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- Find the accumulated amount of the ordinary annuity paying an amortization of 1000P per month at a
rate of 12% compounded monthly for 5 years
GIVEN:
A = 1000P n = 5 years or 60 months
j = 12% or 0.12 i =
- 12
12
i= 0.
n 1
= monthly or 12
FIND: F
FORMULA:
F = A (
( 1 +𝑖)𝑛− 1
𝑖
SOLUTION:
F = 1000P (
( 1 + 0. 01
)
60
− 1
- 01
F = 81670P
FINAL ANSWER: 81,670P will be the accumulated amount of the annuity for five years.
- What present sum is equivalent to a series of 1000P annual end-of-year payments, if a total of 20
payments are made and interest is 12%?
GIVEN:
A = 1000 P
i = 1 2 % = 0.1 2
n = 20
FIND: P
FORMULA:
P = A (
1 − ( 1 + i)
−𝑛
i
SOLUTION:
P = 1000P (
1 −( 1 + 0. 12 )
− 20
- 12
) → 1000P (
1 −( 1. 12 )
− 20
- 12
) → 1000P
P = 7,469.44P
FINAL ANSWER: 7,469.44P is the present sum equivalent to a series of 1000P annual end-of-year
payments.
- A man made ten annual-end-of year purchases of 1000P common stock. At the end of 10
th
year, he sold
all the stock for 12000P. What interest rate did he obtain on his investment?
GIVEN:
F = 12,
A = 1,
n = 10
FIND: i
FORMULA:
F = A (
( 1 +𝑖)
𝑛
− 1
𝑖
SOLUTION:
F = A (
( 1 +𝑖)
𝑛
− 1
𝑖
( 1 +𝑖)
10
− 1
𝑖
i = 0.039 ≈ 4%
FINAL ANSWER: He obtained 4% of interest rate from his investment.
SOLUTION:
P = P150000 + P10000 (
1 −( 1 + 0. 01 )
−( 10 ( 12 )+ 1 )
- 01
−( 5
( 12
) − 1 )
P = 539,170.752P = 539 , 171 P
FINAL ANSWER: The cash price of the condominium unit is 539,171P.
- The owner of the quarry signs a contract to sell his stone on the following basis. The purchaser is to
remove the stone from the certain portion of the pit according to a fixed schedule of volume, price, and
time. The contract is to run 18 years as follows. Eight years excavating a total of 20,000 m per year at 10P
per meter, the remaining ten years, excavating a total of 50,000 m per year at 15P per meter. Based on
equal year end payments during each period by the purchaser, what is the present worth of the pit to the
owner on the basis of 15% interest?
GIVEN:
A
1
= 10 P/meter for 20,000 m/year i = 15% or 0.
A
2
= 15P/meter for 50,000m/year
FIND: A
1
, A
2
, and P
FORMULA:
P = A (
1 −( 1 +𝐼)
−(𝑛− 1 )
𝑖
SOLUTION:
Solving for A 1
and A 2
A
1
10 𝑃
𝑚𝑒𝑡𝑒𝑟
20 , 000 𝑚
𝑦𝑒𝑎𝑟
) A
1
= 200,000P/year
A
2
15 𝑃
𝑚𝑒𝑡𝑒𝑟
50 , 000 𝑚
𝑦𝑒𝑎𝑟
) A
2
= 750,000P/year
Solving for P,
P = A
2
1 −
( 1 +𝐼
)
−(𝑛− 1 )
𝑖
+ 1 } – A
2
1 −
( 1 +𝐼
)
−(𝑛− 1 )
𝑖
P = 750,000{
1 −( 1 + 0. 15 )
−( 10 − 1 )
- 15
1 −( 1 + 0. 15 )
−( 8 − 1 )
- 15
P = 2127948P
FINAL ANSWER: 2,127,948P will be the present worth of the pit to the owner based on 15% interest.
- A wealthy man donated a certain amount of money to provide scholarship grants to deserving students.
The fund will grant 10,000P per year for the first 10 years and 20,000P per year on the years thereafter.
The scholarship grants started one year after the money was donated. How much was donated by the
man if the fund earns 12% interest.
GIVEN:
A
1
= 10,000 A
2
n 1
= 10 years
FIND: P
FORMULA:
1 −( 1 +𝑖)
−𝑛
𝑖
) ; P =
𝐴
𝑖
SOLUTION:
1
1 −( 1 +𝑖)
−𝑛
𝑖
1 −( 1 + 0. 12 )
− 10
- 12
) = 56,502.23P
2
𝐴
𝑖
20 , 000
- 12
= 166,666.67P
3
𝑃
2
( 1 +𝑖)
𝑛
166 , 666. 67
( 1 + 0. 12 )
10
P
Total
= P
1
+ P
3
= 56,502.23P + 53,662.21P = 110,164.
FINAL ANSWER: 110,164.44P was donated by the man.
- What amount of money deposited 40 years ago at 12% interest would now provide a perpetual
payment of 10,000P per annum?
GIVEN:
i = 1 2 % = 0.1 2
A = 10000P/ annum
n = 40 years
FIND: P
SOLUTION:
1 +
- 12 [( 10 )( 12 )]+ 1
12
− 1
- 12
12
= 11 , 616 , 953. 82 ≈ 11 , 616 , 954 P
FINAL ANSWER: The accumulated amount of the rentals is 11 , 616 ,954P.
