Add Math Project Work SPM

ADDITIONAL MATHEMATICS PROJECT WORKFORM 5 YEAR 2017 BY: RAVIND S/O VIJAYANDRAN 5USM IC NUMBER: 000515070797 TEACHER’S NAME: PN.ONG MEEI CHING GROUP MEMBERS: 1. VAITHESVARY KRISHINASAMY 2. ANAND GOPI 3. LEE YI YUN 4. RAVIND VIJAYANDRAN  Ackowledgement The success and the final outcome of this assignment required a lot of guidance and assistance from many people, including my teacher, parents and my fellow friends. The Add Math project work 2017 does not only test my capabilities to just solve an equation but makes me realise the variation in each and every question. Thank you Pn. Ong for giving me this amazing assignment and for all the guidance you have given which made me complete the assignment on time. I am really grateful because I have managed to complete the assignment within the given time by Pn. Ong. This assignment cannot be completed without the cooperation and support of my group members. Last but not least, I would like to express my gratitude to all of those who have lend their support and help upon completing the assignment.  Content Introduction Presentation Assignment 1 Assignment 2 Assignment 3 Assignment 4 Conclusion Reflection References and Attachments  Introduction We students, taking Additional Mathematics are required to carry out a project work while we are in Form 5. This year , the school has prepared a few tasks for us. We are to complete the tasks, assignments to be more specific. We are assigned into groups to complete the given tasks. 1) Apply and adapt a variety of problem solving 2) Enhance acquisition of mathematical knowledge and skills through problem solving 3) Cultivate creativity towards additional mathematics. 4) Prepare ourselves for the demands of the future undertakings 5) Use technology , especially ICT appropriately and properly We are expected to submit the project work within one week from the first day the task is being administered to us. Assignment NO.1 COMPANY A Starting monthly pay: RM900.00 a=900 Monthly increament: RM50.00 d=50 COMPANY B Starting monthly pay: RM750.00 a=750 Monthly increament : RM60.00 d=60 a) Method 1 𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑 Company A Company B 𝑇𝑛 = 900 + n − 1 50 𝑇𝑛 = 750 + 𝑛 − 1 60 900 + 𝑛 − 1 50 = 750 + 𝑛 − 1 60 900 + 50𝑛 − 50 = 750 + 𝑛 − 1 60 850 + 50𝑛 = 690 + 60𝑛 850 − 690 = 60𝑛 − 50𝑛 160 = 10𝑛 𝑛 = 16 Company A(Ali) Company B(Ahmad) 𝑇16 = 900 + 16 − 1 50 𝑇16 = 750 + 16 − 1 60 = RM1650 = RM1650 Monthly salary of Ali and Ahmad will be the same at the 16th month of working when both their salaries reach RM1650 Method 2 Ali Ahmad 900 + 50n = 750 + 60n 900 + 50𝑛 = 750 + 60𝑛 900 − 750 = 60𝑛 − 50𝑛 150 = 10𝑛 150 𝑛= 10 𝑛 = 15 Including first year = 15+1 =16 Company A(Ali) Company B(Ahmad) 𝑇16 = 900 + 16 − 1 50 𝑇16 = 750 + 16 − 1 60 = RM1650 = RM1650 On the 16th month of working , Ali and Ahmad will obtain the same salary which is RM1650 Method 3 MONTHS MONTH;LY SALARY IN COMPANY A MONTHLY SALARY IN COMPANY B 1 900 750 2 950 810 3 1000 870 4 1050 930 5 1100 990 6 1150 1050 7 1200 1110 8 1250 1170 9 1300 1230 10 1350 1290 11 1400 1350 12 1450 1410 13 