Add Maths Project 2007 by Chin Wynn

8/14/2019 Add Maths Project 2007 by Chin Wynn

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SEKOLAH MENENGAH KEBANGSAAN

BUKIT KEMUNING

40460 SHAH ALAM

ADDITIONAL MATHEMATICS PROJECT

WORK 2007

FORM 5

FIXED DEPOSIT ACCOUNT AND

INTEREST RATE

Students Name : CHIN WYNN

Form : 5 SCIENCE 1

I.C. No. : 910715-14-5397

Teachers Name : PN. ONG LIN LIN


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CONTENTS

NO. TITLE PAGE1. Acknowledgement 2

2. Introduction 3

3. Conjecture 9

4. Discussion 9

5. Identifying Information 11

6. Strategy 12

7. Conclusion 37

8. Attachment 44

9. Appendix 46

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ACKNOWLEDGEMENT

Firstly, I would like to thank our AdditionalMathematics Teacher, Madam Ong Lin Lin for guiding us

throughout this project. She explained and showed us every

contents of this project clearly.

Next, I would like to thank my friends for giving

assistance and advice about this project. Besides that, they

also gave me some mental support to doing this project.

Last but not least, I appreciate that my parents fully

believed and supported me. They sacrificed their time to

send me to my friends house in order to complete this

project. They also contributed money for me to carry out

this assignment.

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INTRODUCTION

Our group would like to show our kindness by helping Ang, Bakar and Chandran

to calculate their annual total income as well as the total amount of money theywill receive by saving in the fixed deposit account for the same period of time

but of different interest rate.

The Theory of Interest

Irving Fisher

BiographyIrving Fisher (27 Feb 1867-29 April 1947) was an American economist. his

father was a teacher and congregational minister, and the son was brought up to

believe he must be a useful member of society. Irving had mathematical ability

and a flair for inventing things. A week after he was admitted to Yale

University, his father died at age 53. Irving carried on, however, supporting his

mother, brother, and himself, mainly by tutoring. He graduated from Yale with

B.A. degree in 1888. Irvings best subject was Mathematics. He went on to write

a doctorial thesis combining both subjects, on mathematical economics which

resulted in his being granted the first Yale Ph. D in economics, in 1891.

While most of Fishers energy went into causes and business ventures, and the

better part of his scientific effort was devoted to monetary economics, he isbest remembered today for his theory of interest and capital, studies of an

ideal world from which the real world deviated at its peril. His most enduringintellectual work has been his theory of capital (economics), investment, and

interest rates, first exposited in his 1906 The Nature of Capital and income and1907 The Rate of Interest. His 1930 treatise, The Theory of Interest, summed

up a lifetimes work on capital, capital budgeting, credit markets, and thedeterminants of interest rates, including the rate of inflation.

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Definition of Fixed Deposit

A bank deposit for a fixed period of time.

A fixed-income debt security, usually issued by banks. A Fixed Deposit is like

loaning the bank your money. In return, they pay you interest.

A specific sum of money deposited in a financial institution for a fixed term

earning a pre-agreed interest rate.

A deposit of money, generally with a bank, finance company or large corporation,

repayable on a certain date. Interest may be payable at regular intervals during

the term of the deposit or on withdrawal.

1. Date of Fixed Deposit

The effective date of Fixed Deposit shall be the date on which the bank gets

funds to its account. No Fixed Deposit Receipt is issued in anticipation of

realisation of the cheque.

2. Auto Renewal

Fixed deposits are accepted with auto-renewal facility at periodicity as opted

by the depositors

3. Fixed Deposit Receipt / Memorandum of Deposit

Fixed Deposit Receipt or Memorandum of Deposit will be issued to account

holder as may be decided by the Bank from time to time for customer

convenience.

4. Tax Deducted at Source

Tax at source is deducted as applicable from the interest on fixed deposit, asper Income Tax Act, 1961.

5. Payment of Interest on Fixed Deposit

Interest on fixed deposits is reckoned/paid at quarterly intervals, at the ratesapplicable at the time of acceptance as the case may be, in accordance with RBI

directives.

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6. Premature closure of Fixed Deposits

In the event of the Fixed Deposit being closed before completing the original

term of deposit, interest will be paid at the rate applicable on the date ofdeposit for the period for which the deposit has remained with the Bank. The

deposit may be subject to penal rate of interest as prescribed by the Bank on

the date of deposit.

Definition of Bank

A bank is a commercial or state institution that provides financial services,

including issuing money in form of coins, banknotes or debit cards, receivingdeposits of money, lending money and processing transactions. A commercial

bank accepts deposits from customers and in turn makes loans based on those

deposits. Some banks (called Banks of issue) issue banknotes as legal tender.

Many banks offer ancillary financial services to make additional profit; for most

banks also rent safe deposit boxes in their branches.

Currently in most jurisdictions commercial banks are regulated and require

permission to operate. Operational authority is granted by bank regulation bank

regulatory authorities and provide rights to conduct the most fundamentalbanking services such as accepting deposits and making loans. A commercial bank

is usually defined as an institution that accepts both deposits and makes loans;

there are also financial institutions that provide selected banking services

without meeting the legal definition of a bank. Banks have influenced economies

and politics for centuries. The primary purpose of a bank was to provide loans to

trading companies. Banks provide funds to allow businesses to purchase

inventory, and collected those funds back with interest when the goods weresold. For centuries, the banking industry only dealt with businesses, not

consumers. Commercial lending today is a very intense activity, with bankscarefully analysing the financial condition of its business clients to determine

the level of risk in each loan transaction. Banking services have expanded toinclude services directed at individuals and risk in these much smaller

transactions are pooled.

