Problem 49 - Algebra and Trigonometry

Question

Mixture Problem The radiator in a car is filled with a solution of 60\(\%\) antifreeze and 40\(\%\) water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50\(\%\) antifreeze. If the capacity of the radiator is 3.6 \(\mathrm{L}\) , how much coolant should be drained and replaced with water to reduce the antifreeze concentration to the recommended level?

Step-by-Step Solution

Answer

Drain and replace 0.6 L of the coolant with water.

Step 1: Understand the Problem

We have a radiator with a capacity of 3.6 L, filled with a solution that is 60% antifreeze. We need to obtain a solution that is 50% antifreeze by draining some of the current mixture and replacing it with water.

Step 2: Set up the Equation

Let \( x \) be the amount in liters of the solution to be drained and replaced with water. Initially, the 60% antifreeze solution contains \( 0.6 \times 3.6 = 2.16 \) liters of pure antifreeze. After draining \( x \) liters, \( 3.6 - x \) liters remain, still containing a proportionate amount of antifreeze: \( 0.6 \times (3.6 - x) \). We add \( x \) liters of water to this to maintain a total of 3.6 liters. The new concentration is \( \frac{0.6(3.6-x)}{3.6} = 0.5 \).

Step 3: Solve the Equation

To solve the equation \( \frac{0.6(3.6-x)}{3.6} = 0.5 \), first eliminate the fraction by multiplying both sides by 3.6: \[ 0.6(3.6-x) = 1.8. \]Expand: \[ 2.16 - 0.6x = 1.8. \]Subtract 2.16 from both sides: \[ -0.6x = -0.36. \]Divide by -0.6: \[ x = 0.6. \]

Step 4: Interpret the Solution

We have calculated \( x = 0.6 \), which means that 0.6 liters of the current antifreeze solution should be drained and replaced with water to reduce the concentration of antifreeze to 50%.

Key Concepts

Antifreeze Concentration

In the realm of car maintenance, understanding the right antifreeze concentration is crucial for optimal engine performance. Antifreeze, usually mixed with water, helps prevent the vehicle’s cooling system from freezing in the winter and overheating in the summer. The standard concentration varies with seasons and manufacturers' recommendations.
For this exercise, we started with a 60% antifreeze concentration. However, during the summer, a 50% concentration is suggested for better cooling. It's all about balance: too little antifreeze might lower boiling points and cause the engine to overheat, while too much can reduce overall system effectiveness. Ensuring the correct mixture ensures that the engine runs smoothly all year round.

Radiator Capacity

The radiator is a key component in regulating a car's engine temperature. The radiator's capacity, in this instance, is noted as 3.6 liters. This measurement tells us the total volume available for the antifreeze-water solution.
When considering the radiator's capacity, it is essential to ensure that any adjustments made, whether adding or removing fluids, maintain the total capacity. This exercise's objective is to adjust the mixture while keeping this fixed volume in mind. Calculations are based on not exceeding or falling short of the original radiator's capacity, ensuring no strain is placed on the system.

Algebraic Equation Solving

Solving algebraic equations in mixture problems involves logical steps to reach an accurate solution. First, we define the variable of interest—in this case, the amount of mixture to drain, denoted by \( x \). We then express the initial conditions through algebraic expressions.
To solve the given exercise, we created an equation from the new composition requirements: \( \frac{0.6(3.6-x)}{3.6} = 0.5 \). Begin by simplifying this equation: eliminate fractions through multiplication. It's crucial to maintain balance in the equation by performing identical operations on both sides.

• First, we multiplied through by 3.6 to clear fractions, leading to \( 0.6(3.6-x) = 1.8 \).
• Simplify the expression by expanding and then isolating the variable \( x \).
• Finally, derive \( x = 0.6 \), indicating how much solution should be swapped.

This structured approach allows a straightforward path from an observed problem to a calculated solution, emphasizing clarity and logic in algebraic manipulation.