Burroughs School of Real Estate

#7000
Tom Stepanski
Keymaster

Proration Problems:

1. Seller purchased a 3-year insurance policy on September 1, 1998. Total costs was $2,700.00. Buyer will assume and pay for seller’s insurance. The day of closing is November 1, 1999. Prorate the cost to Buyer. Answer is $1,650
Policy will expire 8/31/2001 and closing is 11/1/1999, so we would subtract the day of closing from the expiration date. It would look like this; (Year-Month-Day)
Yr Mo Day
2001 08 31
2000 20 31
1999 11 01
1 9 30
Remember when we borrow, they are in Years, Months and Days. We had to borrow 1 year or 12 months because 11 was greater than 8.
$2,700 ÷3=$900 per year, $900÷12=$75 per month, $75÷30=$2.50 per day
$900 + $675 + $75 = $1,650

2. Taxes are $480.00 per year. Sale closes April 15. Prorate the taxes. (charge to whom and credit to whom.); Debit the seller $140 and credit the buyer $140
First, we would charge the seller and credit the buyer the same figure. The seller is paying their part of the yearly taxes.
The time that has passed since the first of the year is 3 months and 15 days.
$480÷$12=$40 per month, $40÷30=$1.333 (only carry 3 decimal places, do not round yet)
3 x $40 = $120 $1.333 x 15 = $19.995 or rounded, $20;
$120 + $20 = $140, debit the seller $140 and credit the buyer $140

3. Interest on a loan the buyer is assuming is 8%. On date of closing, June 10, the unpaid balance is $20,000.00. Interest is paid in arrears and is to be prorated. Seller $43.84 Buyer $87.67
10 days for the seller and 20 days for the buyer.
20,000 x .08 = 1,600, 1,600 ÷ 365 = 4.383561643835616 (time 10 for the seller and 20 for the buyer)
Seller $43.84 Buyer $87.67