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http://en.wikipedia.org/wiki/Financial_serviceshttp://en.wikipedia.org/wiki/Coinhttp://en.wikipedia.org/wiki/Banknotehttp://en.wikipedia.org/wiki/Debit_cardhttp://en.wikipedia.org/wiki/Deposit_accounthttp://en.wikipedia.org/wiki/Commercial_bankhttp://en.wikipedia.org/wiki/Commercial_bankhttp://en.wikipedia.org/wiki/Loanhttp://en.wikipedia.org/wiki/Banknotehttp://en.wikipedia.org/wiki/Safe_deposit_boxhttp://en.wikipedia.org/wiki/Financial_serviceshttp://en.wikipedia.org/wiki/Coinhttp://en.wikipedia.org/wiki/Banknotehttp://en.wikipedia.org/wiki/Debit_cardhttp://en.wikipedia.org/wiki/Deposit_accounthttp://en.wikipedia.org/wiki/Commercial_bankhttp://en.wikipedia.org/wiki/Commercial_bankhttp://en.wikipedia.org/wiki/Loanhttp://en.wikipedia.org/wiki/Banknotehttp://en.wikipedia.org/wiki/Safe_deposit_box


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A bank generates a profit from the differential between what level of interest

it pays for deposits and other sources of funds, and what level of interest it

charges in its lending activities. This difference is referred to as the spread

between the cost of funds and the loan interest rate. Historically, profitabilityfrom lending activities has been cyclic and dependent on the needs and

strengths of loan customers. In recent history, investors have demanded a more

stable revenue stream and banks have therefore placed more emphasis on

transaction fees, primarily loan fees but also including service charges on array

of deposit activities and ancillary services (international banking, foreignexchange, insurance, investments, wire transfers, etc.). However, lending

activities still provide the bulk of a commercial bank's income.

The name bankderives from the Italian word banco, desk, used during theRenaissance by Florentines bankers, who used to make their transactions abovea desk covered by a green tablecloth.

Moral Values

Responsibility: The society should keep their money in the bank because the

bank is the safest and the most reliable place to store cash. By keepingmoney in the bank, people will get an interest rate and their money grows.

The money will not get stolen easily and people will feel secure.

Work opportunities: Having a bank will provide more work opportunities for

unemployed people. This will also increase the economic rate of the country.

Sooner or later the value of the ringgit will increase and will stand proudly

next to the dollar and the Euro.

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http://en.wikipedia.org/wiki/Italian_languagehttp://en.wikipedia.org/wiki/Renaissancehttp://en.wikipedia.org/wiki/Florencehttp://en.wikipedia.org/wiki/Italian_languagehttp://en.wikipedia.org/wiki/Renaissancehttp://en.wikipedia.org/wiki/Florence


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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 RM2000 per

month. Bakar is a sales executive whose income depends solely on the

commission he receives. He earns a commission of RM1000 for his first month

and this commission increases by RM100 for each subsequent month. On the

other hand, Chandran decides to go into business. He opens a caf and makes a

profit of RM100 in his first month. For the first year, his profit in each

subsequent month is 50% more than that of the previous month.

In the second year, Ang receives a 10% increment in his monthly pay. Onthe other hand, the commission received by Bakar is reduced by RM50 for each

subsequent month. In addition, the profir made by Chandran is reduced by 10%

for each subsequent month.

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.)

(b) What is the percentage change in their total income for the second

year compared to the first year? Comment on the answers.

(c) Ang, Bakar and Chandran, each decided to open a fixed depositaccount of RM10 000 for three years without any withdrawal.

Ang keeps the amount at an interest rate of 2.5% per annumfor 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 3

years.

(ii) Compare and comment on the difference in the interests

received. If you were to invest RM10 000 for the same

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period of time, which fixed deposit account would you

prefer? Give your reasons.

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Further Exploration

2. (a) When Chandrans first child, Johan is born, Chandran invested

RM300 for him at 8% compound interest per annum. He continues

to invest RM300 on each of Johans birthday, up to and includinghis 18th birthday. What will be the total value of the investment on

Johans 18th birthday?

(b) If Chandran starts his investment with RM500 instead of RM300at the same interest rate, calculate on which birthday will the total

investment be more than RM25 000 for the first time.

Note: Compound interest is the interest earned during a period calculated on

the basis of the original sum together with interest earned from theprevious periods.

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CONJECTURE

Our group predicts that a fixed deposit account of RM10 000 for 3 years

without any withdrawal will earn a higher interest if the investment plan offersa higher interest rate per annum. Hence, Chandran should receive the highest

total amount of money after 3 years.

DISCUSSION

By using mathematical facts, formulae and methods such as arithmetic

progression, geometric progression, listing and counting, different types of

interests such as simple interest and compound interest can be calculated.

Furthermore, diagrams, tables and graphs are also inserted to help presenting

the data and information. Also, all answers in this folio are in at least 4significant figures.

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Mathematical Facts and Formulae that are used

Types of Interest

Simple Interest

Simple interest is the product of the principal, the interest rate (per period),

and the number of time periods.

To find the simple interest: c = prt, where c (simple interest) is found as the

product of p (principal), r (rate) and t (time). For example, Catherin borrows

RM23 000 to buy a new car, and the rate is 5.5% over five years. What is the

resulting simple interest?

RM23 000 x 5.5% x 5 = RM6325

* The simple interest on Cats auto loan is RM6325. If Cat repays his debtin full, he will repay RM29 325.

