BUSANA1 Chapter2 - Compound Interest

Compound Interest Christopher F. Santos

BUSANA1 Chapter 2 De La Salle University 2nd Term 2015-16

Christopher F. Santos

Compound Interest

DLSU BUSANA1 Chapter 2

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Compound Interest

For simple interest, the interest earned is constant every year throughout the term of the investment. For compound interest, the interest earned is added to the principal at regular time intervals and the sum becomes the new principal, in which case we say that the interest is compounded or converted into the principal.


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Conversion Period and Frequency of Conversion

Interest may be compounded/converted into the principal more than once in a year. The length of time between two successive conversion is called the conversion period or interest period. The number of interest periods in one year is called the frequency of conversion m (assumed to be an integer).


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Conversion Period and Frequency of Conversion

Conversion Period Frequency (m) annually 1 year = 12 months 1 semiannually 6 months 2 quarterly 3 months 4 monthly 1 month 12 every 4 months 4 months 3 every 2 months 2 months 6


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Total Number of Conversion

If t is the term of investment, the total number of conversion for the entire term is computed as n = tm where m is the frequency of conversion.


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Example

Frequency of Term Number of Conv (m) Conv (n) annually 1 1 1 semiannually 2 3 6 quarterly 4 2 8 monthly 12 5 60 every 4 months 3 6 18 every 2 months 6 4 21 27


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The Nominal Rate and Interest Rate Per Period i Example: Find the compound amount and interest if 100PhP is invested at 10% compounded semiannually for 3 years. The stated annual rate of interest is called the nominal rate j and the interest rate per period is computed as i=


nominal rate j = frequency of conv m

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The Nominal Rate and Interest Rate Per Period i Example: Find the compound amount and interest if 100PhP is invested at 10% compounded semiannually for 3 years. The stated annual rate of interest is called the nominal rate j and the interest rate per period is computed as i=


nominal rate j = frequency of conv m

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Compound Amount and Interest Formula P = original principal i = interest rate per period n = number of interest periods F = compound amount F = P(1 + i)n I =F −P I = P(1 + i)n − P I = P[(1 + i)n − 1]


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Examples 1. Find the compound amount and interest if 100PhP is invested at 10% compounded semiannually for 3 years. ANS. F = 134.01PhP, I = 34.01PhP 2. If 80,000PhP is deposited at 12 13 % for 6 years, what will be the compound amount and interest if interest is compounded a. semiannually ANS. F = 164, 039.40PhP, I = 84, 039.40PhP b. monthly ANS. F = 167, 042.74PhP, I = 87, 042.74PhP


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Examples

3. On May 6, 2009, Dianne borrowed 6,000PhP and promised to pay the principal and interest at 16% compounded quarterly on August 6, 2016. How much will she repay then? ANS. 18,711.91PhP


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Examples

3. On May 6, 2009, Dianne borrowed 6,000PhP and promised to pay the principal and interest at 16% compounded quarterly on August 6, 2016. How much will she repay then? ANS. 18,711.91PhP


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Present Value at Compound Interest The present value of an amount F due in n periods is the value P(principal) which is invested now at a given rate. F = P(1 + i)n P=

F (1 + i)n

P = F (1 + i)−n


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Present Value at Compound Interest

To discount F due in n periods means to find its present value P at n periods before F is due. P is the discounted value of F at n periods before it is due. The factor (1 + i)−n is called the discount factor.


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Examples 1. Discount 5,500PhP for 7 years at 8% compounded annually. ANS. 3,209.20PhP 2. A man needs 400,000PhP in 3 years to start a small business. How much money should he place in a savings account that gives 4.02% compounded semiannually so he can start the business by then? ANS. 354,979.69PhP Christopher F. Santos

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Examples

3. A laptop computer was bought on installment basis: 10,000PhP downpayment and the balance of 60,000PhP to be paid in two years. What is its cash value if interest rate is 20% compounded quarterly? ANS. 50,610.36PhP


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Examples

3. A laptop computer was bought on installment basis: 10,000PhP downpayment and the balance of 60,000PhP to be paid in two years. What is its cash value if interest rate is 20% compounded quarterly? ANS. 50,610.36PhP


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Assignment 1

2

3

Accumulate 5,000PhP for 7 14 years at 15% compounded quarterly. If 13,000PhP is due on December 2, 2013, find its present value on June 2, 2010 if money was invested at 10% compounded semiannually. The buyer of an automobile pays a 150,000PhP down payment and the balance of 500,000PhP to be paid two years later. What is the cash price of the automobile if money is worth 12% converted annually? Christopher F. Santos

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Finding the Nominal Interest Rate F = P(1 + i)n F P

