2. At the end of the lesson,
the learner is able to
compute maturity value,
interest, and present
value, and solve
problems involving
compound interest
when compound
interest is computed
more than once a year.
3. Lesson Outline:
1.Compounding more than once a year
2.Maturity value, interest, and present value
when compound interest is computed
more than once a year
4. Sometimes, interest may be compounded more than once a year.
Consider the following example.
Example 1. Given a principal of PhP 10,000, which of the
following options will yield greater interest after 5 years:
OPTION A: Earn an annual interest rate of 2% at the end
of the year, or
OPTION B: Earn an annual interest rate of 2% in two
portionsβ1% after 6 months, and 1% after another 6
months?
6. OPTION B: Interest is compounded semi-annually, or every 6 months. Under
this option, the interest rate every six months is 1% (2% divided by 2).
7. Option B will give the
higher interest after 5 years.
If all else is equal, a more
frequent compounding will
result in a higher interest,
which is why Option B gives
a higher interest than
Option A.
8. Definition of Terms:
β’ Frequency of conversion (m) β number of conversion
periods in one year
β’ Conversion or interest period β time between successive
conversions of interest
β’ Total number of conversion periods n
n = mt = (frequency of conversion) x (time in years)
β’ Nominal rate (π(π)
) β annual rate of interest
β’ Rate (j) of interest for each conversion period
9. Examples of nominal rates and the corresponding
frequencies of conversion and interest rate for each period:
11. Example 1:
Find the maturity value and interest
if P10,000 is deposited in a bank at 2%
compounded quarterly for 5 years.
Given: P = 10,000 π(4)
= 0.02
t = 5 years m = 4
Find: a. F b. Ic
12. Example 1:
Given: P = 10,000 π(4)
= 0.02 t = 5 years m = 4
Find: a. F
b. Ic
SOLUTION: π =
π(4)
π
π =
0.02
4
π = 0.005
13. Example 1:
Given: P = 10,000 π(4)
= 0.02 t = 5 years m = 4
Find: a. F
b. P
SOLUTION:
π = ππ‘
π = (4)(5)
π = 20
14. Example 1:
Given: P = 10,000 π(4)
= 0.02 t = 5 years m = 4
Find: a. F
b. Ic
SOLUTION:
πΉ = π(1 + π)π
πΉ = 10,000(1 + 0.005)20
πΉ = 11,048.96
π = 0.005 π = 20
15. Example 1:
Given: P = 10,000 π(4)
= 0.02 t = 5 years m = 4
Find: a. F
b. Ic
SOLUTION: πΌπ = πΉ β π
πΌπ = 11,048.96 β 10,000
πΌπ = 1,048.96
16. Example 2:
Find the maturity value and interest
if P10,000 is deposited in a bank at 2%
compounded monthly for 5 years.
Given: P = 10,000 π(12)
= 0.02 t = 5 years m = 12
Find: a. F
b. Ic
17. Example 2:
Given: P = 10,000 π(12)
= 0.02 t = 5 years m = 12
Find: a. F
b. Ic
SOLUTION:
π =
π(12)
π
π =
0.02
12
π = 0.0166 β¦
18. Example 2:
Given: P = 10,000 π(4)
= 0.02 t = 5 years m = 12
Find: a. F
b. P
SOLUTION:
π = ππ‘
π = (12)(5)
π = 60
19. Example 2:
Given: P = 10,000 π(4)
= 0.02 t = 5 years m = 4
Find: a. F
b. Ic
SOLUTION: πΉ = π(1 + π)π
πΉ = 10,000(1 + 0.0167)60
πΉ = P11,050.79
20. Example 2:
Given: P = 10,000 π(4)
= 0.02 t = 5 years m = 4
Find: a. F
b. Ic
SOLUTION:
πΌπ = πΉ β π
πΌπ = 11,050.79 β 10,000
πΌπ = P1,050.79.
21. Example 3:
Cris borrows P50,000 and promises to
pay the principal and interest at 12%
compounded monthly. How much must
he repay after 6 years?
Given: P = P50,000 π(12)
= 0.12 t = 6 years m = 12
Find: F
22. Example 3:
SOLUTION:
Given: P = P50,000 π(12)
= 0.12 t = 6 years m = 12
Find: F
π = π·(π +
π(ππ)
π
)ππ
πΉ = 50,000(1 +
0.12
12
) 12 (6)
πΉ = 50,000(1.01)72 π = π·πππ, πππ. ππ
Thus, Cris must
pay P102,354.97
after 6 years.
