of C – B is n × 5. next Row # 1 3 2 4 n --- T 12 36 24 48 n × 12 --- A = ( 1 / 4 T) 3 9 6 12 n × 3 --- C 7 21 14 28 n × 7 --- B = ( 1 / 6 T) 2 6 4 8 n × 2 --- C – B 5 15 10 20 n × 5 --- That is, we want the value of n for which 5 × n = 180. Thus, n = 180 ÷ 5, or 36.

23. next © 2007 Herbert I. Gross So looking at the 36th row (that is, n = 36) we see that… next Row # 1 3 2 4 n --- T 12 36 24 48 n × 12 --- A = ( 1 / 4 T) 3 9 6 12 n × 3 --- C 7 21 14 28 n × 7 --- B = ( 1 / 6 T) 2 6 4 8 n × 2 --- C – B 5 15 10 20 36 × 5 --- T = 36 × 12 = 432. 36 …which means that the three girls have a total of $432. 36 × 12 432 180 next

24. © 2007 Herbert I. Gross And we also have the additional, but unasked for, information that… next Column # 1 3 2 4 n --- T 12 36 24 48 n × 12 --- A = ( 1 / 4 T) 3 9 6 12 n × 3 --- C 7 21 14 28 n × 7 --- B = ( 1 / 6 T) 2 6 4 8 n × 2 --- C – B 5 15 10 20 36 × 5 --- A = 36 × 3 = 108 …which means that Alice has $108 36 × 3 36432 108 180 next

25. © 2007 Herbert I. Gross next Column # 1 3 2 4 n --- T 12 36 24 48 n × 12 --- A = ( 1 / 4 T) 3 9 6 12 n × 3 --- C 7 21 14 28 n × 7 --- B = ( 1 / 6 T) 2 6 4 8 n × 2 --- C – B 5 15 10 20 36 × 5 --- B = 36 × 2 = 72 …which means that Betty has $72 36 × 2 36432 108 72 next C = 36 × 7 = 252 …which means that Cathy has $252 36 × 7 252 180 next

26. © 2007 Herbert I. Gross As seen previously, since the total amount Betty and Cathy received was 3 times the amount Alice received, we can deduce that Alice received 1 / 4 of the total sum. next An Arithmetic Solution And since the total amount Alice and Cathy received was 5 times the amount Betty received, we can also deduce that Betty received 1 / 6 of the total sum.

27. next © 2007 Herbert I. Gross Therefore, Alice and Betty received 1 / 4 + 1 / 6 or 5 / 12 of the total sum. And since Cathy was the only other person, she received the rest of the total sum; that is, she received 7 / 12 of the total sum. next An Arithmetic Solution

28. next © 2007 Herbert I. Gross We also know that the difference between Cathy’s share and Betty’s share is $180; but it is also 7 / 12 – 1 / 6 or 5 / 12 of the total sum. next An Arithmetic Solution If 5 twelfths of the total amount is $180, 1 twelfth of the total sum if $180 ÷ 5 or $36. Therefore, 12 twelfths of the total sum (that is, the total sum) must be 12 × $36 or $432. next

29. © 2007 Herbert I. Gross We can obtain a more visual form of our arithmetic solution (and one that replaces fractions by whole numbers) by letting a “corn bread” represent the total amount of money the three girls shared. A Corn Bread Solution Knowing that Betty received 1 / 6 of the total amount and that Alice received 1 / 4 of the total amount suggests that we should divide the corn bread into 12 (i.e., the least common multiple of 4 and 6) equally sized pieces. next

30. © 2007 Herbert I. Gross Since Betty receives 1 / 6 of the 12 pieces, it means that she gets 2 pieces, A Corn Bread Solution next Thus, if we represent the amounts of money that Betty, Alice and Cathy receive by the letters, B, A and C respectively, our corn bread looks like… and the fact that Alice receives of 1 / 4 the 12 pieces means that she get 3 of the pieces. Therefore, Cathy must get the remaining 7 pieces. BBAAACCCCCCC

31. © 2007 Herbert I. Gross In other words, independently of the fact that C is 180 more than B, the ratio B:A:C is 2:3:7. next A Corn Bread Solution Returning to the present problem, we see that since the ratio of C to B is 7:2, the difference between B and C is represented by 7 – 2 or 5 pieces. next BBAAACCCCCCC Thus, 5 pieces represent $180, whereupon each piece represents $36.

32. © 2007 Herbert I. Gross In other words, since each piece represents $36, we see that… A Corn Bread Solution next So Betty has 2 × $36 or $72. 363636363636363636363636 BBAAACCCCCCC Alice has 3 × $36 or $108. Cathy has 7 × $36 or $252. $72$108$252

33. © 2007 Herbert I. Gross All in all the three girls share $72 + $108 + $252 = $432. A Corn Bread Solution next 363636363636363636363636 BBAAACCCCCCC $72$108$252 $432

34. © 2007 Herbert I. Gross Concluding Note It is not possible to anticipate all the different methods that can be used to solve a particular problem. next Thus, our above solutions should be viewed as a cross section of possible approaches rather than an exhaustive search.

35. © 2007 Herbert I. Gross Concluding Note Because the non-algebraic approaches are so highly subjective, we have not included an exercise set for Lesson 23. Rather the the exercise set in Lesson 24 (in which we will use algebra to solve the same problems as we solved in this lesson) will consist of problems that you will be free to solve by any method you chose, algebraic or otherwise.