= (1 + i)n

F P

j = m

n1

=1+i

n1 F −1 i= P

" 1 # n F j = mi = m −1 P Christopher F. Santos

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Examples

1. If 25,000PhP earned an interest of 3,000PhP after 2 years, at what rate compounded annually was the money invested? ANS. 5.83% 2. At what rate compounded quarterly will 3,000PhP grow to 18,000PhP after 108 months? ANS. 20.41%


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Examples



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Examples

3. At what rate converted semiannually will a given principal triple its value in 10 years? ANS. 11.29%


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Examples

3. At what rate converted semiannually will a given principal triple its value in 10 years? ANS. 11.29%


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Logarithm and Its Properties

bx = N ⇐⇒ logb N = x Ex. 23 = 8

⇐⇒

log2 8 = 3

Ex. 3x = 81 ⇐⇒ x = log3 81 = 4 Ex. 2x = 6 ⇐⇒ x = log2 6 = ??


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bx = N ⇐⇒ logb N = x 1

2

common logarithm: 10x = N ⇐⇒ log N = x natural logarithm: ex = N ⇐⇒ ln N = x Ex. 102 = 100

⇐⇒

Ex. e1 = e

⇐⇒


Compound Interest

log 100 = 2 ln e = 1


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Properties: 1

2

3

4

logb (MN) = logb M + logb N logb M N = logb M − logb N logb N r = r logb N logB N log N ln N logb N = = = logB b log b ln b

Ex. 2x = 6 ⇐⇒ x = log2 6 =


Compound Interest

log 6 ln 6 = = 2.58 log 2 ln 2


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Properties: 1

2

3

4


Ex. 2x = 6 ⇐⇒ x = log2 6 =


Compound Interest

log 6 ln 6 = = 2.58 log 2 ln 2


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Properties: 1

2

3

4


Ex. 2x = 6 ⇐⇒ x = log2 6 =


Compound Interest

log 6 ln 6 = = 2.58 log 2 ln 2


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Finding Time F = P(1 + i)n F P

= (1 + i)n

n = log(1+i) PF tm =

log PF n= log (1 + i)

log PF n t= = m m log (1 + i) Christopher F. Santos

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Examples 1. How long will it take 5,000PhP to accumulate to 20,000PhP at 12% compounded semiannually? ANS. 11.90 years 2. On March 15, 2009, a man invested 50,000PhP in a bank that gives 15% converted every 4 months. If he decides to withdraw his money once it accumulates to 60,000PhP, when should he make his withdrawal? ANS. June 15, 2010 Christopher F. Santos

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Equivalent Rates

Two annual rates of interest with different conversion periods are said to be equivalent rates if a given principal produces the same amount at each of these rates for the same period of time. Ex. 12% compounded semiannually is equivalent to 12.36% compounded annually. Here, 12% is the nominal rate while 12.36% is called the effective rate.


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Equivalent Rates



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Equivalent Rates



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Effective Rate The effective rate u is the annual rate that, when compounded annually, is equivalent to the nominal rate j compounded m times a year. Hence, m j P(1 + u) = P 1 + m

(1 + u) = 1 +

u = 1+


j m

m

Compound Interest

j m

m

−1


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Effective Rate

Solving for j, we get

(1 + u) = 1 +

j m

m

(1 + u) m = 1 +

1

j m

1

j m

(1 + u) m − 1 = 1

j = m[(1 + u) m − 1]


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Examples

1. Find the effective rate that is equivalent to 10% compounded a. b. c. d.

annually semiannually quarterly monthly

ANS. 10% ANS. 10.25% ANS. 10.38% ANS. 10.47%

2. The effective rate of 10% is equivalent to what nominal rate compounded a. quarterly b. monthly


ANS. 9.65% ANS. 9.57%

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Examples


ANS. 10% ANS. 10.25% ANS. 10.38% ANS. 10.47%




ANS. 9.65% ANS. 9.57%

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Examples


ANS. 10% ANS. 10.25% ANS. 10.38% ANS. 10.47%




ANS. 9.65% ANS. 9.57%

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Examples


ANS. 10% ANS. 10.25% ANS. 10.38% ANS. 10.47%




ANS. 9.65% ANS. 9.57%

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Examples


ANS. 10% ANS. 10.25% ANS. 10.38% ANS. 10.47%




ANS. 9.65% ANS. 9.57%

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Examples


ANS. 10% ANS. 10.25% ANS. 10.38% ANS. 10.47%




ANS. 9.65% ANS. 9.57%

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Examples


ANS. 10% ANS. 10.25% ANS. 10.38% ANS. 10.47%


2. The effective rate of 10% is equivalent to what nominal rate compounded ANS. 9.65% ANS. 9.57%

a. quarterly b. monthly


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Examples


ANS. 10% ANS. 10.25% ANS. 10.38% ANS. 10.47%


2. The effective rate of 10% is equivalent to what nominal rate compounded ANS. 9.65% ANS. 9.57%

a. quarterly b. monthly


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Other Equivalence of Rates

A. Two Nominal Rates n1 n2 j1 j2 1+ = 1+ m1 m2 B. Nominal Rate and Simple Interest Rate n j 1+ = 1 + rt m


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Other Equivalence of Rates

C. Nominal Rate and Discount Interest Rate n j 1 1+ = m 1 − dt D. Simple Interest Rate and Discount Interest Rate 1 + rt =


