Problem 63 - Algebra for College Students

Question

Solve each problem. Randall wants to mix 50 lb of nuts worth \(\$ 2\) per \(\mathrm{lb}\) with some nuts worth \(\$ 6\) per \(\mathrm{lb}\) to make a mixture worth \(\$ 5\) per \(1 \mathrm{~b}\). How many pounds of \(\$ 6\) nuts must he use?

Step-by-Step Solution

Answer

Randall must use 150 pounds of the \( \$6 \) nuts.

Step 1: Define Variables

Let\( x \) be the number of pounds of \( \$6 \) per pound nuts Randall will use.

Step 2: Set Up the Equation

The total weight of the mixture will be \( 50 + x \) pounds. The total cost of the mixture will be the sum of the costs of the inputs, and it should equal the cost of the final mixture. Set up the equation based on the costs: \[ 2 \times 50 + 6x = 5(50 + x) \]

Step 3: Simplify and Solve the Equation

Distribute and simplify the equation: \[ 100 + 6x = 250 + 5x \] Subtract \( 5x \) from both sides: \[ 100 + x = 250 \] Subtract \( 100 \) from both sides: \[ x = 150 \]

Step 4: Conclusion

Randall must use \( 150 \) pounds of the \( \$6 \) per pound nuts to get the desired mixture.

Key Concepts

Variable Definition

In mixture problems, we need to represent unknown quantities with variables. This helps in setting up and solving equations easily. In this problem, we are asked to find the amount of \(\$6\) per pound nuts needed. Let's define our variable: \(x\), which represents the number of pounds of \(\$6\) per pound nuts Randall will use. Choosing defined variables simplifies the complexity and makes equations more manageable. Always start by identifying what you need to find and let your variable stand for that unknown.

Equation Setup

Next, we set up an equation based on the problem's conditions. We know Randall has 50 pounds of \(\$2\) per pound nuts, and he needs to mix them with \(\$6\) per pound nuts (denoted by \(x\)) to make a mixture worth \(\$5\) per pound.
* The total weight of the mixture: \(50 + x\).
* The cost of \(\$2\) nuts: \(2 \times 50 = 100\).
* The cost of \(\$6\) nuts: \(6x\).
* Total cost of the mixture: \(5(50 + x)\).
By equating the total cost of inputs to the final mixture cost, we get:
\[2 \times 50 + 6x = 5(50 + x)\]

Cost Mixture Calculation

Now, let's break down the cost mixture step-by-step:
1. Calculate the cost for each type of nut.
2. Determine the total cost to ensure the mixture cost matches the desired outcome.
Remember:
* \(2 \times 50\) is the expense of \(\$2\) nuts.
* \(6x\) is the expense of \(\$6\) nuts.
* Final mixture cost: \[5(50 + x)\].
By setting up the equation \[2 \times 50 + 6x = 5(50 + x)\], we balance the cost. This ensures the mixture is worth \(\$5\) per pound. This approach works for any mixture problem—if we account for each component's cost accurately.

Algebraic Simplification

Solving the mixture equation involves standard algebraic techniques.
* Start by distributing and combining like terms.
\[ 2 \times 50 + 6x = 5(50 + x)\]
Simplifies to: \[ 100 + 6x = 250 + 5x\]
* Next, isolate \(x\) by subtracting \(5x\) from both sides:
\[ 100 + x = 250\]
* Finally, solve for \(x\) by subtracting 100:
\[ x = 150\]
This means Randall needs 150 pounds of \(\$6\) nuts. Understanding these algebraic steps is crucial—they are foundational blocks for solving similar problems. Just follow the steps logically and carefully check for errors.