Question

Lily invested $60,000 in an account paying an interest rate of 6 1 2 6 2 1 ​ % compounded continuously. Lydia invested $60,000 in an account paying an interest rate of 6 1 8 6 8 1 ​ % compounded monthly. To the nearest hundredth of a year, how much longer would it take for Lydia's money to triple than for Lily's money to triple?

Answer 1

Using interest equations, it is found that it will take 1.08 year more for Lydia's money to triple than for Lily's.

What is compound interest?

The amount of money earned, in compound interest, after t years, is given by:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

In which:

• A(t) is the amount of money after t years.

• P is the principal(the initial sum of money).

• r is the interest rate(as a decimal value).

• n is the number of times that interest is compounded per year.

• t is the time in years for which the money is invested or borrowed.

What is the amount in continuous compounding?

It is given by:

$A(t) = Pe^{rt}$

For Lily, the parameters are as follows: r = 0.065, with continuous compounding. The time to triple is t for which A(t) = 3P, hence:

$A(t) = Pe^{rt}$

$3P = Pe^{0.065t}$

$e^{0.065t} = 3$

$\ln{e^{0.065t}} = \ln{3}$

$0.065t = \ln{3}$

$t = \frac{\ln{3}}{0.065}$

$t = 16.90$.

For Lydia, using compound interest, the parameters are: r = 0.06125, n = 12, hence:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

$3P = P\left(1 + \frac{0.06125}{12}\right)^{12t}$

$(1.00510416667)^{12t} = 3$

$\log{(1.00510416667)^{12t}} = \log{3}$

$12t\log{(1.00510416667)} = \log{3}$

$t = \frac{\log{3}}{12\log{(1.00510416667)}}$

$t = 17.98$

17.98 - 16.9 = 1.08, hence, it will take 1.08 year more for Lydia's money to triple than for Lily's.

More can be learned about interest equations at https://brainly.com/question/25781328