A year-end bonus of $\$23,000$ will generate how much money at the beginning of each month for the next year, if it can be invested at $6.3\%$, compounded monthly? (Round your answer to the nearest cent.)
This might be a very simple question for this website but I'm having trouble with the language. I'm not being able to figure out that by "at the beginning of each month for the next year" what they are meaning.
If it's asking that : "If I'm investing $\$23,000$ at rate $\%6.3$, compounded monthly then what will be the total amount of money generated at the beginning of the first month after 1 year?" Then I can easily answer as below :
Since, $$ A = P \Big(1+\frac{r}{100\cdot n} \Big)^{nt}$$
And here : $$ P = 23000, r = 6.3, n = 12, t = 1$$
Therefore the answer is $$ A = 24491.5$$.
Please can anyone check if I'm understanding it right?
You are right for the second question. At the first question you invest 12 times $x$ dollars at the beginning of every month, not only once as in the second question. At the sketch you can see that the 9-th payment (at the end of the 8-th month) has to be compounded $4$ times with the monthly interest rate $i_{12}=\frac{0.063}{12}$ to obtain the future value at $t=12$ (end of the year).
So the equation for all 12 payments is
$$x\cdot \sum_{k=1}^{12} \left(1+\frac{0.063}{12}\right)^k=23000$$
At $k=4$ you have the 9-th payment at $t=8$. The sum is a geometric series with a fixed upper bound. It has a closed form. Let´s say that $q_{12}=1+i_{12}$. Due that the payments are made at the beginning of the month the formula for the sum is
$$q_{12}\cdot \frac{q_{12}^{12}-1}{q_{12}-1}$$
$$\left(1+\frac{i}{12}\right)\cdot \frac{\left(1+\frac{i}{12}\right)^{12}-1}{\left(1+\frac{i}{12}\right)-1}$$
So the equation simplifies to
$$x\cdot \left(1+\frac{0.063}{12}\right)\cdot \frac{\left(1+\frac{0.063}{12}\right)^{12}-1}{1+\frac{0.063}{12}-1}=23000$$
$$x\cdot \left(1+\frac{0.063}{12}\right)\cdot \frac{\left(1+\frac{0.063}{12}\right)^{12}-1}{\frac{0.063}{12}}=23000$$
$$x\cdot \left(1+\frac{0.063}{12}\right)\cdot \left(\left(1+\frac{0.063}{12}\right)^{12}-1\right)=23000\cdot \frac{0.063}{12}$$
$$x=\frac{23000\cdot \frac{0.063}{12}}{\left(1+\frac{0.063}{12}\right)\cdot \left(\left(1+\frac{0.063}{12}\right)^{12}-1\right)}=1852.23$$