Problem 41 - Technical Mathematics with Calculus

Question

A landlord owns a house that consumes 2100 gal of heating oil in three winters. He buys another (insulated) house, and the two houses together use 1850 gal of oil in two winters. How many winters would it take the insulated house alone to use 1250 gal of oil?

Step-by-Step Solution

Answer

It would take approximately 6 winters for the insulated house to use 1250 gal of oil.

Step 1: Determine oil usage per winter for the first house

Since the first house uses 2100 gallons of heating oil in three winters, you can find the usage per winter by dividing the total gallons by the number of winters. This is calculated as \(\frac{2100 \text{ gal}}{3 \text{ winters}} = 700 \text{ gal/winter}\).

Step 2: Determine combined oil usage per winter for both houses

To find the combined usage per winter, divide the total usage for the two houses by the number of winters. The calculation is \(\frac{1850 \text{ gal}}{2 \text{ winters}} = 925 \text{ gal/winter}\).

Step 3: Calculate the oil usage per winter for the insulated house

Subtract the first house's usage per winter from the combined usage to get the insulated house's usage per winter. This gives \(925 \text{ gal/winter} - 700 \text{ gal/winter} = 225 \text{ gal/winter}\).

Step 4: Determine number of winters for the insulated house to use 1250 gal of oil

Divide the total oil amount needed by the usage per winter for the insulated house to get the number of winters. This comes out to \(\frac{1250 \text{ gal}}{225 \text{ gal/winter}} \approx 5.56\) winters. Since you can't have a fraction of a winter, you round up to the next whole number, which is 6 winters.

Key Concepts

Mathematical Problem-Solving

Understanding mathematical problem-solving is crucial for tackling real-world issues like the one the landlord faces. It involves identifying the problem, selecting appropriate mathematical tools, and applying them systematically. In essence, it’s about translating a word problem into a mathematical equation that can be solved.

In this scenario, the problem is determining how long it will take for a certain amount of resource (heating oil) to be used by only one of two consumers (the insulated house). The solution process is broken down into steps: defining individual consumption rates, reconciling them with combined consumption, and ultimately using that information to make projections. This structured approach not only finds an answer but also develops skills that can be applied to a variety of mathematical challenges.

Algebraic Equations

An algebraic equation is a mathematical statement that shows the equality of two expressions. In our landlord's problem, setting up the correct equations is pivotal. For instance, to find the insulation house's fuel usage rate, we had to deal with the equation \( 925 \text{ gal/winter} - 700 \text{ gal/winter} = 225 \text{ gal/winter} \).

The beauty of algebraic equations lies in their ability to represent real-world situations in symbolic form, thus simplifying the complexity of the problem. By introducing an unknown variable, equations can be manipulated to solve for that variable, a technique that underpins not only basic math but also advanced scientific theories.

Unit Rate Calculations

Unit rate calculations are a fundamental aspect of problem-solving, especially when dealing with proportions and comparisons. It is the ratio that compares a quantity to one unit of another quantity, such as miles per hour or, in our case, gallons of oil per winter.

To determine how quickly the insulated house goes through heating oil, we had to calculate \( \frac{1250 \text{ gal}}{225 \text{ gal/winter}} \), which represents the unit rate of consumption. Grasping this concept of unit rate is essential for students as it also appears in evaluating speeds, costs, densities, and various other aspects of daily life.