Problem 48 - Algebra 1

Question

You are climbing a 500 -foot cliff. By 1: 00 P.M. you have climbed 125 feet up the cliff. By 4: 00 P.M. you have reached a height of 290 feet. If you continue climbing the cliff at the same rate, at what time will you reach the top?

Step-by-Step Solution

Answer

The climber will reach the top of the 500-foot cliff by approximately 7:49 P.M.

Step 1 - Calculate the rate of climbing

First step is to find out how many feet the climber is able to climb per hour. This can be done by subtracting the initial height (125 feet) from the final height at 4:00 P.M (290 feet) and dividing the result by the number of hours (3 hours). So it's, rate = (290 - 125) / 3 = 55 feet/hour.

Step 2 - Calculate the remaining distance and time to reach the top

The climber has already reached 290 feet at 4:00 P.M. and needs to climb up another (500 - 290) = 210 feet to reach the top. Now that we know the rate, we can calculate the time by dividing the remaining feet by the rate. So the time = 210 / 55 hours. That results to approximately 3.82 hours.

Step 3 - Convert decimal hours to minutes and calculate the final time

The decimal part of the hours (.82) should be converted to minutes by multiplying by 60. Hence .82 hours is approximately equal to 49 minutes. So, the climber will reach the top by around 4:00 P.M. + 3 hours 49 minutes = 7:49 P.M.

Key Concepts

Linear Equations

Linear equations are foundational to understanding rate of change in algebra. They describe a relationship where change occurs at a constant rate. The equation takes the form of y = mx + b, where y represents the dependent variable, x is the independent variable, m is the slope, and b represents the y-intercept.

Connect this to a real-world scenario such as climbing a cliff. Visualize it as a linear graph where the vertical climb (y) depends on the time spent climbing (x). In our exercise, the climber's rate of change, or slope, is the constant speed at which they're ascending. Understanding linear equations allows us to predict outcomes, like forecasting what time they will reach the cliff's summit based on their climbing speed.

Problem-solving Strategies

Effective problem-solving strategies in algebra involve breaking down complex problems into simpler steps. Here, the strategy followed a sequential approach: calculate the climbing rate, determine the remaining distance, then compute the remaining time. This step-by-step strategy helps students tackle challenging assignments by focusing on one component at a time.

Each step involves a clear mathematical operation, such as subtraction to find distance climbed, or division to calculate the hourly rate. The sequence of these steps leads to a solution that is easy to follow and understand, which is key in applying algebra to solve time and distance problems.

Unit Rate Calculation

In algebra, a unit rate is a type of ratio that compares any two separate but related quantities. It's essential when dealing with problems that involve time, rates of speed, or other measures of change. To calculate a unit rate, we divide the first quantity by the second.

In our example, we calculated the climber's rate by dividing the total feet climbed by the hours taken, which gave us feet per hour. Being adept with unit rate calculations is crucial not only in academia but also in real-life scenarios, such as calculating the cost per ounce of a product or miles per gallon for a vehicle.

Time and Distance Problems

Algebra often deals with problems that involve finding the relationship between time and distance. When solving these problems, it's important to understand the concept of rate, which relates the two quantities. We apply these principles to determine how long it will take to cover a certain distance at a given rate, like a climber scaling a cliff.

In the exercise, applying the constant rate found in the unit rate calculation to the remaining distance gave us the additional time needed to complete the climb. By mastering these concepts, students can resolve issues of time management, logistics, and much more in everyday life and professional settings.