- The amount of the perspective investor pay for a bond if he desires an 8% return on his investment
and the bond will return 1000P per year for 20 years and 20,000P after 20 years is
GIVEN:
C = F = 20, 000 P
I = 1000P
i = 8 % = 0. 08
n = 20
FIND: P
FORMULA:
P = I (
1 − ( 1 + i)
−𝑛
i
−𝑛
SOLUTION:
P = 1000P (
1 −
( 1 + 0. 08
)
− 20
- 08
− 20
P = 1000P( 9. 8181 ) + 4 , 290. 9641 𝑃
P = 14 , 10 9. 11 P P = 14 , 10 9P
FINAL ANSWER: The amount of the perspective investor pay for a bond is 14,109P.
- A machine costs 50,000P. Find the capitalized cost if the annual maintenance and operational cost is
5000P and money worth 15% per annum.
GIVEN:
FC = 50,000P
MC = 5,000P
i = 15% → 0.
CR = 0
RC = 0
SV = 0
FIND: CC
FORMULA:
CC = FC +
𝑀𝐶
𝑖
𝐶𝑅
( 1 +𝑖)
𝑘
− 1
𝑅𝐶
( 1 +𝑖)
𝐿
− 1
SOLUTION:
CC = FC +
𝑀𝐶
𝑖
𝐶𝑅
( 1 +𝑖)
𝑘
− 1
𝑅𝐶
( 1 +𝑖)
𝐿
− 1
CC = 50,000 +
5 , 000
- 15
CC = 83,333.33P
FINAL ANSWER: The capitalized cost is 83,333.33P.
- A machine cost 50,000P. Find the capitalized cost if the annual maintenance cost is 5000P and cost
of repair is 4000P every 4 years and money worth 12% per annum
GIVEN:
FC = 50,000P i= 12% or 0.
MC= 5000P k = 4 years
CR= 4000P
FIND: Capitalized Cost (CC)
CC = 10,000,000 +
60 , 000
- 15
200 , 000
( 1 + 0. 15 )
5
− 1
9 , 650 , 000
( 1 + 0. 15 )
25
− 1
CC = 10,900,082.89 P or 10.9M P
FINAL ANSWER: 10.9M P is the capitalized cost if money is worth 15% per annum.
- A salesman earns 1000P on the 1st month, 1500P on the 2nd month, 2000P on the 3rd month and
so on. Find the accumulated amount of his income at the 10th month if money worth 12% compounded
monthly.
GIVEN:
st
month = 1000P G = 500P
nd
month = 1500P j = 12% or 0.12, compounded monthly
rd
month = 2000P n = 10
A = 1000P
FIND: F
FORMULA:
F = F
a
+ F
g
SOLUTION:
i =
𝑗
𝑛
1
- 12
12
F = F
a
+ F
g
= 1000[
( 1 + 0. 01 )
10
− 1
- 01
] + (
500
- 01
) [
( 1 + 0. 01 )
10
− 1
- 01
− 10 ]
F = 33,572.84P or 33,573P
FINAL ANSWER: The accumulated amount of the salesman’s income at the 10
th
month is 33 ,573P.
- A man wishes to accumulate a total of 500,000P at the age of 30. On his 20th birthday, he deposited
a certain amount of money at a rate of 12% per annum. If he increases his deposit by 10% each year until
the 30th birthday, how much should his initial deposit be?
GIVEN:
F = 500,
i = 0.
r = 0.
FIND: A
FORMULA:
W =
1 +𝑟
1 +𝑖
P = F (1+i)
SOLUTION:
W =
1 + 0. 1
1 + 0. 12
P = F ( 1 + 0. 12 )
− 11
− 11
P = 143,738.
A = 143,738.0521( 1 + 0. 12 ){
1 − 0. 982143
1 − 0. 982143
11
A = 15,988.90P
FINAL ANSWER: The initial deposit is 15,988.90P.
SOLUTION:
F = 1000P (
( 1 + 0. 08 )
15
- 08
4
3
2
1
0
F = 21 , 285. 5 1P F = 21 , 286 P
FINAL ANSWER: At the end of 15 years, he will have 21 , 286 P in his account.
- Twenty-five thousand pesos is deposited in a savings account that pays 5% interest, compounded
semi-annually. Equal annual withdrawals are to be made from the account, beginning one year from now
and continuing forever. Find the maximum amount of the equal annual withdrawal.
GIVEN:
P = 25,
j = 5% = 0.05 n 1
= 2 (semi-anually)
FIND: A
FORMULA:
𝐴
𝑖
SOLUTION:
A = P(i e
= 25 , 000 [( 1 +
- 05
2
2
− 1 ]
= 1,265.63P
FINAL ANSWER: The maximum amount of the equal annual withdrawal is 1,265.63P.
- What amount of money deposited 50 years ago at 8% interest would now provide a perpetual
payment of 10000P per year?
GIVEN:
n = 50
i = 8% = 0.
F = 10000P
FIND: P
FORMULA:
SOLUTION: Using 50 as a focal date
10 , 000
- 08
= P(1.08)
50
P = 2665.15P P = 2665P
FINAL ANSWER: The amount of money deposited 50 years ago at 8% interest is 2665P.
i = 8% per year
P P(1.08)
50
A
A = 10,000P per year
A
