Suppose you deposit $1,500 in a savings account that pays interest at an annual rate of 5%. If no money is added or withdrawn from the account, answer the following questions.
a. How much will be in the account after 3 years?
b. How much will be in the account after 18 years?
c. How many years will it take for the account to contain $2,000?
d. How many years will it take for the account to contain $2,500?
To calculate the amount in the account after a certain number of years, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future amount in the account
P = the principal amount (the initial deposit)
r = the annual interest rate (expressed as a decimal)
n = the number of times interest is compounded per year
t = the number of years
In this case, we have:
P = $1,500
r = 5% = 0.05 (as a decimal)
n = 1 (interest is compounded annually)
a. After 3 years:
A = $1,500(1 + 0.05/1)^(1*3)
A = $1,500(1.05)^3
A ≈ $1,500(1.157625)
A ≈ $1,736.44
So, there will be approximately $1,736.44 in the account after 3 years.
b. After 18 years:
A = $1,500(1 + 0.05/1)^(1*18)
A = $1,500(1.05)^18
A ≈ $1,500(1.978049)
A ≈ $2,967.07
So, there will be approximately $2,967.07 in the account after 18 years.
c. To find the number of years it will take for the account to contain $2,000, we can rearrange the formula and solve for t:
A = P(1 + r/n)^(nt)
2,000 = $1,500(1 + 0.05/1)^(1*t)
2,000/1,500 = 1.05^t
1.333 = 1.05^t
To solve for t, we can take the logarithm of both sides:
log(1.333) = log(1.05^t)
t*log(1.05) = log(1.333)
t = log(1.333) / log(1.05)
t ≈ 3.945
So, it will take approximately 3.945 years for the account to contain $2,000.
d. Similarly, to find the number of years it will take for the account to contain $2,500:
A = P(1 + r/n)^(nt)
2,500 = $1,500(1 + 0.05/1)^(1*t)
2,500/1,500 = 1.05^t
1.667 = 1.05^t
Taking the logarithm of both sides:
log(1.667) = log(1.05^t)
t*log(1.05) = log(1.667)
t = log(1.667) / log(1.05)
t ≈ 4.918
So, it will take approximately 4.918 years for the account to contain $2,500.