Add Math Project 2007

ADDITIONAL MATHEMATICS PROJECT WORK 2007 FORM 5

NAME

: LOW YING HAO

I/C

: 901002-01-6021

CLASS : 5 GEMILANG

ADDITIONAL MATHEMATICS PROJECT WORK 2007 FORM 5

Ang, Bakar and Chandran are friends and they have just graduated from a local university. Ang works in a company with a starting pay of RM 2000 per month .Bakar is a sales executive whose income depends solely on the commission he receives. He earns a commission of RM1000 for the first month and this commission increases by RM1000 for each subsequent month. On the other hand, Chandran decides to go into business. He opens a café and makes a profit of RM100 in the first month. For the first year, his profit in each subsequent month is 50% more than that the previous month. In the second year, Ang receives a 10% increment in his monthly pay. On the other hand, the commission received by the Bakar is reduced by RM50 for each subsequent month. In addition, the profit made by Chandran is reduced by 10% for each subsequent month.

Solution Question 1 (a) How much does each of them receive at the end of the first year? (Two or more methods are required for this question.) Method 1: Calculation month by month Answer as shown in table below Year1 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Toatal

Ang(RM) 2000.00 2000.00 2000.00 2000.00 2000.00 2000.00 2000.00 2000.00 2000.00 2000.00 2000.00 2000.00 24000.00

Bakar(RM) 1000.00 1100.00 1200.00 1300.00 1400.00 1500.00 1600.00 1700.00 1800.00 1900.00 2000.00 2100.00 18600.00

Chandran(RM) 100.00 150.00 225.00 337.50 506.25 759.38 1139.06 1708.59 2562.89 3844.34 5766.50 8649.76 25749.27

At the end of the 1st year: Ang Bakar Chandran

: RM24000.00 : RM18600.00 : RM25749.27

Method 2: Arithmetic progression and Geometric progression For Ang used used arithmetic progression. Tn

= a+(n-1)d

Wher e

a = fisrt term d = common difference

where , a = beginning of month n = no. of month d = salary change per month T12

a = 2000 n = 12 d = 2000

= RM2000+(12-1)(RM2000) = RM2000+(11)(RM2000)] = RM 24000

Therefore the total at the end of the 1st year 1 is RM 24000.00 .

For Bakar used arithmetic progression. Sn

=

n 2

a = fisrt term d = common difference

where , a = beginning of month n = no. of month d = salary change per month

S12

Wher e

[2a+(n-1)d]

a = 1000 n = 12 d = 1100 - 1000 = 100

12 [2(RM1000)+(12-1)(RM100)] 2 = 6 [RM2000+(11)(RM100)] = RM 18600 =

Therefore the total at the end of the 1st year 1 is RM 18600.00 .

For Chandran used geometric progression. a (r n-1) Sn

=

r-1

,r> 1

where , a = beginning of month n = no. of month r = ratio of salary change per month

S12

=

(RM100) ((150/100) 12 – 1) (150/100) -1

=

RM25749.26758

Wher e

a = fisrt term r = common ratio

a = 100 n = 12 150 r = 100

Therefore the total at the end of the 1st year 1 is RM 25749.27 .

(b) What is the percentage change in their total income for the second year compared to the first year? Comment on the answers. For 2nd Year the income for Ang, Bakar,Chandran Year 2 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Ang(RM) 2200.00 2200.00 2200.00 2200.00 2200.00 2200.00 2200.00 2200.00 2200.00 2200.00 2200.00 2200.00 26400.00

Bakar(RM) 2050.00 2000.00 1950.00 1900.00 1850.00 1800.00 1750.00 1700.00 1650.00 1600.00 1550.00 1500.00 21300.00

Chandran(RM) 7784.78 7006.30 6305.67 5675.10 5107.59 4596.83 4137.15 3723.44 3351.09 3015.98 2714.39 2442.95 55861.28

For Ang, the % of change is (RM26400.00 – RM24000.00)÷RM24000 X 100% = 10 % For Bakar, the % of change is (RM21300.00 - RM18600.00)÷RM 18600.00 X 100% = 14.52% For Chandran, the % of change is (RM55861.28 – RM25749.27)÷RM 25749.27 X 100% = 116.94% Comment: For the 3 different source of income that are fixed salary for Ang, salary based on commission for Bakar and business profit for Chandran, the highest percentage of income change is Chandran which record a change of 116.94% whereas the lowest percentage change is Ang which is 10% only.