1500 1470 14 1550 1530 15 1600 1590 16 1650 1650 17 1700 1710 18 1750 1770 19 1800 1830 20 1850 1890 Based on the table above, Ali and Ahmad will obtain the same salary at the 16th month of working when both salaries in company A and company B is RM1650 b) Justification Salary scale ofcompany B is a better deal. The monthly increament in company B is higher than in company A. Thus, in a short period of time company A provides a higher salary than comapany B. Yet, in a longer period of time company B offers higher salary than company A and prooving that company B is the better salary deal  Assignment No 2 a) Method 1 𝑇𝑛 𝐴𝑖𝑑𝑖𝑙 > 𝑇𝑛 𝐴𝑧𝑟𝑖𝑛 (300)(1.15)𝑛−1 > (500)(1.1)𝑛−1 1.15 ( )𝑛−1 > 53 1.1 5 (n-1)log10 ( 1.15 1.1 )> log10 ( ) 3 (n-1)> 11.49 𝑛 > 12.49 𝑛 = 13 Thus, this method has proof that Aidil's salary will be more than Azrin's salary in 13th month. b) Method 2 Number Aidil's of Monthly Azrin's Monthly Salary(RM) Months salary (RM) 1 300.00 500.00 2 345.00 550.00 3 396.75 605.00 4 456.26 665.50 5 524.70 732.05 6 603.41 805.26 7 693.92 885.78 8 798.01 974.36 9 917.71 1071.79 10 1055.37 1178.97 11 1213.67 1296.87 12 1395.72 1426.56 13 1605.08 1569.21 14 1845.84 1726.14 15 2122.72 1898.75 Thus, this table proves that Aidil's salary will be more than Azrin's salary in the 13th month. Thus, this table proves that Aidil's salary will be more than Azrin's salary in the 13th month. alary in 13th month. n the 13th month. n the 13th month. (a) Assignment 3 Number of Years .n 1 2 3 4 Yearly Salary Increment(RM) 11280 12240 13200 14160 12240-11280=960 Difference Between 13200-12240=960 Each Salary Increment 14160-13200=960 Graph of Yearly Salary Increment Against Number of Years 16000 14000 Yearly Salary Increment 12000 10000 8000 6000 4000 2000 0 0 1 2 3 4 Number of years The graph above shows a gradual increase in the increment of the salary. this proves that the increment of the salary is an arithmetic progression with the common difference d=960 (b) Number of Monthly years,n Salary 3 1003.52 4 1123.94  5 1258.82 6 1409.87 7 1579.06 Number Of Month 1 2 3 4 Monthly Salary Increment 120.42 134.87 151.05 169.18 Difference Between Each Salary Increment 134.87/120.42=1.12 151.05/134.87=1.12 169.18/151.05=1.12 Graph Of Monthly Increment Against the Month 180 Monthly Salary Increment 160 140 120 100 80 60 40 20 0 0 1 2 3 4 5 Number Of Month The increment in this salary increment is a goemetric progression whereby the increament i got here is a common ratio of r=1.12. Then the graph shows a gradual increase which that this is a geometric progression  Assignment 4 (a) 𝑇10 55 𝑇11 82 𝑇12 144 𝑇13 233 𝑇14 377 𝑇15 610 (b) 𝑇9 34 𝑎 𝑆𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦 =1−𝑟 = 𝑇8 21 = 1−1.6 1 =1.619047619 =-1..666666667 Conjecture: The ratio is recurring decimal Numbe r of Value of Tn d © Terms 1 1 1 2 1 2 3 2 1.5 4 3 1.667 5 5 1.6 6 8 1.625 7 13 1.615 8 21 1.619 9 34 1.618 10 55 1.618 11 89 1.618 12 144 1.618 13 233 1.618 14 377 1.618 15 610 1.618 ⬚ 𝑇𝑛 Let y= lim 𝑛→∞ 𝑇𝑛−1 𝑇𝑛 1 lim = y=1+ 𝑛→∞ 𝑇𝑛−1 𝑦 𝑇𝑛 +𝑇𝑛 −1 𝑦2 = 𝑦 + 1 = lim ( 𝑛→∞ 𝑇𝑛 𝑦2 − 𝑦 − 1 = 0 𝑇𝑛 𝑇𝑛 −1 = lim + 𝑛→∞ 𝑇𝑛 𝑇𝑛 𝑏 ± 𝑏2 4  2 y= −𝑏 ± 𝑏 − 4𝑎𝑐 𝑇𝑛−1 2𝑎 = lim 1 + 𝑛→∞ 𝑇𝑛 1± 12 +4 = 2(1) − −1 + (−1)2 −4(1)(−1) − −1 − (−1)2 −4(1)(−1) 𝑦1 = 𝑦2 = 2(1) 2(1) = 1.618 = −0.618 (d) phonesType of phoneslength width length/width value iphone 5 4.87 2.31 4.87/2.31 2.108 iphone 5s 4.89 2.31 4.89/2.31 2.116 iphone iphone 6 15.8 7.71 15.8/7.71 2.031 iphone 6s 15.8 7.78 15.8/7.78 2.031 iphone 7 13.8 6.71 13.8/6.71 3.719 huawei y8 15.43 7.7 15.43/7.7 2.003 huawei nexus 6p15.93 7.78 15.93/7.78 2.048 huawei huawei p8 14.49 7.41 14.49/7.41 1.955 huawei p9 14.5 7 14.5/7 2.071 huawei p10 14.5 6.9 14.5/6.9 2.101 oppo A37 14.3 7 14.3/7 2.042 oppo A57 14.9 7.3 14.9/7 2.129 oppo oppoF3 15.3 7 15.3/7 2.156 oppo neo 7 14.27 7.17 14.27/7 2.028 oppo F1s 7.1 14.6 7.1/14 0.5071 sumsung j7 15.2 7.8 15.2/7.8 1.948 sunsung corby 10.3 6.36 10.3/6.36 1.6184 sumsung sunsung y 10.4 5.8 10.4/5.8 1.79 sumsung s8 14.8 6.8 14.8/6.8 2.176 sunsung s8+ 15.9 7.3 15.9/7.3 2.178 honor 4X 15.29 7.7 15.29/7.7 1.985 honor 8 14.5 7.1 14.5/7.1 2..042 honor honor 6x 15 7.6 15/7.6 1.973 honor 7 14 8.65 14/8.65 1.61849 honor 8lite 14.7 7.2 14.7/7.2 2.042 e) (i) 𝒳 𝑐𝑚 𝒴 𝑐𝑚 𝒳 𝒳+1 = 1 𝒳 𝑥2 − 𝓍 − 1 = 0 −(−1)± (−1)2 −4(−1)(1) 𝒳= 2(1) 𝓍1 = 1.618 𝓍2= 0.618 1.618𝑐𝑚 1 𝑐𝑚 1𝑐𝑚 0.618𝑐𝑚 For small rectangle: 1 𝜙= 0.618 (iii) Method 1: 𝑦 = 1.618 𝑥 When x=8, 𝑦 = 1.618 8 y=12.94  Method 2: 𝒴 𝒳−𝒴 = 𝒳 𝒴 8−𝒴 1.618= 𝒴 1.618𝒴 = 8 − 𝒴 8 𝒴= 0.618 = 12.94 Thus, the length of the golden rectangle is 12.94. Method 2: 𝒳𝑐𝑚 8𝑐𝑚 8𝑐𝑚 𝒳 − 8𝑐𝑚 8 𝒳 = 𝒳−8 8 𝑥 2 − 8𝒳 − 64 = 0 −(−8) ± (−8)2 −4(1)(−64) 𝒳= 2(1) 𝒳1 = 12.94 𝒳2 = −4.94 Method 2: 2𝒳 + 2𝒴 = 20 𝒳 = 10 − 𝒴........................(1) 𝐴 = 𝒳𝒴 … … … … … … (2) Sub (1) into (2), 𝐴 = 10 − 𝒴 𝒴 =10𝒴 − 𝒴 2 𝑑𝐴 𝑊ℎ𝑒𝑛 = 0, 𝑑𝑥 10 − 2𝒴 = 0 𝒴 = 5 … … (3) Sub (3) into (1),  𝒳 = 10 − 5 =5 Length(cm) Width(cm),y=10-x Perimeter(cm) Area 1 9 20 9 2 8 20 16 3 7 20 21 4 6 20 24 5 5 20 25 6 4 20 24 7 3 20 21 8 2 20 16 Method 1: 𝑦 =1.618 𝑥 𝑦 = 1.618𝑥 … … … … … … . 1 2y+2x=20.........................(2) Sub (1) into (2), 2(1.618x)-2x=20 x=3.82..................(3) Sub (3) into (1), y= (1.618)(3.82) =6.181 Hence, it is shown that the golden rectangle has a width of 3.82cm and a lenghof 6.181cm  Conclusion After doing research, answering the questions, planning a table , drawing graphs and some problem solving, I realised that progressions is important in our daily lives and business activities. It is not only used in the business facility but also widely used in banking account skills. I learnt a lot through the Additional Mathematics project work such as the golden ratio and the fibronacci sequence which are widely used on the products in the market today.  Reflection Add Math You are a killer subject Yet with hard work, Youre nothing else than just Satisfying Add Math, Intergration, Diffrentiation, You Roar, Yet I can tame, The almighty roar you claim, Progression. Arithmetic, Geometric, All in my veins, Giuding me with knowledge, More to attain Add Math, You’re a companion, you’re a friend Above all, you’re the knowledge I crave, To attain  References 1) addmathsprojectwork.blogspot.com/ 2) https://www.scribd.com/doc/34117115/Add-Maths-Folio- Form-5