Melanie puts $$\$ 1,100$$ in an investment account that she expects will make \(5 \%\) interest for each three-month period. However, after a year she realizes she was wrong about the interest rate and she has $$\$ 50$$ less than she expected. Assuming the interest rate the account earns is constant, which of the following equations expresses the total amount of money, \(x\), she will have after \(t\) years using the actual rate? A) \(x=1,100(1.04)^{4 t}\) B) \(x=1,100(1.05)^{4 t-50}\) C) \(x=1,100(1.04)^{t / 3}\) D) \(x=1,100(1.035)^{4 t}\)

Short Answer

Expert verified
The short answer is: The correct equation is \(x = 1100(1.035)^{4 t}\).

Step by step solution

01

Find the expected amount after one year with the given interest rate

Using the given interest rate of $$5\%$$ per quarter, we can calculate the expected amount Melanie should have after one year. Since there are four quarters in a year, the interest rate will be applied four times. The formula for compound interest is: \[A=P(1+\dfrac{r}{n})^{nt}\] Where: - A is the future value of the investment - P is the initial investment - r is the annual interest rate (as a decimal) - n is the number of times the interest rate is compounded per year - t is the time in years In this case, P = $$1100$$, r = $$0.05$$, n = 4 (four quarters per year), t = 1 (one year). Applying these values to the formula: \[A = 1100(1+0.05/4)^{4(1)}\]
02

Calculate the expected amount

Now, we can plug in the values and calculate the expected amount: \[A = 1100(1+0.0125)^{4}\] \[A = 1100(1.0125)^{4}\] \[A \approx 1100(1.050946)\] \[A \approx 1156.04\]
03

Calculate the amount Melanie actually had after one year

We are given that Melanie had $$50$$ less than expected after one year. Therefore, the actual amount she had after one year is $$1106.04 - 50$$: \[A_{actual} = 1106.04 - 50\] \[A_{actual} = 1056.04\]
04

Find the actual interest rate

As we want to find a formula for the total amount of money, x, she will have after t years using the actual rate, we can arrange the formula from Step 1: \[x = P(1+\dfrac{r}{n})^{nt}\] We know that at the end of one year, the actual amount she had was $$1056.04$$. Plug in the values for P, n and t: \[1056.04 = 1100(1+\dfrac{r}{4})^{4(1)}\] Now we need to calculate the actual r.
05

Compare the given equations with the actual situation

Looking at the given options, we can identify which equation fits the problem: A) \(x = 1100(1.04)^{4t}\) doesn't match our equation. B) \(x = 1100(1.05)^{4t-50}\) doesn't make sense, as the exponent should not have $$-50$$. C) \(x = 1100(1.04)^{t/3}\) the exponent should be multiplied by the number of quarters in a year. D) \(x = 1100(1.035)^{4t}\) is very close to our derived equation, so it's plausible and we can consider this as the best option, but still we don't have the interest rate $$r$$. Checking option D, plugging the actual amount ($$1056.04$$) in place of x: \[1056.04 = 1100(1.035)^{4(1)}\] Solving for this equation, we can see that it holds true, so the correct equation is: \(x = 1100(1.035)^{4 t}\)

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Key Concepts

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

Understanding Compound Interest
When it comes to understanding the growth of investments over time, compound interest plays a central role. Compound interest is often introduced in SAT math practice because it exemplifies real-world financial scenarios and involves exponential functions, which are a key topic in algebra.

Compound interest refers to earning or paying interest on both the initial principal and the accumulated interest from previous periods. Unlike simple interest, which only earns interest on the principal amount, compound interest allows your balance to grow at an accelerated rate since each interest calculation is based on the increasing amount.

Mathematically, compound interest is represented by the formula:
\[A = P(1 + \frac{r}{n})^{nt}\]
where:
  • \(A\) is the accumulated amount after time \(t\).
  • \(P\) is the principal amount.
  • \(r\) is the annual interest rate in decimal form.
  • \(n\) is the number of times interest is compounded per year.
  • \(t\) is the time the money is invested for, in years.

Regular practice with compound interest problems can significantly improve students' understanding of exponential growth and its applications, reinforcing their algebra skills along the way.
Exponential Functions in Financial Mathematics
Exponential functions are central to many SAT math problems, particularly those involving financial contexts like compound interest. An exponential function is a mathematical expression in which a constant base is raised to a variable exponent. This concept is crucial for tracking the growth of investments over time.

Most exponential growth scenarios, including compound interest, are represented by functions like:
\[f(x) = a(1 + r)^{nx}\]
where \(a\) is the initial amount, \(r\) is the rate of growth (or decay if negative), and \(nx\) denotes the number of times the growth is compounded over the period \(x\).

In the context of the SAT problem about Melanie's investment, we see that the function representing her investment's growth takes on the exponential form, with her principal amount forming the base and the number of compounding periods forming the exponent.
Recognizing the structure of an exponential function is key for students to efficiently approach these problems and solve them without confusion.
Algebraic Expressions and Their Application
The ability to work with algebraic expressions is essential for solving various SAT math problems. An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like \(x\) or \(y\)), and operations between them, such as addition, subtraction, multiplication, and division.

In our example, the task was to find the algebraic expression that correctly represents Melanie's financial situation. Students must understand how to manipulate algebraic expressions to isolate variables, combine like terms, and apply the correct algebraic structure to describe a real-world situation.

Building a strong foundation in algebraic manipulation allows students to approach and solve problems systematically. Whether it's simplifying expressions, factoring, or working with exponents, the ability to confidently handle algebra underpins success in SAT math and beyond. Regularly practicing algebra problems helps students develop fluidity in converting word problems into algebraic expressions, a skill that is invaluable both on standardized tests and in real-life problem-solving scenarios.

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Most popular questions from this chapter

Students in a physics class are studying how the angle at which a projectile is launched on level ground affects the projectile's hang time and horizontal range. Hang time can be calculated using the formula \(t=\frac{2 v \cdot \sin (\theta)}{g}\), where \(t\) is the hang time in seconds, \(v\) is the initial launch velocity, \(\theta\) is the projectile angle with respect to level ground, and \(g\) is the acceleration due to gravity, defined as \(9.8 \mathrm{~m} / \mathrm{s}^2\). Horizontal range can be calculated using the formula \(R=\frac{v^2 \sin (2 \theta)}{g}\), where \(R\) is the distance the projectile travels from the launch site, in feet. Which of the following gives the value of \(v\), in terms of \(R, t\), and \(\theta\) ? A) \(v=\frac{t \sin (\theta)}{2 R \sin (\theta)}\) B) \(v=\frac{2 t \sin (\theta)}{R \sin (\theta)}\) C) \(v=\frac{2 R \sin (\theta)}{t \sin (2 \theta)}\) D) \(v=\frac{2 R \sin (2 \theta)}{t \sin (\theta)}\)

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Luciano measured the amount of water that evaporated over a period of time from a container holding \(w\) ounces of water, where \(w\) is greater than 12. By the end of the first day, the cup had lost 2 ounces of water. By the end of the 7th day, the cup had lost an additional 8 ounces of water. By the end of the 11th day, the cup had lost half of the water that remained after the 7th day. Which of the following represents the remaining amount of water, in ounces, in Luciano's container at the end of the 11th day? A) \(\frac{w-2}{8}\) B) \(\frac{w-2}{2}-10\) C) \(\frac{1}{2} w-10\) D) \(\frac{w-10}{2}\)

A) NO CHANGE B) ameliorated C) gone down D) subsided

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