add math project 2007

ADDITIONAL MATHEMATICS

PROJECT WORK 2007FORM 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 Ang(RM) Bakar(RM) Chandran(RM)Jan 2000.00 1000.00 100.00Feb 2000.00 1100.00 150.00Mar 2000.00 1200.00 225.00Apr 2000.00 1300.00 337.50May 2000.00 1400.00 506.25Jun 2000.00 1500.00 759.38Jul 2000.00 1600.00 1139.06

Aug 2000.00 1700.00 1708.59Sep 2000.00 1800.00 2562.89Oct 2000.00 1900.00 3844.34Nov 2000.00 2000.00 5766.50Dec 2000.00 2100.00 8649.76

Toatal 24000.00 18600.00 25749.27

At the end of the 1st year:

Ang : RM24000.00Bakar : RM18600.00Chandran : RM25749.27

Method 2:Arithmetic progression and Geometric progression

For Ang used used arithmetic progression.

Tn

=a+(n-1)d

Where

a = fisrt term

d = common difference

where ,                        a = beginning of month          a = 2000n =  no. of month                  n = 12d = salary change per month         d = 2000

T1

2=

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

[2a+(n-1)d]

Where

a = fisrt term

2d = common difference

where ,                        a = beginning of month          a = 1000n =  no. of month                  n = 12d = salary change per month         d = 1100 - 1000                            = 100

S1

2 =

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.

Sn

=

a (r n-1) , r >

1

Where

a = fisrt term

r-1r = common ratio

                where ,                                        a = beginning of month                  a = 100        n =  no. of month                         n = 12        r = ratio of salary change per month  r

=

150100

S1

2 =

(RM100) ((150/100) 12 – 1)

(150/100) -1

= RM25749.26758

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 Ang(RM) Bakar(RM) Chandran(RM)Jan 2200.00 2050.00 7784.78Feb 2200.00 2000.00 7006.30Mar 2200.00 1950.00 6305.67Apr 2200.00 1900.00 5675.10May 2200.00 1850.00 5107.59Jun 2200.00 1800.00 4596.83Jul 2200.00 1750.00 4137.15

Aug 2200.00 1700.00 3723.44Sep 2200.00 1650.00 3351.09Oct 2200.00 1600.00 3015.98Nov 2200.00 1550.00 2714.39Dec 2200.00 1500.00 2442.95Total 26400.00 21300.00 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 Ang(RM) Bakar(RM) Chandran(RM)Investment 10000.00 10000.00 10000.00

Year 1Jan 10020.83Feb 10041.71Mar 10062.63 10075.00Apr 10083.59May 10104.60Jun 10125.65 10150.56 10175.00Jul 10146.75

Aug 10167.89Sep 10189.07 10226.69Oct 10210.30Nov 10234.57Dec 10252.88 10303.39 10353.06

Year 2Jan 10274.24Feb 10295.65Mar 10317.10 10380.67Apr 10338.59May 10360.13Jun 10381.71 10458.52 10534.24Jul 10403.34

Aug 10425.02

Sep 10446.74 10536.96Oct 10468.50Nov 10490.31Dec 10512.16 10615.99 10718.59

Year 3Jan 10534.06Feb 10556.01Mar 10578.00 10695.61Apr 10600.04May 10622.12Jun 10644.25 10775.83 10906.17Jul 10666.43

Aug 10688.65Sep 10710.92 10856.64Oct 10733.23Nov 10755.59

Dec(total amount for 3years)

10778.00 10938.07 11097.02

Method 2: Geometric progression method

For Ang,

T  = arn-1                  where ,                         a = beginning of the month          a = 10000r = common ratio of interest rate        r = 1 + 0.025÷12n = no. of period                n = 37

Total amount = 10000(1 + 0.025÷12)37 – 1

                  =  RM10778.00

For Bakar,

T  = arn-1                  where ,                         a = beginning of the month          a = 10000r = common ratio of interest rate        r = 1 + 0.03÷4n = no. of period                n = 13

Total amount = 10000(1 + 0.03÷4)13 – 1

                   =  RM10938.07

For Chandran

T  = arn-1                  where ,                         a = beginning of the month         a = 10000r = common ratio of interest rate         r = 1 + 0.035 ÷ 2n = no. of period                n = 7

Total amount = 10000(1 + 0.035÷2)7 – 1

                 =  RM11097.02

(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 forhim 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 Chandran’s Son,Johan(RM)0 300.001 324.002 673.923 1051.834 1459.985 1900.786 2376.847 2890.998 3446.279 4045.97

10 4693.6511 5393.1412 6148.5913 6964.4814 7845.6315 8797.2816 9825.0717 10935.0718 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 18T18 = (RM10935.07+RM300) X 1.08%      = RM12133.88

(b) If Chandran starts his investment with RM500 instead of RM300 at the sameinterest rate, calculate on which birthday will the total investment be more thanRM25 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 Chandran’s Son,Johan(RM)0 500.001 540.002 907.203 1,303.784 1,732.085 2,194.646 2,694.227 3,233.758 3,816.459 4,445.77

10 5,125.4311 5,859.4712 6,652.2213 7,508.4014 8,433.0715 9,431.7216 10,510.2617 11,675.0818 12,933.0819 14,291.7320 15,759.07

21 17,343.7922 19,055.3023 20,903.7224 22,900.0225 25,056.02

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