Compound Interest

Compound interest is very similar to simple interest. The difference is that theprincipal changes with every time period, unlike simple interest, where the

principal remains the same. The new principal at the end of every time period isessentially the simple interest on the principal at the beginning of the time

period, added to the principal.

Progression

Arithmetic Progression (A.P.)

Tn = a + (n - 1) d

Sn = n [ 2a + (n 1) d ]2

Geometric Progression (G.P.)

Tn = arn - 1

Sn = a ( rn 1 )

r 1

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IDENTIFYING INFORMATION

To find the interest rate.

To find out the best fixed deposit account. The importance of fixed deposit account.

Question 1(a)To determine how much each of them receive at the end of the first year.

Question 1(b)

To investigate the percentage change in their total income for the second yearcompared to the first year.

Question 1(c)(i)

To determine the amount each person will receive after three years.

Question 1(c)(ii)

To identify the difference in the interest received and the best fixed deposit

account.

Further Exploration

Question 2(a)

To determine the total value of the investment on Johans 18th birthday.

Question 2(b)

To investigate on which birthday the total investment will be more than

RM25 000.

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STRATEGY

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: Listing and Manual Calculation

Angs salary is constant throughout the first year, which is RM2000 per month.Bakar earns RM1000 for the first month and it increases by RM100 for each

subsequent month.Chandran makes a profit of RM100 in the first month and it increases by 50%

compared to the previous month.

Year 1 Ang (RM) Bakar (RM) Chandran (RM)

Jan 2000.0000 1000.0000 100.0000

Feb 2000.0000 1100.0000 150.0000

Mar 2000.0000 1200.0000 225.0000

Apr 2000.0000 1300.0000 337.5000

May 2000.0000 1400.0000 506.2500

Jun 2000.0000 1500.0000 759.3750Jul 2000.0000 1600.0000 1139.0625

Aug 2000.0000 1700.0000 1708.5938

Sep 2000.0000 1800.0000 2562.8907

Oct 2000.0000 1900.0000 3844.3361

Nov 2000.0000 2000.0000 5766.5042

Dec 2000.0000 2100.0000 8649.7563

Total (2 d.p.) 24000.00 18600.00 25749.27

Table 1: Income received by Ang, Bakar and Chandran in the first year

Method 2: Arithmetic Progression (A.P.) and Geometric Progression (G.P.)

Total income received by Ang at the end of the first year

The salary is same throughout the year, therefore the total income at the end

of the first year = RM2000 x 12 = RM24 000.00

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Total income received by Bakar at the end of the first year

Since Bakar earns a commission of RM1000 for the first month and his

commission increases by RM100 for each subsequent month, his monthly

commission could be arranged in a number sequence according to the ascendingorder of the months.

RM1000, RM1100, RM1200, ...... (A.P.)

Hence, his total commission at the end of the first year could becalculated using arithmetic progression (A.P.).

a = RM1000

d = RM1100 RM1000= RM100

Sn = n [ 2a + ( n- 1 ) d ], where n = number of months

2

S12 = 12 [ 2 (1000) + ( 12 1 ) ( 100 ) ]2

= 6 [ 2000 + 11 ( 100 ) ]= 6 ( 2000 + 1100 )

= 6 ( 3100 )

= RM18 600.00

Nevertheless, the calculations above have shown that Bakar receives a

total of RM18 600.00 at the end of the first year.

Total income received by Chandran at the end of the first year

Since Chandran makes a profit of RM100 in the first month and his profit

increases by 50% more than that of the previous month in each

subsequent month, the profits he made in the second, third or the

following months could be calculated using the following formulae.

Profit in nth month = Profit in ( n 1 )th month + [ 50% x Profit in ( n 1 )th month]

Profit made in 1

st

month = RM100 (stated in question)

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Profit made in 2nd month = RM100 + ( 50% x RM100 )

= RM100 + RM50

= RM150

Profit made in 3rd month = RM150 + ( 50% x RM150 )= RM150 + RM75

= RM225

From the calculations above, a number sequence of profits made by Chandranaccording to the ascending order of the months could be formed.

RM100, RM150, RM225, ...... (G.P.)

Hence, his total profit at the end of the first year could be calculatedusing geometric progression (G.P.).

a = RM100

r = RM225

RM150

= 1.5

Sn = a ( rn

1 ), where n = number of monthsr 1

S12 = 100 [ ( 1.5 )12 1 ]

1.5 1

= 100 [ ( 1.5 )12 1 ]

0.5

= RM25 749.27 (2 d.p.)

Henceforth, the calculations above have shown that Chandran receives a

total of RM25749.27 at the end of the first year.

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1. (b) What is the percentage change in their total income for the second

year compared to the first year? Comment on the answers.


Angs income increases by 10% in the second year.Bakars income is reduced by RM50 for each subsequent month.

Chandrans income is reduced by 10% for each subsequent month.


Jan 2200.0000 2050.0000 7784.7807

Feb 2200.0000 2000.0000 7006.3026

Mar 2200.0000 1950.0000 6305.6723Apr 2200.0000 1900.0000 5675.1051

May 2200.0000 1850.0000 5107.5946

Jun 2200.0000 1800.0000 4596.8351

Jul 2200.0000 1750.0000 4137.1516

Aug 2200.0000 1700.0000 3723.4364

Sep 2200.0000 1650.0000 3351.0928

Oct 2200.0000 1600.0000 3015.9835

Nov 2200.0000 1550.0000 2714.3852Dec 2200.0000 1500.0000 2442.9467

Total (2 d.p.) 26400.00 21300.00 55861.29

Table 2: Income received by Ang, Bakar and Chandran in the 2nd year

For Ang, the percentage change

= RM26 400.00 RM24 000.00 x 100%

RM24 000.00= +10%

For Bakar, the percentage change

= RM21 300.00 RM18 600.00 x 100%

RM18 600.00

= +14.52%

For Chandran, the percentage change is

= RM55 861.29 RM25 749.27 x 100%

RM25 749.27= +116.94%

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Method 2: Arithmetic Progression (A.P.) and Geometric Progression (G.P.)

