Problem 228 - Elementary Algebra

Question

A motor boat travels 60 miles down a river in three hours but takes five hours to return upstream. Find the rate of the boat in still water and the rate of the current.

Step-by-Step Solution

Answer

The rate of the boat in still water is 16 mph and the rate of the current is 4 mph.

Step 1: Define Variables

Let the rate of the boat in still water be denoted as \(b\) (in miles per hour) and the rate of the current be denoted as \(c\) (in miles per hour).

Step 2: Write the Downstream Equation

The rate downstream is \(b + c\). Given that the boat travels 60 miles downstream in 3 hours, use the relation distance = rate × time to form the equation: \[ 60 = (b + c) \times 3 \] Simplify to: \[ b + c = 20 \]

Step 3: Write the Upstream Equation

The rate upstream is \(b - c\). Given that the boat travels 60 miles upstream in 5 hours, use the relation distance = rate × time to form the equation: \[ 60 = (b - c) \times 5 \] Simplify to: \[ b - c = 12 \]

Step 4: Solve the System of Equations

We have two equations: \[ b + c = 20 \] \[ b - c = 12 \] Add these two equations to eliminate c: \[ (b + c) + (b - c) = 20 + 12 \] This simplifies to: \[ 2b = 32 \] Solve for b: \[ b = 16 \] Now, substitute \(b = 16\) back into the first equation: \[ 16 + c = 20 \] Solve for c: \[ c = 4 \]

Key Concepts

Rate Problems

Rate problems are a common type of algebra word problem. These problems often involve finding the speed or rate at which something moves. In this motorboat problem, the rates we need to find are the speed of the boat in still water and the speed of the current. When solving rate problems, a useful strategy is to define variables for each unknown rate. This helps you organize the information and form equations. In our example, we defined the boat's speed in still water as \(b\) and the current's speed as \(c\). Identifying the relationships between different rates and distances, we can use equations to solve for these rates.

System of Equations

A system of equations involves two or more equations with two or more variables that we solve together. In this problem, we formed a system of linear equations using the downstream and upstream information.
The downstream rate is the boat's speed plus the current's speed, so our first equation became \(b + c = 20\). The upstream rate is the boat's speed minus the current's speed, leading to our second equation: \(b - c = 12\). By adding these two equations, we eliminated the variable \(c\) and solved for \(b\). This method is called the elimination method. Once we found \(b\), we substituted it back into one of the original equations to find \(c\). The process showed how solving a system of equations can effectively determine rates in word problems.

Distance-Rate-Time Relationship

The distance-rate-time relationship is a fundamental concept in algebra that we often use to tackle rate problems. This relationship is expressed by the formula \( \text{Distance} = \text{Rate} \times \text{Time} \). It tells us that the distance traveled is equal to the speed of travel multiplied by the time taken. In our motorboat problem, we used this relationship twice. First for the downstream trip: \(60 = (b + c) \times 3\), and then for the upstream trip: \(60 = (b - c) \times 5\). These equations were derived by knowing the distances, times, and using the rates as unknowns, showing the versatility of the distance-rate-time relationship in solving various real-world problems.