Time Payments Question (1 Viewer)

[Amleops]

An amount of $280000 is borrowed. Interest is calculated monthly at 12% pa on the balance owing. The loan is to be repaid over 25 years by equal monthly installments. If the repayment were doubled, how long would it then take to repay the loan.

I got up to finding the original monthly repayment which was $2949.03 (therefore double of that is $5898.06), but I'm not too sure where to go next.

 

[Carrotsticks]

1. Let the new amount $5898.06 be K, for simplicity of calculation later on.

2. Work your way backwards. Find the amount owing after 1 month, after 2 months, after 3 months etc until you establish a pattern.

3. Once you find the pattern, write down the general 'formula' for the amount owing after the nth month.

4. We want to set the amount owing to become 0 (since the loan is fully repaid), so then we can make an equation to solve for n.

5. Once you solve for n, that is the number of months it takes to repay.

6. Your answer should be less than 25 years. If it is more, then you did something wrong.

 

[Amleops]

OK, I was doing something similar to that, it was just Step 5 I was having a problem with.

2800(1.01^n) = K(1.01^n - 1)

I've tried a couple of ways to solve that and it hasn't worked so far.

 

[Carrotsticks]

OK, I was doing something similar to that, it was just Step 5 I was having a problem with.

2800(1.01^n) = K(1.01^n - 1)

I've tried a couple of ways to solve that and it hasn't worked so far.

Make 1.01^n the subject, then use logarithms to solve for n.

 

[Amleops]

Ah OK, I think I've got it now then:

2800(1.01^n) = K(1.01^n - 1)
2800 x 1.01^n = K1.01^n - K
2800 x 1.01^n - K1.01^n + K = 0
1.01^n(2800-K) + K = 0
1.01^n(2800-K) = -K
1.01^n = -K/(2800-K)
log1.01(-K/2800-K) = n
n = log(-K/2800-K) / log(1.01)
n = log(-5898.06/2800-5898.06) / log(1.01)
n = 64.7061273
Therefore, the loan is repaid after 65 months.

Thanks for your help