Understanding Boat and Stream Problems

Boat and stream problems are common in quantitative aptitude. They involve calculating speeds of a boat or a person rowing in water, considering the effect of the water's current (stream). There are a few key terms and concepts to understand:

• Speed in Still Water: The speed of the boat or person if there were no current.
• Speed of the Stream: The speed of the water current.
• Downstream Speed: The speed of the boat when moving in the same direction as the stream. It is the sum of the speed in still water and the speed of the stream. \(\text{Downstream Speed} = \text{Speed in Still Water} + \text{Speed of Stream}\)
• Upstream Speed: The speed of the boat when moving against the direction of the stream. It is the difference between the speed in still water and the speed of the stream. \(\text{Upstream Speed} = \text{Speed in Still Water} - \text{Speed of Stream}\)

In this problem, we are given information about the boat's speed in still water and its travel times for specific distances downstream and upstream. We need to find the speed of the stream.

Calculating Boat Speed in Still Water

The question states that the boat can row 24 km in 6 hours in still water. We can use the basic formula: Speed = Distance / Time.

Speed of boat in still water \( = \frac{\text{Distance}}{\text{Time}} = \frac{24 \text{ km}}{6 \text{ hours}}\)

Speed of boat in still water \( = 4 \text{ km/hr}\)

Let's denote the speed of the boat in still water as \(b\) and the speed of the stream as \(s\). From the first piece of information, we know \(b = 4 \text{ km/hr}\).

Setting Up the Equations for Downstream and Upstream Travel

The boat travels 56 km downstream and 30 km upstream in a total of 38 hours. We need to express the time taken for each part of the journey using the speeds we defined.

• Downstream speed \( = b + s = 4 + s\) km/hr
• Upstream speed \( = b - s = 4 - s\) km/hr

The time taken to travel a certain distance is given by Time = Distance / Speed.

• Time taken to travel 56 km downstream \( = \frac{56}{\text{Downstream Speed}} = \frac{56}{4 + s}\) hours
• Time taken to travel 30 km upstream \( = \frac{30}{\text{Upstream Speed}} = \frac{30}{4 - s}\) hours

The total time for both journeys is 38 hours. So, we can write the equation:

\(\text{Time Downstream} + \text{Time Upstream} = \text{Total Time}\)

\(\frac{56}{4 + s} + \frac{30}{4 - s} = 38\)

Solving for the Stream Speed

Now we need to solve the equation \(\frac{56}{4 + s} + \frac{30}{4 - s} = 38\) for \(s\).

We can solve this equation algebraically or by testing the given options. Since the options are simple values, testing might be faster.

Let's test the options:

• Option 1: \(s = 3.5\) km/hr
  Downstream speed = \(4 + 3.5 = 7.5\) km/hr
  Upstream speed = \(4 - 3.5 = 0.5\) km/hr
  Total time = \(\frac{56}{7.5} + \frac{30}{0.5} = \frac{560}{75} + 60 = \frac{112}{15} + 60 \approx 7.47 + 60 = 67.47\) hours. This is not 38 hours.
• Option 2: \(s = 5\) km/hr
  Downstream speed = \(4 + 5 = 9\) km/hr
  Upstream speed = \(4 - 5 = -1\) km/hr. Speed cannot be negative, so this option is invalid. The speed of the stream must be less than the speed of the boat in still water for upstream travel to be possible. \(s < b\).
• Option 3: \(s = 3\) km/hr
  Downstream speed = \(4 + 3 = 7\) km/hr
  Upstream speed = \(4 - 3 = 1\) km/hr
  Total time = \(\frac{56}{7} + \frac{30}{1} = 8 + 30 = 38\) hours. This matches the given total time.
• Option 4: \(s = 4\) km/hr
  Downstream speed = \(4 + 4 = 8\) km/hr
  Upstream speed = \(4 - 4 = 0\) km/hr. Upstream speed cannot be zero (unless the boat makes no progress), and division by zero is undefined. This option implies the boat cannot move upstream against the stream. \(s < b\).

The only option that satisfies the conditions is \(s = 3\) km/hr.

Alternatively, let's solve the equation algebraically:

\(\frac{56}{4 + s} + \frac{30}{4 - s} = 38\)

Multiply by \((4 + s)(4 - s)\) to clear the denominators:

\(56(4 - s) + 30(4 + s) = 38(4 + s)(4 - s)\)

\(224 - 56s + 120 + 30s = 38(16 - s^2)\)

\(344 - 26s = 608 - 38s^2\)

Rearrange into a quadratic equation:

\(38s^2 - 26s + 344 - 608 = 0\)

\(38s^2 - 26s - 264 = 0\)

Divide by 2:

\(19s^2 - 13s - 132 = 0\)

We can solve this quadratic equation using the quadratic formula or by factoring. Since we found that \(s=3\) is a solution by checking the options, \((s-3)\) must be a factor of the quadratic. We can factor it as:

\((s - 3)(19s + 44) = 0\)

This gives two possible solutions for \(s\):

• \(s - 3 = 0 \implies s = 3\)
• \(19s + 44 = 0 \implies 19s = -44 \implies s = -\frac{44}{19}\)

Since the speed of the stream cannot be negative, the only valid solution is \(s = 3\) km/hr.

Final Answer

The speed of the stream is 3 km/hr.

Revision Table: Boat and Stream Concepts

Additional Information on Boat and Stream Problems

Boat and stream questions often require setting up equations based on the speeds and times for downstream and upstream journeys. Remember that the speed of the stream always helps downstream movement and hinders upstream movement.

• If the boat's speed in still water is less than the stream's speed (\(b < s\)), the boat cannot travel upstream. However, problems usually give scenarios where upstream travel is possible, implying \(b > s\).
• Sometimes, the question might ask for the distance traveled or the speed of the boat in still water or the stream, given other parameters. Always identify what is known and what needs to be found, then use the formulas relating speeds, distances, and times.
• Practice solving different variations of these problems to become comfortable with setting up the equations and solving them.