Percentage change in Angs total income for the second year compared to the

first year

Angs new monthly salary = RM2 000 + 10% increase

= RM2 000 + ( RM2 000 x100

10)

= RM2 000 + RM200= RM2 200

Total salaries in the 2nd year = RM2 200 x 12

= RM26 400

From the calculations above, Ang earns a total of RM26 400.00 in his secondyear where he receives an increment of 10% in his monthly pay.

The following formulae could be used to calculate the percentage change in his

total income for the second year compared to the first year.

Percentage change

= Total income in the 2nd year Total income in the 1st year x 100%

Total income in the 1st year

= RM26 400 RM24 000 x 100%

RM24 000

= RM2 400 x 100%

RM24 000

= +10%

As a result, the result above has shown that the total salary earned by

Ang in the second year increased by 10% more than the total salaries inthe first year. Specifically, Ang earns an additional RM2400.00 in the

second year compared to the first year.

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Percentage change in Bakar s total income for the second year compared

to the first year

The income at the end of the first year:

1000, 1100, 1200, ...... (A.P.)

a = 1000

d = 100n = 12

T12 = 1000 + ( 12 1 ) ( 100 )

= 1000 + 1100

= RM2100.00

Income for the second year:

2050, 2000, 1950, ......

a = 2050

d = -50

n = 12

Sn = 12 [ 2 ( 2050 ) + ( 12 1 ) ( -50) ]

2= RM21 300.00

From the calculations above, Bakar earns a total of RM21 300.00 in his second

year.

Percentage change= Total income in the 2

nd year Total income in the 1st year x 100%


= RM21 300 RM18 600 x 100%

RM18 600

= RM2 700 x 100%

RM18 600

= +14.52%

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As a result, the result above has shown that the total income earned by

Bakar in the second year increased by 14.52% more than the total

income in the first year. Specifically, Bakar earns an additional

RM2700.00 in the second year compared to the first year.

Percentage change in Chandrans total income for the second yearcompared to the first year

The income at the end of the first year:

100, 100 x 1.5, 100 x ( 1.5 )2, 100 x ( 1.5 )11 (G.P.)

a = 100r = 1.5n = 12

T12 = 100 ( 1.5 )11

= RM8649.76

Income for the second year:

T1 = 8649.76 x 0.9 = RM7784.78

7784.78, 7784.78 x 0.9, 7784.78 x ( 0.9 )2, 7784.78 x ( 0.9 )11

a = 7784.78

r = 0.9

n = 12

S12 = 7784.78 [ 1 ( 0.9)12 ]

1 0.9

= RM55 861.28

From the calculations above, Chandran earns a total of RM55 861.28 in hissecond year.

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Percentage change

= Total income in the 2nd year Total income in the 1st year x 100%


=

RM55 861.28 RM25 749.27

x 100%RM25 749.27

= RM30 112.01 x 100%

RM25 749.27= +116.94%

As a result, the result above has shown that the total income earned by

Chandran in the second year increased by 116.94% more than the totalincome in the first year. Specifically, Chandran earns an additional

RM30 112.01 in the second year compared to the first year. In general, hehas earned more than twice the amount of the total profit he made in the

first year.

Name 1st year income

(RM)

2nd year income

(RM)

Percentage change

( % )

Ang 24 000.00 26 400.00 + 10

Bakar 18 600.00 21 300.00 + 14.52

Chandran 25 749.27 55 861.29 + 116.94

Table 3: Percentage change in the total income received by Ang, Bakar

and Chandran respectively

1. The change is positive for all 3 of them.

2. Ang has the least increase by a fixed amount of 10%.

3. Chandrans income increase higher than Bakars income because

his increase is by geometric progression, whereas Bakars

income increase is by arithmetic progression.

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1. (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 aduration of 3 months renewable at the end of every 3 months.

Chandran keeps the amount at an interest rate of 3.5% per annumfor a duration of 6 months renewable at the end of every 6

months.

(i) Find the total amount each of them will receive after 3 years.


For Ang: Interest rate per annum = 2.5%

Renewal period = At the end of each month

For Bakar: Interest rate per annum = 3%

Renewal period = At the end of every 3 months

For Chandran: Interest rate per annum = 3.5%

Renewal period = At the end of every 6 months


Investment 10000.0000 10000.0000 10000.0000

Jan 10020.8333

Feb 10041.7100

Mar 10062.6302 10075.0000

Apr 10083.5940May 10104.6015

Jun 10125.6528 10150.5625 10175.0000

Jul 10146.7479

Aug 10167.8870

Sep 10189.0701 10226.6917

Oct 10210.2973

Nov 10231.5688

Dec 10252.8846 10303.3919 10353.0625

Table 4: Interest received by Ang, Bakar and Chandran in the 1st

year

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Jan 10274.2448

Feb 10295.6495

Mar 10317.0988 10380.6673Apr 10338.5928

May 10360.1315

Jun 10381.7151 10458.5223 10534.2411

Jul 10403.3437

Aug 10425.0173

Sep 10446.7361 10536.9612

Oct 10468.5001

Nov 10490.3095

Dec 10512.1643 10615.9884 10718.5903

Table 5: Interest received by Ang, Bakar and Chandran in the 2nd year


Jan 10534.0646

Feb 10556.0106

Mar 10578.0023 10695.6083

Apr 10600.0398May 10622.1232

Jun 10644.2526 10775.8254 10906.1656

Jul 10666.4281

Aug 10688.6498

Sep 10710.9178 10856.6441

Oct 10733.2322

Nov 10755.5931

Dec (2 d.p.) 10778.00 10938.07 11097.02

Table 6: Interest received by Ang, Bakar and Chandran in the 3rd year

The total amount received by Ang after three years is RM10 778.00.

