Last payment =$61.59 #### Explanation: $\textcolor{b r o w n}{\frac{1}{12} \text{ of a month is an unexpected value. Could you mean years?}}$Making assumptions about the distribution of payment values. There will be other ways of doing this. The use of 'unit payment' is just a trick to solve this problem type. It is a (constant of proportionallity) way of measuring the distribution of payment. Nothing else! $4 \frac{1}{12} \text{ months"= 49/12" months}$So 49xx"unit payment" =$232.20

=>1 unit payment =($232.20)/49~~$4.738.....

There are 12 units of payments in 1 month $\left(\frac{12}{12} = 1\right)$

So we have 3 months at 12xx4.738... = $56.87 per month rounded to 2 decimal places This gives a total payment at the end of 3 months of 3xx$56.87= $170.61 So the payment for the last month is: $232.20-$170.61=$61.59

Jan 22, 2017

An interpretation of the question that is different to my first one.

4 payments at $56.86 to 2 decimal places Last payment at$4.74 to 2 decimal places

#### Explanation:

You should read through my other solution first to see where the numbers come from.

There are 5 payments

$4 \frac{1}{12} \text{ months"= 49/12" months}$

So 49xx"unit payment" = $232.20  =>1 unit payment =($232.20)/49~~$4.738..... There are 12 units of payments in 1 month $\left(\frac{12}{12} = 1\right)$4 payments at 12xx$4.738.. ->4xx$56.8553... =$227.46
Closing payment at $4.738 " "->" "ul(" "$4.74)
" "\$232.20