To determine the simple interest rate required for \($15,000 to double in 8 years, we can use the simple interest formula:
\[ A = P(1 + rt) \]
where:
- \( A \) is the final amount
- \( P \) is the principal amount
- \( r \) is the interest rate
- \( t \) is the time in years
Given:
- The principal amount \( P = \$\)15,000 \)
- The final amount \(\( A = 2 \times 15000 = \$30,000 \)\)
- The time \(\( t = 8 \)\) years
Rearranging the formula to solve for \(\( r \)\):
\(\[ A = P(1 + rt) \]\)
\(\[ 30000 = 15000(1 + r \times 8) \]\)
\(\[ 30000 = 15000 + 120000r \]\)
\(\[ 15000 = 120000r \]\)
\(\[ r = \frac{15000}{120000} \]\)
\(\[ r = 0.125 \]\)
To convert this to a percentage, we multiply by 100:
\(\[ r = 0.125 \times 100 = 12.5\% \]\)
Therefore, the simple interest rate needed for $15,000 to double in 8 years is 12.5%.
For the second part of your question, we need to arrange these fractions of a year in ascending order: 190 days, 5 months, 160 days, 6 months, 200 days.
First, we convert all the given time units to a common unit, fractions of a year. Assuming a year has 365 days:
- 190 days: \(\( \frac{190}{365} \approx 0.5205 \)\) of a year
- 5 months: \(\( 5 \times \frac{30}{365} = \frac{150}{365} \approx 0.410 \)\)
- 160 days: \(\( \frac{160}{365} \approx 0.4384 \)\) of a year
- 6 months: \(\( 6 \times \frac{30}{365} = \frac{180}{365} \approx 0.4932 \)\)
- 200 days: \(\( \frac{200}{365} \approx 0.5479 \)\) of a year
Now, we arrange these fractions in ascending order:
- 5 months: \(\( 0.410 \)\) of a year
- 160 days: \(\( 0.4384 \)\) of a year
- 6 months: \(\( 0.4932 \)\) of a year
- 190 days: \(\( 0.5205 \)\) of a year
- 200 days: \(\( 0.5479 \)\) of a year
In ascending order, these fractions of a year are:
5 months, 160 days, 6 months, 190 days, 200 days.
