1 4 compound-interest
TRANSCRIPT
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1.4 Compound Interest1.4 Compound Interest
Compound Interest
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1.4 Compound Interest1.4 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.
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1.4 Compound Interest1.4 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
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
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.
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
More Formulas:
€
j = mF
P
⎛
⎝ ⎜
⎞
⎠ ⎟
1
n−1
⎡
⎣
⎢ ⎢
⎤
⎦
⎥ ⎥€
P = F(1+ i)−n
€
n =log F
P( )
log(1+ i)
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1.4 Compound Interest1.4 Compound Interest
1. Given P = Php25,200, i = 3%, n = 16, find F.
€
F = P(1+ i)n
€
=25,200(1.03)16
€
=Php40,438.60
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
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%
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
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%
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
• 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.
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
€
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
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
€
100,000
€
=50,000(1.04)5
€
+ x
€
x = Php39,167.35
€
100,000
€
50,000
€
x
Obligation(s)
Payment(s) 2.5 5
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
€
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
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1.4 Compound Interest1.4 Compound Interest
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
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1.4 Compound Interest1.4 Compound Interest
€
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