my car cost 16,758.90 dollars, my annual interest rate is 25.99% for 72 months. my monthly payments are 464.38 dollars. at the end of those 72 monte ill have paid 33,435.36 dollars in total. my question is what if i pay 10,000 dollars every year plus my monthly payments, how much interest am i going to be paying and in how many months ill be done paying my car off?

Guest May 17, 2014

#1**+5 **

First I am going to check if your original figures are true

The car costs $16758.90, the loan is for 72 months, interest is 25.99% pa (monthly reducible)

How much will each (end of month payment) need to be.

I am going to use this formula

$$\boxed{A=R\times \frac{1-(1+i)^{-n}}{i}}$$ ==> $$R = A\times \frac{i}{1-(1+i)^{-n}}$$

A=16758.90

n=72

i=0.2599/12=0.0216583333333

R is the regular payment.

$${\frac{{\mathtt{16\,758.9}}{\mathtt{\,\times\,}}{\mathtt{0.021\: \!658\: \!333\: \!333\: \!3}}}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\left({\mathtt{1.021\: \!658\: \!333\: \!333\: \!3}}\right)}^{\left(-{\mathtt{72}}\right)}\right)}} = {\mathtt{461.671\: \!365\: \!148\: \!752\: \!942\: \!6}}$$

So from my calculations if you want the loan finished in 72 months you will need to pay $461.67 Which is just a little less than what your question stated - near enough - the difference is most likely a rounding error.

If I can work out how to do it at all I will use the **$464.38** that was given in your question.

Now you want to know how much difference it would make if you make an extra $10 000 payment at the end of each year.

Ummm Let 1+i= I It will make the maths easier.

Amount owing after 1 month = 16758.90*1.021658333333- 464.38

Amount owing after 1 month =AI-R

Amount owing after 2 months = (AI-R)I-R = AI^{2}-RI-R = AI^{2}-R(I+1) = AI^{2}-R(1+I)

Amount owing after 3 months = [AI^{2}-R(1+I)]I-R = AI^{3}-RI(1+I)-R = AI^{3}-RI-RI^{2}-R=AI^{3}-R(1+I+I^{2})

Amount owing after n months = AI^{n}-R(1+I+I^{2}+ ....I^{n-1})

The brackets house the sum of a GP

$$\boxed{S_n=\frac{a(r^n-1)}{r-1}}$$ where a=1, r=I, n=n

$$S_n=\frac{I^n-1}{I-1}$$

Amount owing after n months = $$AI^n-\frac{R(I^n-1)}{I-1}$$

$${\mathtt{16\,758.9}}{\mathtt{\,\times\,}}{{\mathtt{1.021\: \!658\: \!333\: \!333\: \!3}}}^{{\mathtt{12}}}{\mathtt{\,-\,}}{\frac{{\mathtt{464.38}}{\mathtt{\,\times\,}}\left({{\mathtt{1.021\: \!658\: \!333\: \!333\: \!3}}}^{{\mathtt{12}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{0.021\: \!658\: \!333\: \!333\: \!3}}}} = {\mathtt{15\,386.023\: \!881\: \!635\: \!638\: \!058}}$$

Amont owing after 12 months = $15386.02

Now a $10000 payment is made do the amount owing immediately drops to $5386.02

Now the amount owing after another year will be

$${\mathtt{5\,386.02}}{\mathtt{\,\times\,}}{{\mathtt{1.021\: \!658\: \!333\: \!333\: \!3}}}^{{\mathtt{12}}}{\mathtt{\,-\,}}{\frac{{\mathtt{464.38}}{\mathtt{\,\times\,}}\left({{\mathtt{1.021\: \!658\: \!333\: \!333\: \!3}}}^{{\mathtt{12}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{0.021\: \!658\: \!333\: \!333\: \!3}}}} = {\mathtt{678.531\: \!308\: \!282\: \!506\: \!542\: \!7}}$$

this gets knocked on the head with another one off payment.

So With the extra one off payments the loan has taken 24 months to pay off and the cost of the amount repaid has become $21823.65

$${\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{464.38}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10\,678.53}} = {\mathtt{21\,823.65}}$$

the total amount of interest paid is $${\mathtt{21\,823.65}}{\mathtt{\,-\,}}{\mathtt{16\,758.9}} = {\mathtt{5\,064.75}}$$

I think that about covers it.

