Question

1.Find the amount to which $ 6000 will grow if invested at 4% per year compounded quarterly for 5 years. Amount = $

2 How many years does it take a principle to triple in value at 2% p.a. compounded monthly? ___ years.
3 Solve for the number of days needed for $3000 compounded 3 times per year at 5% per annum to become $3100. ___ days.
4 An amount of $2,340 is deposited into a bank paying an annual interest rate of 3.1% compounded continuously. Fill in the blanks: The balance at the end of 3 years is __. It takes approximately __ years for the money to double.
5 Suppose that $700 is compounded continuously for 122 days and the account has grown to $715. The interest rate must have been __ % per year

Answer 1

The compound interest problems can be solved using the general compound interest formula for periodic compounding and a different formula for continuous compounding. The first question's answer is approximately $7401.20, the second is about 55.45 years, the third is 81.37 days, the fourth balance is $2568.33, with money doubling time of nearly 22.45 years, and the fifth question's interest rate is approximately 4.39% per year.

Step-by-step explanation:

To solve the various compound interest problems, the general formula A = P(1 + r/n)^(nt) is used for compounding interest, where A is the amount of money accumulated after n years, including interest, P is the principal amount, r is the annual interest rate (decimal), n is the number of times that interest is compounded per year, and t is the time the money is invested for in years. Another formula used when interest is compounded continuously is A = Pe^(rt), where e is the base of the natural logarithm, approximately equal to 2.71828.

For the first problem, using the compound interest formula for quarterly compounding we find:

A = $6000(1 + 0.04/4)^(4*5) = $6000(1.01)^20 ≈ $7401.20

For the second problem, we need to find how many years it takes for the principal to triple using the formula a bit differently:

3P = P(1 + 0.02/12)^(12t)

Solving for t, we find it takes approximately 55.45 years.

For the third problem, we look for the number of days needed for $3000 to become $3100:

$3100 = $3000(1 + 0.05/3)^(3*(t/365))

Solving for t, we find approximately 81.37 days are needed.

For the fourth problem, we'll use the continuous compounding formula:

The balance at the end of 3 years is A = $2340 * e^(0.031*3) ≈ $2568.33.

Regarding the time required for the money to double, we will set the amount A to $4680 (which is double of $2340) and solve for t:

$4680 = $2340 * e^(0.031*t)

Solving for t, the money takes approximately 22.45 years to double.

For the fifth problem, to find the interest rate given the account growth over 122 days, we use:

$715 = $700 * e^(r*(122/365))

Solving for r, the rate is approximately 4.39% per year.