1 1 − dt

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Examples 1. If a debtor pays interest to his creditor at 16% compounded semiannually, at what rate compounded monthly could he just as well borrow money? ANS. 15.49% 2. John Paul borrows 50,000PhP with agreement to pay the principal and simple interest at the rate of 8% at the end of 4 years. At what nominal rate compounded monthly could he just as well pay the interest? ANS. 6.96% Christopher F. Santos

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Examples

3. What simple discount rate is equivalent to 15% compounded every 4 months in a 1-year term of investment? ANS. 13.62% 4. Find the simple interest rate that is equivalent to 6% simple discount rate in a 3-month transaction. ANS. 6.09%


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Assignment 1

2

3

4

At what rate converted semiannually will an investment of 40,000PhP earn interest of 12,000PhP in 5 12 years? Elaine borrowed 100,000PhP with interest at 12% compounded quarterly and agreed to pay 305,000PhP to settle this debt. For how long was the money borrowed? What effective rate is equivalent to 13% compounded quarterly? Find the simple discount rate equivalent to 8.6% converted quarterly in an 8-year transaction. Christopher F. Santos

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Quiz 1

The coverage of Quiz 1 ends here.


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Values of Obligation Ex. A 4,000PhP obligation due in 18 months bears interest of 15% compounded semiannually. If money is worth 12% compounded monthly, find the value of this obligation a. at the end of 2 years ANS. 5,274.89PhP

b. now ANS. 4,154.33PhP

To find the value of an obligation at any date, we first compute its maturity value on the due date before we accumulate/discount it to the desired date. Christopher F. Santos

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Comparison of Values

Ex. Which of the following two amounts of money is more valuable? A. 15,000PhP due at the end of 3 years B. 19,000PhP due at the end of 6 years Assume that money is worth 8% compounded quarterly.


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Note that we CANNOT compare the values of two amounts of money which are due on different dates, UNLESS both amounts are brought on the same date by either accumulating/discounting. This date is called the comparison date.


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In the previous example, suppose we set the comparison date to be at the end of 5 years. (a) Accumulate 15,000PhP to CD by 2 years: F = 17,574.89PhP (b) Discount 19,000PhP to CD by 1 year: P = 17, 553.06 Hence, 15,000PhP due in 3 years is more valuable than 19,000PhP due in 6 years.


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NOTE: 1 Even if a different CD were used, the conclusion will still be the same. 2 If we are free to choose the CD, then we use any of the given due dates to make comparisons faster.


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In the previous example, if we set the CD to be at the end of 6 years, then we only need to accumulate 15,000PhP by 3 years before comparing the resulting amount (19,023.63PhP) to 19,000PhP. Hence, we arrive the same conclusion.


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Comparison of Values of Obligations Ex. Which obligation is more valuable? A. 27,000PhP due in 4 years with interest at 5% compounded quarterly B. 26,000PhP due in 3 years with interest at 6% compounded semiannually Money is worth 7 21 % compounded annually. Use the end of 5 years as comparison date. ANS. At CD = 5yrs, OA = 35,407.29PhP while OB = 35,876.79PhP. Hence, obligation B is more valuable than obligation A. Christopher F. Santos

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Comparison of Values of Obligations Ex. Which obligation is more valuable? A. 27,000PhP due in 4 years with interest at 5% compounded quarterly B. 26,000PhP due in 3 years with interest at 6% compounded semiannually Money is worth 7 21 % compounded annually. Use the end of 5 years as comparison date. ANS. At CD = 5yrs, OA = 35,407.29PhP while OB = 35,876.79PhP. Hence, obligation B is more valuable than obligation A. Christopher F. Santos

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Equation of Values

An equation of values is an equation that shows that one set of values is equal to another set on the same comparison date.

