Simple and Compound Interest

Simple and Compound Interest 

Simple Interest  Suppose you won P10,000 and you plan to invest it for 5 years. A cooperative group offers 2% simple interest rate per year. A bank offers 2% compounded annually. Which will you choose and why?

Solution: Simple interest, with annual rate r  Tim ime( e(t) t) Pr Prin inci cipa pall

Simple Interest

Amount after t years

(P)

(Maturity Value Solution

Answer

(10,000) (0.02) (1)

200

10,000 + 200 = 10,200

2

(10,000) (0.02) (2)

400

10,000 + 400 = 10,400

3

(10,000) (0.02) (3)

600

10,000 + 600 = 10,600

4

(10,000) (0.02) (4)

800

10,000 + 800 = 10,800

5

(10,000) (0.02) (5)

1000

10,000 + 1000 = 11,000

1

10,000

Solution: Compound Interest, with annual rate r  Time (t)

Amount at the start of year t 

Compound Interest

Solution

Answer

Amount at the end of year t (Maturity Value)

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

Compare the interests gained in two 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.

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 =

problem

 365

depending on the requirement in the

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=

     

Example 1: Issa deposited Php 10,000 in a bank that offers a simple interest rate of 2.5%. If she placed the money for 3 years, how much interest will she earn at the end of 3 years? Given: P = Php 10,000 r = 2.5% t = 3 years Solution: I = Prt = (10,000) (0.025) (3) = Php 750 Thus, Issa will earn Php 1,750 worth of interest after 3 years.

Example 2: JB borrowed Php100 000 from Yani to have his car repaired. If Yani charges a simple interest of 3% per annum and JB plans to pay his loan in 6 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 Thus, JB should pay Php 1500 to Yani for interest alon

Example 3: How much interest will be earned when Php 520 000 is invested at the rate of 5.75% per year for 90 days. Assume 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 Thus, Php 520 000 should be repaid including interest that amounts to

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 t = number of years the amount is deposited or borrowed for

Example 1: An amount of Php 1500 is deposited in a bank paying an annual interest rate of 4.3%, compounded quarterly. What is the balance after 6 years? Given: P = Php 1500 r = 4.3% t = 6 years n=4 Solution: A = 1500 (1 +

0.043 (4)(6) ) 4

= Php 1 938.84 The balance after 6 years is approximately Php 1 938.84

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?

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

Therefore, Andrew borrowed Php 360 000

Example 2: At what rate did Simon invest his money amounting to Php 175 000 if he received Php 2843.75 worth of interest after 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% Thus, Simon invested his money at the rate of 3.25% per year.

Example 3: How long will it take for an investment of Php 250 000 to 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

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

Maturity (Future Value) of 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 t = term/ time in years

Example 1: Find the maturity value if 1 million pesos is deposited in a bank at an annual simple interest rate of 0.25% after (a) 1 year and (b) 5 years. Given: P = Php 1 000 000 r = 0.25% (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

(a) After 5 years Given: P = Php 1 000 000 r = 0.25% 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

Maturity (Future Value) of 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

Ic = F - P

Example1: Find the maturity value and the compound interest if P10 000 is compounded annually at an interest rate of 2% in 5 years. Given: P = P10 000 r = 2% t = 5 years

Solution: F = P (1 + r)t = 10000 (1 + 0.02) 5 = 11, 040.081 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.

Present Value Simple Interest P = Future Value (1 + rt)

Compound Interest P = Future Value 

(1 + )nt 

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

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?

Formulas: Simple Interest I = Prt

Compound Interest 

A = P (1 + )nt 

Ic = F - P

Future Value / Maturity Value F = P + Is

Future Value / Maturity Value F = P (1 + r)t

F = P (1 + rt)

Present Value P= F__ (1 + rt)

Present Value P= F___ 

(1 + )nt 

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?

Quiz 3. Find the maturity value and interest if Php 50,000 is invested a. at an annual simple interest rate of 5% after 8 years b. at 5% compounded annually for 8 years

Question to Ponder:

Where people pay by installment?

Insurance Companies

Car Loans

Home Appliances Installments

House and Lot Installment 

Payments by installment are done periodically, and in equal amounts. These scheme is called annuity.

Annuity A sequence of payment made at equal (fixed) intervals or periods of time. Payment interval = the time between successive payments Annuities may be classified in different ways, as follows:

According to payment interval and interest period Simple Annuity = an annuity where the payment interval is the same as the interest period General Annuity = an annuity where the payment interval is not the same as the interest period

According to Time of Payment  Ordinary Annuity ( or annuity immediate) = a type of annuity in which the payments are made at the end of each payment interval Annuity Due = a type of annuity in which the payments are made at beginning of each payment interval

According to Duration Annuity Certain = an annuity in which payments begin and end at definite times. Ex. installment basis of paying a car, appliances, house and lot, tuition fees, etc. Contingent Annuity = an annuity in which the payment extend over an indefinite (or indeterminate) length of time. Ex. Life insurance, pension payments

Term of Annuity, t = time between the first payment interval and the last payment interval

Regular or Periodic Payment, R = the amount of each payment Amount (Future Value) of an annuity, F = sum of future values of all the payments to be made during the entire term of the annuity Present value of an annuity, P = sum of present values of all the payments to be made during the entire term of the annuity

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

Future Value Value of Simple Simple Annuities

To find j:  j = _i_ m where i = interest m = number of conversion per year

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:

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:

Future Value of General 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.

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:

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

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

Example2: 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? Given: R = P5000 n = 2(10) = 20 i12 = 0.25% = 0.0025 m = 12 Find F:

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

Present Value of Simple Annuities The present value of a simple annuity is given by:

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:

Example2: Find the present value if quarterly payments of P2000 for 5 years with interest rate of 8% compounded quarterly.

Given:

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

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

0.02

P = 32, 702.87

Present Value of General 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.

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

Thus, Ken borrowed P100,000 from Kat.

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: