Elementary School, Primary School, Junior High School, Middle School, High School.
Apply the rule of three for a direct proportionality relationship and fill in missing values in a table.
A rule of three problem for a direct proportionality relationship has to be solved. It is presented in the form of a table. To solve the problem, complete the table by filling in the missing values.
The table is provided with values that will be sufficient to solve the problem. The difficulty of the problem is depending on the values that are provided and the values that have to be calculated and derived from others.
The values of the tables are as follows:
argument x1>1 value y1
divisor on left divisor on right
argument x2=1 value y2
factor on left = factor on right
argument x3≠x1 value y3
The level of difficulty can be chosen from five levels. The level of difficulty determined what values will be provided with the table:
Level 1: Argument x1, value y1, divisor on left, argument x2, argument x3 are given. The solution is straightforward.
Level 2: Argument x1 and value y1 plus others are given as necessary so that values have to be derived in a way that the table can be filled from from top to bottom down the table rows.
Level 3: Argument x1, value y1 are provided togther with x3 on the left or value y3 on the right, so value will have to be derived in a bottom to top way too.
Level 4: Only one of argument x1 or value y1 are given, plus others as necessary to solve the problem.
Level 5: Neither argument x1 nor value y1 are given, the derivation of values from others will have to be performed in a bottom up fashion.
The number of problems is selectable. The first problem may be presented as a sample problem, showing the complete table with all values. A number range can be chosen for the numbers appearing in the table.
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