# Simple and Compound Interest

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## Compound Interest

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### Solution:

Simple interest, with annual rate Simple interest, with annual rate r r

T Timime(e(t)t) PrPrinincicipapall (P) (P) S Siimmpplle e IInntteerreesstt AAmmoouunnt t aafftteer r t t yyeeaarrss (Maturity

(Maturity VValuealue

S Soolluuttiioonn AAnnsswweerr 1 1 1100,,000000 ((1100,,000000) ) ((00..0022) ) ((11)) 220000 1100,,00000 0 + + 22000 0 = = 1100,,220000 2 2 ((1100,,000000) ) ((00..0022) ) ((22)) 440000 1100,,00000 0 + + 44000 0 = = 1100,,440000 3 3 ((1100,,000000) ) ((00..0022) ) ((33)) 660000 1100,,00000 0 + + 66000 0 = = 1100,,660000 4 4 ((1100,,000000) ) ((00..0022) ) ((44)) 880000 1100,,00000 0 + + 88000 0 = = 1100,,880000 5 5 ((1100,,000000) ) ((00..0022) ) ((55)) 11000000 1100,,00000 0 + + 1100000 0 = = 1111,,000000

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### Solution:

Compound Interest, with annual rate r

Time (t) Amount at the start of year t

Compound Interest Amount at the end of year t (Maturity Value) Solution Answer 1 10,000 (10,000)(0.02)(1) 200 10,000 + 200 = 10,200 2 10,200 (10,200)(0.02)(1) 204 10,200 + 204 = 10,404 3 10,404 (10,404)(0.02)(1) 208.08 10,404 + 208.08 = 10,612.08 4 10,612.08 (10,612.08)(0.02)(1) 212.24 10,612.08+ 212.24 = 10,824.32 5 10,824.32 (10,824.32)(0.02)(1) 216.49 10,824.32+ 216.49= 11,040.81

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### investments.

Simple Interest (in pesos) 11,000 – 10, 000 = 1,000

Compound Interest (in pesos) 11,040.81 – 10,000 = 1, 040.81

What are your observations in simple and compound interest?

Simple interest remains constant throughout the

investment term. In compound interest, the interest from the previous year also earns interest. Thus, the interest grows every year.

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Note that the rate r is expressed in percent. Thus, do not forget to convert it to decimals when used in the computations. The time t is normally expressed in years. In cases that time is

expressed in months or days, do not forget to convert it to its equivalent number of years using the following conversion formulas:

t = n months means that

t = 

12 years

t = m days means that

t = 

360 years or t = 

365 depending on the requirement in the

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### Simple Interest

To compute for the simple interest, we apply the formula:

I = Prt

There are also instances when P, r or t might be unknown. We use the following formulas to compute for P, r, and t

respectively. P =   r =   t =  

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### of 3 years?

Given: P = Php 10,000 r = 2.5% t = 3 years Solution: I = Prt = (10,000) (0.025) (3) = Php 750

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### months, how much interest should JB pay?

Given: P = Php 100 000 r = 3% t = 6 months or 6 12 years or 0.5 years Solution: I = Prt = (100 000) (0.03) (0.5) = Php 1500

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### that there are 360 days in a year.

Given: P = Php 520 000 r = 5.75% t = 90 days or 90 360 year or 0.25 year Solution: I = Prt = (520 000) (0.0575) (0.25) = Php 7 475

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### Compound Interest

To compute for compound interest, we apply the formula,

A = P (1 + 

)

nt

A = amount of money accumulated after n years, including interest

P = principal r = rate

n = number of times the interest is compounded per year

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### is the balance after 6 years?

Solution:

A = 1500 (1 + 0.043

4 )

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= Php 1 938.84

The balance after 6 years is approximately Php 1 938.84

Given:

P = Php 1500 r = 4.3%

t = 6 years n = 4

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### Assignment

1. Andy borrowed Php 50 000 from Banco Matatag to pay for

her daughter’s tuition. If the bank charges 4.25% interest rate,

how much interest should Andy pay after 3 years?

2. Ronald invested Php 1 000 000 using an investment

instrument that promises to pay a simple interest rate of 1.75% per annum. If Ronald withdraws the money after 9 months, how much interest will he earn by then?

