Problem 80 - Precalculus

Question

A family is coloring Easter eggs and wants to make 2 shades of purple, "light purple" and "deep purple." They have 30 tablespoons of deep red solution and 2 tablespoons of blue solution. If "light purple" consists of \(2 \%\) blue solution and "deep purple" consists of \(10 \%\) blue solution, how much of each version of purple solution can be created?

Step-by-Step Solution

Answer

0 tbsp of 'light purple' and 10 tbsp of 'deep purple' can be created.

Step 1: Set Up Equations

Let the amount of 'light purple' solution be represented by \( x \) tablespoons and the 'deep purple' solution be represented by \( y \) tablespoons. We know that the total amount of blue solution used in both solutions equals the total blue solution available.

Step 2: Equation for Blue Solution

The equation for the blue solution is given by the percentages: \( 0.02x + 0.10y = 2 \). Here, \( 0.02x \) represents the blue solution in 'light purple' and \( 0.10y \) represents the blue solution in 'deep purple'.

Step 3: Equation for Red Solution

Similarly, we set up an equation for the red solution. The equation for the red solution is: \( 0.98x + 0.90y = 30 \). Here, \( 0.98x \) and \( 0.90y \) represent the red content in 'light purple' and 'deep purple', respectively.

Step 4: Solve Simultaneous Equations

First, solve the blue solution equation: \( 0.02x + 0.10y = 2 \). Simplifying gives \( x = 50 - 5y \) by multiplying through by 50. Substituting this into the red solution equation: \( 0.98(50 - 5y) + 0.90y = 30 \). This gives \( 49 - 4.9y + 0.90y = 30 \) \( y = 10 \).

Step 5: Substitute Back to Find x

Substitute \( y = 10 \) back into the equation \( x = 50 - 5y \): \( x = 50 - 5(10) \), resulting in \( x = 0 \). This means all 10 tablespoons are allocated to 'deep purple'.

Key Concepts

Simultaneous Equations

When dealing with multiple simultaneous equations, the goal is to find values that satisfy all given equations at the same time. These equations often contain two or more unknown variables. This is common in problems involving mixtures, where different amounts of ingredients must add up to a predetermined total.

In the given exercise, two simultaneous equations are needed. One equation captures the total amount of each color (blue and red) in the mixture. Solving these requires understanding how each solution contributes to the amount used.

The first step is to represent the unknowns. Letting \( x \) be the amount of 'light purple' and \( y \) the amount of 'deep purple', allows us to form equations based on the percentage of blue in each mixture. By setting up and solving these equations, you can find the correct amounts of each solution, ensuring that the total does not exceed the given quantities.

Percentages in Mixture Problems

Mixture problems often involve working with percentages to determine how much of each ingredient is required to meet certain criteria. This requires a clear understanding of how percentages represent parts of a whole.

For instance, in our example, 'light purple' and 'deep purple' each contain different percentages of blue solution. 'Light purple' consists of \(2\%\) of blue, while 'deep purple' consists of \(10\%\).

• This means that in a 100 tablespoon solution of 'light purple', 2 tablespoons would be blue.
• Similarly, in 'deep purple', 10 tablespoons of every 100 are blue.

Using percentages, we set up equations that reflect these compositions. By converting the percentages to decimals and multiplying by the quantities of 'light purple' (\( x \)) and 'deep purple' (\( y \)), we accurately calculate the needed proportion of blue solution.

Linear Equations

Linear equations form the backbone of algebra and play a crucial role in solving mixture problems. These equations typically take the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) and \( y \) are variables.

In our scenario, linear equations are used to express the relationship between the amounts of blue and red solutions. The exercise illustrates setting up two linear equations:

• One for the blue solution: \( 0.02x + 0.10y = 2 \)
• Another for the red solution: \( 0.98x + 0.90y = 30 \)

By solving these equations simultaneously, we determine the feasible amounts of 'light purple' and 'deep purple'.
The simplification of these equations often involves substitution or elimination methods. Solving them precisely is essential because it directly impacts the resulting mixture quantities. By substituting the known solution back into another equation, all variable values can be found, demonstrating the utility and reliability of linear methods in problem-solving.