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Chapter 3: Problem 28

Nenuphar wants to invest a total of \(\$ 30,000\) into two savings accounts, one paying \(6 \%\) per year in interest and the other paying \(9 \%\) per year in interest (a more risky investment). If after 1 year she wants the total interest from both accounts to be \(\$ 2100\), how much should she invest in each account?

Short Answer

Expert verified
Invest \$20000 in the 6% account and \$10000 in the 9% account.

Step by step solution

01

Define Variables

Let \( x \) be the amount invested in the account paying 6% interest, and \( y \) be the amount invested in the account paying 9% interest.
02

Set up Equations

We have two main equations based on the problem. The total investment equation: \( x + y = 30000 \). The total interest equation: \( 0.06x + 0.09y = 2100 \).
03

Solve for One Variable

From the total investment equation: \( y = 30000 - x \).
04

Substitute and Solve

Substitute \( y \) into the total interest equation: \[ \begin{array}{rcl} 0.06x + 0.09(30000 - x) & = & 2100 \end{array} \] Simplify the equation: \[ 0.06x + 2700 - 0.09x = 2100 \] \[ -0.03x + 2700 = 2100 \] \[ -0.03x = -600 \] \[ x = 20000 \]
05

Find y

Using the value of \( x \) in the total investment equation, \[ y = 30000 - 20000 = 10000 \]

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Key Concepts

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

linear equations
In mathematics, a linear equation is any equation that can be written in the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. They graph as straight lines and typically have one or more variables.
Linear equations are used in real-world scenarios like Nenuphar's investment problem. Here, we have two linear equations: \( x + y = 30000 \) and \( 0.06x + 0.09y = 2100 \).
Each equation represents a condition. The first equation represents the total amount of money invested, whereas the second represents the interest earned from both accounts.
Solving these equations simultaneously allows us to determine how much money has to be invested in each type of account.
interest calculation
Interest calculation is crucial in finance as it helps determine how much earnings or cost will be accrued over time. Simple interest is often calculated using the formula: \( I = P \times R \times T \).
In the formula, \( I \) is the interest, \( P \) is the principal amount (initial investment), \( R \) is the rate of interest, and \( T \) is the time period.
In Nenuphar's problem, the interest from the 6% account can be calculated as \( 0.06x \) and from the 9% account as \( 0.09y \). The total interest earned from both accounts is \( 2100 \). Using these simple interest formulas allows us to derive the necessary equations to solve the problem.
substitution method
The substitution method is a common technique for solving systems of linear equations. It involves solving one equation for one variable and then substituting that result into the other equation.
In Nenuphar's problem, we start with two equations: \( x + y = 30000 \) and \( 0.06x + 0.09y = 2100 \).
Step 1: Solve the total investment equation for \( y \): \( y = 30000 - x \).
Step 2: Substitute this expression into the total interest equation: \( 0.06x + 0.09(30000 - x) = 2100 \).
Step 3: Simplify and solve for \( x \): \[ \begin{align*} 0.06x + 2700 - 0.09x &= 2100 \ -0.03x + 2700 &= 2100 \ -0.03x &= -600 \ x &= 20000 \end{align*} \].
Step 4: Substitute \( x = 20000 \) back into the total investment equation to find \( y \): \[ y = 30000 - 20000 = 10000 \].
This method is particularly useful for solving problems where you have multiple linear equations.

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Most popular questions from this chapter

A husband drives a heavily loaded truck that can go only 55 mph on a 650 -mile turnpike trip. His wife leaves on the same trip 2 hours later in the family car averaging \(70 \mathrm{mph}\). Recall that distance traveled \(=\) speed \(\cdot\) time traveled. a. Derive an expression for the distance, \(D_{h},\) the husband travels in \(t\) hours since he started. b. How many hours has the wife been traveling if the husband has traveled \(t\) hours \((t \geq 2) ?\) c. Derive an expression for the distance, \(D_{w}\), that the wife will have traveled while the husband has been traveling for \(t\) hours \((t \geq 2)\) d. Graph distance vs. time for husband and wife on the same axes. e. Calculate when and where the wife will overtake the husband. f. Suppose the husband and wife wanted to arrive at a restaurant at the same time, and the restaurant is 325 miles from home. How much later should she leave, assuming he still travels at \(55 \mathrm{mph}\) and she at \(70 \mathrm{mph} ?\)

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