Appendix The Model-Drawing Method with Algebra (Reproduced from Teaching Secondary School Mathematics: ‘A Resource Book 2007. pp. 393-412) Introduction ‘One of the main objectives of mathematics education in Singapore is to enable students to develop their abilities in problem solving. The model-drawing method was introduced in the 1980s by the Primary Mathematics Project of the Ministry of Education. It involves the construction of pictorial models, namely the part-whole model and the comparison model, to help students visualise abstract mathematical relationships and various problem structures through pictorial representations (Kho, 1987). It is a powerful visual aid for solving complex problems involving fraction, ratio and percentage. Above all, it is closely related to the algebraic method for solving algebra word problems (Kho, 1987; Fong, 1994; Ng, 2001: Cheong, 2002; Beckmann, 2004). This article demonstrates how the model-drawing method can help students visualise and conceptualise a problem so that they can formulate an algebraic equation to solve it Mathematical Models ‘The Part-Whole Model ‘The part-whole model (also known as the ‘part-part-whole’ model) shows the relationship between a whole and its parts, e.g The pictorial model shows that the whole w is divided into two parts a and b. That is, weatb. In the part-whole model, we may divide the whole into a number of equal parts, e.g ‘The pictorial model shows that the whole w is divided into 3 equal parts, and each part is p. Thatis, w = 3p.The Comparison Model The comparison model shows the relationship between two quantities when they are compared, e.g. a 7 The pictorial model shows that the quantity a is more than the quantity b, and their difference is d. That is, d= a—b. Also, The sum of a and b is s. Thatis, s In the comparison model, we may express one quantity as a multiple of the other, e.g, The pictorial model shows that the quantity a is 3 times as much as the quantity B. That is, a=3b, Example 1 There are 50 children in a dance group. If there are 10 more boys than girls, how many girls are there? Method 1. Let the number of girls be x. We form the equation: x+(x410)= 50 The solution is. = 20. There are 20 girls.‘The same problem can be solved by formulating a different equation as shown im the alternative method below Method 2 Here we form the equation: (0-1) -x=10 The solution is x= 20. ‘There are 20 girl. Example 2 $120 is shared among 3 persons A, B and C. If receives $20 less than B, and B receives 3 times as much money as C, how much money does C receive? Method 1 Let C receive Sx. We form the equation: (Bx -20) + 3x4 2= 120 ‘The solution is x= 20. C receives $20. Method 2 120 Here we form the equation: 3x — (120 — 4x) = 20 The solution is x= 20. C receives $20. a7Example 3 A has 3 times as much money as B. B has $200 less than C. C has $50 more than A. Find the roral amount of money that A, B and C have. Method 1 Let B have Sx, We form the equation: Gx + 50)—x= 200 The solution is x = 75, Total amount of money = S(3x + .x-+ 3x +50) ‘Method 2. Here we form the equation: (x +200) -3r= 50 5. ‘The solution is 4 Total amount of money = S(3x + .x+. + 200 ) = $575 118Here we form the equation: 2x +50 = 200 ‘The solution is x = 75. ‘Total amount of money = S(3x +.x+x+ 200) = $575 ‘More examples are given as follows. Example 4 John is 4 times as old as his son. If their total age 10 years ago was 60, find their present ages. [16, 64] Joba «« (a 10 years ago: John 10 is’ (0 [(4x ~ 10) + (x= 10) = 60] ngExample 5 Raju had 3 times as much money as Gopal. After spending $60 each, Raju had 4 times as much money as Gopal. How much money did Raju have at first? [S540] Raju After spending S60 each: Raju pat $2 o 3x — 60 = 4(x-60)] Example 6 A box contains a total of 200 blue, yellow and orange beads. Twice the number of blue beads is 10 more than the number of yellow beads. There are 50 more yellow beads than orange beads: How many blue beads are there? Blue [x-+ 2x10) + (2x- 60) Fraction, Ratio and Percentage The Part-Whole Model for Fraction In the part-whole model, when we divide the whole into equal parts, we may use some of the parts to represent a fraction of the whole, e.g. —EE_ Ee The pictorial model shows that a is Sof w,In this case, w comprises 4 units, and a is equal to 3 units. We write: a= 3 w= St rite: @ “ 4 For example, the following model shows 3 of a whole. If we let 3 of the whole be-x, then the whole is 4(5) If we let the whole be 4x, 48) then 3 of the whole is 3x ay ae The Comparison Model for Fraction and Ratio In the comparison model, we may express one quantity’ as a fraction of the other, e.g. ‘The pictorial model shows that two quantities @ and b are 3 units and 2 units respectively. We say: aisdofd. bisdora. We write: 121The relationship between a and b can also be expressed as a ratio. We say: We write F a3 The ratio of a 10 b is 3:2 aad The ratio of b to ais 2:3. bai For example, the following model shows that one quantity is + of the other. The same model shows that the two quantities are in the ratio 3 : 2 3Example 7 Susan had 2 as much money as Mary at frst. After receiving + of Susan's money, Mary had $210, How much money did Susan have at first? Susan |= Let Susan's money be 2x and Mary's money be 3 at first. We form the equation ax 210 ‘The solution is x Susan’s money at first = S(2x) = $105 Example 8 Susan had > as much money as Mary at first. After receiving = of Mary’s money, Susan had $210, How much money did Susan have at first? Before Susan Afr Susan 20 60. ve The solution is x at first = $(2x) = $120 Susan’s mone’ 123,Example 9 3 ofthe beads in a box are yellow beads. The rest are orange and blue beads. The ratio of the number of orange beads to the nunber of blue beads is 4 : 5, If there are 30 more blue beads than orange beads, how many yellow beads are there? yellow orange & blue Let the number of orange beads and blue beads be 4x and 5x respectively. We form the equation Sx-4x=30 The solution is x= 30. Number of yellow beads = 3( More examples are given as follows. Example 10 A box contains total of 200 blue and yellow beads. 