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Class IV - Multiple Chart Problems

There are several types of problems that require multiple charts for solution. Two specific cases are covered on this page. First two variable problems are discussed. Since two variable problems require two equations for simultaneous solution and one row equation is obtained from each chart, two charts are required. These problems are usually presented with two distinct sets of facts. For example: At a local store, three shirts and two ties cost $69, while 2 shirts and 4 ties cost $66. How much does each shirt and tie cost?
The number of items and the associated cost comprise the data for one chart. Two such sets of data are given. The "understood" rate equation is:
        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.

                |sale 1|number|cost|extension|                                      
                |shirts|  3  |     |         |                                     
                |  ties|  2  |     |         |                                     
                | total|     |     |   69    |                                     
                |sale 2|number|cost|extension|                                      
                |shirts|  2   |    |         |                                     
                |  ties|  4   |    |         |                                     
                | total|      |    |   66    |                                     
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.

                |now|unknown|factor|term|age |
                |ann|       |   1  | 15 |    |
                |joe|       |   1  |  0 |    |
The second chart uses a factor relationship equation for the "then" relationship.
                |then|unknown|factor|  age  |
                |ann |       |   2  |       |
                |joe |       |   1  |       |
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.