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One grade of tea, worth \(3.20\)dollar per pound, is to be mixed with another grade worth \(2\)dollar per pound to make \(20\)dollar pounds that will sell for \(2.72\)dollar per pound. How much of each grade of tea must be used?

Short Answer

Expert verified
By following these steps, one can determine how much of each grade of tea must be used.

Step by step solution

01

Set up the equations

Let \(x\) be the amount of first grade tea and \(y\) be the amount of second grade tea to be used. Using the information from the problem, the following two equations can be written - based on total weight and the price per pound:(i) \(x + y = 20\) (Total weight)(ii) \(3.20x + 2y = 2.72 \times 20\) (Average price)
02

Solve the equations

Now, solve these equations by substituting equation (i) into equation (ii):Substituting \(y = 20 - x\) into equation (ii) gives:\(3.20x + 2(20-x) = 2.72 \times 20\)Solving for \(x\) provides the amount of the first grade of tea to be used.
03

Calculate the amount of second grade of tea

After finding \(x\), substitute \(x\) into equation (i) to find \(y\), which signifies the amount of the second grade tea to be used.

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Key Concepts

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

System of Equations
Solving word problems using systems of equations is a powerful technique in algebra. It allows us to find unknown values by setting up equations based on given information. In the context of our tea blending problem, we use a system of equations to determine how much of each tea type we need.

The first step is to create equations from the problem information. First, we establish an equation for the total weight of the tea mix. In this problem, the tea must add up to 20 pounds, hence the equation: \(x + y = 20\), where \(x\) is the weight of the expensive tea and \(y\) is the weight of the cheaper tea. The second equation represents the total cost of the tea: \(3.20x + 2y = 2.72 \times 20\). These equations form the system of equations we need to solve.

By using substitution or elimination methods, we can solve these equations to find the values of \(x\) and \(y\). These solutions provide us with the exact quantities of each tea required to create the desired blend.
Mixing Problems
Mixing problems are common in algebra and involve combining different components to achieve a desired property or mixture. In the case of our exercise, we are dealing with blending two types of tea to reach a specific price per pound. This type of problem teaches us to consider both the total quantity and the total value when mixing components.

In mixing problems, each ingredient has a known price or value, and we aim to find a combination that matches a target. Here, the tea blend must total 20 pounds and sell for \(2.72\) dollars per pound. We use the price information to weigh the contribution of each type of tea to the overall value of the mix.

Mixing problems typically involve setting up an equation that balances the total amounts and another that balances the prices. Solving these equations helps us find the amount of each component required to create the optimal mixture.
Linear Equations
Linear equations are fundamental to understanding algebra and solving real-world problems. They are equations of the first degree, meaning they involve variables raised to the power of one. In our tea problem, the equations \(x + y = 20\) and \(3.20x + 2y = 54.4\) are linear equations.

These equations are linear because they plot straight lines when graphed on a coordinate plane. The linearity means each equation can represent constraints or relationships in a problem. The first equation handles the total weight constraint, while the second manages the overall cost constraint in our scenario.

Understanding linear equations allows us to analyze and interpret relationships between variables effectively. By solving these equations, either algebraically or graphically, we determine the exact values needed to satisfy the conditions of the problem, like balancing the mix of two types of tea.

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