To find the interest rate charged given that \($680 is worth $\)698.70 after three months, we can utilize the simple interest formula:
\(\[ A = P(1 + rt) \]\)
Where:
- \(\( A \)\) is the final amount.
- \(\( P \)\) is the initial amount.
- \(\( r \)\) is the interest rate.
- \(\( t \)\) is the time period in years.
We are given:
- \(\( P = 680 \)\)
- \(\( A = 698.70 \)\)
- \(\( t = 3 \)\) months, which is equivalent to \(\( \frac{3}{12} \)\) years.
First, set up the equation with the known values:
\(\[ 698.70 = 680 \left(1 + r \cdot \frac{3}{12}\right) \]\)
Next, simplify the time period:
\(\[ 698.70 = 680 \left(1 + r \cdot 0.25\right) \]\)
Then, solve for \(\( r \)\):
\(\[ 698.70 = 680 (1 + 0.25r) \]\)
Divide both sides by 680:
\(\[ \frac{698.70}{680} = 1 + 0.25r \]\)
This simplifies to:
\(\[ 1.0275 = 1 + 0.25r \]\)
Subtract 1 from both sides to isolate the term with \(\( r \)\):
\(\[ 0.0275 = 0.25r \]\)
Divide both sides by 0.25 to solve for \(\( r \)\):
\(\[ r = \frac{0.0275}{0.25} = 0.11 \]\)
The interest rate \(\( r \)\) is 0.11, which can be expressed as a percentage by multiplying by 100:
\(\[ r \times 100 = 0.11 \times 100 = 11 \% \]\)
Thus, the interest rate charged is:
\(\[ 11.0\% \]\)
Rounded to the nearest cent as required.
