1. ## Loan repayment

Joshua borrows $94500 to purchase his house. The loan is over 25 years at 10% p.a interest compounded on the balance owing after each monthly repayment i) Calculate the value of his monthly repayments. I got$856 but that is not the answer >< using the principal that:
94500 x (1.0083^300-1) = a(1.0083^300 -1)/(1.0083-1)

Thank you for help given.

2. Originally Posted by christina
Joshua borrows $94500 to purchase his house. The loan is over 25 years at 10% p.a interest compounded on the balance owing after each monthly repayment Calculate the value of his monthly repayments. I got$856 but that is not the answer >< using the principal that:
94500 x (1.0083^300-1) = a(1.0083^300 -1)/(1.0083-1)
.10/12 = .0083333......; NOT .0083
This results in correct payment of 858.722...; 858.72 rounded.
Use (1 + .10/12)

Your formula is not correct; should be:

payment = ai / [1 - 1/(1 + i)^n]
a = amount borrowed (94500)
i = interest (.10/12)
n = number of payments (300)

payment = 94500(.10/12) / [1 - 1/(1 + .10/12)^300] = 858.722...

OK?

3. Hello, christina!

Why are these problems assigned
. . without teaching the Amortization Formula?
Is everyone expected to derive it?

Joshua borrows $94,500 to purchase his house. The loan is over 25 years at 10% p.a interest compounded on the balance owing after each monthly repayment i) Calculate the value of his monthly repayments. The necessary formula is: .$\displaystyle A \;=\;P\,\frac{i(1+i)^n}{(1+i)^n-1}$. . where: .$\displaystyle \begin{Bmatrix} A & = & \text{periodic payment} \\ P &=& \text{principal amount of loan} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}$We have: .$\displaystyle \begin{Bmatrix}P &=& 94,\!500 \\ i &=& \frac{0.10}{12} &=& \frac{1}{120} \\ n &=& 25\cdot12 &=& 300 \end{Bmatrix}$Hence: .$\displaystyle A \;=\;94,\!500\cdot\frac{\left(\frac{1}{120})(1 + \frac{1}{120}\right)^{300}} {\left(1 + \frac{1}{120}\right)^{300} - 1} \;=\;858.7222047$The monthly payment is: .$\displaystyle \$858.72$

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Actually, lending institutions always round up.

At $858.7222047 per month, they would receive: . . .$\displaystyle 300 \times 858.7222047 \:\approx\;\$257,\!616.66$

At "only" $858.72 per month, they would receive: . . .$\displaystyle 300 \times 858.72 \:=\:\$257,\!616.00$

They would lose 66¢ to which they are entitled . . . horrors!

Although the loss is spread over 25 years,
. . you just have to feel sorry for them, don't you?

4. Thank you for your help Soroban and Wilmer !
Hmm, I have not seen that formula, I will go searching for it in my textbook - thank you especially Soroban for your tips and interesting information ^^

5. Sorry to ask again, but this question has two more parts to it and I'm stuck ><""

ii) If Joshua decides to pay $1092 a month off his loan instead, how much does owe after one repayment? The answers say$94500 x 1.008333 - 1092 = $94195.50 but what does this mean exactly? iii) If he continues to pay$1092 a month until the loan is paid off, how many years and months does he save on his loan?
Does this mean we have to calculate how long it takes him to pay off the loan with a monthly payment of $1092 because I am unsure >< THANK YOU AGAIN. 6. Originally Posted by christina ii) If Joshua decides to pay$1092 a month off his loan instead, how much does owe after one repayment?
The answers say $94500 x 1.008333 - 1092 =$ 94195.50
but what does this mean exactly?

iii) If he continues to pay $1092 a month until the loan is paid off, how many years and months does he save on his loan? Does this mean we have to calculate how long it takes him to pay off the loan with a monthly payment of$1092 ?
ii) 1st month interest cost = 94500 * .008333... = 787.50
94500 + 787.50 - 1092 = 94195.50

iii) Yes: calculate length of loan assuming 1092 payment.
So solve for n in the equation I or Soroban gave you.