Question

review the following note receivable transactions in the journal of mediterranean services: view the transactions. mediterranean services has a december 31 fiscal year - end. read the requirements. requirement 1. record the preceding note receivable transactions in the journal of mediterranean services. use a 365 - day year for interest computations, and round interest amounts to the nearest dollar. (record debits first, then credits. exclude explanations from any journal entries.) october 1: loaned $21,000 cash to brandy shields on a one - year, 10% note. date accounts debit credit october 1 transactions october 1 loaned $21,000 cash to brandy shields on a one - year, 10% note. december 6 performed service for lawn pro, receiving a 90 - day, 9% note for $5,000. december 16 received a $4,000, six - month, 11% note on account from peabody company. december 31 accrued interest revenue for the year.

Explanation:

Step1: Record the loan on October 1

When a company loans cash and receives a note receivable, the Notes Receivable account is debited (increase in asset) and the Cash account is credited (decrease in asset).

$$\begin{align*} \text{Debit: Notes Receivable - Brandy Shields}&= 21000\\ \text{Credit: Cash}&=21000 \end{align*}$$

Step2: Calculate interest for each note on December 31 (accrued interest)

For Brandy Shields' note:

The formula for simple interest is $I = P\times r\times t$, where $P$ (principal) = $21000$, $r$ (annual interest rate) = $10\%=0.10$, and $t$ (time in years). From October 1 to December 31 is $t=\frac{92}{365}$ years.

$$\begin{align*} I_{1}&=21000\times0.10\times\frac{92}{365}\\ &=21000\times0.10\times0.252055\\ &\approx 529 \end{align*}$$

For Lawn Pro's note:

$P = 5000$, $r=9\% = 0.09$, from December 6 to December 31 is $t=\frac{25}{365}$ years

$$\begin{align*} I_{2}&=5000\times0.09\times\frac{25}{365}\\ &=5000\times0.09\times0.068493\\ &\approx 31 \end{align*}$$

For Peabody Company's note:

$P = 4000$, $r = 11\%=0.11$, from December 16 to December 31 is $t=\frac{15}{365}$ years

$$\begin{align*} I_{3}&=4000\times0.11\times\frac{15}{365}\\ &=4000\times0.11\times0.041096\\ &\approx 18 \end{align*}$$

Total accrued interest $I=I_{1}+I_{2}+I_{3}=529 + 31+18=578$
The journal entry for accrued interest:
Debit: Interest Receivable = 578
Credit: Interest Revenue = 578

Answer:

Date	Accounts	Debit	Credit
	Cash		21000
December 31	Interest Receivable	578
	Interest Revenue		578