Answer to Question #161170 in Financial Math for Bdll

Question #161170

Mario deposits $25 at the end of each month for 4 years into an account that pays 9.6%/a compounded monthly. He then makes no further deposits and no withdrawals. Determine the balance 10 years after his last deposit.


1
Expert's answer
2021-02-24T06:42:40-0500

This problem is related to annuity problem.


The formula for compound interest, including principle sum is:


A=P×[(1+rk)Nk1]rkA=\frac{P\times[(1+\frac{r}{k})^{Nk}-1]}{\frac{r}{k}}


where,

A = the balance in the account after N years.

P = the regular deposit (the amount you deposit each year, each month, etc.)

r = the annual interest rate in decimal form =9.6100=0.096\frac{9.6}{100}=0.096


k = the number of compounding period in one year

N = the number of years that interest is compounded


so, A=25×[(1+0.09612)4×121]0.09612A=\frac{25\times[(1+\frac{0.096}{12})^{4\times12}-1]} {\frac{0.096}{12}}


A=25×[(1.008)481]0.008=25×[1.465904041]0.008=1,455.95012A=\frac{25\times[(1.008)^{48}-1]}{0.008}=\frac{25\times[1.46590404-1]}{0.008}=1,455.95012


Thus, the amount after 4 years = $1,455.95012

but the there is no deposit and withdrawal after this and the amount 10 years after the last deposit is :


A=P×(1+R100)TA=P\times(1+\frac{R}{100})^T


where,

A = total amount after the compounded interest on the principal amount

P = Principal amount

R = rate of interest compounded annually =9.612=\frac{9.6}{12} %

T = time in years

A=1455.95012×(1+9.612×100)10×12A=1455.95012\times(1+\frac{9.6}{12\times100})^{10\times12}


A=1455.95012×(1.008)120=1455.95012×2.60173976A=1455.95012\times(1.008)^{120}=1455.95012\times2.60173976


A = $3,788.00332


Thus, the amount 10 years after the last deposit is : A = $3,788.00332



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