Chapter 1
Money and Interest
1. A man wishes to accumulate P3,722 after 5 years, 8 months and 28 days. How much should
be deposited by the man in a bank if the ordinary simple interest is 15% per annum?
Given:
F = 3,722.00
I = 15% or 0.15
n = 5 years, 8 months, 28 days
Required:
P=?
Solution:
𝑛 = 5 + 8 𝑚𝑜𝑛𝑡ℎ𝑠 𝑥
1 𝑦𝑒𝑎𝑟
12 𝑚𝑜𝑛𝑡ℎ𝑠
1 𝑦𝑒𝑎𝑟
+ 28 𝑑𝑎𝑦𝑠𝑥 360 𝑑𝑎𝑦𝑠
n = 5 + 0.67 + 0.078
𝑃=
𝐹
1+𝑖𝑛
3,722.00
= 1+(0.15)(5.745)
𝑃 = ₱2,000.00
2. A man deposited P2,000 in a bank at the rate of 15% per annum from March 21, 1996 to
October 25, 1997. Find the exact simple interest.
Given:
P = 2,000.00
i = 15% or 0.15
n = March 21, 1996 – October 25, 1997
(Mar 22, 1996 – Mar 21, 2997 = 365 days)
March – 10 (excluding Mar 21)
April – 30
May – 31
June – 30
July – 31
Aug – 31
Sept – 30
Oct -
Required:
P=?
Solution:
𝑛 = 5 + 8 𝑚𝑜𝑛𝑡ℎ𝑠 𝑥
n = 5 + 0.67 + 0.078
1 𝑦𝑒𝑎𝑟
12 𝑚𝑜𝑛𝑡ℎ𝑠
1 𝑦𝑒𝑎𝑟
+ 28 𝑑𝑎𝑦𝑠𝑥 360 𝑑𝑎𝑦𝑠
𝑃=
𝐹
1+𝑖𝑛
3,722.00
= 1+(0.15)(5.745)
𝑃 = ₱2,000.00
3. A bank charges 1.5% per month on a loan. Find the equivalent nominal rate of interest.
Given:
I = 1.5% or 0.015
n = 12
Required:
j=?
Solution:
𝑗
𝑖 = 𝑛1 𝑜𝑟 𝑗 = (𝑖)(𝑛1)
j = 0.015 x 12
j = 0.18 or 18%
4. A financing company charges 1.5% per month on a loan. Find the equivalent effective rate
of interest.
Given:
i = 1.5% or 0.015
n = 12
Required:
𝐼𝑒 = ?
Solution:
𝐼𝑒 = (1 + 𝐼)𝑛1 − 1
𝐼𝑒 = (1 + 0.015)12 − 1
𝐼𝑒 = 0.1964 𝑜𝑟 19.56%
5. A nominal rate of 12% compounded monthly is equal to an effective rate of ___________.
Given:
j = 12%
n = monthly or 12
Required:
𝐼𝑒 = ?
Solution:
𝑗
𝐼𝑒 = (1 + 𝑛)𝑛 − 1
𝐼𝑒 = (1 +
12% 12
12
)
−1
= (1 + 0.01)12 − 1
= 1.1268 − 1
𝐼𝑒 = 0.1268 𝑜𝑟 12.68% compounded monthly
6. Convert 16% compounded semi-annually to equivalent nominal rate which is compounded
daily.
Given:
𝑗
(1 + 365)365 − 1 = (1 +
𝑗
365
(1 + 365)
365
√(1 +
𝑗
)
365
= (1 +
365
=
𝑗
𝑗
= 0.000421793
2
0.16 2
365
√
1+365 = 1.000421793
365
0.16 2
2
)
729
625
) −1
j = 0.153954
j = 15.39%
7. Find the accumulated amount of P1,000 after 5 years when deposited in a bank at the rate
of 16% compounded monthly.
Given:
P = 1,000.00
j = 16%
𝑛1 = 12
n = 5 years or 60 months
Required:
F=?
Solution:
F = P (1 +
𝑗 𝑛
𝑛1
)
0.16 60
)
12
F = 1000 (1 +
F = 1000 (2.2138)
F = ₱2, 213.81
8. How long in years will a certain sum of money to triple its amount when deposited at a rate
of 12% compounded monthly.
