Problem 62 - Elementary Algebra

Question

If Kent can mow the entire lawn in \(m\) minutes, what fractional part of the lawn has he mowed at the end of 20 minutes?

Step-by-Step Solution

Answer

Kent has mowed \(\frac{20}{m}\) of the lawn.

Step 1: Understanding the Problem

Kent can mow the entire lawn in \(m\) minutes. We need to determine what fraction of the lawn he has completed at the end of 20 minutes.

Step 2: Determine Rate of Mowing

Since Kent can mow the whole lawn in \(m\) minutes, he mows at a rate of \(1\) full lawn per \(m\) minutes. Therefore, his mowing rate is \(\frac{1}{m}\) of the lawn per minute.

Step 3: Calculate Amount Mowed in 20 Minutes

Now, we need to determine how much Kent has mowed in 20 minutes. Since he mows at \(\frac{1}{m}\) of the lawn per minute, in 20 minutes he would mow \(20 \times \frac{1}{m}\) of the lawn.

Step 4: Simplify the Fraction

The expression \(20 \times \frac{1}{m}\) simplifies to \(\frac{20}{m}\). So Kent has mowed \(\frac{20}{m}\) of the lawn at the end of 20 minutes.

Key Concepts

Rate of Work

The rate of work is a central concept for solving problems involving tasks done over time. It refers to how much of a task can be completed in a certain period. In this context, Kent's rate of mowing the lawn is crucial. You can think of this rate as the amount of lawn Kent can mow in one minute.

• Kent completes the entire lawn in \(m\) minutes. Thus, in one minute, he finishes a fractional part of the lawn, specifically \(\frac{1}{m}\) of it.
• This fraction \(\frac{1}{m}\) establishes his rate of work per minute, meaning every minute, Kent mows \(\frac{1}{m}\) of the lawn.

Using this rate, we can then determine how much he has accomplished in any other amount of time, which leads us to calculate how much of the lawn Kent mows in 20 minutes.

Simplifying Fractions

Simplifying fractions is a fundamental skill in algebra that makes problems easier to solve and interpret. In the problem, we eventually end up with the fraction \(\frac{20}{m}\), which represents the part of the lawn Kent mows in 20 minutes. Simplifying a fraction can make this information more approachable and concise.

• The fraction \(\frac{20}{m}\) comes from multiplying Kent's rate of work over 20 minutes: \(20 \times \frac{1}{m} = \frac{20}{m}\).
• Ensuring the fraction is in its simplest form can help in understanding and further calculations, especially when dealing with multiple steps or more complex problems.

This approach of simplifying fractions is not just limited to this exercise but is widely used across various algebraic problems to ease calculations.

Word Problems in Algebra

Word problems in algebra are designed to translate real-world situations into mathematical equations, which once solved, give insights about the scenario. The key skill here is to interpret the words correctly into mathematical terms.

• Begin by identifying the known and unknown variables, like Kent’s total mowing time \(m\) and partial time spent, 20 minutes in this problem.
• Establish relationships using these variables, such as Kent’s rate of mowing \(\frac{1}{m}\) in our case.

By setting up equations based on this information, as done by finding \(20 \times \frac{1}{m}\), we solve to find the fractional part of the task, making such problems less daunting and more solvable.