Questions: Bob took out a loan for 225 days and was charged simple interest at an annual rate of 7.3%. The total interest he paid on the loan was 117. How much money did Bob borrow? Assume that there are 365 days in a year, and do not round any intermediate computations.

Solution Steps

To determine how much money Bob borrowed, we can use the simple interest formula:

\[ I = P \times r \times t \]

where:

• \( I \) is the interest paid (\$117),
• \( P \) is the principal amount (the amount borrowed),
• \( r \) is the annual interest rate (7.3\% or 0.073),
• \( t \) is the time the money is borrowed for, in years (225 days / 365 days).

We need to solve for \( P \):

\[ P = \frac{I}{r \times t} \]

We are given:

• Interest paid, \( I = 117 \)
• Annual interest rate, \( r = 0.073 \)
• Time borrowed in days, \( t = 225 \)
• Days in a year, \( 365 \)

Convert the time borrowed from days to years: \[ t_{\text{years}} = \frac{225}{365} \approx 0.6164 \]

The simple interest formula is: \[ I = P \times r \times t \]

Rearrange to solve for \( P \): \[ P = \frac{I}{r \times t} \]

Substitute the given values into the formula: \[ P = \frac{117}{0.073 \times 0.6164} \]

Perform the calculation: \[ P \approx \frac{117}{0.0459972} \approx 2600.0 \]

Final Answer