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

Key Stone Problem…Key Stone Problem…

next

Set 23© 2007 Herbert I. Gross

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

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


next


PrefacePrefacenext

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.

next


PrefacePrefacenext

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.

next


PrefacePrefacenext

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 boththeir computational skills and their

ability to discover patterns.

next


PrefacePreface

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.

next


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 23Keystone Problem for Lesson 23next

What was the sum of money shared by the three girls?

next


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

next


A Trial-and-Error SolutionA 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.

next

next



For example, just by reading the problemwe should see that…

next

A + B + C = Tand..

B + C = 3A

Hence, we may replace B + C in A + B + C = T

by its value in B + C = 3A .

next

next



A + B + C = T

A + (B + C) = T

We see that…

A + 3A = T

4A = T

A = 1/4T

From the equation, A = 1/4T, we see that T has to be divisible by 4.

next

next


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 = T6B = T

From the equation, B = 1/6T, we see that T has to be divisible by 6.

next

(A + C) + B = T

B = 1/6T


next


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


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

next


In this case we see that…

next


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

TT A = (A = (11//44T)T) A + BA + B C = (T – [A + B]) C = (T – [A + B]) B = (B = (11//66T)T) C – B C – B

240 60 100 140 40 100

next


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


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


240 60 100 140 40 100

--- --- --- ----- --- (180)

480 120 200 280 80 200

next

next


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


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.

next


So if we choose T to be 372, we see that…

next


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


240 60 100 140 40 100

--- --- --- ----- --- (180)

372 93 155 21762 155

480 120 200 280 80 200

next


For example, starting with T = 372 our chart would look like…



372 93 155 217 62 155

396 99 165 231 66 165

384 96 160 22464 160

408 102 170 238 68 170

432 108 180 252 72 180

420 105 175 24570 175

next


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 RefinementsSome Possible Refinements

A increases by 3

TT A = (A = (11//44T)T) A + BA + B C = (T – [A + B]) C = (T – [A + B]) B = (B = (11//66T)T) C – B C – B 372 93 155 217 62 155

396 99 165 231 66 165384 96 160 22464 160

408 102 170 238 68 170

432 108 180 252 72 180420 105 175 24570 175

B increases by 2

C increases by 7 B – C increases by 5

next


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 #Row #

1

3

2

4

n

---

TT

12

36

24

48

n × 12

---

A = (A = (11//44T)T)

3

9

6

12

n × 3

---

CC

7

21

14

28

n × 7

---

B = (B = (11//66T)T)

2

6

4

8

n × 2

---

C – B C – B

5

15

10

20

n × 5

---

next


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 #Row #1

3 2

4

n ---

TT12

3624

48

n × 12---

A = (A = (11//44T)T)3

96

12

n × 3---

CC7

2114

28

n × 7---

B = (B = (11//66T)T)2

64

8

n × 2---

C – B C – B 5

1510

20

n × 5---

That is, we want the value of n forwhich 5 × n = 180. Thus, n = 180 ÷ 5, or 36.

next


So looking at the 36th row (that is, n = 36) we see that…

next

Row #Row #1

3 2

4

n ---

TT12

3624

48

n × 12---

A = (A = (11//44T)T)3

96

12

n × 3---

CC7

2114

28

n × 7---

B = (B = (11//66T)T)2

64

8

n × 2---

C – B C – B 5

1510

20

36 × 5---

T = 36 × 12 = 432.

36

…which means that the three girls have a total of $432.

36 × 12432 180

next

next


And we also have the additional, but unasked for, information that…

next

Column #Column #1

3 2

4

n ---

TT12

3624

48

n × 12---

A = (A = (11//44T)T)3

96

12

n × 3---

CC7

2114

28

n × 7---

B = (B = (11//66T)T)2

64

8

n × 2---

C – B C – B 5

1510

20

36 × 5---

A = 36 × 3 = 108

…which means that Alice has $108

36 × 336 432 108 180

next

next


next

Column #Column #1

3 2

4

n ---

TT12

3624

48

n × 12---

A = (A = (11//44T)T)3

96

12

n × 3---

CC7

2114

28

n × 7---

B = (B = (11//66T)T)2

64

8

n × 2---

C – B C – B 5

1510

20

36 × 5---

B = 36 × 2 = 72

…which means that Betty has $72

36 × 236 432 108 72

next

C = 36 × 7 = 252

…which means that Cathy has $252

36 × 7 252 180

nextnext

next


As seen previously, since the total amount Betty and Cathy received was 3 times the

amount Alice received, we can deducethat Alice received 1/4 of the total sum.

next

An Arithmetic SolutionAn 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.

next


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


next


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


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

next


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 SolutionA 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

next


Since Betty receives 1/6 of the 12 pieces, it means that she gets 2 pieces,

A Corn Bread SolutionA Corn Bread Solutionnext

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.

B B A A A C C C C C C C

nextnext

next


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 SolutionA 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

B B A A A C C C C C C C

Thus, 5 pieces represent $180, whereupon each piece represents $36.

next


In other words, since each piece represents $36, we see that…


So Betty has 2 × $36 or $72.

3636 3636 3636 3636 3636 3636 3636 3636 3636 3636 3636 3636BB BB AA AA AA CC CC CC CC CC CC CC

nextnext

Alice has 3 × $36 or $108.

Cathy has 7 × $36 or $252.

$72$72 $108$108 $252$252

next


All in all the three girls share $72 + $108 + $252 = $432.


3636 3636 3636 3636 3636 3636 3636 3636 3636 3636 3636 3636BB BB AA AA AA CC CC CC CC CC CC CC

$72$72 $108$108 $252$252$432$432

next


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.

next


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

orotherwise.

Documents

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