Posttest items

The post-equation problem was similar, in the nature of the mathematical concept it depicts, to the pre-equation problem. It could be solved by using equations. However, this problem was of higher difficulty level, since it required the use of a system of equations of four unknowns. Students are not familiar with this concept; however, the problem could be solved by using other strategies. The problem presented a situation where different numbers of cows were found in pastures connected by bridges. The given data was represented by a given figure, on which the number of cows in two adjoining pastures was provided. Students were asked to interpret the data presented by the given figure, and would be asked to check for any unnecessary given data. This would help in evaluating how students handle given information. A sophisticated solution strategy would start by interpreting the given data into the following equations: A+B=29; A+C=25; B+D=32; C+D=28; B+C=26, where A is the number of cows in the top left pasture, B in the top right pasture, C in the bottom left pasture, and D in the bottom right pasture. After examining the first two equations, a student may notice that the difference between the number of cows in pastures B and C is the difference between 29 and 25 which is 4. Thus, B and C are two numbers whose difference is 4 and sum is 26, according to the last equation. Students can use guess and check to find that B=15 and C=11, or they can find this answer by solving the equations B+C=26 and B-C=4. Once B and C are found, students can then use the other equations to find that A=14 and D=17. Some students may not be able to

48 interpret the given data into the above equations, and will rely only on guess and check from the very beginning.

The post-pattern problem was similar in nature to the second item in the pretest, yet higher in its difficulty level. It could be solved using a similar strategy; however, the given number of cells (30) was intentionally high in order to force students to try and come up with an effective solving strategy. Instead of listing all the possibilities, which is a tedious job, they should start to notice the pattern that this problem presents. To connect cell1 to every other cell, 29 passages are needed, to connect cell2 with every other cell, except to itself and to cell1 again, 28 passages are needed, and so on…. Adding up all these cells gives 29+28+27+…+2+1=435 passages. Part ‘a’ of this item, which is to explain why no three cells are collinear, will help in evaluating students’ understanding of problem conditions.

The post-reasoning problem also presented a situation that would need mathematical logical reasoning to be solved. Students were asked to justify their work which helped the researcher in evaluating students’ level of understanding the problem situation and solving processes. The word problem stated that there were a total of 4 dozens dogs that were distributed among six differently colored kennels: blue, red, orange, yellow, green and purple. To find the number of dogs in each kennel, some clues were given. In order to solve this problem effectively, the given data needs to be properly represented. Students might resort to using a diagram that shows the six kennels, with the given clues written next to each kennel. The easiest to start with is the orange kennel, because it is clearly given that it is the largest kennel with ten dogs in it. Since the orange kennel has the largest number of dogs, students should keep in mind that no other kennel should

49 contain ten or more dogs. The second clue states that the smallest kennel has six dogs in it, but the color is not known. The yellow and green kennels has the same number of dogs, so students may use a variable to represent this number, for example D. A third clue states that there are 13 dogs in both the red and blue kennels which should contain the smallest number of dogs. So now, students should reason that either the red or the blue is the smallest kennel and thus one of them should contain 6 dogs, as stated in the given information, and the other should has 7 in order to add up to 13. However, it is not yet known which color is for each kennel. The last clue should help in determining this. The purple kennel has 2 more dogs than the blue kennel, so it can either have 6+2=8 dogs or 7+2=9 dogs. The students should then consider each of the two possibilities in order to determine what the correct answer is. If there were 8 dogs in the purple kennel then the sum of all the dogs in all kennels would be represented by this equation: 6+7+10+D+D=48, which when solved would give D=8.5 dogs which is not a possible answer. However, if there were 9 dogs in the purple kennel, the solution would be D= 8 dogs in each of the yellow and green kennels. This is a valid answer. Thus, the red kennel has 6 dogs, and the blue has 7 dogs, the orange has 10, the yellow and green has 8 in each, and the purple has 9 dogs. It has to be mentioned here that students do not need to use an equation to solve this problem. A less advanced strategy would be to add up the known number of dogs which is 32 dogs, and then subtract this from the total of 48 dogs which would give 16 dogs to be distributed equally among the yellow and green kennels and so 8 dogs in each.