Learning of Mathematics - Solving a Maths - Math Problem Example

Problem Solving Name: Institution: Tutor: Date: Problem Three tired and hungry men went to sleep with a bag of apples. One man woke up, ate 1/3 of the apples, then went back to sleep. Later a second man woke up and ate 1/3 of the remaining apples, then went back to sleep. Finally, the third man woke up and ate 1/3 of the remaining apples. Solution A (My solution) Let the number of apples in the bag be represented by X. Let the first tired man be represented by A, the second by B and the third by C. The first man to wake up will be person A, second B, and third C. Taking the total to be X, when the first man wakes, he eats 1/3 of X (a third of what the bag contains.) This brings 1/3 x X =( 1/3)X The remaining apples are, total apples (X) minus the apples eaten by first person (1/3)X. That is,   X – (1/3)X  X - 1/3X =  (2/3)X (this is what remains after the first person has eaten)  When the second person wakes he eats 1/3 of the remaining. B eats 1/3 of  (2/3)X 1/3 x 9 2/3)X = 2/9X (what is eaten by second person B) After the second person has eaten the remainder becomes, the number of apples left by the first person (A) minus the number eaten by the second (B). (2/3)X – (2/9)X = (4/9)X.(what remains after B has eaten)  The third person eats a third of what the second left. He eats 1/3 of 4/9X 1/3 x (4/90) X = (4/27)X. The remainder after he has eaten is the number that was left by the second by the number that he C has eaten. Number of apples left by B is (4/9)X Number of apples eaten by C is (4/27)X. The remaining after the third has eaten is number left by B minus number eaten by C (4/9)X – (4/27)X =( 8/27)X (8/27)X can be simplified to be (1/3)X. (The common divisor is three; divide both numerator and denominator by three)  We are informed that after they have eaten then there are eight apples that remain in the bag. The number of apples that remain at the end is (8/27)X. Now, we equate (8/27)X  to 8 apples (8 that remained at last and the number that remains after all the  three men have eaten) (8/27)X = 8 27  x  (8/27)X = 8 x 27  (multiplying both sides by 27) 8X= 216 (only the numerator contains X and therefore X remains on the right hand side) X = 216/8 (diving both sides by 9 to remain with X on the subject side) X = 27 X was the number of apples that the bag contained initially.  So, the number of apples in the bag originally was 27 apples. Solution B The following table was created by another student who is not taking the unit. Apples A B C W 1/3 x W 1/3 x 2/3W 1/3 x 4/9W Fraction consumed 1/3W 2/9W 4/27W Remainder After eating by each individual 2/3W 2/3W – 2/9W = 4/9W 4/9W–4/27W = 8/27W Then he equated (8/27)W to the number of apples that remained. (8/27)W = 8 W = 8 x 27/8 W = 27. Solution C The student assumed that the total number of apples in the bag are T He multiplied T by 1/3 and obtained 1/3T as the number that was consumed by person A The remainder he gets 2/3T He further multiplies this by 1/3 to get 2/9T (the number consumed by second person) The remainder he obtains by subtracting (2/9)T from (2/3)T and gets (4/9)T Multiplies remainder by 1/3 and obtains (4/27)T (what the third person consumes) He then argues that when at last, there were 8 left, after they had eaten some, then it is true to deduce that; The number eaten by the first plus number by second plus number by third plus 8 should bring T, the initial number. (1/3)T + (2/9)T + (4/27)T + 8 = T T + 8= T. He terminates there. No solution! Solution D Possible Solution by grade 5 student. In Grade five, I may use trial and error method. I takes let’s say 12 objects. At each time I removes a third of the objects and place them aside. The remaining I again removes a third and places aside. This will lead to gradual decrease in the number. If 12 do not work, I consider other numbers. Taking 1/3 aside result to 8 left. This solution is ignored since the second and the third has not picked but the remaining is already nine. If 12 gives a remainder of 8 after the first person had picked, I multiply the number chosen initially by three. This gives 24. I try it out and if does not work still, I continue with other numbers. I then try a very close numbers to 24 and see the outcome. I try both the upper numbers and the lower numbers. The numbers that are greater than 24. I observe the remainder very keenly. It is the remainder that will aid in getting the correct solution.  There are different strategies that are employed in these three different solutions. The forth method (possible solution by grade 5) involves gauze work. The solver is never sure of the answer. It is both time consuming and discouraging. The solvers level of understanding this problem is low and cannot choose a direct and formula method for solving it. No concept is used. The third method seems to be correct. Even the arguments are understandable. The method terminates at some point and nothing can be made further. The solver has some understanding of the problem. He used algebra but this does not work out. The second involves use of table. Though it gives a correct answer, but the method of explanation cannot be understood by students of lower grades. It is self explanatory method. There is very little formulation used. It is just a presentation of figures. This method works best with the first learners or students who have perfect knowledge of the problem. Concepts are never used. The first solution that uses algebra is more elaborate and understandable at all levels. The solver has in mind both the first and the slow learners. The method is step after the other and full understanding is enhanced. The method is clear and simple. It is the most effective method.  The problem has numerous ways of solving it. The end result is one. This indicates that the students ability were different. The ability of grade five and other upper grades are different. In solving the problem and observing others solve it, I have learnt that when students are faced with such problems, it becomes very difficult to argue out clearly the procedure. This is as per the work of the second student. The students know the solution but cannot argue out the procedure. He only places figure without any explanations attached to it. In assistance of other people in learning of mathematics, and student centered method of learning should be encouraged. It gives the students opportunity to explain for themselves and conduct further research in all the problems. In learning of mathematics, I use research method. It enhances the expansion of the unit covered. Discussion should as well be encouraged in learning of mathematics. It means the bringing of new ideas and methods as well as confirming and evaluating the ones used. It gives weaker students to learn from their fellow. Discussion also provides an opportunity for the students who are shy in class to participate. This is more convenient when a very small group is used. The most appropriate group should compos of 5-6 students. Reflection is a method that should not be used. It means the presentation of the educators’ ideas and views as they were given. This narrows down the thinking capacity of students. As I do the research, I bump into new and very complex problems that need further research and consultations. After learning and completion of every unit we use both the past papers and revision in assessment. In the solution of such problems-especially by algebra- the understanding of fractions are enhanced. This takes place at the time that subtraction and multiplication are involved. Reference; Haylock, P. et al (2009), ‘Mathematics Explained For Primary Teachers’, Oxford publishers.   Read More