The total amount received by Bakar after three years is RM10 938.07.

The total amount received by Chandran after three years is

RM11 097.02.

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Method 2: Geometric Progression (G.P.)

The total amount Ang will receive after three years

On top of constructing a table to determine the amount of deposit inAngs fixed deposit account, the geometric progression method could also

be used.

Before that, the amount of deposit in the first few months will have tobe calculated beforehand. Here, the amount of deposit in the first three

months (January, February and March in the first year) is calculated.

Principal amount = RM10 000.00000

Amount of deposit in the 1st month= RM10 000.00000 + Interest received in the month

= RM10 000.00000 + ( RM10 000.00000 x 2.5% x12

1)

= RM10 000.00000 + ( RM10 000.00000 x480

1)

= RM10 020.83333

Amount of deposit in the 2nd

month= RM10 020.83333 + Interest received

= RM10 020.83333 + ( RM10 020.83333 x480

1)

= RM10 041.71007

With the principal amount, the amount of deposit in the first and second

month, a number sequence could be formed as shown in the following.

RM10 000.00000, RM10 020.83333, RM10 041.71007, (G.P.)

To calculate the amount Ang will receive after three years in his fixed

deposit account, geometric progression can now be used.

a = RM 10 000.00000

r = RM 10 020.83333RM 10 000.00000

= 1.002083333

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T37 (Amount at the end of the 36th month)

Tn = arn 1, where n = sequence of month

T37 = 10 000 x ( 1.002083333 )36= RM10 778.00 (2 d.p.)

The result from the calculation above has proven that the total amount

Ang will receive after 3 years is RM10 778.00. He earns an additionalRM778.00 from the interest after keeping RM10 000.00 in his fixed

deposit account without any withdrawal for three years.

The total amount Bakar will receive after three years

The amount of deposit in the first few months will have to be calculated

beforehand. Here, the amount of deposit in the first three months

(March, June and September in the first year) is determined.


Amount of deposit in March (1st year)

= RM10 000.00 + Interest received in the month

= RM10 000.00 + ( RM10 000.00 x 3% x12

3)

= RM10 000.00 + ( RM10 000.00 x400

3)

= RM10 075.00

Amount of deposit in June (1st year)

= RM10 075.00 + Interest received

= RM10 075.00 + ( RM10 075.00 x400

3 )

= RM10 150.56

With the principal amount, the amount of deposit in March and June of

the 1st year, a number sequence could be formed as shown in the following.

RM10 000.00, RM10 075.00, RM10 150.56, (G.P.)

In order to calculate the amount Bakar will receive after three years inhis fixed deposit account, geometric progression can now be used.

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a = RM 10 000.00

r = RM10 075.00

RM10 000.00

= 1.0075

T13 (T1 is the principal amount, T2 is the amount after March of the 1st year, T3 is

the amount after June of the 1st year, T13 is the amount after December of

the 3rd year)

T13 = 10 000.00 x ( 1.0075 )12

= RM10 938.07

As a result, the calculation above has shown that the total amount Bakarwill receive after 3 years is RM10 938.07. On top of that, Bakar earnsan additional RM938.07 from the interest after keeping RM10 000 in his

fixed deposit account without any withdrawal for three years.

The total amount Chandran will receive after three years

The amount of deposit in the first few months will have to be calculated.

Here, the amount of deposit in the first three months (June andDecember in the first year, and June in the second year) is determined.


Amount of deposit in June (1st year)

= RM10 000.00 + Interest received in the month

= RM10 000.00 + ( RM10 000.00 x 3.5% x12

6)

= RM10 000.00 + ( RM10 000.00 x4007 )

= RM10 175.00

Amount of deposit in December (1st year)= RM10 175.00 + Interest received

= RM10 175.00 + ( RM10 175.00 x400

7)

= RM10 353.06

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With the principal amount, the amount of deposit in June and December

of the 1st year and June of the 2nd year, a number sequence could be

formed as shown in the following.

RM10 000.00, RM10 175.00, RM10 353.06, (G.P.)

In order to calculate the amount Chandran will receive after three yearsin his fixed deposit account, geometric progression can now be used.

a = RM 10 000.00

r = RM10 175.00

RM10 000.00

= 1.0175

T7 (T1 is the principal amount, T2 is the amount after June of the 1st year, T3 is

the amount after December of the 1st year, T7 is the amount after December

of the 3rd year)

T7 = 10 000 x ( 1.0175 )6

= RM11 097.02

Therefore, the result above has shown that the total amount Chandran willreceive after 3 years is RM11 097.02. On top of that, Bakar earns an

additional RM1 097.02 from the interest after keeping RM10 000 in his fixed

deposit account without any withdrawal for three years.

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1. (c) (ii) Compare and comment on the difference in 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.

Comparisons between the interests received by Ang, Bakar and Chandran after

three years

The tables and calculations in the question 1 (c)(i) have shown that Ang, Bakarand Chandran received different amount of interests after three years under

different rates of interest and renewal period of their fixed deposit accounts.

