Learn on PengiAoPS: Introduction to Algebra (AMC 8 & 10)Chapter 19: Exponents and Logarithms

Lesson 19.3: Interest-ing Problems

In this Grade 4 AoPS Introduction to Algebra lesson, students apply compound interest formulas to solve real-world problems involving future value, present value, and unknown interest rates. Using the compound interest formula and its inverse, learners practice isolating variables with exponents to find quantities such as the present value of a future sum or the annual percentage rate on a loan. This lesson is part of Chapter 19's coverage of exponents and their practical applications within the AMC 8 and AMC 10 competition math context.

Section 1

Solving Complex Interest Rate Equations

Property

To find unknown interest rates in complex scenarios, set up equations using the compound interest formula A=P(1+r)nA = P(1 + r)^n and solve by isolating the interest rate term: (1+r)n=AP(1 + r)^n = \frac{A}{P}, then r=(AP)1n1r = \left(\frac{A}{P}\right)^{\frac{1}{n}} - 1

Examples

Section 2

Compound Interest with Compounding Periods

Property

The future value AA of a principal amount PP (present value) after tt years with an annual interest rate rr compounded nn times per year is given by the formula:

A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}

To find the present value PP needed to achieve a future amount AA, the formula is rearranged:

P=A(1+rn)ntP = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}

Examples

  • To find the future value of a 1,000 dollar investment after 5 years at a 6% annual rate compounded semi-annually (n=2n=2), you calculate:
A=1000(1+0.062)2×5=1000(1.03)101343.92 dollarsA = 1000\left(1 + \frac{0.06}{2}\right)^{2 \times 5} = 1000(1.03)^{10} \approx 1343.92 \text{ dollars}
  • To find the present value you must invest to have 5,000 dollars in 3 years at a 4% annual rate compounded quarterly (n=4n=4), you calculate:
P=5000(1+0.044)4×3=5000(1.01)124437.04 dollarsP = \frac{5000}{\left(1 + \frac{0.04}{4}\right)^{4 \times 3}} = \frac{5000}{(1.01)^{12}} \approx 4437.04 \text{ dollars}

Explanation

This formula is an extension of simple compound interest, accounting for interest being calculated more than once a year. The annual rate rr is divided by the number of compounding periods nn to get the rate per period. The exponent ntnt represents the total number of times interest is compounded over the investment''s lifetime. This skill allows for more precise calculations of future and present value in real-world financial scenarios where interest is often compounded semi-annually, quarterly, or even monthly.

Section 3

Present Value Calculations

Property

Present value is the current worth of a future amount of money, calculated using:

PV=FV(1+r)nPV = \frac{FV}{(1 + r)^n}
where PVPV is present value, FVFV is future value, rr is the interest rate (as a decimal), and nn is the number of compounding periods.

Examples

Book overview

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Chapter 19: Exponents and Logarithms

  1. Lesson 1

    Lesson 19.1: Exponential Functions

  2. Lesson 2

    Lesson 19.2: Show Me the Money

  3. Lesson 3Current

    Lesson 19.3: Interest-ing Problems

  4. Lesson 4

    Lesson 19.4: What is a Logarithm?

Lesson overview

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Section 1

Solving Complex Interest Rate Equations

Property

To find unknown interest rates in complex scenarios, set up equations using the compound interest formula A=P(1+r)nA = P(1 + r)^n and solve by isolating the interest rate term: (1+r)n=AP(1 + r)^n = \frac{A}{P}, then r=(AP)1n1r = \left(\frac{A}{P}\right)^{\frac{1}{n}} - 1

Examples

Section 2

Compound Interest with Compounding Periods

Property

The future value AA of a principal amount PP (present value) after tt years with an annual interest rate rr compounded nn times per year is given by the formula:

A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}

To find the present value PP needed to achieve a future amount AA, the formula is rearranged:

P=A(1+rn)ntP = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}

Examples

  • To find the future value of a 1,000 dollar investment after 5 years at a 6% annual rate compounded semi-annually (n=2n=2), you calculate:
A=1000(1+0.062)2×5=1000(1.03)101343.92 dollarsA = 1000\left(1 + \frac{0.06}{2}\right)^{2 \times 5} = 1000(1.03)^{10} \approx 1343.92 \text{ dollars}
  • To find the present value you must invest to have 5,000 dollars in 3 years at a 4% annual rate compounded quarterly (n=4n=4), you calculate:
P=5000(1+0.044)4×3=5000(1.01)124437.04 dollarsP = \frac{5000}{\left(1 + \frac{0.04}{4}\right)^{4 \times 3}} = \frac{5000}{(1.01)^{12}} \approx 4437.04 \text{ dollars}

Explanation

This formula is an extension of simple compound interest, accounting for interest being calculated more than once a year. The annual rate rr is divided by the number of compounding periods nn to get the rate per period. The exponent ntnt represents the total number of times interest is compounded over the investment''s lifetime. This skill allows for more precise calculations of future and present value in real-world financial scenarios where interest is often compounded semi-annually, quarterly, or even monthly.

Section 3

Present Value Calculations

Property

Present value is the current worth of a future amount of money, calculated using:

PV=FV(1+r)nPV = \frac{FV}{(1 + r)^n}
where PVPV is present value, FVFV is future value, rr is the interest rate (as a decimal), and nn is the number of compounding periods.

Examples

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 19: Exponents and Logarithms

  1. Lesson 1

    Lesson 19.1: Exponential Functions

  2. Lesson 2

    Lesson 19.2: Show Me the Money

  3. Lesson 3Current

    Lesson 19.3: Interest-ing Problems

  4. Lesson 4

    Lesson 19.4: What is a Logarithm?