3. If you have a bank account whose principal is Php 1000 and your bank compounds the interest twice a year at an interest

rate of 5%, how much money do you have in your account at the end of the year?

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There are certain situations when the unknown could be the P, r and t . Let’s have the following examples:

Example1: Andrew borrowed money from Magnificent Lending Services at 2.75% simple interest for 120 days to add to his

funds for house repair. If he was charged Php 3 300 for interest, how much did Andrew borrow? (Assume that there are 360

days) Given: r = 2.75% t = 120 days or 120 360 years or 0.3333 years I = Php 3 300 Solution: P =   = 3300 (0.0275)(0.3333) = Php 360 000

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### 6 months?

Given: P = Php 175 000 I = Php 2843.75 T = 6 months or 0.5 years Solution: r =   = 2843.75 (175000)(0.5) = 0.0325 or 3.25%

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### double itself if it is invested at the rate of 8%?

Given: P = Php250 000 r = 8% I = Php 250 000 Solution: t =   = 250000 (250000)(0.08)

= 12.5 years or 12 years and 6 months

Therefore, it will require 12 years and 6 months to double the investment at the given rate of interest

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### Group Activity:

Complete the entries in the given table: Assume 360 days

Principal (P) Rate (r) Time (t) Interest (I)

1 Php 120 000 1% 5 years (1) 2 Php 50 000 2.8% (2) Php 1 250 3 Php 10 000 (3) 3 months Php 525 4 (4) 3.5% 120 days Php 4 130.75 5 Php 500 000 2.5% 8 months (5) 6 Php 25 000 6.25% (6) Php 1 200 7 Php 1 200 000 (7) 1.5 years Php 20 000 8 (8) 4.20% 90 days Php 18 065.25 9 Php 25 000 (9) 45 days Php 375 10 (10) 2.05% 8 months Php 5 250

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### Simple Interest

F = P + Is

Where

F = maturity or future value P = Principal

Is = simple interest

F = P (1 + rt)

Where

F = maturity or future value P = Principal

r = rate

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### (a) After 1 year

Method 1: Is = Prt = (1000000)(0.0025) (1) = 2 500 F = P + Is = 1 000 000 + 2 500 = 1 002 500 Method 2: F = P (1 + rt) = (1000000) [1 + (0.0025) (1)] = 1 002 500

The future or maturity value after 1 year is Php 1 002 500

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### (a) After 5 years

Method1: Is = Prt = (1000000)(0.0025) (5) = 12 500 F = P + Is = 1 000 000 + 12 500 = 1 012 500 Method2: F = P (1 + rt) = (1000000) [1 + (0.0025) (5)] = 1 012 500

The future or maturity value after 5 years is Php 1 012 500

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### Compound Interest

F = P (1 + r)t

Where

P = Principal or present value F = maturity or future value r = interest rate

t = term/ time in years

The compound Interest Ic is given by

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### is compounded annually at an interest rate of 2% in 5 years.

Given: P = P10 000 r = 2% t = 5 years Ic = F - P = Php 11 040.081 – Php 10 000 = 1, 040.081

The future value is Php 11 040.081 and the compound interest is Php 1 040.081.

Solution:

F = P (1 + r)t

= 10000 (1 + 0.02) 5

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### Present Value

Simple Interest P = Future Value (1 + rt) Compound Interest P = Future Value (1 +  ) nt

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### Example1:

What is the present value of Php 50,000 due in 7 years if money is worth 10% compounded annually?

Given: F = 50,000 t = 7 years r = 10% P = ? Solution: P = 50 000 (1 + 0.10)7 = 25, 657.91

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### Group Activity:

1. Complete the table by finding the unknown.

Principal (P) Rate (r) Time (t) Interest (I) Maturity Value (F) 60 000 4% 15 (1) (2) (3) 12% 5 15000 (4) 50 000 (5) 2 (6) 59 500 (7) 10.5% (8) 157 500 457 500 1 000 000 0.25% 6.5 (9) (10)

2. Find the maturity value and interest if P50 000 is invested at 5% compounded annually for 8 years?

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### Formulas:

Simple Interest

I = Prt

Future Value / Maturity Value

F = P + Is F = P (1 + rt) Present Value P = F__ (1 + rt) Compound Interest A = P (1 +  ) nt Ic = F - P

Future Value / Maturity Value

F = P (1 + r)t Present Value P = F___ (1 +  ) nt

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### Quiz:

1. You want to start saving money. You placed a time deposit in Bank A for Php 50 000 with an interest rate of 5%. How much will your saving be after 5 years if:

a) Invested following simple interest rate? b) If compounded twice a year.

