Question

You deposit 2000 in account A, which pays 2.25% annual interest compounded monthly. You deposit another 2000 in account b, which pays 3% annual interest compounded monthly. when is the sum of the balance in both accounts at least 5000

Answer 1

To model this situation, we are going to use the compound interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where
$A$ is the final amount after $t$ years 
$P$ is the initial deposit 
$r$ is the interest rate in decimal form 
$n$ is the number of times the interest is compounded per year
$t$ is the time in years 

For account A: 
We know for our problem that $P=2000$ and $r= \frac{2.25}{100} =0.0225$. Since the interest is compounded monthly, it is compounded 12 times per year; therefore, $n=12$. Lets replace those values in our formula:
$A=2000(1+ \frac{0.0225}{12} )^{12t}$

For account B:
$P=2000$, $r= \frac{3}{100} =0.03$, $n=12$. Lest replace those values in our formula:
$A=2000(1+ \frac{0.03}{12} )^{12t}$

Since we want to find the time, $t$, when  the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
$2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000$

Now that we have our equation, we just need to solve for $t$:
$2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000$
$(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000} {2000}$
$(1.001875)^{12t}+(1.0025 )^{12t}= \frac{5} {2}$
$ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})$
$12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})$
$t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})$
$t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )}$
$17.47$

We can conclude that after 17.47 years the sum of the balance in both accounts will be at least 5000.