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Chapter 4: Problem 34

Udit received \(\$ 1200\) from his parents as a graduation present. He invested part of it at \(4 \%\) interest, and he invested the remainder at \(6 \%\). If the total yearly interest amounted to \(\$ 62\), how much did he invest at each rate?

Short Answer

Expert verified
Udit invested \$500 at 4% and \$700 at 6%.

Step by step solution

01

Define Variables

Let's define the variables for the amounts invested. Let \( x \) be the amount Udit invested at \( 4\% \) interest, and \( y \) be the amount he invested at \( 6\% \) interest.
02

Set Up Equations

Now, we have two pieces of information:1. Udit invested a total of \( \\(1200 \), so the equation is: \[ x + y = 1200 \]2. The total yearly interest from both investments is \( \\)62 \). Thus, the interest from the \( 4\% \) investment \( (0.04x) \) plus the interest from the \( 6\% \) investment \( (0.06y) \) equals \( \$62 \):\[ 0.04x + 0.06y = 62 \]
03

Solve the System of Equations

Using the first equation \( x + y = 1200 \), we can express one variable in terms of the other. Let's express \( y \):\[ y = 1200 - x \]Now substitute \( y \) in the second equation:\[ 0.04x + 0.06(1200 - x) = 62 \]This simplifies to:\[ 0.04x + 72 - 0.06x = 62 \]
04

Simplify and Solve for x

Combine the terms in the equation:\[ 72 - 0.02x = 62 \]Isolate the variable \( x \) by subtracting \( 72 \) from both sides:\[ -0.02x = -10 \]Divide both sides by \( -0.02 \):\[ x = 500 \]
05

Solve for y

Now that we have \( x = 500 \), use the equation \( y = 1200 - x \) to find \( y \):\[ y = 1200 - 500 = 700 \]
06

Check Your Work

We calculated that \( x = 500 \) and \( y = 700 \). Verify by checking if the interests add up to \( \$62 \):For the \( 4\% \) investment: \( 0.04 \times 500 = 20 \).For the \( 6\% \) investment: \( 0.06 \times 700 = 42 \).Total interest: \( 20 + 42 = 62 \), which matches the given total.

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Key Concepts

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

System of Equations
A system of equations is a set of two or more equations with the same variables. Solving a system of equations involves finding values for the variables that satisfy all equations simultaneously.
The most common methods to solve a system of equations include substitution, elimination, and graphical representation.
  • Substitution: This method involves solving one equation for a variable and substituting the result into another equation.
  • Elimination: By adding or subtracting equations, you eliminate one variable, allowing you to solve for the other.
  • Graphical Representation: Plotting each equation on a graph to find where they intersect can also solve a system of equations. The coordinates of the intersection point provide the solution.
In the given exercise, we used substitution. By expressing one variable in terms of another (i.e., \( y = 1200 - x \)), we substituted the expression into the second equation (\( 0.04x + 0.06y = 62 \)) to find the solution.
Interest Calculation
Interest calculation helps understand the growth of borrowed or invested money over time at a specified rate. Interest can be calculated through a simple formula:
  • Simple Interest: Calculated using the formula \( I = P \times r \times t \), where \( I \) is the interest, \( P \) the principal amount, \( r \) the interest rate, and \( t \) the time period.
In our problem, we had two investments with different interest rates, \(4\%\) and \(6\%\). The total interest from these two investments equals \$62. To find this, we calculated as follows:
  • Interest from the first investment: \( 0.04x \)
  • Interest from the second investment: \( 0.06y \)
By adding these two interest amounts, we equaled them to the total interest to solve the problem.
Variable Definition
Defining variables is key in translating real-world scenarios into mathematical statements. By defining variables, we can create equations that represent situations and solve them.
  • Choosing clear and meaningful variable names or symbols helps keep track of what each variable represents.
  • Variables should be defined based on what needs to be determined in the problem being solved.
In the given exercise, two variables were defined: \( x \) for the amount invested at \(4\%\) and \( y \) for the amount at \(6\%\). Defining these variables allowed for setting up equations that describe the situation and finding the solution by using algebraic methods.

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