Problem 27 - Algebra for College Students

Question

A tank contains 20 gallons of water. One-half of the water is removed and replaced with antifreeze. Then one-half of this mixture is removed and replaced with antifreeze. This process is continued eight times. How much water remains in the tank after the eighth replacement process? \(\frac{5}{64}\) gallon

Step-by-Step Solution

Answer

After the eighth replacement process, \( \frac{5}{64} \) gallon of water remains in the tank.

Step 1: Initial State

The tank initially contains 20 gallons of pure water. This is our starting point before any replacements occur.

Step 2: First Replacement

In the first replacement, half of the 20 gallons of water is removed. This means 10 gallons of water are removed and replaced with 10 gallons of antifreeze. The tank now contains 10 gallons of water and 10 gallons of antifreeze.

Step 3: General Replacement Process

For each subsequent replacement, half of the tank's current contents are removed and replaced with antifreeze. This means that each time, only half of the remaining water is kept.

Step 4: Mathematical Representation

Let the amount of water remaining after each replacement be represented by a sequence, with the initial amount as 20 gallons of water. The formula to calculate the amount of water after each replacement is given by: \[ W_n = W_{n-1} \times \frac{1}{2} \] where \( W_0 = 20 \). This formula is applied repeatedly for each replacement step.

Step 5: Applying the Replacement 8 Times

Starting with 20 gallons of water, apply the formula: - After 1st replacement: \( W_1 = 20 \times \frac{1}{2} = 10 \) gallons - After 2nd replacement: \( W_2 = 10 \times \frac{1}{2} = 5 \) gallons - After 3rd replacement: \( W_3 = 5 \times \frac{1}{2} = 2.5 \) gallons - After 4th replacement: \( W_4 = 2.5 \times \frac{1}{2} = 1.25 \) gallons - After 5th replacement: \( W_5 = 1.25 \times \frac{1}{2} = 0.625 \) gallons - After 6th replacement: \( W_6 = 0.625 \times \frac{1}{2} = 0.3125 \) gallons - After 7th replacement: \( W_7 = 0.3125 \times \frac{1}{2} = 0.15625 \) gallons - After 8th replacement: \( W_8 = 0.15625 \times \frac{1}{2} = 0.078125 \) gallons

Step 6: Convert Decimal to Fraction

Convert 0.078125 gallons to a fraction: 0.078125 can be expressed as \( \frac{5}{64} \) gallons since decimal places represent powers of 2 (binary fractions).

Key Concepts

Mathematical Modeling

Mathematical modeling is a tool that helps us represent real-world scenarios in mathematical form. In this problem, we use sequence and series to predict how much water remains in a tank after repeated actions.

• A sequence is a list of numbers that typically follows a specific mathematical rule.
• Here, we have a decreasing geometric sequence because each term is half of the previous one.

The challenge is to find the amount of water left after a series of operations. So, we create a formula for calculating the amount iteratively. We start with an initial value of 20 gallons, and for each step, the amount of water is reduced by half. Hence, modeling such scenarios mathematically helps predict outcomes efficiently even after complex operations.

Mixture Problems

Mixture problems involve combining different substances and examining the properties of the resulting mixture. In this exercise, we alternately remove water and add antifreeze to a tank.

• Initially, the tank contains 20 gallons of water.
• In each step, half of what is in the tank is replaced by a different substance, in this case, antifreeze.

After each operation, the problem transitions into analyzing how much pure water remains. Such problems often require understanding proportions and how each step affects the overall composition of the mixture. By treating each step mathematically, we can solve for reduced quantities systematically.

Exponential Decay

Exponential decay describes processes where quantities diminish over time, at a rate proportional to their current value. In our sequence problem, the amount of water in the tank diminishes exponentially, halving with each replacement.

• The general formula applied is: \( W_n = W_{n-1} \times \frac{1}{2} \).
• This results in each term being half of the previous term, showing exponential decay in water volume.

As you continue the process of replacement, the quantity gets very small, approaching zero but never quite getting there. Such decay is similar to radioactive decay or population decrease where the amount reduces significantly over time. Understanding this principle helps in visualizing how quickly substances can diminish under certain operations.