Problem 75 - Algebra and Trigonometry

Question

Mixing Coolant A truck radiator holds 5 gal and is filled with water. A gallon of water is removed from the radiator and replaced with a gallon of antifreeze; then a gallon of the mixture is removed from the radiator and again replaced by a gallon of antifreeze. This process is repeated indefinitely. How much water remains in the tank after this process is repeated 3 times? 5 times? \(n\) times?

Step-by-Step Solution

Answer

After 3 times, 3.2 gallons remain; after 5 times, about 2.05 gallons. For \(n\) times, use \(\left(\frac{4}{5}\right)^n\times5\).

Step 1: Understand the process of replacement

Initially, the tank contains 5 gallons of water. Each operation involves removing 1 gallon of the current mixture and replacing it with 1 gallon of antifreeze. This means after every replacement, the composition of the mixture changes, with the percentage of water decreasing each time.

Step 2: Calculate the water remaining after 1 replacement

When the first gallon of water is removed, the tank remains with 4 gallons of water and 1 gallon of antifreeze is added. This means the tank now has \(4\) gallons of water and \(1\) gallon of antifreeze.

Step 3: Calculate the water remaining after 2 replacements

For the second replacement, 1 gallon of the 4:1 water to antifreeze mixture is removed. This is equivalent to removing \(\frac{4}{5}\) gallon of water. Thus, \(\frac{1}{5}\) gallon of antifreeze remains. Then 1 gallon of antifreeze is added again. The new ratio of water is \(\frac{4}{5}\times4\), reducing it to \(\frac{16}{25}\) gallons.

Step 4: Calculate the water remaining after 3 replacements

Repeating the process: remove 1 gallon \(\Rightarrow\) remaining water is \(\frac{16}{25}\times\frac{4}{5}\) = \(\frac{64}{125}\). Add 1 gallon of antifreeze.

Step 5: Calculate the water remaining after 5 replacements

Continuing this process, by the third replacement, \[\text{Remaining water = } \left(\frac{4}{5}\right)^3\times5 = \frac{64}{125}\times5 = 3.2 \text{ gallons}\]After five replacements you continue: \[\left(\frac{4}{5}\right)^5\times5 = 2.05 \text{ gallons}\]

Step 6: General formula for n replacements

For any number of replacements \(n\), the formula for the amount of water remaining is: \[\left(\frac{4}{5}\right)^n\times5\]This formula comes from the consistent reduction factor \(\frac{4}{5}\) applied to each gallon of water removed and replaced.

Key Concepts

Algebraic Expressions

In mixture problems, algebraic expressions help us quantify changes in the mixture, like the reduction of water in this radiator exercise. Initially, the radiator is filled entirely with water. As the antifreeze replaces water, the solution becomes a combination of both liquids. Algebraic expressions describe the reduction in water. For instance, if the radiator starts with 5 gallons of water, removing one gallon and replacing it with antifreeze involves computing the water percentage left. Each repetition shrinks this water percentage, forming a geometric pattern.

• Express the water percentage with each replacement
• Track how much water is left using expressions like \( \frac{4}{5} \,times\, x \)

These expressions simplify the ongoing calculations of water reduction in formulaic terms.

Geometric Sequences

A geometric sequence appears when the number of elements changes by a constant factor in each step, like in this radiator example where the amount of water reduces regularly. The formula \( \left(\frac{4}{5}\right)^n \,times\, 5 \) showcases how the geometric sequence is used here. The 5 represents the initial total gallons of water, and the fraction \( \frac{4}{5} \) is the reduction factor applied every time a replacement occurs. Each replacement is a term in our geometric sequence.

• Initial value: 5 gallons of water
• Reduction factor: \( \frac{4}{5} \)
• General formula for water remaining after n replacements

This enables easy forecasting of the water left after any number of replacements:

Radiator

The radiator in this scenario is a closed system where the exercise models the physical process of replacing water with antifreeze through a sequence of adjustments. To understand the dynamics of water replacement, one must consider the radiator's constant total volume. With each procedure, one gallon is extracted to be substituted with antifreeze, making the radiator

• Capable of maintaining a consistent liquid volume
• An applicable space for monitoring fluid change rates
• Ensuring the total volume always amounts to 5 gallons

The radiator, therefore, acts as a controlled environment for studying mixture problems, turning practical processes into mathematical equations.

Replacement Method

The replacement method involves systematically swapping fluids in a container, which introduces a pattern of change over multiple iterations. In this exercise, water is replaced with antifreeze to model continuous dilution. After each replacement, the quantity of water is reduced by the fraction \( \frac{4}{5} \). The first cycle replaces pure water with antifreeze, while subsequent iterations further dilute the water content.

• 1 gallon of current mixture is removed
• 1 gallon of antifreeze is added again
• Successful in demonstrating gradual reductions

This method contextualizes how the repetitive action impacts the system, especially for predicting future quantities accurately using mathematical expressions.