That is one of CPhill's pairs of glasses - I 'acquired' them, no one deserves 3 pairs especially when they are soooo cool!

Melody
May 17, 2014

#1**+5 **

Best Answer

First I am going to check if your original figures are true

The car costs $16758.90, the loan is for 72 months, interest is 25.99% pa (monthly reducible)

How much will each (end of month payment) need to be.

I am going to use this formula

$$\boxed{A=R\times \frac{1-(1+i)^{-n}}{i}}$$ ==> $$R = A\times \frac{i}{1-(1+i)^{-n}}$$

A=16758.90

n=72

i=0.2599/12=0.0216583333333

R is the regular payment.

$${\frac{{\mathtt{16\,758.9}}{\mathtt{\,\times\,}}{\mathtt{0.021\: \!658\: \!333\: \!333\: \!3}}}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\left({\mathtt{1.021\: \!658\: \!333\: \!333\: \!3}}\right)}^{\left(-{\mathtt{72}}\right)}\right)}} = {\mathtt{461.671\: \!365\: \!148\: \!752\: \!942\: \!6}}$$

So from my calculations if you want the loan finished in 72 months you will need to pay $461.67 Which is just a little less than what your question stated - near enough - the difference is most likely a rounding error.

If I can work out how to do it at all I will use the **$464.38** that was given in your question.

Now you want to know how much difference it would make if you make an extra $10 000 payment at the end of each year.

Ummm Let 1+i= I It will make the maths easier.

Amount owing after 1 month = 16758.90*1.021658333333- 464.38

Amount owing after 1 month =AI-R

Amount owing after 2 months = (AI-R)I-R = AI^{2}-RI-R = AI^{2}-R(I+1) = AI^{2}-R(1+I)

Amount owing after 3 months = [AI^{2}-R(1+I)]I-R = AI^{3}-RI(1+I)-R = AI^{3}-RI-RI^{2}-R=AI^{3}-R(1+I+I^{2})

Amount owing after n months = AI^{n}-R(1+I+I^{2}+ ....I^{n-1})

The brackets house the sum of a GP

$$\boxed{S_n=\frac{a(r^n-1)}{r-1}}$$ where a=1, r=I, n=n

$$S_n=\frac{I^n-1}{I-1}$$

Amount owing after n months = $$AI^n-\frac{R(I^n-1)}{I-1}$$

$${\mathtt{16\,758.9}}{\mathtt{\,\times\,}}{{\mathtt{1.021\: \!658\: \!333\: \!333\: \!3}}}^{{\mathtt{12}}}{\mathtt{\,-\,}}{\frac{{\mathtt{464.38}}{\mathtt{\,\times\,}}\left({{\mathtt{1.021\: \!658\: \!333\: \!333\: \!3}}}^{{\mathtt{12}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{0.021\: \!658\: \!333\: \!333\: \!3}}}} = {\mathtt{15\,386.023\: \!881\: \!635\: \!638\: \!058}}$$

Amont owing after 12 months = $15386.02

Now a $10000 payment is made do the amount owing immediately drops to $5386.02

Now the amount owing after another year will be

$${\mathtt{5\,386.02}}{\mathtt{\,\times\,}}{{\mathtt{1.021\: \!658\: \!333\: \!333\: \!3}}}^{{\mathtt{12}}}{\mathtt{\,-\,}}{\frac{{\mathtt{464.38}}{\mathtt{\,\times\,}}\left({{\mathtt{1.021\: \!658\: \!333\: \!333\: \!3}}}^{{\mathtt{12}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{0.021\: \!658\: \!333\: \!333\: \!3}}}} = {\mathtt{678.531\: \!308\: \!282\: \!506\: \!542\: \!7}}$$

this gets knocked on the head with another one off payment.

So With the extra one off payments the loan has taken 24 months to pay off and the cost of the amount repaid has become $21823.65

$${\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{464.38}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10\,678.53}} = {\mathtt{21\,823.65}}$$

the total amount of interest paid is $${\mathtt{21\,823.65}}{\mathtt{\,-\,}}{\mathtt{16\,758.9}} = {\mathtt{5\,064.75}}$$

I think that about covers it.

That is one of CPhill's pairs of glasses - I 'acquired' them, no one deserves 3 pairs especially when they are soooo cool!

Melody
May 17, 2014