X


Payment/s =

X

Compound Interest

Obligation/s


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Examples

1. Ricky owes Janet 10,000PhP due in 4 12 years. On the third year, he pays 3,000PhP. How much would he have to pay on the sixth year to discharge the rest of his obligation if money is worth 9% compounded semiannually? ANS. 7,504.88PhP


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Examples

1. Ricky owes Janet 10,000PhP due in 4 12 years. On the third year, he pays 3,000PhP. How much would he have to pay on the sixth year to discharge the rest of his obligation if money is worth 9% compounded semiannually? ANS. 7,504.88PhP


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Examples 2. Mark owes Michelle the following debts: A. 5,000PhP due in 1 year B. 7,000PhP due in 9 months with accumulated simple interest from today at 9% C. 4,000PhP due in 18 months with accumulated interest from today at 15%, m = 2

Mark wishes to replace these debts with two equal payments at the end of 10 months and 2 years, respectively. Find the amount of each payment if money is worth 12% compounded monthly. ANS. 9,111.23PhP


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Examples 2. Mark owes Michelle the following debts: A. 5,000PhP due in 1 year B. 7,000PhP due in 9 months with accumulated simple interest from today at 9% C. 4,000PhP due in 18 months with accumulated interest from today at 15%, m = 2

Mark wishes to replace these debts with two equal payments at the end of 10 months and 2 years, respectively. Find the amount of each payment if money is worth 12% compounded monthly. ANS. 9,111.23PhP


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Examples

3. What payment at the end of 5 years and 6 months, in addition to 12,000PhP at the end of 3 years, will discharge the following obligations: A. 9,500PhP due now B. 25,000PhP due at the end of 4 12 years with accumulated interest from today at 10% compounded semiannually

if money is worth 12% compounded semiannually? Use the end of 5 12 years as comparison date. ANS. 45,551.94PhP


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Examples

3. What payment at the end of 5 years and 6 months, in addition to 12,000PhP at the end of 3 years, will discharge the following obligations: A. 9,500PhP due now B. 25,000PhP due at the end of 4 12 years with accumulated interest from today at 10% compounded semiannually

if money is worth 12% compounded semiannually? Use the end of 5 12 years as comparison date. ANS. 45,551.94PhP


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Assignment 1

2

Joey owes 2,000PhP due in 2 years and 6,000PhP due in 5 years. What single payment on the 4th year will settle these debts if money is worth 10% compounded semiannually? A man owes 12,000PhP due in 3 years and 18,000PhP due in 7 years with accumulated interest from today at 10%, m = 4. He decides to pay by making a payment of 8,000PhP at the end of 1 year, 5,000PhP at the end of 5 years, and another payment at the end of 9 years. If money is worth 14%, m = 2, find the size of the payment at the end of 9 years. Christopher F. Santos

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Varying Interest Rates

If the interest rate changes during the investment term, the amount at the previous rate is computed first before applying the new rate.


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Examples

1. Tin Tin invested 120,000PhP in a fund that pays interest at 9%, m = 4, for the first 6 years; 8%, m = 4 for the next 8 years; and 7%, m = 12 thereafter. If the money is invested for 20 years, how much will it be then in the fund? ANS. 586,381.25PhP


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Examples

1. Tin Tin invested 120,000PhP in a fund that pays interest at 9%, m = 4, for the first 6 years; 8%, m = 4 for the next 8 years; and 7%, m = 12 thereafter. If the money is invested for 20 years, how much will it be then in the fund? ANS. 586,381.25PhP


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Examples

2. On June 21, 2006, Dylan invested 40,000PhP in a fund. How much money will he have on September 21, 2015 if the rate of interest is 9% compounded every 4 months until June 21, 2011, and then change to 10% compounded every 3 months in the remaining years? ANS. 94, 825.27PhP


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Examples

2. On June 21, 2006, Dylan invested 40,000PhP in a fund. How much money will he have on September 21, 2015 if the rate of interest is 9% compounded every 4 months until June 21, 2011, and then change to 10% compounded every 3 months in the remaining years? ANS. 94, 825.27PhP


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Continuous Compounding - interest are compounded very frequently The amount when principal P is invested for t years at the nominal rate j compounded continuously is F = Pejt and, consequently, we can compute the present value by using the formula P = Fe−jt


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Examples

1. Find the amount in 4 years if 6,900PhP is invested at 7 21 % compounded continuously. ANS. 9,314.03PhP 2. How much should be invested now in order to have 50,000PhP in 3 41 years if it is invested at 6 23 % compounded continuously? ANS. 40,259.92PhP


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3. Find the effective rate of interest that is equivalent to 11% converted continuously. ANS. 11.63% 4. Find the amount due at the end of 8 years if 35,000PhP is invested at 12% converted continuously for the first 4 years and 14% converted quarterly for the last 4 years. ANS. 98,078.77PhP


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Examples



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Assignment 1

2

3

A certain amount P was invested in a fund at a rate of 7.5% simple interest for the first 5 years, at 18% compounded monthly for the next 5 years, and at 14% compounded semiannually for the remaining years. If the total amount in the fund after 13 years is 89,560PhP, find the initial amount P invested in the fund. Find the amount due if 8,000PhP is invested for 6 years at 8% converted (a) monthly (b) continuously. Find the rate compounded continuously that is equivalent to 8% compounded quarterly. Christopher F. Santos

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BUSANA1 Chapter2 - Compound Interest

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