Question

A person invests 4500 dollars in a bank. The bank pays 5.25% interest compounded

monthly. To the nearest tenth of a year, how long must the person leave the money
in the bank until it reaches 6400 dollars?

Answer 1

To solve this problem, we can use the formula for compound interest. In this case, the person must leave the money in the bank for approximately 2.3 years to reach $6400.

Step-by-step explanation:

To solve this problem, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:

• A is the final amount of money
• P is the principal amount (initial investment)
• r is the annual interest rate (in decimal form)
• n is the number of times the interest is compounded per year
• t is the number of years

In this case, the principal amount (P) is $4500, the annual interest rate (r) is 5.25% or 0.0525, the interest is compounded monthly (n = 12), and we want to find the number of years (t) it will take for the amount to reach $6400.

Plugging in the values:

$6400 = $4500(1 + 0.0525/12)^(12t)

Dividing both sides by $4500:

1.42222222222 = (1 + 0.0525/12)^(12t)

Taking the natural logarithm of both sides:

ln(1.42222222222) = ln((1 + 0.0525/12)^(12t))

Using logarithmic properties to bring down the exponent:

ln(1.42222222222) = 12t * ln(1 + 0.0525/12)

Dividing both sides by 12 * ln(1 + 0.0525/12):

t = ln(1.42222222222) / (12 * ln(1 + 0.0525/12))

Using a calculator to evaluate the expression:

t ≈ 2.3

Therefore, the person must leave the money in the bank for approximately 2.3 years to reach $6400.

Answer 2

Step-by-step explanation:

We can use the formula for compound interest to solve this problem:

A = P(1 + r/n)^(nt)

where A is the amount of money after t years, P is the principal amount, r is the annual interest rate (as a decimal), n is the number of times the interest is compounded per year, and t is the time in years.

Using this formula, we can find the time t required for the investment to grow from $4500 to $6400.

First, we need to convert the annual interest rate of 5.25% to a monthly rate by dividing it by 12:

r = 0.0525 / 12 ≈ 0.004375

Next, we can substitute the given values into the formula and solve for t:

6400 = 4500(1 + 0.004375/12)^(12t)

6400/4500 = (1 + 0.004375/12)^(12t)

1.4222 = (1 + 0.004375/12)^(12t)

ln(1.4222) = 12t ln(1 + 0.004375/12)

t = ln(1.4222) / (12 ln(1 + 0.004375/12))

t ≈ 3.3

Rounding the answer to the nearest tenth of a year, we get:

t ≈ 3.3 years

Therefore, the person must leave the money in the bank for approximately 3.3 years until it reaches $6400.