While building a medieval cathedral, it cost 37 guilders (monetary unit) to hire 4 artists and 3 stonemasons or 33 guilders for 3 artists and 4 stonemasons. What would be the expense of just 1 of each worker?

From Burns, S. (Ed.). (2003). September’’s Menu of Problems. Mathematics Teaching in the Middle School, 9, 32-36.

 1-5 E Their artifacts are shown in Figures 1-5 in Appendix B. The first group, Figure 1, argued that if one artist and one stonemason together made 11 dollars, then the total for three of each would be 33 dollars. However, they reasoned that because we know that 33 dollars is enough to pay those six workers plus another stonemason, then one stonemason and one artist must together make less than 11 dollars. This group then used the guess-and-check method. They first assumed that the total for one artist and one stonemason was 8 dollars. They tried the combination of 1 dollar for the cost of one artist and 7 dollars for the cost of one stonemason (8 dollars total); however, they discovered that the total for 3 artists and 4 stonemasons was less than the needed 33 dollars. They tried other combinations but saw that the total cost was decreased; so, they abandoned the idea of an 8-dollar total. They then assumed that the total for one artist and one stonemason was 9 dollars. However, their starting guess for the cost of one artist was 2 dollars, not 1 dollar. They found that the combination of three artists at 3 dollars each and four stonemasons at 6 dollars did total the needed 33 dollars. However, when they used these amounts in the second scenario, they found that it did not work: four artists at 3 dollars each and three stonemasons at 6 dollars did not total the needed 37 dollars. They then assumed that the total for one artist and one stonemason was 10 dollars. Using the same logic, starting at 1 dollar per artist and 9 dollars per stonemason, then 2 dollars per artist and 8 dollars per stonemason, etc., they arrived at a solution of 3 dollars per artist and 7 dollars per stonemason, which they demonstrated would satisfy both requirements. In all three guess-and-check calculations, they assumed that the artists would earn less than the stonemasons would; so, they arrived at the correct figures for the solutions but had the assignments to the two types of workers backwards. They showed that four artists at 3 dollars each and three stonemasons at 7 dollars each would total 33 dollars. However, the original question stated that the cost of 33 dollars applied to three artists and four stonemasons. So, although their logic was correct, they made a minor error in the interpretation.

The second group, Figure 2, presented a tabular representation of the two scenarios. There are also several indications that they chose the values of seven and three dollars for the costs of the two types of workers, but there is no clear explanation of how they arrived at that conclusion. At the top left of the poster, four rows of seven marks each are made to represent the artists; each group of seven is circled, showing that the cost of each of four artists is 7 dollars. To the right, there are three rows of three marks each, representing that each of three stonemasons earns three dollars each. There is no indication that any values other than the correct solution were considered. There is also no indication of exactly how the correct values were calculated. However, at the bottom of the poster, two expressions are written:                These appear to indicate that an algebraic solution using two simultaneous equations was employed to arrive at the solution.

The third group, Figure 3, presented an addition solution using symbols to balance two equations. The top of the poster shows the two scenarios and the bottom of the poster shows the addition of these two. The top left of their poster shows three ten’s (squares) and seven units (circles) to represent thirty-seven dollars. On the right of the equal sign, there are four A’s, for artists, and three S’s, for stonemasons. Directly below that is a similar configuration to represent a thirty-three dollar cost for three artists and four stonemasons. The lower left portion of the poster shows six ten’s (squares) and ten units (circles) to represent seventy dollars. This is the addition of the ten’s (squares) and units (circles) from the two equations on the top of the poster. On the right side, there are seven A’s and seven S’s, which are the sum of the A’s and S’s from the two equations. This group reasoned that they now had a total of seven artists and seven stonemasons and a total of seventy dollars. They circled one artist, one stonemason, and the group of ten units (circles) to show that one of each worker would cost ten dollars. This was one of the few groups who answered the question as written. They made no attempt to determine the individual costs for one artist or one stonemason. Members of this group were not unanimous about whether or not they should do so; however, several members of this group were confident that the question merely asked for the cost for one artist and one stonemason together and that individual costs were not required.

The fourth group lists the two scenarios and then depicts the artists making 7 dollars each and the stonemasons making 3 dollars each. Below that, the group lists their check work. This seems to be backwards. They reason that 33 dollars and 37 dollars added together equals 70 dollars; simultaneously, they reason that three artists and four stonemasons added to four artists and three stonemasons results in seven of each type of worker. If seven artists and seven stonemasons cost 70 dollars, the group reasons that one artist and one stonemason cost 10 dollars. An interesting approach to finding the individual cost for each type of workers follows. First, the group realizes that both scenarios have three artists and three stonemasons. Once scenario has an extra artist and the other scenario has an extra stonemason. Based on their conclusion that one artist and one stonemason cost 10 dollars, they derive that three artists and three stonemasons cost 30 dollars. Using this baseline, they argue that the scenario which has the extra artist is 37 dollars, which is 7 dollars more than their baseline. Therefore, the artist must cost 7 dollars. And, the scenario which has an extra stonemason costs 33 dollars, which is 3 dollars more than their baseline. Therefore, the stonemason must cost 3 dollars.

The fifth group, Figure 5, drew 37 hash marks on the left side of the paper and 33 hash marks on the right side of the paper. On the left side, there are three boxes drawn, each around three hash marks, and marked with an “s,” for stonemason. And, the are remaining hash marks are separated into four groups of seven by being encircled; each is marked with an “a,” for artist. On the right side, there are four boxes drawn, each around three of the hash marks, and marked with an “s,” for stonemason. And, there are remaining hash marks are separated into three groups of seven by being encircled; each is marked with an “a,” for artist. The poster, of course, only represents their final product and does not give insight to their thinking processes.

A review of the artifacts from the other three groups reveals a similar variety of approaches. Some groups used a strictly algebraic strategy with the simultaneous equations and  One group started off with finding ways to arrive at 37, including 25 + 12, 26 + 11, and 27 + 10, even though none of these contain one value which is divisible by 7 and another value which is divisible by 3. Their last attempt, though, 28 + 9, does factor correctly. They then used the same strategy to arrive at 33, using 19 + 14, 20 + 13, and finally arriving at the correct 21 + 12. Another group first wanted to determine who made more. They reasoned that the cost with an extra artist was greater than the cost with an extra stonemason, concluding that artists cost more. Using algebra, they found that the cost for one artist was 4 dollars greater than the cost of a stonemason. They then drew four boxes to represent four artists and wrote a “5” inside each one. They also drew three circles to represent three stonemasons and drew one hash mark in each one, because stonemasons make 4 dollars less than artists do. They computed the total, 23 dollars and saw that it was short of the required 37 dollars. They added a hash mark to each square and circle and added the four 6’s and three 2’s to arrive at a total of 30, again too low. Adding one hash mark to each square and circle again, they added the four 7’s and three 3’s to get the required total of 37. From their picture, it was clear to see that an artist earns 7 dollars and a stonemason earns 3 dollars. Appendices 1-5 E