Question

A person invests 7500 dollars in a bank. The bank pays 4% interest compounded annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 12800 dollars?

Answer 1

Answer:

13.6 years

Step-by-step explanation:

From the question;

• Money invested (Principal) is 7500 dollars
• Rate of interest is 4% compounded annually
• Amount, (money accrued ) is 12800 dollars

We are needed to determine the time it took for the money to reach the given amount;

• To solve the question we need to know the compound interest formula, that is;

$A=P(1-\frac{r}{100})^n$

Where n is the interest periods;

• Therefore, substituting the variables with the corresponding values, we can determine n.

Therefore;

$12800=7500(1-\frac{4}{100})^n$

Dividing both sides by 7500, we get;

$1.7067=(1-\frac{4}{100})^n$

$1.7067=(1.04)^n$

Introducing logarithms on both sides;

$log1.7067=log(1.04)^n\\nlog1.04=log1.7067\\n=\frac{log1.7067}{log1.04} \\n=13.63$

Thus, n=13.6 years

Thus, it would take 13.6 years for the invested money to accumulate to 12800 dollars