| List of Formulas | Anglish Features |
| Types of Story Problems | Examples of Story Problems |
| Index | Inveractive Anglish Programing |
number of items X cost per item = cost extension for item.
Two unknowns are required for each chart - one for the cost of ties and one
for the cost of shirts. However, since the cost of ties and shirts are the
same in both charts, the one set of two unknowns can be used for both charts.
In fact, the two charts have the same row and column names also. Therefore, they must be labeled with different names in the upper left hand corners so that chart, row and column name combinations can be used to uniquely identify any chart, row, column or cell.
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|sale 1|number|cost|extension|
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|shirts| 3 | | |
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| ties| 2 | | |
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| total| | | 69 |
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|sale 2|number|cost|extension|
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|shirts| 2 | | |
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| ties| 4 | | |
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| total| | | 66 |
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The two charts give two equations written using two unknowns, which
can be solved by substitution. The approach is to
subtract one equation from the other eliminating one of the variables. The
result is one equation with one unknown which can be solved. The solved
unknown is substituted into an equation to get the value of the second unknown.
To start by eliminating the first variable, the ratio of the variable
between the two equations must be deterimed so that the amount by which the equation
should be multiplied is known. The ratio is calculated by:
ratio = sale2's number of ties / sale1's number of ties.
Multiplying the sale1's equation by the ratio and solving for the shirt's cost gives:
sale1's shirts's cost = (ratio X sale1's total extension - sale2's total extension)
/ (ratio X sale1's number of shirt's - sale2's number of shirts)
The second unknown (the ties' cost) is given by:
sale1's ties's cost = (sale1's total extension - sale1's number of shirts
X sale1's shirts's cost / sale1's number of ties
The second type of multiple chart problems discussed on this page requires
a three dimensional chart to organize the data. These problems are characterized by
giving two sets of relational facts such that each set requires its own chart. However, a
solution can be obtained only by relating the two charts. Since each chart
can be considered a plane in the
three dimensional problem, the equation that relates the charts is called a "plane equation."
These problems are called relational relational problems because they use one
relational equation to relate a second set of related entities.
As an example of this type of problem consider: Ann
is fifteen years older than Joe. Ten years from now, Ann will be twice
as old as Joe. How old is each now?
One chart is set up for the current or "now" age relationship using a factor
term relationship equation.
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|now|unknown|factor|term|age |
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|ann| | 1 | 15 | |
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|joe| | 1 | 0 | |
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The second chart uses a factor relationship equation for the "then" relationship.
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|then|unknown|factor| age |
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|ann | | 2 | |
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|joe | | 1 | |
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The problem is solved by relating the age of Ann now to the age of Ann in the
future (then). These ages differ by 10 years. The plane equation is:
age of ann now + 10 = age of ann then
Where the "age of ann now" is given by:
now unknown X factor + term or unknownX1+15
and the "age of ann then" is:
then unknown X factor or (unknown+10)X2.
Since the age of Ann then is 10 years more than the age of Ann now.
It is easier to equate Ann's ages than Joe's because her age is given in terms of Joe's and therfore contian all of the relevant factor and term information. Setting up the plane equation and solving for the (now) unknown gives
unknown = (ann's now term + 10 - ann's future factor X 10)
/ (ann's future factor - ann' now factor)
To get the now age, the now term must be added in:
now's age = unknown + now's term.
In the above, the possessive form of the chart names can be used to identify specific values in the problem. For example, the value for "ann's factor" is ambiguous since applies to in both charts. However, "now's Ann's factor" and "then's Ann's factor" are unique (The alternative notation "Ann's now factor" and "Ann's then factor" are a little more readable in this case. Either form is acceptable.)
Some problems can be combined and solved at one time using Multi-dimensional Scopes - another type of problem that uses multiple tables.