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Compound Interest Calculation: $1000 Investment at 5% Monthly Compounding for 5, 10, 20 Years
Mathematics
Grade 11 (Senior High School)
Question Content
The equation for compound interest is $A = P\\left(1+\\frac{r}{n}\\right)^{nt}$ where $P$ is the principal amount invested at interest rate $r$ compounded $n$ times per year. In $t$ years $P$ will grow to the amount $A$. The product $nt$ is the total number is payment periods. Let's say grandma gave you $1,000 (P=1000) as a gift for your thirteenth birthday and you decide to invest it. You put the money in an account with an interest rate of 5% ($r=0.05$) compounded monthly ($n=12$). How much money would you have after 5 years? 10 years? 20 years? Make a graph of the amount relative to the time it's invested.
Correct Answer
After 5 years: ≈$1283.36; After 10 years: ≈$1647.01; After 20 years: ≈$2712.64
Detailed Solution Steps
1
Step 1: Identify all given values: $P=1000$, $r=0.05$, $n=12$
2
Step 2: Calculate the amount after 5 years ($t=5$): Substitute values into the formula: $A = 1000\\left(1+\\frac{0.05}{12}\\right)^{12*5} = 1000\\left(1+0.0041667\\right)^{60} ≈ 1000*1.28336 ≈ 1283.36$
3
Step 3: Calculate the amount after 10 years ($t=10$): Substitute values into the formula: $A = 1000\\left(1+\\frac{0.05}{12}\\right)^{12*10} = 1000\\left(1+0.0041667\\right)^{120} ≈ 1000*1.64701 ≈ 1647.01$
4
Step 4: Calculate the amount after 20 years ($t=20$): Substitute values into the formula: $A = 1000\\left(1+\\frac{0.05}{12}\\right)^{12*20} = 1000\\left(1+0.0041667\\right)^{240} ≈ 1000*2.71264 ≈ 2712.64$
5
Step 5: For the graph: Plot time $t$ (in years, x-axis: 0 to 20) against amount $A$ (y-axis: 1000 to 2800). The graph will be an increasing exponential curve starting at (0, 1000), passing through (5, 1283.36), (10, 1647.01), and (20, 2712.64).
Knowledge Points Involved
1
Compound Interest Formula
The formula $A = P\\left(1+\\frac{r}{n}\\right)^{nt}$ calculates the future value $A$ of a principal investment $P$, where $r$ is the annual interest rate (decimal), $n$ is the number of compounding periods per year, and $t$ is the time in years. It is used for investments that earn interest on both the principal and accumulated interest over time, unlike simple interest which only applies to the principal.
2
Exponential Growth
Compound interest follows an exponential growth model, where the value of the investment increases at a rate proportional to its current value. This results in a curved, upward-sloping graph, as the growth accelerates over time. Exponential growth is common in finance, population studies, and natural sciences.
3
Substitution and Evaluation of Algebraic Formulas
This skill involves replacing variables in a formula with given numerical values and computing the result using order of operations (PEMDAS/BODMAS). It is a foundational algebraic skill used to solve real-world problems with predefined mathematical relationships.
4
Graphing Exponential Functions
When graphing the compound interest function $A(t) = 1000\\left(1+\\frac{0.05}{12}\\right)^{12t}$, the x-axis represents time $t$ and the y-axis represents the accumulated amount $A$. The graph starts at the principal value (when $t=0$) and increases exponentially, showing the growth of the investment over time. Key points (t, A) are plotted and connected to form the curve.
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