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To solve this problem, we use the formula for simple interest:

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

where:
- $ I $ is the interest paid, which is $146.
- $ r $ is the annual interest rate, 2.5% or 0.025 as a decimal.
- $ t $ is the time in years. Since the loan was for 292 days, we convert this to years: 
  \[
  t = \frac{292}{365}
  \]

We need to find $ P $, the principal amount (the amount Charlie borrowed). Rearranging the formula, we get:

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

### Step-by-Step Calculation:

1. **Convert the time to years:**
   \[
   t = \frac{292}{365} \approx 0.8 \text{ years}
   \]

2. **Set up the formula with known values:**
   \[
   P = \frac{146}{0.025 \times 0.8}
   \]

3. **Calculate the denominator:**
   \[
   0.025 \times 0.8 = 0.02
   \]

4. **Divide the interest by the result:**
   \[
   P = \frac{146}{0.02} = 7300
   \]

So, Charlie borrowed **$7,300**.

Would you like more details or have any questions?

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Here are some related questions for further exploration:

1. What is the formula for calculating compound interest?
2. How would the answer change if the interest rate were compounded monthly?
3. How can you convert days to years in general interest calculations?
4. How would this calculation differ if the interest were compounded daily?
5. What would be the total amount Charlie needs to repay after 292 days?

**Tip:** Always double-check your unit conversions, especially when dealing with time, as it can impact the accuracy of your financial calculations.

Charlie took out a loan for 292 days and was charged simple interest at an annual rate of 2.5%. The total interest he paid on the loan was $146. How much money did Charlie borrow? Assume that there are 365 days in a year, and do not round any intermediate computations.

Learn how to calculate the principal amount Charlie borrowed using simple interest. With a loan duration of 292 days and an interest rate of 2.5%, this problem includes detailed calculations on converting days to years and applying the simple interest formula. Perfect for Grade 10 students learning interest calculation.

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