Lucky Champ owes $231.20O interest on a 8% loan he took out on his March 17 birthday to upgrade an oven in his Irish restaurant, Lucky's Pub and Grub. The loan is due on August 17. What is the principal? (Use Days in a year table.) (Use 360 days a year. Do not round intermediate calculations.) Principal

FINANCIAL ACCOUNTING
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Chapter1: Financial Statements And Business Decisions
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**Problem Statement:**

Lucky Champ owes $231.20 interest on an 8% loan he took out on his March 17 birthday to upgrade an oven in his Irish restaurant, Lucky’s Pub and Grub. The loan is due on August 17. What is the principal? 

(Use *Days in a year* table.) *(Use 360 days a year. Do not round intermediate calculations.)*

**Solution:**

To find the principal, you can use the formula for simple interest:
\[ \text{Interest} = \text{Principal} \times \text{Rate} \times \text{Time} \]

- Interest (I) is $231.20.
- Rate (R) is 8%, or 0.08 as a decimal.
- Time (T) is the period from March 17 to August 17. 

To calculate Time in terms of the 360-day year:
- March 17 to August 17 covers approximately 5 months.
- With 30 days per month convention for accounting purposes: 5 months × 30 days/month = 150 days.
- Time = 150/360 in terms of the year.

Now, substitute in the formula and solve for Principal (P):

\[ 231.20 = P \times 0.08 \times \frac{150}{360} \]

\[ P = \frac{231.20}{0.08 \times \frac{150}{360}} \]

Calculate P to find the principal amount.
Transcribed Image Text:**Problem Statement:** Lucky Champ owes $231.20 interest on an 8% loan he took out on his March 17 birthday to upgrade an oven in his Irish restaurant, Lucky’s Pub and Grub. The loan is due on August 17. What is the principal? (Use *Days in a year* table.) *(Use 360 days a year. Do not round intermediate calculations.)* **Solution:** To find the principal, you can use the formula for simple interest: \[ \text{Interest} = \text{Principal} \times \text{Rate} \times \text{Time} \] - Interest (I) is $231.20. - Rate (R) is 8%, or 0.08 as a decimal. - Time (T) is the period from March 17 to August 17. To calculate Time in terms of the 360-day year: - March 17 to August 17 covers approximately 5 months. - With 30 days per month convention for accounting purposes: 5 months × 30 days/month = 150 days. - Time = 150/360 in terms of the year. Now, substitute in the formula and solve for Principal (P): \[ 231.20 = P \times 0.08 \times \frac{150}{360} \] \[ P = \frac{231.20}{0.08 \times \frac{150}{360}} \] Calculate P to find the principal amount.
The image displays a detailed table often referred to as a Julian calendar. The table is structured to align days of the month (indicated on the far left) with cumulative day counts across each month from January to December. Each month column represents the number of days that have passed in the year by the end of that day in the specified month.

### Table Explanation

- **Columns:**
  - The table consists of columns representing the months from January (31 days) to December (31 days), with a special consideration for February (28 days).
  - Each column provides cumulative day counts as the year progresses, starting from 1 for January 1st to 365 for December 31st.

- **Rows:**
  - Rows are labeled with the day of the month, ranging from 1 to 31.

- **Cumulative Days:**
  - For instance, the value under the column for March (31) against row 10 indicates the 69th day of the year.
  - July 15th is noted as day 196.

- **Anomalies:** 
  - February is marked with 28 days, adhering to a non-leap year format.
  - Certain months are highlighted in grey for emphasis at the top of the columns: January (31), February (28), March (31), April (30), May (31), June (30), July (31), August (31), September (30), October (31), November (30), and December (31).

- **Notes:**
  - The bottom note mentions, "*Often referred to as a Julian calendar.*" indicating traditional use similar to the Julian system.

This table is useful for understanding day sequences within the year and aligning events or schedules against a fixed calendar framework.
Transcribed Image Text:The image displays a detailed table often referred to as a Julian calendar. The table is structured to align days of the month (indicated on the far left) with cumulative day counts across each month from January to December. Each month column represents the number of days that have passed in the year by the end of that day in the specified month. ### Table Explanation - **Columns:** - The table consists of columns representing the months from January (31 days) to December (31 days), with a special consideration for February (28 days). - Each column provides cumulative day counts as the year progresses, starting from 1 for January 1st to 365 for December 31st. - **Rows:** - Rows are labeled with the day of the month, ranging from 1 to 31. - **Cumulative Days:** - For instance, the value under the column for March (31) against row 10 indicates the 69th day of the year. - July 15th is noted as day 196. - **Anomalies:** - February is marked with 28 days, adhering to a non-leap year format. - Certain months are highlighted in grey for emphasis at the top of the columns: January (31), February (28), March (31), April (30), May (31), June (30), July (31), August (31), September (30), October (31), November (30), and December (31). - **Notes:** - The bottom note mentions, "*Often referred to as a Julian calendar.*" indicating traditional use similar to the Julian system. This table is useful for understanding day sequences within the year and aligning events or schedules against a fixed calendar framework.
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