To sum everything up, the following table is designed to compare the amount of

interests received by Ang, Bakar and Chandran after three years.

Accountowner

Interest

rate perannum

( % )

Period ofrenewal

Initial

deposit

(RM)

Amount of

deposit

after 3

years

(RM)

Total

interestreceived

(RM)

Ang 2.5

At the end of

every month

Bakar 3.0At the end of

every 3 months

Chandran 3.5At the end of

every 6 months

10 000

10 778.00 778.00

10 938.07 938.07

11 097.02 1 097.02

Table 7: Comparisons between Angs, Bakars and Chandrans fixed deposit

accounts after 3 years

By reading and analysing the data in the table above, it can be proven that

Chandran receives the most amount of interest in his fixed deposit accountafter three years compared to the fixed deposit accounts owned by Ang and

Bakar.

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Choice of fixed deposit account and reasoning

Without a doubt, the preferred fixed deposit account of mine would be

Chandrans fixed deposit account if I were to invest RM10 000 for 3 years.

One of the main reasons I would choose Chandrans fixed deposit account is

that the percentage of interest received after 3 years in Chandrans account ishigher than both Angs and Bakars fixed deposit account. To prove that the

percentage of interest received after 3 years in Chandrans account is thehighest, lets do some calculations.

Percentage increase = Amount of deposit after 3 years Initial deposit x 100%

Initial deposit

Percentage increase in Angs account after 3 years

=00.10000

00.1000000.10778

RM

RM x 100%

=00.10000

00.778x 100%

= 0.0778 x 100%

= 7.78%

Percentage increase in Bakars account after 3 years

=00.10000

00.1000007.10938

RM

RMRM x 100%

=00.10000

07.938

RM

RMx 100%

= 0.093807 x 100%

= 9.93807%

Percentage increase in Chandrans account after 3 years

=00.10000

00.1000002.11097RM

RMRM x 100%

=00.10000

02.1097

RM

RMx 100%

= 0.109702 x 100%

= 10.9702%

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As seen above, the percentage increase in Angs account after 3 years is the

lowest among all the three accounts. On the other hand, the percentage

increase in Bakars fixed deposit account is lower when compared with

Chandrans fixed account, but higher than Angs account.

This is because the interest earned is more when the fixed deposit period is

longer. The advantage of short period of renewable interest is that the moneycan be withdrawn during emergencies. The disadvantage is that less interest is

earned. On the other hand, the advantage pf long period of renewable interestis that more interest is earned. However, the disadvantage is that if there is a

need of money, interest will be lost if the fixed amount is withdrawn.

Hence, by comparing the difference between the percentage increase in all the

three accounts after three years, it is very clear that the Chandrans fixeddeposit account receives the highest percentage increase compared to theother two accounts.

This is why I have chosen Chandrans fixed deposit account over the other two

accounts because of its high percentage of interest after three years despite

the disadvantage since my objective is to gain the interest from my savings.

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Further Exploration

2. (a) When Chandrans first child, Johan is born, Chandran invested RM300 for

him at 8% compound interest per annum. He continues to invest RM300 on

each of Johans birthday, up to and including his 18th birthday. What willbe the total value of the investment in Johans 18th birthday?


Year Value of Investment (RM)

0 300.0000

1 624.0000

2 973.92003 1351.8336

4 1759.9803

5 2200.7787

6 2676.8410

7 3190.9883

8 3746.2674

9 4345.9688

10 4993.646311 5693.1380

12 6448.5890

13 7264.4761

14 8145.6342

15 9097.2849

16 10125.0677

17 11235.0731

18 (2 d.p.) 12433.88

Table 8: Value of investment on Johan

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Value of Investment against Year

624.0000973.9200

1351.83361759.9803

2200.7787

2676.8410

3190.9883

3746.2674

4345.9688

4993.6463

5693.1380

6448.5890

7264.4761

8145.6342

9097.2849

10125.0677

11235.0731

12433.8800

0

2000

4000

6000

8000

10000

12000

14000

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

ea

a

o

en

CHIN WYNN

Graph 1:

Value of

Investment

on Johan

Thus, thetotal value of investment on Johans 18th birthday is RM12 433.88.

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To fulfil the requirement of the question, we are to calculate the total value of

the investment on Johans 18th birthday.

The solution of this problem is by performing repetitive calculations.

1st year = RM300 + Interest received

= RM300 + ( RM300 x 8% )= RM300 + ( RM300 x 0.08 )

= RM300 ( 1 + 0.08 )= RM300 x ( 1.08 )

2nd year = RM300 x ( 1.08 )2

3rd year = RM300 x ( 1.08 )3

4th year = RM300 x ( 1.08 )4

5th year = RM300 x ( 1.08 )5

6th year = RM300 x ( 1.08 )6

7th year = RM300 x ( 1.08 )7

8th year = RM300 x ( 1.08 )8

9th year = RM300 x ( 1.08 )9

10th year = RM300 x ( 1.08 )10

11th year = RM300 x ( 1.08 )11

12th year = RM300 x ( 1.08 )12

13th year = RM300 x ( 1.08 )13

14th year = RM300 x ( 1.08 )14

15th year = RM300 x ( 1.08 )15

16th year = RM300 x ( 1.08 )16

17th year = RM300 x ( 1.08 )17

18th year = RM300 x ( 1.08 )18 + Investment when Johan was 0 year old

= RM300 x ( 1.08 )18 + RM300

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Thus, to find the total value of the investment on Johans 18th birthday, it is

necessary to use the geometric progression method.

a = Investment when Johan was 0 year old = RM300

r = Investment in the 2nd year

Investment in the 1st year

= RM300 x ( 1.08 )2

RM300 x ( 1.08 )

= 1.08

Sn = a ( rn 1 )

r - 1

S19 = 300 [ ( 1.0819 ) 1 ]

1.08 1

= 300 [ ( 1.0819 ) 1 ]

0.08

= RM12 433.88 (2 d.p.)

