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Chapter 19: Problem 469

The I. N. Vestor Company loaned out a total of \(\$ 30,000\), part at \(6 \%\) and the rest at \(9 \%\). The annual dividends from both investments was the same amount as that earned by the total loan if invested at \(7 \%\). Find the amount loaned out at each rate.

Short Answer

Expert verified
The I. N. Vestor Company loaned out \(20,000\) at 6% and \(10,000\) at 9%.

Step by step solution

01

Define the variables

Let x be the amount loaned at 6% and y be the amount loaned at 9%.
02

Write the equations

We have two equations to describe the problem. The first equation represents the total amount loaned and can be written as: \(x + y = 30,000\) The second equation represents the equivalence of annual dividends earned from both investments and the amount earned by the total loan if invested at 7%. The equation can be written as: \(0.06x + 0.09y = 0.07 \times 30,000\)
03

Solve the system of equations

We can solve the system of equations using the substitution or elimination method. We will use the substitution method. First, solve for x in the first equation: \(x = 30,000 - y\) Now, substitute this value for x into the second equation and solve for y: \(0.06(30,000 - y) + 0.09y = 0.07 \times 30,000\) \(1,800 - 0.06y + 0.09y = 2,100\) Combine like terms: \(0.03y = 300\) Now, divide by 0.03 to solve for y: \(y = 10,000\) Now that we have the value of y, we can find the value of x by substituting y back into the equation we derived from the first equation: \(x = 30,000 - 10,000\) \(x = 20,000\)
04

State the answer

The I. N. Vestor Company loaned out \(20,000\) at 6% and \(10,000\) at 9%.

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Key Concepts

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

Linear Algebra
Linear algebra is an essential field within mathematics, dealing with vector spaces and linear mappings between these spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces. When we talk about systems of equations, such as the one presented in our investment problem, we're tapping into linear algebra's ability to describe and solve linear equations and systems. This problem, made up of two variables and two algebraic equations, is a perfect example of a linear system that can be solved using methods taught in linear algebra, namely substitution or elimination.
Substitution Method
The substitution method is a technique used in solving a system of equations where you solve one equation for one variable and then substitute that solution into the other equation. This method is especially useful when the equations are not easily aligned for elimination. In our investment problem, once we defined the variables as the amounts loaned at different interest rates, we solved the first equation for one variable, and then we 'substituted' this expression into the second equation. By doing so, we reduced the system of equations from two variables to one, which allowed us to find the definitive values for both variables, solving the investment problem at hand.
Investment Problem
Investment problems often involve finding the best way to divide money among different investment opportunities to maximize returns or meet certain criteria. In our exercise, we tackled an investment problem where money was loaned out at two different interest rates, and the goal was to determine how much was loaned at each rate given that the annual returns were equivalent to a third rate. This type of problem can often be modeled using a system of algebraic equations, demonstrating the real-world application of linear algebra in financial contexts.
Algebraic Equations
Algebraic equations are mathematical statements that assert the equality of two expressions. They contain one or more variables, which are symbols representing unknown quantities that we aim to solve for. In the context of our exercise, we formulated two algebraic equations to represent the relationships between the amounts loaned at different interest rates and their respective returns. By solving these equations using the substitution method, we are able to find the numerical values of the variables. This is a skill set that underpins much of algebra and provides a basis for solving a wide variety of real-world problems.

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Most popular questions from this chapter

Consider a market that is controlled by two brands \(B_{1}\) and \(\mathrm{B}_{2} .\) Suppose it is known that \(10 \%\) of the buyers of each brand switch to the other brand during any given month. At the beginning of a given month 500 buyers are divided so that 300 purchase \(\mathrm{B}_{1}\) and 200 purchase \(\mathrm{B}_{2} .\) What will be the number of buyers who will purchase \(B_{1}\) and \(B_{2}\) during the given month?

1) Suppose we want to fit an exponential curve to a set of data. Explain how we could use the least squares method. 2) Fit an exponential curve to the following data: \((1,1.00)\), \((2,1.20),(3,1.80),(4,2.50),(5,3.60),(6,4.70),(7,6.60)\) \((8,9.10)\)

Ms. Wong has a total of \(\$ 4200\) invested in securities \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). The rates of annual dividends are \(4 \%, 6 \%\) and \(5 \%\) respectively, yielding total annual dividends of \(\$ 214 .\) If the sum of \(\mathrm{A}\) and \(\mathrm{B}\) is twice \(\mathrm{C}\), find the amount invested in each security.

Consider a buyer who at the beginning of each month decides whether to buy brand 1 or brand 2 that month. Each month the buyer will select either brand 1 or 2 . The selection he makes at the beginning of any one month depends at most on the selection he made in the immediately preceding month and not on any other previous selections. Let \(\mathrm{p}_{\mathrm{ij}}\) denote the probability that at the beginning of any month the buyer selects brand \(j\) given that he bought brand i the preceding month. Suppose we know that \(p_{11}=3 / 4, p_{12}=1 / 4\), \(\mathrm{p}_{21}=1 / 2\) and \(\mathrm{p}_{22}=1 / 2\) 1) First draw the transition diagram. 2) Assuming that the first month under consideration is January, determine the distribution of probabilities for April.

1) Generalize the least squares method of curve-fitting to a quadratic function. 2) Fit the following data points to a quadratic function: \((-5,2)\), \((-4,7),(-3,9),(-2,12),(-1,13),(0,14),(1,14),(2,13),(3,10)\) \((4,8),(5,4)\)
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