Chapter 5.7, Problem 55AYU

Comparing Bank Accounts Two bank accounts are opened at the same time. The first has a principal of $\$1000$ in an account earning $5\%$ compounded monthly. The second has a principal of $\$2000$ in an account earning $4\%$ interest compounded annually. Determine the number of years, to the nearest tenth, at which the account balances will be equal.

To determine

To calculate: The number of years to the nearest tenth at which the account balances of two accounts will be equal, where the first has a principal of $\$1000$ earning $5\%$ compounded monthly and the second one has a principal of $\$2000$ earning $4\%$ compounded annually.

Answer to Problem 55AYU

Solution:

The account balances of two accounts will be equal at $64.9$ years, where the first has a principal of $\$1000$ earning $5\%$ compounded monthly and the second one has a principal of $\$2000$ earning $4\%$ compounded annually.

Explanation of Solution

Given information:

Invested amount in the first account $P_1=1000$

Compounded interest rate in the first account per annum $r_1=0.05$

Number of times the interest compounded in the first account per annum $n_1=12$

Invested amount in the second account $P_2=2000$

Compounded interest rate in the first account per annum $r_2=0.04$

Number of times the interest compounded in the first account per annum $n_2=1$

Formula used:

The compound interest formula is $A=P\cdot {\left(1+\frac{r}{n}\right)}^{nt}$

Calculation:

As per the account, balances in the two accounts will be equal.

Use the compound interest formula with $P_1=1000$, $r_1=0.05$, $n_1=12$ and $P_2=2000$, $r_2=0.04$, $n_2=1$ to find the number of years.

$P_1{\left(1+\frac{r_1}{n_1}\right)}^{n_{1}t}=P_2{\left(1+\frac{r_2}{n_2}\right)}^{n_{2}t}$

$\Rightarrow 1000{\left(1+\frac{0.05}{12}\right)}^{12t}=2000{\left(1+\frac{0.04}{1}\right)}^t$

$\Rightarrow {(1.00416667)}^{12t}=2{(1.004)}^t$

$\Rightarrow 12t\ln (1.00416667)=\ln 2+t\ln (1.004)$

$\Rightarrow 12t\times 0.004158010481=0.6931471806+t\left(0.3922071315\right)$

$\Rightarrow 0.04989612577t-\left(0.3922071315\right)t=0.6931471806$

$\Rightarrow t=64.9$.

Hence, the account balances in both the accounts will be the same at $64.9$ years.