Financial Mathematics, Simple interest question. Help. Laurie deposits $\$60,000$ 
in a bank at $5\%$ interest per annum. 
Andrew deposits $\$40,000$ in bank at $8\%$ per annum.
How long wil it take, by simple interest, for Andrew to have more money than Laurie?
What I have done so far:
I calculated the simple interest earned by both in $5$ years:
Laurie:
$\$60,000 \cdot 5 \cdot 5/ 100 =\$15000$
Andrew:
$\$40,000 \cdot 8 \cdot 5 / 100 = \$16000$
 A: Hint: The simple interest per year for Laruie is $0.05\cdot 60000=3000$, while the simple interest for Andrew is $0.08\cdot 40000=3200$. So the total balance of Andrew has to be greater than Laurie's, i.e. solve
$$40000+3200x>60000+3000x$$
A: Simple Interest is calculated using the formula: $I=PRT$
where $P$ is the starting princple, $R$ is the interest rate in decimal form, and $T$ is time in years.
Thus, the final balance after after adding interest is: 
$P + PRT = P(1 + RT) $
Let $P_{A} $  represent Andrew's principle. Let $P_{L}$ represent Laurie's principle. We can start by finding when their balances will be equal. Essentially, any time after that Andrew will have more money since he has a higher interest. So we set their equations equal to each other and solve for $T$:
$P_{A}(1 + .08T) = P_{L}(1 + .05T) $
$40,000(1 + .08T) = 60,000(1 + .05T) $
$1 + .08T = 1.5(1 + .05T) $
$1 + .08T = 1.5 + .075T $
$ .005T = .5 $
$ T = 100 $
Thus, after 100 years Andrew will have more money. (However, banks typically use compound interest).
A: Let $L(n)$ and $A(n)$ be the amount of money Laurie and Andrew have after $n$ years, respectively. Then
$$\begin{align*}
L(n) &= 60000\left(1+\frac1{20}\right)^n=60000\left(\frac{21}{20}\right)^n\\
A(n) &= 40000\left(1+\frac2{25}\right)^n=40000\left(\frac{27}{25}\right)^n\\
\end{align*}$$
We want to find $$n^* = \operatorname{argmin}_n\{A(n) > L(n) \}.$$
Since $L$ and $A$ are positive, increasing functions, this is equivalent to $\frac{A(n)}{L(n)}>1$. We have
$$ \frac{A(n)}{L(n)} = \frac{40000}{60000}\left(\frac{27}{25}\right)^n\left(\frac{21}{20}\right)^{-n}= \frac23\left(\frac{36}{35}\right)^n$$
Since $\frac{A(n)}{L(n)}>1$ if and only if $\log\left(\frac{A(n)}{L(n)}\right) >0$,
$$ \log\left(\frac{A(n)}{L(n)}\right) = \log\frac23 + n\log\frac{36}{35}>0\iff n>\frac{\log\frac23}{\log\frac{36}{35}}\approx 14.393.$$
So Andrew will have more money after $$\lceil 14.393\rceil = 15$$ years.