By using the method of calculation above, which is the sum of the first nth

terms, Sn of a geometric progression method, we can finally determine the

total value of investment on Johans 18th birthday. Noticeably, the result fromthe calculation has shown that the total value of investment on Johans 18th

birthday is RM12 433.88.

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2. (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 investment of Chandran has changed from RM300 to

RM500.

Year Value of Investment (RM)

0 500.0000

1 1040.0000

2 1623.20003 2253.0560

4 2933.3005

5 3667.9645

6 4461.4017

7 5318.3138

8 6243.7789

9 7243.2812

10 8322.743711 9488.5632

12 10747.6483

13 12107.4601

14 13576.0569

15 15162.1415

16 16875.1128

17 18725.1218

18 20723.1316

19 22880.982120 25211.4607

Table 9: Value of investment on Johan

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Graph 2: Value of Investment on Johan

Hence, the total investment will be more than RM25 000.00 for the first timeon Johans 20th birthday and this includes the RM500.00 invested on his 20th

birthday.

36

500.00001040.0000

1623.20002253.0560

7.9645

4461.4017

5318.3138

6243.7789

724 .

9488.5632

10747.6483

12107.4601

13576.0569

15162.1415

16875.1128

18725.1218

20723.1316

22880.98

25211.4607

000

15000

20000

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CHIN WYNN


The question requires us to calculate on which birthday will the total investment

be more than RM25 000 for the first time if Chandran starts his investment

with RM500 instead of RM300 at the same rate of interest.

Since the only change in the investment in this question compared to the ones inthe previous question is the value of investment on every Johans birthday, we

can still use the sum of the first nth terms, Sn of a geometric progressionmethod (refer to the solution of the previous question) to calculate the total

value of investment on a specific year. However, the only change to theformulae is to modify the value of ato RM500 instead of RM300 because this

question states that Chandran starts his investment with RM500 instead of

RM300. The value of rremains the same because there is no difference in therate of interest between both the questions.

a = RM500

r = 1.08

Therefore, the new formula for calculating the total value of investment on a

specific year in this question is shown in the following.

Sn = 500 [ ( 1.08)n 1 ]

1.08 - 1

To fulfil the requirement of the question, we are supposed to calculate on whichbirthday will the total investment be more than RM25 000 for the first time if

Chandran starts his investment with RM500 instead of RM300 at the same rateof interest.

500 [ ( 1.08)n 1 ] 25 000

1.08 1

500 [ ( 1.08)n 1 ] 25 000

0.08500 [ ( 1.08 )n 1 ] 2000

( 1.08 )n 1 4( 1.08 )n 5

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Until this stage, the value of nneeds to be determined to find out on which

birthday will have the total investment of more than RM25 000 for the first

time. For that reason, logarithm method is used to work out the value of n.

lg 1.08n lg 5n lg 1.08 lg 5

n08.1lg

5lg

n 20.91 (4 s.f.)

n = 21

Although n = 21, the first year is when Johan is 0 year old. Therefore, the total

investment will be more than RM25 000.00 for the first time on Johans 20th

birthday.

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Question 1(b)

Percentage change in Angs total incomes = +10%

Angs total income at the end of the 1st year = RM24 000.00

Let Angs total income at the end of the 2nd year be x.x 24 000.00 = 10%

24 000.00

x= ( 0.1 x 24 000.00 ) + 24 000.00

x= RM26 400.00

Angs total income at the end of the 2nd year = RM26 400.00

Angs salary per month =12

00.26400RM

= RM2 200.00

Percentage change in Bakars total incomes = +14.52%

Bakars total income at the end of the 1st year = RM18 600.00

Let Bakars total income at the end of the 2nd year be y.

y 18 600.00 = 14.52%

18 600.00

y= ( 0.1452 x 18 600.00 ) + 18 600.00

y= RM21 300.72

Bakars total income at the end of the 2nd year = RM21 300.72, which is

approximately same as the answer in the strategy, that is RM21 300.00.Using the arithmetic progression method,

Let a be the commission Bakar received for the 1st month.d = - RM50.00

n = 12

Sn =2

n[ 2a + ( n 1 ) d ]

21 300.00 =2

12[ 2a + 11 ( - 50.00 ) ]

3550.00 = 2a 550.00

2a = 4100.00a = RM2 050.00

The answer is the same as the commission Bakar received for the 1st month,which is also RM2 050.00. Thus, it is proven.

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Percentage change in Chandrans total incomes = +116.94%

Chandrans total income at the end of the 1st year = RM25 749.27

Let Chandrans total income at the end of the 2nd year be z.

z 25 749.27 = 116.94%

25 749.27z= ( 1.1694 x 25 749.27 ) + 25 749.27

z= RM55 860.47

Chandrans total income at the end of the 2nd year = RM55 860.47, which isapproximately same as the answer in the strategy, that is RM55 861.29.

Using the geometric progression method,Let a be the profit Chandran earned for the 1st month.

r = 90%

= 0.9n = 12

Sn = a ( 1 - rn )

1 - r

55 861.29 = a [ 1 - ( 0.9 )12 ]

1 0.9

5 586.129 = a [ 1 ( 0.9 )12 ]

a = RM7 784.78

The answer is the same as the profit Chandran earned in the 1st month, which is

also RM7 784.78. Thus, it is proven.

