A person invested $ 2, 200 $2,200 in an account growing at a rate allowing the money to double every 11 years. How long, to the nearest tenth of a year would it take for the value of the account to reach $ 7, 600 $7,600?

Question

Mathematics

Leonel

22 July, 01:27

A person invested $ 2, 200 $2,200 in an account growing at a rate allowing the money to double every 11 years. How long, to the nearest tenth of a year would it take for the value of the account to reach $ 7, 600 $7,600?

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Answers (1)

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Kayden Woods · Answer 1

It takes 38 years for the invested amount to reach a value of $7600.

Step-by-step explanation:

It is given that,

The invested amount in the account is $2,200. The money gets doubled every 11 years.

Therefore, it is determined that $2,200 will become $4,400 at the end of 11 years.

The increase of money in one year must be calculated in order to find how long it takes to reach the value of $7,600.

Money increased in 1 year = Invested amount in the account / 11 years

⇒ 2200 / 11

⇒ 200 dollars.

In 1 year, the amount the is increased by $200.

To find the number of years it takes to reach $7,600:

Number of years = Total amount / money increasing each year.

⇒ 7,600 / 200

⇒ 38 years.

It takes 38 years for the invested amount to reach a value of $7600.