Understanding Boat and Stream Problems

This problem involves the concept of relative speed in the context of boat and stream motion. When a boat moves downstream, the speed of the current adds to the speed of the boat in still water. When the boat moves upstream, the speed of the current opposes the boat's speed, reducing its effective speed.

We can define the key speeds involved:

• Speed of the boat in still water: Let's denote this as \(v_b\).
• Speed of the current: Let's denote this as \(v_c\).

Based on these, the effective speeds are:

• Speed downstream: \(v_d = v_b + v_c\) (The boat and current move in the same direction)
• Speed upstream: \(v_u = v_b - v_c\) (The boat moves against the current)

The relationship between distance, speed, and time is given by:

Time \(=\) Distance \(/\) Speed

Setting Up Equations for Sailor's Journey

The problem provides two scenarios with different distances traveled downstream and upstream, and the total time taken for each scenario. We can use the formula Time \(=\) Distance \(/\) Speed to set up a system of equations.

Let's analyze the information given:

Scenario 1:

• Distance downstream \(=\) 36 km
• Distance upstream \(=\) 24 km
• Total time \(=\) 6 hours

Time taken downstream in Scenario 1 \(=\) \(\frac{36}{v_b + v_c}\)

Time taken upstream in Scenario 1 \(=\) \(\frac{24}{v_b - v_c}\)

Equation 1: \(\frac{36}{v_b + v_c} + \frac{24}{v_b - v_c} = 6\)

Scenario 2:

• Distance downstream \(=\) 24 km
• Distance upstream \(=\) 36 km
• Total time \(=\) 6.5 hours (or \(6 \frac{1}{2}\) hours)

Time taken downstream in Scenario 2 \(=\) \(\frac{24}{v_b + v_c}\)

Time taken upstream in Scenario 2 \(=\) \(\frac{36}{v_b - v_c}\)

Equation 2: \(\frac{24}{v_b + v_c} + \frac{36}{v_b - v_c} = 6.5\)

Solving the System of Equations for Speeds

We have a system of two equations with two variables, \(v_b\) and \(v_c\). To make it easier to solve, let's use substitution.

Let \(x = \frac{1}{v_b + v_c}\) and \(y = \frac{1}{v_b - v_c}\).

The equations become:

Equation 1: \(36x + 24y = 6\)

Equation 2: \(24x + 36y = 6.5\)

We can simplify Equation 1 by dividing by 6:

\(6x + 4y = 1\) (Equation 3)

Now, we can solve this system of linear equations. Let's use the elimination method. Multiply Equation 3 by 6 and Equation 2 by 1:

• \(6 \times (6x + 4y) = 6 \times 1 \Rightarrow 36x + 24y = 6\)
• \(1 \times (24x + 36y) = 1 \times 6.5 \Rightarrow 24x + 36y = 6.5\)

This brings us back to the original equations, which wasn't the most helpful approach. Let's try eliminating one variable using suitable multipliers. Multiply Equation 3 by 6 and Equation 2 by 1:

• Multiply Equation 1 (\(36x + 24y = 6\)) by 2: \(72x + 48y = 12\)
• Multiply Equation 2 (\(24x + 36y = 6.5\)) by 3: \(72x + 108y = 19.5\)

Now, subtract the first new equation from the second new equation:

\((72x + 108y) - (72x + 48y) = 19.5 - 12\)

\(60y = 7.5\)

\(y = \frac{7.5}{60} = \frac{75}{600} = \frac{1}{8}\)

Now substitute the value of \(y\) into Equation 3 (\(6x + 4y = 1\)):

\(6x + 4\left(\frac{1}{8}\right) = 1\)

\(6x + \frac{1}{2} = 1\)

\(6x = 1 - \frac{1}{2} = \frac{1}{2}\)

\(x = \frac{1}{2 \times 6} = \frac{1}{12}\)

So, we have \(x = \frac{1}{12}\) and \(y = \frac{1}{8}\).

Recall that \(x = \frac{1}{v_b + v_c}\) and \(y = \frac{1}{v_b - v_c}\).

• \(\frac{1}{v_b + v_c} = \frac{1}{12} \Rightarrow v_b + v_c = 12\) (Downstream speed)
• \(\frac{1}{v_b - v_c} = \frac{1}{8} \Rightarrow v_b - v_c = 8\) (Upstream speed)

Now we have a simpler system of linear equations:

1) \(v_b + v_c = 12\)

2) \(v_b - v_c = 8\)

Add the two equations:

\((v_b + v_c) + (v_b - v_c) = 12 + 8\)

\(2v_b = 20\)

\(v_b = \frac{20}{2} = 10\) km/h (Speed of the boat in still water)

Subtract the second equation from the first equation:

\((v_b + v_c) - (v_b - v_c) = 12 - 8\)

\(v_b + v_c - v_b + v_c = 4\)

\(2v_c = 4\)

\(v_c = \frac{4}{2} = 2\) km/h (Speed of the current)

Calculating the Speed of the Current

From the calculations above, the speed of the current (\(v_c\)) is found to be 2 km/h.

Verification of the Solution

Let's check if \(v_b = 10\) km/h and \(v_c = 2\) km/h satisfy the original conditions.

Downstream speed \(=\) \(v_b + v_c = 10 + 2 = 12\) km/h

Upstream speed \(=\) \(v_b - v_c = 10 - 2 = 8\) km/h

Scenario 1: 36 km downstream and 24 km upstream.

Time \(=\) \(\frac{36}{12} + \frac{24}{8} = 3 + 3 = 6\) hours. This matches the given total time.

Scenario 2: 24 km downstream and 36 km upstream.

Time \(=\) \(\frac{24}{12} + \frac{36}{8} = 2 + 4.5 = 6.5\) hours. This matches the given total time.

Since both scenarios are satisfied, our calculated speeds are correct. The speed of the current is 2 km/h.

Revision Table: Key Boat and Stream Concepts

Additional Information on Boat and Stream Problems

Boat and stream problems are a common type of question in quantitative aptitude tests. They test your understanding of relative speeds. It is crucial to correctly identify whether the boat is moving with the current (downstream) or against the current (upstream) to apply the correct effective speed formula.

Setting up equations based on the given time or distance information is usually the next step. Often, these problems lead to a system of linear equations, which can be solved using methods like substitution or elimination, as demonstrated in this solution.

Remember that downstream speed is always greater than the speed of the boat in still water, and upstream speed is always less than the speed of the boat in still water. Also, upstream speed must be a positive value, implying the boat's speed must be greater than the speed of the current (\(v_b > v_c\)) for the boat to make progress upstream.