Question

An avid collector wants to purchase a signed basketball from a particular playoff game. He plans to put away​ 4% more money each​ year, in a safe at his​ home, to save up for the basketball. In the sixth​ year, he puts​ $580 in the safe and realizes that he has exactly enough money to purchase the basketball.

Answer 1

The collector started with $284.31, saving 4% more each year. After 6 years, he had $580, enough to buy the basketball.

Let's denote the initial amount of money the collector puts in the safe as
`\( P \)`. Since he plans to put away 4% more money each year, the amount he puts away in the sixth year is
`\( P + 0.04P \)`, which is equivalent to
`\( 1.04P \)` The total amount of money in the safe after the sixth year is
`\( P + 1.04P = 2.04P \).`

Given that he has
`\( $580 \)` in the safe after the sixth year, we can set up the equation:

`\[ 2.04P = 580 \]`

Now, solve for \( P \):

`\[ P = (580)/(2.04) \]`

`\[ P \approx 284.31 \]`

So, the initial amount of money the collector put in the safe in the first year was approximately
`\( $284.31 \)`. Since he puts away 4% more each year, the amount he puts away in the sixth year is
`\( 1.04 * 284.31 \` approx 296.43 \).