EXERCISE 1.1
1. How Much will be the future worth of money after 12 months if the sum of ₱35,000 is invested
today at a simple interest rate of 3% per month?
Solution:
P = 35,000
r = 3%
t = 12
F=P(1+rt)
= 35,000 ( 1 + 0.03 * 12 )
= 35,000 ( 1 + 0.36 )
= 35,000 (1.36)
F = 47,600
3. How long will it take the money to triple itself if invested at 9.5% compounded semi-annually?
Solution:
F = 3P
P
F=P(1+i)n
i=j/m
3P = P ( 1 + 0.095 / 2 ) n
3 = ( 1 + 0.0475 ) n
ln (3) = n ln (1.0475)
n = ln (3) / ln (1.0475)
n = 23.67373751
j = 0.095
m=2
Number of years is
t=n/m
t = 23.67373751 / 2
t = 11. 84 years
5. What is the effective rate corresponding to 18% compounded daily using 360 days in one year?
Solution:
j = 0.18
m = 365
ER = ( 1 + j / m ) m – 1
= ( 1 + 0.18 / 365 ) 365 – 1
= ( 1.000493151 ) 365 – 1
= 1.197164245 – 1
= 0.1972 or 19.72%
7. A certain bank advertises 9.25% account that yields 9.58% annually. Find how often the interest is
compounded.
Solution:
ER = 0.0958
j = 0.0925
ER = ( 1 + j / m ) m – 1
if m = 4
0.0958 = ( 1 + 0.0925 / m ) m – 1
1.0958 = ( 1 + 0.0925 / 4 ) 4
1.0958 = ( 1 + 0.0925 / m ) m
1.0958 = (1 + 0.023125)4
Trial and Error
if m = 2
1.0958 = 1.002139063 (X)
1.0958 = (1.023125)4
1.0958 = 1. 095758346 or 1.0958
therefore, m = 4.