24. Example 1:
Find the present value of P50,000
due in 4 years if money is invested at
12% compounded semi-annually.
Given: F = 50,000 π(2)
= 0.12 t = 4 years m = 2
Find: P
26. Example 1:
SOLUTION:
π = ππ‘
π = (4)(2)
π = 8
Given: F = 50,000 π(2)
= 0.12 t = 4 years m = 2
Find: P
27. Example 1:
SOLUTION:
π· =
π
(π + π)π
Given: F = 50,000 π(2)
= 0.12 t = 4 years m = 2
Find: P
π =
50,000
(1 + 0.06)8
π =
50,000
(1.06)8 π· = π·ππ, πππ. ππ
28. Example 2:
What is the present value of P25,000
due in 2 years and 6 months if money
is worth 10% compounded quarterly?
Given: F = 25,000 π(4)
= 0.10 t = 2 Β½ years m = 4
Find: P
30. Example 2:
SOLUTION:
π = ππ‘
π = (4)(2
1
2
)
π = 10
Given: F = 25,000 π(4)
= 0.10 t = 2 Β½ years m = 4
Find: P
31. Example 2:
SOLUTION:
π· =
π
(π + π)π
π =
25,000
(1 + 0.025)10
π =
25,000
(1.025)10 π· = P19,529.96
Given: F = 25,000 π(4)
= 0.10 t = 2 Β½ years m = 4
Find: P
32. SEATWORK: (1 whole)
Solve the following problems on compound interest.
1. Cian lends P45,000 for 3 years at 5%
compounded semi-annually. Find the future
value and interest of this amount.
2.Tenten deposited P10,000 in bank which gives
1% compounded quarterly and let it stay there
for 5 years. Find the maturity value and
interest.
33. SEATWORK: (1 whole)
Solve the following problems on compound interest.
3. How much should you set aside and invest in a
fund earning 9% compounded quarterly if you
want to accumulate P200,000 in 3 years and 3
months?
4. How much should you deposit in a bank paying 2%
compounded quarterly to accumulate an amount
of P80,000 in 5 years and 9 months?
34. 5. Miko has P250,000 to invest at 6% compounded
monthly. Find the maturity value if he invests for
(a) 2 years? (b) 12 years? (c) How much is the
additional interest earned due to the longer
time.
6. Yohan sold his car and invested P300,000 at 8.5%
compounded quarterly. Find the maturity value
if he invests for (a) 3 years? (b) 6 years? (c) How
much is the additional interest earned due to the
longer time.
35. 7. Maryam is planning to invest P150,000. Bank A is
offering 7.5% compounded semi-annually while
Bank B is offering 7% compounded monthly. If she
plans to invest this amount for 5 years, in which
bank should she invest?
8. Yani has a choice to make short term investments
for her excess cash P60,000. She can invest at 6%
compounded quarterly for 6 months or (b) 5%
compounded semi-annually fo 1 year. Which is
larger?
Editor's Notes
The investment scheme in Option B introduces new concepts because interest is compounded twice a year, the conversion period is 6 months, and the frequency of conversion is 2. As the investment runs for 5 years, the total number of conversion periods is 10. The nominal rate is 2% and the rate of interest for each conversion period is 1%.
Frequency of conversion (m) β number of conversion periods in one year
Conversion or interest period β time between successive conversions of interest
Total number of conversion periods n
n = mt = (frequency of conversion) x (time in years)
Nominal rate ( π (π) ) β annual rate of interest
Rate (j) of interest for each conversion period
J = Interest rate in a conversion period
Nominal rate of interest / frequency of conversion
N = total number of conversion periods
Frequency conversion x time
Frequency of conversion (m) β number of conversion periods in one year
Conversion or interest period β time between successive conversions of interest
Total number of conversion periods n
n = mt = (frequency of conversion) x (time in years)
Nominal rate ( π (π) ) β annual rate of interest
Rate (j) of interest for each conversion period
J = Interest rate in a conversion period
Nominal rate of interest / frequency of conversion
N = total number of conversion periods
Frequency conversion x time
J = Interest rate in a conversion period
Nominal rate of interest / frequency of conversion
J = Interest rate in a conversion period
Nominal rate of interest / frequency of conversion
N = total number of conversion periods
Frequency conversion x time
J = Interest rate in a conversion period
Nominal rate of interest / frequency of conversion
N = total number of conversion periods
Frequency conversion x time