Problem 52 - Intermediate Algebra

Question

Gasoline. \(\quad\) A truck owner drove his pickup to a service station to fill the nearly empty 24 -gallon gas tank. If the truck runs on 89 -octane gasoline, but the station only sells 87 -octane and 93 octane gas, how many gallons of catch should be pumped to fill the tank with an 89 -octane blend?

Step-by-Step Solution

Answer

Mix 16 gallons of 87-octane with 8 gallons of 93-octane.

Step 1: Understanding the Problem

The problem is asking us to determine how much of each type of gasoline (87-octane and 93-octane) needs to be mixed to achieve a total of 24 gallons of 89-octane gasoline.

Step 2: Assign Variables

Let \( x \) be the number of gallons of 87-octane gasoline and \( y \) be the number of gallons of 93-octane gasoline. We know that the total amount of gasoline is 24 gallons, so we have the equation:\[ x + y = 24 \]

Step 3: Set Up Another Equation Using Octane Ratings

We need the mixture to be 89 octane. So, the octane condition can be written using the equation:\[ 87x + 93y = 89 \times 24 \]where 89 is the octane rating of the desired mixture and 24 is the total number of gallons.

Step 4: Substitute to Solve the System

From the first equation, solve for one variable, say \( y \):\[ y = 24 - x \]Substitute \( y \) in the second equation:\[ 87x + 93(24 - x) = 2136 \]

Step 5: Simplify and Solve for x

Distribute and simplify:\[ 87x + 2232 - 93x = 2136 \]Combine like terms:\[ -6x + 2232 = 2136 \]Subtract 2232 from both sides:\[ -6x = -96 \]Solve for \( x \):\[ x = 16 \]

Step 6: Calculate y Using x

Substitute \( x = 16 \) back into the equation \( y = 24 - x \):\[ y = 24 - 16 \ y = 8 \]

Step 7: Conclusion

Thus, 16 gallons of 87-octane gasoline and 8 gallons of 93-octane gasoline should be used to create 24 gallons of 89-octane gasoline.

Key Concepts

Octane Ratings

Gasoline comes with different octane ratings, which is a measure of a fuel’s ability to resist knocking or pinging during combustion caused by the air/fuel mixture detonating prematurely. The higher the octane rating, the more stable the fuel.
For vehicles designed to run on a specific octane rating, like the truck in our problem that requires 89-octane gasoline, using the correct octane ensures optimal performance and efficiency.
The challenge arises when the exact octane is unavailable at the pump, which is why understanding mixtures becomes crucial. Mixing different octane fuels, 87 and 93 in this scenario, can create the desired octane level needed for the truck.

System of Equations

A system of equations is a set of equations with multiple variables, which allows us to find a comprehensive solution that satisfies all the given conditions.
In mixture problems, like achieving a specific octane level, we often use a system of equations to simultaneously consider different aspects of the problem.

Equation 1: It represents the total volume of the mixture (e.g., \( x + y = 24 \) for 24 total gallons).

Equation 2: It represents the specific octane level condition (e.g., \( 87x + 93y = 89 \times 24 \)).

By solving this system, we find the exact amounts needed of each type of fuel to meet the octane requirement.

Variable Substitution

Variable substitution is a method used to solve systems of equations algebraically. This involves expressing one variable in terms of another and substituting it into a different equation.
In our gasoline problem, after establishing the equations, we isolate one variable.
For example, let's express \( y \) from the total volume equation \( x + y = 24 \) as \( y = 24 - x \).

Using Substitution

We then substitute this expression for \( y \) in the octane condition equation:
\[ 87x + 93(24 - x) = 2136 \]This reduces the problem to a single equation with one variable, making it simpler to solve.
Once we find \( x \), we can quickly determine \( y \) by substituting \( x \) back into the equation \( y = 24 - x \), giving us our final solution.