1 4 compound-interest

1.4 Compound Interest1.4 Compound Interest

Compound Interest


• Compound interest – a type of interest which results from the periodic addition of simple interest to the principal.

• This type of interest often applies to savings accounts, loans, and credit cards.

• Compound amount – the amount at the end of the term (after several compounding).

• It is the sum of the original principal and its compound interest.


• Formula for the compound amount F:

€

F = P(1+ i)n

€

P - original principal

j - rate of interest per year

m - frequency of conversion

i - interest rate per priod; i = jm

t - length of term in years

n - total number of conversion periods; n = tm


Example. Find the compound amount at the end of 12 periods if the principal is Php25,000 and the interest per period is 10%.

€

F = P(1+ i)n

€

=25,000(1+ .10)12

€

F = Php78,460.71


Example. What is the maturity value of a 75,000-peso, three-year investment earning 5% compounded monthly?

€

F = P(1+ i)n

€

=75,000 1+ .0512( )

12

€

F = Php87,110.42

€

Do this ifi is not exact.


Example. Find the compound amount after 5 years and 9 months if the principal is Php150,500 and the rate is 7% compounded annually.

€

F = P(1+ i)n

€

=150,000(1.07)5 912⋅1

€

=Php221,333.92


More Formulas:

€

j = mF

P

⎛

⎝ ⎜

⎞

⎠ ⎟

1

n−1

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥€

P = F(1+ i)−n

€

n =log F

P( )

log(1+ i)


1. Given P = Php25,200, i = 3%, n = 16, find F.

€

F = P(1+ i)n

€

=25,200(1.03)16

€

=Php40,438.60


3. Given P = Php1.8M, j = 11%, t = 7.5 years, m = 2, find F.

€

F = P(1+ i)n

€

=1,800,000(1.055)15

€

=Php4,018,457.69


5. Given F = Php46,000, j = 12%, t = 6.25 years, m = 12, find P.

€

P = F(1+ i)−n

€

=46,000(1.01)−75

€

=Php21,809.96


7. Given F = Php56,471.27, P = Php25,000, t = 8 years 3 months, m = 4, find j.

€

j = mF

P

⎛

⎝ ⎜

⎞

⎠ ⎟

1

n−1

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

€

=456,471.27

25,000

⎛

⎝ ⎜

⎞

⎠ ⎟

1

33

−1

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

€

=10%


11. Given F = Php34,500, P = Php30,000, j = 15%, m = 12, find n.

€

n =log F

P( )

log(1+ i)

€

=log 34,500

30,000( )

log(1.0125)

€

=11.25 periods


13. Given F = Php72,157.25, P = Php48,200, j = 9%, m = 12, find n.

€

n =log F

P( )

log(1+ i)

€

=log 72,157.25

48,200( )

log(1.0075)

€

=54 periods


23. Find the compound amount due in 6 years and 2 months if Php350,000 is invested at 12% compounded monthly.

€

F = P(1+ i)n

€

=350,000(1.01)74

€

=Php730,886.10


27. How much must Ella deposit in a bank that pays 11% compounded quarterly so that she will have Php400,000 after 4 years?

€

P = F(1+ i)−n

€

=400,000(1.0275)−16

€

=Php259,149.70


28. A personal computer was bought on installments – Php5,000 downpayment and the balance of Php22,000 in 2 years. What is the cash price if the interest rate is 20% compounded quarterly?

€

P = F(1+ i)−n

€

=22,000(1.05)−8

€

=Php14,890.47

€

CP = DP + P

€

=5,000 +14,890.47

€

=Php19,890.47


30. On April 15, 2011, Justin borrowed Php1.4M. He agreed to pay the principal and the interest at 8% compounded semi-annually on July 15, 2016. How much will he pay then?

€

F = P(1+ i)n

€

=1,400,000(1.04)10 12

€

=Php2,113,382.46


33. At what rate converted quarterly will Php30,000 become Php40,000 in 7 years?

€

j = mF

P

⎛

⎝ ⎜

⎞

⎠ ⎟

1

n−1

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

€

=440,000

30,000

⎛

⎝ ⎜

⎞

⎠ ⎟

1

28

−1

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

€

=4.13%


37. If Php80,000 is invested at the rate of 6 ½% compounded annually, when will it earn interest of Php15,000?

€

t =log F

P( )

m log(1+ i)

€

=log 95,000

80,000( )

log(1.065)

€

=2.73 years


• Equation of values – a mathematical statement which says that the dated values of two sets of amounts are equal when brought to a particular point in time (the comparison date).

• In the context of borrowing, the equation of values says that

obligations = payments• These sums are obtained by either accumulating

or discounting the debts incurred or the payments made toward the comparison date.


45. What single payment at the end of 6 years would replace the following debts?

a) Php29,000 due in 1 year without interestb) Php690,000 due in 8 years at 14% compounded

quarterly Money is worth 8.5% effective.

€

29,000

€

690,000(1.035)32

€

x

Obligation(s)

Payment(s) 1 6 8


€

29,000(1.085)5

€

+ 690,000(1.035)32(1.085)−2

€

=x

€

x = Php1,805,909.97

€

29,000

€

690,000(1.035)32

€

x

Obligation(s)

Payment(s) 1 6 8


47. For an amount borrowed from a credit cooperative, Janice needs to pay Php100,000 in 5 years. After 2 ½ years , she made a Php50,000 payment. If money is worth 8% compounded semi-annually, how much would she have to pay on the 5th year to fully settle the loan?

€

100,000

€

50,000

€

x

Obligation(s)

Payment(s) 2.5 5


€

100,000

€

=50,000(1.04)5

€

+ x

€

x = Php39,167.35

€

100,000

€

50,000

€

x

Obligation(s)

Payment(s) 2.5 5


49. If money is worth 8% effective, what single payment in 5 years will repay the following two debts:

a) Php125,000 due at onceb) Php500,000 due in 8 years

€

500,000

€

125,000

€

x

Obligation(s)

Payment(s) 1 5 8


€

125,000(1.08)5

€

+ 500,000(1.08)−3

€

=x

€

x = Php580,582.13

€

500,000

€

125,000

€

x

Obligation(s)

Payment(s) 1 5 8


51. As payments for debts of Php300,000 due at the end of 4 years and Php485,000 at the end of 8 years, Jane agrees to pay Php50,000 at once and Php250,000 at the end of 5 years. She will make a third and final payment at the end of 10 years. How much would it be if money is worth 14% compounded semi-annually.

€

485,000

€

300,000

€

50,000

Obligation(s)

Payment(s) 1 4 5 8 10

€

250,000

€

x


€

300,000(1.07)12

€

+ 485,000(1.07)4

€

=

€

x = Php626,121.48

€

50,000(1.07)20

€

+ 250,000(1.07)10

€

+ x

€

485,000

€

300,000

€

50,000

Obligation(s)

Payment(s) 1 4 5 8 10

€

250,000

€

x

1 4 compound-interest

Technology

Transcript of 1 4 compound-interest