Jerry invested $2500 into an account with a 1.5% annual interest rate. Assuming that Jerry did not make any deposits or withdrawals, find the following:
a) Amount in Jerry's account if Jerry invested the amount for 18 years and interest compounded: (I) each second (II) each minute (III) hourly (IV) daily (V) weekly (VI) monthly (VII) quarterly (VIII) semi-annually (IX) yearly (X) continuously
b) number of years of investment in Jerry wanted to have: (I) $5000 in his account and chose interest to be compounded quarterly (II) $5000 in his account and chose interest to be compounded quarterly
Answers:
a)
I) Amount in Jerry's account if interest is compounded each second:
Number of times interest is compounded in a year (n) = 365 x 24 x 60 x 60 = 31,536,000
Annual interest rate (r) = 1.5%
Time in years (t) = 18 years
Amount in account (A) = P(1 + r/n)^(nt) = 2500(1 + 0.015/31,536,000)^(31,536,000 x 18)
A = $3026.22
II) Amount in Jerry's account if interest is compounded each minute:
Number of times interest is compounded in a year (n) = 365 x 24 x 60 = 525,600
Annual interest rate (r) = 1.5%
Time in years (t) = 18 years
Amount in account (A) = P(1 + r/n)^(nt) = 2500(1 + 0.015/525,600)^(525,600 x 18)
A = $3054.24
III) Amount in Jerry's account if interest is compounded hourly:
Number of times interest is compounded in a year (n) = 365 x 24 = 8,760
Annual interest rate (r) = 1.5%
Time in years (t) = 18 years
Amount in account (A) = P(1 + r/n)^(nt) = 2500(1 + 0.015/8,760)^(8,760 x 18)
A = $3061.80
IV) Amount in Jerry's account if interest is compounded daily:
Number of times interest is compounded in a year (n) = 365
Annual interest rate (r) = 1.5%
Time in years (t) = 18 years
Amount in account (A) = P(1 + r/n)^(nt) = 2500(1 + 0.015/365)^(365 x 18)
A = $3071.26
V) Amount in Jerry's account if interest is compounded weekly:
Number of times interest is compounded in a year (n) = 52
Annual interest rate (r) = 1.5%
Time in years (t) = 18 years
Amount in account (A) = P(1 + r/n)^(nt) = 2500(1 + 0.015/52)^(52 x 18)
A = $3081.10
VI) Amount in Jerry's account if interest is compounded monthly:
Number of times interest is compounded in a year (n) = 12
Annual interest rate (r) = 1.5%
Time in years (t) = 18 years
Amount in account (A) = P(1 + r/n)^(nt) = 2500(1 + 0.015/12)^(12 x 18)
A = $3096.99
VII) Amount in Jerry's account if interest is compounded quarterly:
Number of times interest is compounded in a year (n) = 4
Annual interest rate (r) = 1.5%
Time in years (t) = 18 years
Amount in account (A) = P(1 + r/n)^(nt) = 2500(1 + 0.015/4)^(4 x 18)
A = $3120.48
VIII) Amount in Jerry's account if interest is compounded semi-annually:
Number of times interest is compounded in a year (n) = 2
Annual interest rate (r) = 1.5%
Time in years (t) = 18 years
Amount in account (A) = P(1 + r/n)^(nt) = 2500(1 + 0.015/2)^(2 x 18)
A = $3146.43
IX) Amount in Jerry's account if interest is compounded yearly:
Number of times interest is compounded in a year (n) = 1
Annual interest rate (r) = 1.5%
Time in years (t) = 18 years
Amount in account (A) = P(1 + r/n)^(nt) = 2500(1 + 0.015/1)^(1 x 18)
A = $3221.29
X) Amount in Jerry's account if interest is compounded continuously:
Number of times interest is compounded in a year (n) = infinity
Annual interest rate (r) = 1.5%
Time in years (t) = 18 years
Amount in account (A) = Pe^(rt) = 2500e^(0.015 x 18)
A = $3239.33
b)
I) Number of years of investment in Jerry wanted to have $5000 in his account and chose interest to be compounded quarterly:
Let A be the amount that Jerry wants to have in his account, P be the initial amount invested, r be the annual interest rate, and n be the number of times interest is compounded in a year. Then, we can use the formula:
A = P(1 + r/n)^(nt)
Substituting the given values, we get:
5000 = 2500(1 + 0.015/4)^(4t)
Solving for t, we get:
t = (ln(2) / (4 ln(1 + 0.015/4)))
t = 28.75 years
Therefore, Jerry needs to invest his money for about 29 years to have $5000 in his account if he chooses quarterly compounding.
II) Number of years of investment in Jerry wanted to have $10,000 in his account and chose interest to be compounded continuously:
Using the formula for continuous compounding:
A = Pe^(rt)
Substituting the given values, we get:
10000 = 2500e^(0.015t)
Solving for t, we get:
t = (ln(4) / 0.015)
t = 184.07 years
Therefore, Jerry needs to invest his money for about 184 years to have $10,000 in his account if he chooses continuous compounding.