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Chapter 5: Problem 51

Compound Interest If \(\$ 10,000\) is invested at an interest rate of 3\(\%\) per year, compounded semiannually, find the value of the investment after the given number of years. \(\begin{array}{llll}{\text { (a) } 5 \text { years }} & {\text { (b) } 10 \text { years }} & {\text { (c) } 15 \text { years }}\end{array}\)

Short Answer

Expert verified
After 5 years: $11,604.41; after 10 years: $13,439.16; after 15 years: $15,561.12.

Step by step solution

01

Understand the Problem

We know the principal amount is $10,000, the interest rate is 3% per annum, compounded semiannually. We need to find the value of the investment for different years.
02

Identify the Compound Interest Formula

The formula for compound interest is \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where \( A \) is the amount, \( P \) is the principal amount, \( r \) is the annual interest rate (as a decimal), \( n \) is the number of times interest is compounded per year, and \( t \) is the number of years.
03

Calculate for 5 Years

Using the formula, - \( P = 10000 \), \( r = \frac{3}{100} = 0.03 \), \( n = 2 \), \( t = 5 \).- Substitute these into the formula: \[A = 10000 \left(1 + \frac{0.03}{2}\right)^{2 \times 5}\]- Calculate:\[ A = 10000 \times \left(1 + 0.015\right)^{10} = 10000 \times 1.015^{10} \approx 11604.41 \]
04

Calculate for 10 Years

Now for 10 years, - \( P = 10000 \), \( r = 0.03 \), \( n = 2 \), \( t = 10 \).- Use the formula: \[A = 10000 \left(1 + \frac{0.03}{2}\right)^{2 \times 10}\]- Calculate:\[ A = 10000 \times 1.015^{20} \approx 13439.16 \]
05

Calculate for 15 Years

Finally for 15 years, - \( P = 10000 \), \( r = 0.03 \), \( n = 2 \), \( t = 15 \).- Use the formula: \[A = 10000 \left(1 + \frac{0.03}{2}\right)^{2 \times 15}\]- Calculate:\[ A = 10000 \times 1.015^{30} \approx 15561.12 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Investment
When you decide to grow your money, the initial amount you set aside is your investment. In financial terms, this is often called the "principal." For example, if you start with $10,000, this amount is your starting point. Investments can grow over time with the help of interest.
This initial sum of money is crucial because it determines how much you might earn in the future. Choosing the right kind of investment, like saving accounts, stocks, or bonds, depends on your financial goals and risk tolerance.
Understanding the power of an investment is vital as it sets the foundation for future wealth growth.
Interest Rate
The interest rate is a percentage that shows how much you earn on your investment over a year. In our example, the interest rate is 3% annually. This rate determines how fast your money will grow.
Interest can be simple, calculated only on the principal, or compound, calculated on both the principal and previously earned interest. Small changes in the interest rate can significantly affect your investment's outcome.
  • A higher interest rate leads to more earnings.
  • A lower interest rate means slower growth potential.
As you can see, selecting the right interest rate is crucial when planning your finances.
Compounded Semiannually
Compounding refers to the process where your investment earns interest on both the initial amount and the accumulated interest over time. When an investment is compounded semiannually, it means the interest is calculated and added to the principal twice a year.
This process accelerates growth because as your investment grows, the interest per period increases due to the higher total value. Here's how it works:
  • After six months, interest is added to the principal.
  • After another six months, interest is calculated on the new total (principal + interest).
This compounding strategy can lead to higher returns compared to annual compounding.
Financial Mathematics
Financial mathematics is the study of applying mathematical methods to solve financial problems. It helps calculate interest, investments, loans, and other financial matters. In our context, we use the compound interest formula.
This formula, \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \], helps determine the future value of an investment given it is invested at a particular interest rate and compounding frequency.
  • \( A \) is the amount after time \( t \).
  • \( P \) is the principal investment.
  • \( r \) is the annual interest rate.
  • \( n \) is the number of times interest is compounded per year.
  • \( t \) is the number of years the money is invested.
Mastering financial mathematics equips you to make informed investment decisions, thus optimizing your financial growth.

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Most popular questions from this chapter

Suppose you鈥檙e driving your car on a cold winter day \(\left(20^{\circ} \mathrm{F} \text { outside) and the engine overheats (at }\right.\) about \(220^{\circ} \mathrm{F}\) ). When you park, the engine begins to cool down. The temperature \(T\) of the engine \(t\) minutes after you park satisfies the equation $$\ln \left(\frac{T-20}{200}\right)=-0.11 t$$ (a) Solve the equation for \(T\) . (b) Use part (a) to find the temperature of the engine after \(20 \min (t=20) .\)

\(29-43\) . These exercises deal with logarithmic scales. Comparing Decibel Levels The noise from a power mower was measured at 106 \(\mathrm{dB}\) . The noise level at a rock concert was measured at 120 \(\mathrm{dB}\) . Find the ratio of the intensity of the rock music to that of the power mower.

Find the functions \(f \circ g\) and \(g \circ f\) and their domains. $$ f(x)=\log _{2} x, \quad g(x)=x-2 $$

\(17-24\) . These exercises use the radioactive decay model. Finding Half-life If 250 \(\mathrm{mg}\) of a radioactive element de- cays to 200 \(\mathrm{mg}\) in 48 hours, find the half-life of the element.

\(29-43\) . These exercises deal with logarithmic scales. Finding pH The hydrogen ion concentration of a sample of each substance is given. Calculate the pH of the substance. (a) Lemon juice: \(\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-3} \mathrm{M}\) (b) Tomato juice: \(\left[\mathrm{H}^{+}\right]=3.2 \times 10^{-4} \mathrm{M}\) (c) Seawater: \(\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-9} \mathrm{M}\)
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