(c) Ang, Bakar and Chandran, each decided to open a fixed deposit account of RM10 000 for three years without any withdrawal. Ang keeps the amount at an interest rate of 2.5% per annum for a duration of 1 month renewable at the end of each month. Bakar keeps the amount at an interest rate of 3% per annum for a duration of 3 months renewable at the end of every 3 months. Chandran keeps the amount at an interest rate of 3.5% per annum for a duration of 6 months renewable at the end of every 6 months. (i) Find the total amount each of them will receive after three years. Method 1 : Month by month calculation Fixed Deposit Investment Year 1 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Year 2 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Year 3 Jan Feb Mar Apr May Jun Jul Aug Sep Oct

Ang(RM) 10000.00 10020.83 10041.71 10062.63 10083.59 10104.60 10125.65 10146.75 10167.89 10189.07 10210.30 10234.57 10252.88 10274.24 10295.65 10317.10 10338.59 10360.13 10381.71 10403.34 10425.02 10446.74 10468.50 10490.31 10512.16 10534.06 10556.01 10578.00 10600.04 10622.12 10644.25 10666.43 10688.65 10710.92 10733.23

Bakar(RM) 10000.00

Chandran(RM) 10000.00

10075.00 10150.56

10175.00

10226.69 10303.39

10353.06

10380.67 10458.52

10534.24

10536.96 10615.99

10718.59

10695.61 10775.83 10856.64

10906.17

Nov Dec(total amount for 3years)

10755.59 10778.00

10938.07

Method 2: Geometric progression method For Ang, T = arn-1 where , a = beginning of the month r = common ratio of interest rate n = no. of period

a = 10000 r = 1 + 0.025÷12 n = 37

Total amount = 10000(1 + 0.025÷12)37 – 1 = RM10778.00 For Bakar, T = arn-1 where , a = beginning of the month r = common ratio of interest rate n = no. of period

a = 10000 r = 1 + 0.03÷4 n = 13

Total amount = 10000(1 + 0.03÷4)13 – 1 = RM10938.07 For Chandran T = arn-1 where , a = beginning of the month r = common ratio of interest rate n = no. of period Total amount = 10000(1 + 0.035÷2)7 – 1 = RM11097.02

a = 10000 r = 1 + 0.035 ÷ 2 n=7

11097.02

(ii) Compare and comment on the difference in the interests received. If you were to invest RM10 000 for the same period of time, which fixed deposit account would you prefer? Give your reasons. The difference in the interest receive for Ang, RM 10778.00 – RM 10000.00 = RM 778.00 Bakar, RM 10938.07 – RM 10000.00 = RM 938.07 Chandra, RM 11097.02 – RM 10000 = RM 1097.02

Thus, comparing the above values for the difference in interest receive Chandran has the highest value of interest receive at the interest rate of 3.5% per annum 6 months renewable whereas Ang has the lowest value of interest receive at the interest rate of 2.5% per annum renewable monthly. If I were to invest RM10000 for a period a time, I definitely choose Chandran method because it yield the highest interest earn among the three.

Further Exploration Question 2 (a)When Chandran's first child, Johan is born, Chandran invested RM300 for him at 8% compound interest per annum. He continues to invest RM300 on each of Johan's birthday, up to and including his 18th birthday. What will be the total value of the investment on Johan's 18th birthday? Year Saving 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Chandran’s Son,Johan(RM) 300.00 324.00 673.92 1051.83 1459.98 1900.78 2376.84 2890.99 3446.27 4045.97 4693.65 5393.14 6148.59 6964.48 7845.63 8797.28 9825.07 10935.07 12133.88

Tn = (Tn-1 X 1.08%) + RM300 For example, Year 1 , T1 = RM300 X 1.08% = RM324 As year 1 is the starting year thus Chandran has not invested the additional RM300. Year 2, T2 = (RM324 +RM300) X 1.08% = RM673.92 Year 17, T17 = (RM9825.07+RM300) X 1.08% = RM10935.07 Thus, Year 18

T18 = (RM10935.07+RM300) X 1.08% = RM12133.88 (b) If Chandran starts his investment with RM500 instead of RM300 at the same interest rate, calculate on which birthday will the total investment be more than RM25 000 for the first time. Based on the question , the initial investment of Chandran has changed from RM300 to RM500 but remain RM300 in the subsequent years of investment. Year Saving 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Chandran’s Son,Johan(RM) 500.00 540.00 907.20 1,303.78 1,732.08 2,194.64 2,694.22 3,233.75 3,816.45 4,445.77 5,125.43 5,859.47 6,652.22 7,508.40 8,433.07 9,431.72 10,510.26 11,675.08 12,933.08 14,291.73 15,759.07 17,343.79 19,055.30 20,903.72 22,900.02 25,056.02

Based on the table constructed, the total investment will be more than RM25000 the first time is at year 21.