Given:
F = 3X
P=X
j = 12% or 0.12
𝑛1 = 𝑚𝑜𝑛𝑡ℎ𝑙𝑦 𝑜𝑟 12
Required:
n=?
Solution:
F = P (1 +
3𝑋
𝑋
=X
(1+
𝑗 𝑛
𝑛1
)
0.12 𝑛
)
12
𝑋
3 = (1 + 0.01)𝑛
Ln 3
ln 1.01
=𝑛
ln 1.01
ln 1.01
n = 110.41 months or 9.2 years
9. How much should be deposited in a bank at a rate of 12% compounded continuously for 5
years if its accumulated amount is P9,110.60?
Given:
j = 12% or 0.12
n = 5 years
F = ₱ 9,110.60
Required:
P=?
Solution:
F = P𝑒 𝑗𝑛
𝐹
P = 𝑒 𝑗𝑛 =
₱ 9,110.60
𝑒 (0.12)(5)
= ₱ 5,000.00
10. An effective rate of interest which is 12.75% is equivalent to what percent if compounded
continuously.
Given:
𝑖𝑒 = 12.75% = 0.1275
Required:
j=?
Solution:
𝑖𝑒 = 𝑒 𝑗 − 1
𝑒 𝑗 = 𝑖𝑒 + 1
= 0.1275 + 1 = 1.1275
j = 0.12 or 12%
11. How much is expected to be received by a man that makes a loan of P851.06 which is
payable at once if the bank gave him a discount of 6%?
Required: 𝑅𝑝 = ?
Given:
F = ₱851.06
d = 6% or 0.06
Solution:
d=
𝐷
𝐹
=
0.06 =
𝐹−𝐷
𝐹
851.06−𝑃
851.06
P = ₱800.00
12. Find the cash price of a generator which was bought in installment basis that requires a
down payment of P50,000 and payment of P30,000 after one year, P40,000 after two years,
and a final payment of P76,374.34 after four years at a rate of 15% per annum.
Given:
i = 15% or 0.15
Using 4 as focal date:
Solution:
P(1 + 𝑖)4 = P(1 + 𝑖)3 + P(1 + 𝑖)2 + 𝑃
X – 50,000(1.15)4 = 30,000(1.15)3 + 40,000(1.15)2 + 74,374.38
1.74900625X – 87,450,3125 = 45,626.25 + 52,900 + 74,374.38
1.74900625X = 174,900.59 + 87,450,3125
262,350.4025
1.74900625X = 1.74900625X
X = ₱150,000.00
13. A man made a loan of P100,000 at a rate of 15% per annum and promise to pay it according
to the following manner, P30,000 at the end of first year, unknown payment at the end of
second year and a final payment of P76,374.38 at the end of fourth year. Find the unknown
payment made by the man.
Given:
i = 15% or 0.15
Using 4 as focal date
Solution:
P(1 + 𝑖)4 = P(1 + 𝑖)3 + P(1 + 𝑖)2 + 𝑃
100,000(1.15)4 = 30,000(1.15)3 + X(1.15)2 + 76,374.38
174,900.625 = 45,625.25 + 1.3225X + 76,374.38
1.3225X = 174,900.625 - 45,625.25 - 76,374.38
1.3225X =
52,899.62
1.3225
X = ₱40,000.00
14. Find the present worth of the following payments, P5,000 after 1 year, P4,000 after 2 years,
P8,000 after 4 years at the rate of 12% per annum.
Given:
i = 12 or 0.12
Using 4 focal date
Solution:
P(1 + 𝑖)4 = P(1 + 𝑖)3 + P(1 + 𝑖)2 + 𝑃
X(1.12)4 = 5,000(1.12)3 + 4,000(1.12)2 + 8000
1.57351936X = 7,024.64 + 5,017.6 + 8000
20,042.24
1.57351936X = 1.57351936
X = ₱12,737.21
15. Find the amount of the following payments at the end of 5th year. P3,000 at the end of 1st
year, P4,500 at the end of 2nd year, and P6,000 at the end of 4th year if money worth 12%
per annum.
Given:
i = 12% or 0.12
Using 5 focal date
Solution:
F = P(1 + 𝑖)4 + P(1 + 𝑖)3 + P(1 + i)
F = 3,000(1.12)4 + 4,500(1.12)3 + 6,000(1.12)
F = 4,720.55808 + 6,322.176 + 6,720
F = ₱17,762.73