money to pay the 1st semester bill on time and still have the remainder for the second semester? John has savings account earning 3% simple interest. He has to pay 1300 for first-semester tuition by september 1st and 1300 for second semster tuition by january 1st. How much does he need to earn by september 1st to have enough money to pay the 1st semester bill on time and still have the remainder for the second semester?
My attempt:
P = unknown
r = 3 percent
t = 4/12
I = 1300+1300 = 2600
But this doesnt make sense
 A: The setup here is that he's going to have a certain amount of money on 9/1. Whatever he doesn't use to pay fall semester tuition, he's going to throw into his savings account, let it make him some extra money, and then withdraw three months later to pay his spring semester tuition.
Let's reason. He'd better have at least $1300 to pay fall semester tuition, but he doesn't need the full \$1300 for spring semester because he'll earn some interest. So let's figure out how much he needs to put in his account after he's paid the \$1300 in order to have \$1300 three months later.
Since the interest is simple, he'll be earning 3% annually on 4/12 of a year, so whatever he saves, he'll have (that much times interest) in January. IN symbols:
$$ (\mbox{amount saved in Sept})\bigg(1 + \frac{4}{12}0.03\bigg) = (\mbox{result in January}) $$
(The 0.03 comes from the fact that it's 3% for a year, the 4/12 comes from the fact that he's only getting 3% for 4/12 of the year.) 
Well we know he needs \$1300 in January, so the result in January had better be 1300:
$$ (\mbox{amount saved in Sept})\bigg(1 + \frac{4}{12}0.03\bigg) = 1300 $$
How much does he need in September? He needs fall semester 1300 plus the amount he needs to save: in other words,
$$ \mbox{need in Sept} = (1300 \mbox{ for fall semester}) + (\mbox{amt to save for spring}) $$
Can you do the arithmetic to take it from here?
A: Let $T_1=0$ refer to September $1^{st}$, and $T_2=4$ refer to January $1^{st}$. The difference $T_2-T_1 = 4 \text{ months}$. 
Let $M(T)$ refer to the money John needs at time $T$. 
As given, the problem stipulates $M(T_1)=1300$ and $M(T_2)=1300$
With an interest rate $U=0.03$ we can define $M_{n+1}=rM_n$, where $n$ is the number of months that have elapsed since the beginning of measurement and $r=1+U=1.03$ is the multiplier for the money you have after taking interest one time. 
Setting $k=n+1$ we have $M_{k}=rM_{k-1}=r*(r*...(r*M_1))$. As you can see, we multiply by $r$ once the in the first term and $k-1$ times in the parentheses. Thus, we multiply by $r$ a total of $k$ times, which is $r^k$. 
Solving for $M_k$ we obtain $M_k=M_1r^k$. Thus $$M(T)=M_1r^k$$
The total money John needs is really a two part problem. First, we determine how much John needs by $T_1$ and then assume he has no money left after he pays then. We then solve for $M(T_2)$ assuming he begins with no money. 
Therefore, $$M(T_1)=1300=xr^0=1300*1$$ 
And
$$M(T_2)=1300=xr^4$$
Solving for $x$, we obtain $$x=1300r^{-4}$$ Now plugging in $r=1.03$, $x \approx 455$
Thus $M_{\text{total}}= M(T_1)+M(T_2)=1300+1155=2455\$$ is the amount of money John must have by September first. 