2. You are planning to invest in a high-risk mutual fund of Bank A which will give you 5% interest in 5 years. Your target is to earn a total amount of Php 1,000,000. How much should you invest if the investment is:

a) Under simple interest scheme? b) Compounded monthly?

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### Time Diagram

An installment payment of an appliance of P3000 every month An installment payment of an appliance of P3000 every month for 6 months

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### Annuities

To find j: To find j:  j =  j = _i__i_ m m where i = interest where i = interest

m = number of conversion per year m = number of conversion per year

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### Example1:

In order to save for her high school graduation, Marie decided to save P200 at the end of each month. If the bank pays 0.25%

compounded monthly, how much will her money be at the end of 6 years? Given: R = P200 m = 12 i(12) = 0.25% = 0.0025  j = 0.0025 12 = 0.0002083 t = 6 years n = tm = 6(12) = 72 periods Find F:

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### Example2:

Suppose Mrs. Remoto would like to save P3,000 at the end of each month, for six months, in a fund that gives 9% compounded monthly . How much is the amount of future value of her savings after 6 months? Given: R = P3,000 t = 6 months i(12) = 9% = 0.09 m = 12  j = 0.09 12 = 0.0075 n = 6 periods Find F:

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### Annuities

The formula for the future value of general annuity is the same as that for a simple annuity. The extra step occurs in finding j; the given interest rate per period must be converted to an equivalent rate per payment interval.

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### Example1:

Mel started to deposit P1000 monthly in a fund that pays 6%

compounded quarterly . How much will be in the fund after 15 years? Given: R = Php 1000 n = 12 (15) = 180 payments i4 = 6% = 0.06 m = 4 Find F:

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Since the payments are monthly, the interest rate of 6% compounded quarterly must be converted to its equivalent interest rate that is compounded monthly.

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Thus, the interest rate per monthly payment interval is 0.004975 or 0.4975%

Apply the formula in finding the future value of an ordinary annuity using the computed equivalent rate

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### Example2:

Given: R = P5000 n = 2(10) = 20 i12 = 0.25% = 0.0025 m = 12 Find F:

A teacher saves P5000 every 6 months in a bank that pays 0.25%

compounded monthly. How much will be her savings after 10 years?

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Convert 0.25% compounded monthly to its equivalent interest rate for each semi –annual payment interval.

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### Example1:

Suppose Mrs. Remoto would like to know the present value of her monthly deposit of P3000 when interest is 9% compounded monthly. How much is the present value of her savings at the end of 6 months? Given: R = P3000 i = 9% = 0.09  j = 0.09/12 = 0.0075 n = 6 months m = 12 Find P:

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### Example2:

Given: R = P2000 i = 8% = 0.08  j = 0.08/4 = 0.02 t = 5 years n = 5(4) = 20 payments m = 4

Find the present value if quarterly payments of P2000 for 5 years with interest rate of 8% compounded quarterly.

P = 2000 * 1- (1+0.02)-20

0.02

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### Annuities

The formula for the future value of general annuity is the same as that for a simple annuity. The extra step occurs in finding j; the given interest rate per period must be converted to an equivalent rate per payment interval.

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### Example1:

Ken borrowed an amount of money from Kat. He agrees to pay the principal plus interest by paying P38, 973.76 each year for 3 years. How much money did he borrow if interest is 8%

compounded quarterly ? Given: R = P38,973.76 i4 = 8% = 0.08 m = 4 n = 3 payments Find: P

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### Example2:

Mrs. Reyes would like to buy a television set payable monthly for 6 months starting at the end of the month. How much is the

cost of the TV set if her monthly payment is P3000 and interest is 9% compounded semi-annually ? Given: R = 3000 i(2) = 9% = 0.09 n = 6 payments m = 2 Find P:

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## References

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