Question 1(c)(i)

Total amount received by Ang = RM10 778.00

Using the geometric progression method,

Let a be the initial principal amount.

r = 1 + ( 2.5% x 12

1

)

=480

481

n = 37 (36 months and the first term, which is the initial principal amount)Tn = ar

n-1

10 778.00 = a (480

481)36

a = RM10 000.00

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Total amount received by Bakar = RM10 938.07


Let a be the initial principal amount.

r = 1 + ( 3% x12

3)

=400

403

n = 13 (12 months and the first term, which is the initial principal amount)

Tn = arn-1

10 938.07 = a (400

403)12

a = RM10 000.00

Total amount received by Chandran = RM11 097.02Using the geometric progression method,Let a be the initial principal amount.

r = 1 + ( 3.5% x12

6)

=400

407

n = 7 (6 months and the first term, which is the initial principal amount)Tn = ar

n-1

11 097.02 = a ( 400

407

)6

a = RM10 000.00

Further Exploration

Question 2(a)

Total value of investment on Johans 18th birthday = RM12 433.88


Let a be the money invested when Johan was born.r = 108%

= 1.08

n = 19 ( Johans 18th

birthday, including the year when Johan was born)Sn = a ( r

n 1 )

r - 112 433.88 = a [ ( 1.08 )19 1 ]

1.08 - 1

994.7104 = a [ ( 1.08 )

19

1 ]a = RM300.00

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CHIN WYNN

Question 2(b)

The total investment will be more than RM25 000.00 for the first time on

Johans 20thbirthday.

Using the geometric progression method,Let a be the money invested when Johan was born.

r = 108%

= 1.08

n = 21 ( Johans 20th

birthday, including the year when Johan was born)

a = RM500.00

Sn = a ( r

n

1 )r 1

S21 = 500.00 [ ( 1.08 )21 1 ]

1.08 1

= RM25 211.46

If n = 20,

S20 = 500.00 [ ( 1.08 )20 1 ]

1.08 1

= RM22 880.98

Thus, the total investment will be more than RM25 000.00 for the first time on

Johans 20th birthday.

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CHIN WYNN

Results

Question 1(a)

Angs income at the of the 1st year = RM24 000.00

Bakars income at the end of the 1st year = RM18 600.00Chandrans income at the end of the 1st year = RM25 749.27

Question 1(b)

Angs income at the end of the 2nd year = RM26 400.00Bakars income at the end of the 2nd year = RM21 300.00

Chandrans income at the end of the 2nd year = RM55 861.29

Percentage change in Angs total incomes = +10%

Percentage change in Bakars total incomes = +14.52%Percentage change in Chandrans total incomes = +116.94%

The change is positive for all 3 of them. Ang has the least increase by a

fixed amount of 10%. Chandrans income increase higher than Bakars

income because his increase is by geometric progression, whereas Bakars

income increase is by arithmetic progression.

Question 1(c)(i)

Total amount received by Ang = RM10 778.00Total amount received by Bakar = RM10 938.07

Total amount received by Chandran = RM11 097.02

Question 1(c)(ii)

I have chosen Chandrans fixed deposit account over the other two accounts

because of its high percentage of interest after three years despite the

disadvantage of losing the interest if the fixed deposit is withdrawn since my

objective is to gain the interest from my savings.

Further Exploration

Question 2(a)

Total value of investment on Johans 18th birthday = RM12 433.88

Question 2(b)

The total investment will be more than RM25 000.00 for the first time on

Johans 20thbirthday.

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CHIN WYNN

Generalisation

A fixed deposit account of RM10 000 for 3 years without any withdrawal will

earn a higher interest if the investment plan offers a higher interest rate per

annum. The best example is Chandran, who received the highest total amount ofmoney after 3 years.

Besides that, income which increases by geometric progression experience

greater increase compared to arithmetic progression.

Finally, parents should invest some money when their children are born, likeChandran. From the example, although Chandran invested a small amount of

money only (RM300.00), the outcome after 19 years is RM12 433.88. If he

invested more, the outcome would be greater some more. Hence, the moneycould be used for childrens education like furthering their studies in overseas.

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ATTACHMENT

Document 1: Rubric for Additional Mathematics Project Work 2007

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CHIN WYNN

Document 2: Guidelines for the Implementation of Additional Mathematics

Project Work

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APPENDIX

1.

http://www.tuitionplaza.com/bb/forum_posts.asp?TID=7180&PN=0&TPN=2&dbqp=56921

2. http://geminigeek.com/blog/archives/2004/08/addmath-

project-tips-1/

3. http://nextlevel.com.sg/question/10949
http://www.tuitionplaza.com/bb/forum_posts.asp?TID=7180&PN=0&TPN=2&dbqp=56921http://www.tuitionplaza.com/bb/forum_posts.asp?TID=7180&PN=0&TPN=2&dbqp=56921http://geminigeek.com/blog/archives/2004/08/addmath-project-tips-1/http://geminigeek.com/blog/archives/2004/08/addmath-project-tips-1/http://nextlevel.com.sg/question/10949http://www.tuitionplaza.com/bb/forum_posts.asp?TID=7180&PN=0&TPN=2&dbqp=56921http://www.tuitionplaza.com/bb/forum_posts.asp?TID=7180&PN=0&TPN=2&dbqp=56921http://geminigeek.com/blog/archives/2004/08/addmath-project-tips-1/http://geminigeek.com/blog/archives/2004/08/addmath-project-tips-1/http://nextlevel.com.sg/question/10949

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