Problem 51
Question
You want to invest \(\$ 5000\) in a certificate of deposit for 12 months. You are given the options below. Which would you choose? Explain. (a) \(r=5.25 \%\), quarterly compounding (b) \(r=5 \%\), monthly compounding (c) \(r=4.75 \%\), continuous compounding
Step-by-Step Solution
Verified Answer
The best choice would be option (a) which yields 5263.30 dollars after 1 year.
1Step 1: Calculate the Amount for Option (a)
The given interest rate is 5.25%, or 0.0525 in decimal form, the time is 1 year, and it's quarterly compounding which means \(n=4\). Substituting these values into the formula, we get \(A = 5000 * (1+0.0525/4)^(4*1) = 5000 *(1.013125)^4 = 5000 * 1.052661 = 5263.30\) dollars.
2Step 2: Calculate the Amount for Option (b)
The interest rate for option (b) is 5.00% or 0.05 in decimal form, and it's compounded monthly which implies \(n=12\). Inserting these values into the formula, we have \(A = 5000* (1+0.05/12)^(12*1) = 5000*(1.004167)^{12} = 5000 *1.051267= 5256.34\) dollars.
3Step 3: Calculate the Amount for Option (c)
The interest rate for option (c) is 4.75% or 0.0475, and it's compounded continuously. Utilizing the continuous compounding formula, \(A = 5000*e^{0.0475*1} = 5000* 1.048812 = 5244.06\) dollars.
4Step 4: Choose the Best Option
Now, comparing the Future value of money after 1 year in all three options, we can see that option (a) yields the most amount after 1 year. Thus, the best choice would be option (a): 5.25% with quarterly compounding.
Key Concepts
Quarterly CompoundingMonthly CompoundingContinuous Compounding
Quarterly Compounding
When we talk about quarterly compounding, we're referring to the process of earning interest on investment four times a year. This means every quarter, the interest earned is added to the initial principal, and then this new total becomes the next quarter's principal. Over time, this can lead to more significant growth compared to less frequent compounding.
To calculate the amount with quarterly compounding, you use the formula:
For instance, with an investment of $5000 at 5.25% interest over one year, you calculate as follows:
To calculate the amount with quarterly compounding, you use the formula:
- \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
For instance, with an investment of $5000 at 5.25% interest over one year, you calculate as follows:
- \[ A = 5000 \left(1 + \frac{0.0525}{4}\right)^{4 \times 1} = 5263.30 \]
Monthly Compounding
Monthly compounding takes the compounding process a step further by calculating and adding interest twelve times a year. This frequent updating of balance results in slightly higher returns compared to quarterly compounding.
The formula used is the same as for other discrete compounding methods, adjusted for the frequency:
To illustrate, by investing $5000 at a 5% annual interest rate with monthly compounding for one year, we calculate:
The formula used is the same as for other discrete compounding methods, adjusted for the frequency:
- \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
To illustrate, by investing $5000 at a 5% annual interest rate with monthly compounding for one year, we calculate:
- \[ A = 5000 \left(1 + \frac{0.05}{12}\right)^{12 \times 1} = 5256.34 \]
Continuous Compounding
Continuous compounding represents a theoretical limit of compounding interest where interest is calculated and added to the principal an infinite number of times per year. This uses the natural base "e" to evaluate the future value. The formula is given by:
For instance, with $5000 at a 4.75% rate, continuous compounding is calculated as:
- \[ A = Pe^{rt} \]
For instance, with $5000 at a 4.75% rate, continuous compounding is calculated as:
- \[ A = 5000 \times e^{0.0475 \times 1} = 5244.06 \]
Other exercises in this chapter
Problem 51
In Exercises, solve for \(x\) or \(t\). $$ e^{\ln x}=4 $$
View solution Problem 51
Use a graphing utility to graph the normal probability density function with \(\mu=0\) and \(\sigma=2,3\), and 4 in the same viewing window. What effect does th
View solution Problem 52
In Exercises, determine an equation of the tangent line to the function at the given point. $$ g(x)=\log _{10} 2 x $$ $$ (5,1) $$
View solution Problem 52
In Exercises, solve for \(x\) or \(t\). $$ e^{\ln x^{2}}-9=0 $$
View solution