Problem 28 - Algebra for College Students

Question

The radiator of a truck contains 10 gallons of water. Suppose we remove 1 gallon of water and replace it with antifreeze. Then we remove 1 gallon of this mixture and replace it with antifreeze. This process is carried out seven times. To the nearest tenth of a gallon, how much antifreeze is in the final mixture? \(5.2\) gallons

Step-by-Step Solution

Answer

The radiator ends up with 5.2 gallons of antifreeze after seven iterations.

Step 1: Determine Initial Conditions

The radiator initially contains 10 gallons of pure water and 0 gallons of antifreeze.

Step 2: First Removal and Replacement

When we remove 1 gallon of water and replace it with 1 gallon of antifreeze, the radiator now contains 9 gallons of water and 1 gallon of antifreeze.

Step 3: Second Mixture Adjustment

Remove 1 gallon of the current mixture (9/10 water and 1/10 antifreeze): - Water removed = \( \frac{9}{10} \times 1 = 0.9 \) gallons - Antifreeze removed = \( \frac{1}{10} \times 1 = 0.1 \) gallons Replace with 1 gallon of antifreeze, resulting in: - Water = 9 - 0.9 = 8.1 gallons - Antifreeze = 1 - 0.1 + 1 = 1.9 gallons

Step 4: Third Mixture Adjustment

Repeat the removal of 1 gallon of the new mixture (8.1/10 water, 1.9/10 antifreeze). Calculate removed amounts:- Water removed = \( \frac{8.1}{10} \times 1 = 0.81 \) gallons - Antifreeze removed = \( \frac{1.9}{10} \times 1 = 0.19 \) gallons After adding a gallon of antifreeze, we have:- Water = 8.1 - 0.81 = 7.29 gallons - Antifreeze = 1.9 - 0.19 + 1 = 2.71 gallons

Step 5: Calculate Continuation Pattern

Continue this process, each time:- Removing 1 gallon of the current mixture (fractions of water and antifreeze calculated each step)- Replace with 1 gallon of pure antifreezeFor each step, use the formula:- New water volume = Previous water volume \( \times \frac{9}{10} \)- Antifreeze is the total volume = 10 minus new water volume after each step

Step 6: Final Calculation (Seventh Adjustment)

After seven iterations, using the above operations, we calculate the volume of water and antifreeze. By compound reduction, the water's volume becomes:\[ 10 \times \left( \frac{9}{10} \right)^7 \approx 4.8 \text{ gallons} \] Thus, antifreeze = total capacity - water volume = \(10 - 4.8 = 5.2\) gallons.

Key Concepts

Volume Replacement

Volume replacement is a fundamental concept in algebraic mixture problems and relates directly to transforming the composition of a mixture by replacing one liquid with another. Imagine you have a fixed container, like the truck's radiator, initially filled with water. You decide to gradually change the composition by continually removing a small portion of the existing liquid mixture and replacing it with a new component, such as antifreeze.

The key here is consistency in volume: after each removal of 1 gallon, a gallon of antifreeze takes its place. This constant exchange maintains the original total volume of 10 gallons inside the radiator. Understanding this ensures you grasp how even changing small parts of a mixture can alter the overall make-up of the contents over time. This is foundational when solving similar mixture problems in algebra.

Iterative Process

The iterative process is how the mixture transformation takes place over several steps. An iterative process means repeating the same procedure over and over until a desired outcome is achieved. In this truck radiator example, the specific iterative process involves:

• Removing 1 gallon of the current mixture.
• Replacing it with 1 gallon of antifreeze.

By doing this repeatedly, seven times according to the exercise, you slowly increase the concentration of antifreeze inside the radiator. Each small step nudges the mixture closer to being predominantly antifreeze instead of water. An iterative approach is beneficial in breaking complex transformations into manageable increments, allowing gradual change and making the entire calculation straightforward to follow.

Compound Reduction

Compound reduction is a powerful tool used to calculate changes applied step by step, especially in situations where an iterative process is involved. It specifically refers to how the water's volume decreases in logarithmic steps.

Each removal reduces the water by its fraction, which is calculated by multiplying the current volume of water by 9/10. This fraction is obtained based on removing 1/10 of the water each time because there are 10 parts in 1 gallon. Thus, applying this reduction over each iteration signifies that less and less water remains.

Mathematically, this can be expressed as a geometric sequence, where after each step, the new water volume becomes: \( volume ext{ of water} = Previous ext{ water volume} imes \frac{9}{10} \).

This reduction process compounds over the entire number of steps, allowing us to predict the water volume after multiple iterations, which, in this problem, is after 7 steps.

Antifreeze Mixture Calculation

Calculating the final amount of antifreeze in the radiator involves putting all these concepts together. Since the total volume remains constant at 10 gallons during the entire exchange, the antifreeze quantity can be determined by understanding how much water is left after all iterations.

Using the compound reduction formula, we calculate the water's volume after seven steps:\[ 10 \times \left( \frac{9}{10} \right)^7 \approx 4.8 \text{ gallons} \]

Hence, the remaining volume which the water does not occupy becomes the antifreeze volume:
Antifreeze = Total Volume - Water Volume = \(10 - 4.8 = 5.2\) gallons.

This calculation illustrates the gradual shift from water to antifreeze and highlights the use of iterative analysis and compound reduction principles in solving practical algebraic mixture problems such as the one described here.