# Math Help - Finance word problem help

1. ## Finance word problem help

Can someone please tell me how i solve this ? thanx

A couple wants to buy a house and can afford to pay $1400 per month. a) if they can get a loan for 30 years with the interest at 8% per year on the unpaid balance, how much can they pay for the house? b) what is the total amount paid over the life of the loan? c) what is the interest paid on the loan? I think the answer to part A is$463,680. is that right ? but i dont know how to do parts b and c. thanks

2. a) $A = P\left( 1+\frac{r}{n} \right)^{n\times t}$

$1400\times 12\times 30 = P\left( 1+\frac{8}{12} \right)^{12\times 30}$

Solve for $P$

b) $1400\times 12\times 30$

c) $I= A-P$

3. Hello, Stephen!

This is an Amortization problem.
You should have been given this formula:

$A \;=\;P\cdot\dfrac{i(1+i)^n}{(1+i)^n-1}$

. . where: . $\begin{Bmatrix}A &=& \text{periodic payment} \\ P &=& \text{principal borrowed} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}$

A couple wants to buy a house and can afford to pay $1400 per month. a) If they get a loan for 30 years with 8% interest per year on the unpaid balance, how much can they pay for the house? We have: . $\begin{Bmatrix}A &=& 1400 \\ i &=& \frac{0.08}{12} &=& \frac{1}{150} \\ n &=& 12\cdot30 &=& 360 \end{Bmatrix}$ $\text{Then: }\;1400 \;=\;P\cdot\dfrac{\frac{1}{150}\left(1 + \frac{1}{150}\right)^{360}} {\left(1+\frac{1}{150}\right)^{360}-1}$ Solve for $P\!:\;\;P \;=\;190,\!796.8927$ They can pay up to $\190,\!796.89$ for the house. b) What is the total amount paid over the life of the loan?$1400 per month for 360 months: . $\504,\!000$

c) What is the interest paid on the loan?

Total interest is: . $\504,\!000 - 190,\!796.89 \;=\;\313,\!203.11$

4. Thanks so much for the help. I was actually shown how to do this kind of problem on my calculator but couldn't remember how. thanks again