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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
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© 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.
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© 2007 Herbert I. Gross
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© 2007 Herbert I. Gross
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.
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© 2007 Herbert I. Gross
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.
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© 2007 Herbert I. Gross
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.
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© 2007 Herbert I. Gross
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.
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© 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 23Keystone Problem for Lesson 23next
What was the sum of money shared by the three girls?
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© 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.
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Note
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© 2007 Herbert I. Gross
A Trial-and-Error SolutionA Trial-and-Error Solution
To this end, as an abbreviation for our table we’ll let
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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.
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© 2007 Herbert I. Gross
A Trial-and-Error SolutionA Trial-and-Error Solution
For example, just by reading the problemwe should see that…
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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 .
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© 2007 Herbert I. Gross
A Trial-and-Error SolutionA Trial-and-Error Solution
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.
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A Trial-and-Error SolutionA 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 = T6B = T
From the equation, B = 1/6T, we see that T has to be divisible by 6.
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(A + C) + B = T
B = 1/6T
© 2007 Herbert I. Gross
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© 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.
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A Trial-and-Error SolutionA 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.
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© 2007 Herbert I. Gross
In this case we see that…
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A Trial-and-Error SolutionA 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.
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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
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© 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…
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A Trial-and-Error SolutionA 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.
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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
--- --- --- ----- --- (180)
480 120 200 280 80 200
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© 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.
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A Trial-and-Error SolutionA 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.
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© 2007 Herbert I. Gross
So if we choose T to be 372, we see that…
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A Trial-and-Error SolutionA 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.
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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
--- --- --- ----- --- (180)
372 93 155 21762 155
480 120 200 280 80 200
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© 2007 Herbert I. Gross
For example, starting with T = 372 our chart would look like…
A Trial-and-Error SolutionA Trial-and-Error Solution
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 66 165
384 96 160 22464 160
408 102 170 238 68 170
432 108 180 252 72 180
420 105 175 24570 175
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© 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 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
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© 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…
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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
---
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© 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.
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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.
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© 2007 Herbert I. Gross
So looking at the 36th row (that is, n = 36) we see that…
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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
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© 2007 Herbert I. Gross
And we also have the additional, but unasked for, information that…
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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
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© 2007 Herbert I. Gross
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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
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C = 36 × 7 = 252
…which means that Cathy has $252
36 × 7 252 180
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© 2007 Herbert I. Gross
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.
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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.
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© 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.
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An Arithmetic SolutionAn Arithmetic Solution
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© 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.
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An Arithmetic SolutionAn 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.
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© 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 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.
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© 2007 Herbert I. Gross
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
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© 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.
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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.
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B B A A A C C C C C C C
Thus, 5 pieces represent $180, whereupon each piece represents $36.
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© 2007 Herbert I. Gross
In other words, since each piece represents $36, we see that…
A Corn Bread SolutionA Corn Bread Solutionnext
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
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Alice has 3 × $36 or $108.
Cathy has 7 × $36 or $252.
$72$72 $108$108 $252$252
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© 2007 Herbert I. Gross
All in all the three girls share $72 + $108 + $252 = $432.
A Corn Bread SolutionA Corn Bread Solutionnext
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
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© 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.
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Thus, our above solutions should be viewed as a cross section of possible approaches rather than an exhaustive
search.
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© 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
orotherwise.
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