2 of the blue beads and + of the yellow beads are taken out and used 0 make a necklace. If there are 30 fewer blue beads than yellow beads in the necklace, how many beads are left in the box? (90) sue [= | Yellow “0 [Bx + 22x +30) = 200]is 1 0) Example 11 3 of the students in Sec 1A and > of the students in Sec 1B are girls, Both classes have the Jame number of girls. Sec TA has 4 more boys than Sec 1B. How many students are there in Sec 1A? (40) sirls boys sec 1 I 4] et Be a ot Example 12 The number of fifiy-cent coins and twenty-cent coins are in the ratio 2 : 3. If 4 of the fifty- cent coins are exchanged for twenty-cent coins, the ratio will become 2 : 7. What is the total value ($) of she set of coins? [89.60] ‘Method 1 Before: 50e 206 fier soe [ 10 125‘The Part-Whole Model for Percentage e of the whole. We take the In the part-whole model, we may express a part as a percenta; whole as the base (100%). e.g. 0% 608 008 ‘The pictorial model shows that b is 60% of a. In this case, a comprises 100 units, and is equal to 60 units. We write b= 0.64. ‘The part-whole model can be used to show the relationship between the new value of & {quantity and its original value after an increase or a decrease Let the original value be x (a) The following model shows an increase of 20%. ' 0% Here the increase is 0.2x, and the new value is 1.2. () The following model shows a decrease of 40% oa Here the decrease is 0.4x, and the new value is 0.6s.The Comparison Model for Percentage of the other. For In the comparison model, we may express one quantity as a percent example, to compare b relative to a, we take a as the base (100%). €.g. 08 100% 125% ee | ze The model shows that b is 125% of a. That is, b is 25% more than a. In this case, a comprises 100 units, and b is equal to 125 units. We write b = 1.254. Here b is 25% more than a, but @ is not 25% less than b. Indeed, 1 0.8b This shows that a is 80% of b. That is, a is 20% less than b. Example 13 ‘Mrs Tan has a VIP card that entitles her to a discount of 30% off her dinner bill. If she pay’ for her dinner, how much is the original bill? 0% 10% Let the original bill be Sx. We form the equation O.7x= 133 The solution is x= 190. The original bill is $190. 127Example 14 Jane saved 20% more in January than in February. If she saved a toral of $330 in the two ‘months, how much did she save in February? os 100% 120% j 330 Let Jane save Sv in February. We form the equation: 330 The solution is.x xe 0, Jane saved $150 in February. Example 13 Mrs Tan spent 2 of her money on a handbag. She spent 40% ofthe remaining money on «dress. If the handbag cost $80 more than the dres, how much money did she have at first? handbag Let Mrs Tan have $31 at first. We form the equation: 2x-0.4r= 80 The solution is x = 50. Mrs Tan’s money = $(3x) = $150Example 16 A sum of money is shared among three persons A, B and C so that A receives 10% more than B, and B receives 10% more than C. If A receives $525 more than C, find the sum of money. Let C receive Sx. We form the equation: 25 Lid.Ly- 2500. (1.2Ly + LL +x) = $8275 The solution is Sum of money Example 17 There are twice as many boys as girls in a choir. If the number of boys is decreased by 30%, by what percentage must the number of girls be increased (or decreased) so that there will be an equal number of boys and girls in the choir? 0% 70% 100% rr | (ax) ——4 4 = 40% 129Example 18 The ratio of the number of boys to the number of girls is 5 : 6. Itis given that 30% of the boys wear glasses, and there is an equal number of boys and girls who wear glasses. What percentage of the Is wear glasses? Boys J ins [) Boys [030 Gils Percentage Conclusion Algebra is the key to the learning of higher mathematics. Nevertheless it is a worldwide concern that many students have difficulties making the transition from arithmetic to algebra, especially when they attempt to formulate algebraic equations for solving problems. The integration of the model-drawing method and the algebraic method provides an enriching opportunity for students to engage in the construction and interpretation of algebraic equations through meaningful and active learning. We hope that this approach will help more students develop their competence and confidence in using the algebraic method. References Beckmann, S. (2004), Solving alg demonstrated in grade 4—6 texts used in Si ebra and other story problems with simple diagrams: A method pore. The Mathematics Educator, 14(1).42~46. primary mathematics and the model approach to problem Cheong. N. P. (2002), The teaching solving. Marhematies Newsletter (Issue No. 4). Singapore: Ministry of Edu H. K, (1994). Brid, ning, 14( yematies. Teaching and Fon, the gap between secondary and primary 1 Kho, T. H, (1987). Mathematical models for solving arithmetic problems. Proceedings of the Fourth Southeast Asian Conference on Mathematics Educarion (ICMI-SEAMS), 345-351 ngapore: Institute of Education 1), Secondary school students" perceptions of the relationship between the model method and algebra. Proceedings of the Tweljih International Commission of Mathematical Instruction (ICMD Study Conference (pp. 468-474). Melbourne, Australia: The University of Melbourne.i 1 1 Appendix D Solving Challenging Algebra Word Problems In this article, we illustrate how the Model Method can be integrated with the algebraic method to help students solve challenging algebra word problems in secondary school mathematics. The following examples, taken from a Singapore Mathematics textbook for Secondary One’, illustrate the approach Example 1 A certain amount of water is poured from a jug into an empty mug so that the amount of water in the mug is E the amount of water eft in the jing. If 50 ml of water i further povred from te tothe mug, the amount ef weer athe mag wil bef tha eft nthe jug. Find the original amount of water in the jug. Students may draw a pictorial model to represent the problem situation, + 50 aid (a) After Ist 7 ib) After nd Mug Tug Mug Part (a) shows the amounts of water in the two containers after the first transfer, and part (b) shows the amounts after the second transfer: Mathematics Matters Express p. 136. Published by EPB Pan Pacific, 2007 131‘The model can also be presented as shown below After Ist ransfer: Js a | ci After 2nd transfer: Jug Mug From the model, students can formulate an algebraic equation to solve the problem, Let the amount of water in the mug and the jug after the first transfer be x ml and 6x ml respectively After Ist ranser us| I aa Mug [¥ After 2nd aransfer — Nariatioy After the second transfé (6x - 50) ml respectively the amount of water in the mug and the jug are (x + 50) ml and After Ist rransfer:From the model, students obtain the equation: 6x ~ 50 = Six +50) ‘The solution of the equation is 7x = 2100 The original amount of water is 2100 ml. Variation 2 The amount of water in the mug and the jug after the second transfer can also be expressed as (x + 50) ml and 5(x + 50) ml respectively: After Ist ransfer: After 2nd transfer: Jug Mug From the model, students obtain the equation: 6x-Siv45 30 The solution of the equation is x= 300. 7x=2100 The original amount of water is 2100 ml The following are alternative solutions to the problem when students let x ml be a different unknown quantity Variation 3 Let the amount of water in the mug after the second transfer be x ml. After Ist transfer: Sr+50 ue I vie [ES] After 2nd transfer: mg | vos 133,From the model, students obtain the equation: Sy +50 = 6(x~ 50) The solution of the equation is x 6c = 2100 The original amount of water is 2100 ml Nariation 4 Let the amount of water in the jug after the first transfer be x ml After Lt ranger ee ee From the model, students obtain the following equation to solve the problem: = +50) 6 ‘The solution of the equation is x= 1800. 100 er is 2100 ml. ‘The original amount of wa Example 2 Ara musical concert, class A tickets were sold at $4 each, class B tickets at $2 each, sit class A tickets and and souvenir programmes at $1 each. zo the audience who bout 2 of the audience who bought class B tickets also bought the programmes. The total amount of money collected from both types of tickets was $1400 and the amount of money collected from the programmes was $350. Find the total number of people who attended the concert.Let x and y be the number of people who bought class A and class B tickets respectively ‘The problem can be solved by formulating a pair of simultaneous linear equations: y+ 2y= 1400 xt The total number of people is 500. The problem can also be solved by a pictorial method as follows Let 4x and 3y be the number of people who bought class A and class B tickets respectively. Students can draw a pictorial model to represent the problem situation as follows: ] 400 330 By eliminating 3 groups of 350 from 1400 as shown, students find the value of x as follows: 7x = 1400-350 x 3 0 —> x=50 By eliminating 3x from 350, students find the value of y as follows: 2y =350-50x3 200 — y=100 Total number of people = 4x + 3y = 500 jing problem before students learn simultaneous: jinear equation from the The problem is regarded as a challet linear equations. They may solve the problem by formulating following mode!: The model shows that + of A and = of B bought the programmes. 135Var Let | unit in B be x. Then the total number of people who bought the programmes in B is 2x, and that in A is 350 ~ 2x From the model, students obtain the following equation to solve the problem: 4 4(5)(350 - 2x) + 2(3x) = 1400 The solution of the equation is x= 100, s 500, = 2x) + 3x ‘The total number of people is 500. ‘Variation 2 Let | unit in A be x. Then the total number of people who bought the programmes in A is 0 ~ 3. 3x, and that in B is Be x4 $1400 350-38 From the model, students obtain the following equation to solve the problem: 3) (350-38) = 1400 (ax) + The solution of the equation is x = 50. 500 4x43 (350-3) ‘The total number of people is 500. 136