Diane owes $232.20 for a gown for a special occasion. She pay for the gown in `4 1/12` months. What will her monthly payment be? What is her last payment?

`color(brown)(1/12" 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 1/12 " 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 `(12/12=1)`

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`

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

There are 5 payments

`4 1/12 " 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 `(12/12=1)`

4 payments at `12xx$4.738.. ->4xx$56.8553... =$227.46`
Closing payment at `$4.738 " "->" "ul(" "$4.74)`
`" "$232.20`