compound interest

**Babs0201** · April 26th 2008, 12:28 PM

Mary wants to take out a loan. Suppose she can afford to make monthly payments of 200 dollars and the bank charges interest at an annual rate of 5 percent, compounded monthly. What is the maximum amount that Mary could afford to borrow if the loan is to be paid off eventually?

I am not sure how to do a problem with "eventually."
I have tried to approach this like a regular compound interest problem, but it is not working.

Yn = Yo (1 + i )^n

b / (1-a) = 200 / (1 - 1.041667) = -4799.96

i = 0.5 / 12 = 0.041667
n = ????

-4799.96 + (Yo - (-4799.96)(1.041667^n) = ????

How can I do this without knowing how much she took out, or how long it will take to pay off?

**icemanfan** · April 26th 2008, 12:50 PM

Originally Posted by Babs0201

Mary wants to take out a loan. Suppose she can afford to make monthly payments of 200 dollars and the bank charges interest at an annual rate of 5 percent, compounded monthly. What is the maximum amount that Mary could afford to borrow if the loan is to be paid off eventually?

I am not sure how to do a problem with "eventually."
I have tried to approach this like a regular compound interest problem, but it is not working.

Yn = Yo (1 + i )^n

b / (1-a) = 200 / (1 - 1.041667) = -4799.96

i = 0.5 / 12 = 0.041667
n = ????

-4799.96 + (Yo - (-4799.96)(1.041667^n) = ????

How can I do this without knowing how much she took out, or how long it will take to pay off?

$i = \frac{0.05}{12}$

What you want is to figure out what loan generates $200 of interest over the course of a month. So $L + 200 = L(1 + \frac{0.05}{12})$ .

**Babs0201** · April 26th 2008, 01:01 PM

Thank you for your time!

so I am getting 47961.6 for the amount she can borrow.
I solved for L.
L + 200 = L + 0.00417L

200 / 0.00417 = 47961.6

**icemanfan** · April 26th 2008, 01:20 PM

Originally Posted by Babs0201

Thank you for your time!

so I am getting 47961.6 for the amount she can borrow.
I solved for L.
L + 200 = L + 0.00417L

200 / 0.00417 = 47961.6

Now, the minor technicality is that if she borrows exactly that much money, she'll never pay it off. Anything less will suffice (assuming she lives long enough)!

**Soroban** · April 26th 2008, 02:30 PM

Hello, Babs!

Icemanfan has an elegant solution . . . impressive!

I deliberately did this "the long way" . . .

Mary wants to take out a loan.
Suppose she can afford to make monthly payments of $200 dollars
and the bank charges 5% annually, compounded monthly.
What is the maximum amount that Mary could afford to borrow
if the loan is to be paid off eventually?

This is an Amortization problem.

Formula: . $A \;=\;P\,\frac{i(1+i)^n}{(1+i)^n-1}\quad \text{where: }\;\begin{Bmatrix}A & = & \text{periodic payment} \\ P &=& \text{principal borrowed} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}$

We have: . $A = \$200,\;i \,=\,\frac{5\%}{12} \,=\,\frac{1}{240}$

Then:. . $200 \;=\;P\,\frac{\frac{1}{240}\left(1 + \frac{1}{240}\right)^n}{\left(1 + \frac{1}{240}\right)^n - 1} \quad\Rightarrow\quad 200\left[\left(1 + \frac{1}{240}\right)^n - 1\right] \;=\;\frac{P}{240}\left(1 + \frac{1}{240}\right)^n$

. . . . $48,\!000\left(1 + \frac{1}{240}\right)^n - 48,000 \;=\;P\left(1 + \frac{1}{240}\right)^n$

. . . . $48,\!000\left(1 + \frac{1}{240}\right)^n - P\left(1 + \frac{1}{240}\right)^n \;=\;48,000$

Factor: . $\left(1 + \frac{1}{240}\right)^n(48,\!000 - P) \;=\;48,\!000 \quad\Rightarrow\quad \left(1 + \frac{1}{240}\right)^n \;=\;\frac{48,\!000}{48,\!000-P}$

We have the form: . $X^n \;=\;\frac{48,\!000}{48,\!000-P}$

Since the left side is positive, we see that $P < 48,\!000$

Therefore, she can borrow any amount less than $48,000.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Of course, as $P$ approaches $48,000, the problem gets silly.

If she borrowed $47,990, we have: . $\left(\frac{241}{240}\right)^n \:=\:\frac{48,\!000}{10} \:=\:4,\!800$

Take logs: . $\ln\left(\frac{241}{240}\right)^n \:=\:\ln(4,\!800) \quad\Rightarrow\quad n\!\cdot\!\ln\left(\frac{241}{240}\right) \;=\;\ln(4,\!800)$

Hence: . $n \;=\;\frac{\ln(4,\!800)}{\ln\left(\frac{241}{240}\ right)} \;=\;2038.564336\text{ months}\;\approx\;170\text{ years}$

**Babs0201** · April 26th 2008, 04:40 PM

Thank you so much. It really helped.

By looking at your solutions I figured out a shorter way to do it.

simply take the interest 0.05 and divid it by the months (12) and multiply it by the unknown amout she is borrowing K. Set it equal to 200. so...

(0.05/12)*k = 200

k = 48000, which is the amount she will never be able to pay off with the amount she is paying now!!!!

Thanks again for helping to explain this!!!

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