Understanding Boat and Stream Problems

Boat and stream problems are a common type in quantitative aptitude, dealing with relative speeds. When a person or boat moves in water, its speed is affected by the speed of the water current (stream).

• Speed Downstream: When the boat moves in the same direction as the stream, the speed of the stream adds to the speed of the boat. This is the speed downstream.
  Speed downstream = Speed of boat in still water + Speed of stream
• Speed Upstream: When the boat moves against the direction of the stream, the speed of the stream opposes the speed of the boat. This is the speed upstream.
  Speed upstream = Speed of boat in still water - Speed of stream

In this specific boat and stream problem, we are given the speed of the man in still water and a relationship between the time taken to travel upstream and downstream. We need to find the velocity (speed) of the stream.

Step-by-Step Solution for Stream Velocity

Let's denote the given quantities:

• Speed of the man in still water, $V_{man} = \frac{28}{3}$ km/h.
• Let the velocity of the stream be $V_{stream}$ km/h.

Now, we can express the speed downstream and speed upstream in terms of $V_{man}$ and $V_{stream}$:

• Speed downstream ($S_{down}$) = $V_{man} + V_{stream} = \left(\frac{28}{3} + V_{stream}\right)$ km/h.
• Speed upstream ($S_{up}$) = $V_{man} - V_{stream} = \left(\frac{28}{3} - V_{stream}\right)$ km/h.

We are told that it takes the man thrice as much time to row upstream as it takes to row downstream. Let's assume the distance covered in both cases is the same, say $D$ km.

The formula relating distance, speed, and time is: Time = Distance / Speed.

• Time taken downstream, $T_{down} = \frac{D}{S_{down}} = \frac{D}{\frac{28}{3} + V_{stream}}$.
• Time taken upstream, $T_{up} = \frac{D}{S_{up}} = \frac{D}{\frac{28}{3} - V_{stream}}$.

According to the problem statement, $T_{up} = 3 \times T_{down}$. Substituting the expressions for $T_{up}$ and $T_{down}$:

$$ \frac{D}{\frac{28}{3} - V_{stream}} = 3 \times \frac{D}{\frac{28}{3} + V_{stream}} $$

Assuming $D \neq 0$, we can cancel $D$ from both sides:

$$ \frac{1}{\frac{28}{3} - V_{stream}} = \frac{3}{\frac{28}{3} + V_{stream}} $$

Now, we can cross-multiply to solve for $V_{stream}$:

$$ 1 \times \left(\frac{28}{3} + V_{stream}\right) = 3 \times \left(\frac{28}{3} - V_{stream}\right) $$ $$ \frac{28}{3} + V_{stream} = 3 \times \frac{28}{3} - 3 \times V_{stream} $$ $$ \frac{28}{3} + V_{stream} = 28 - 3 V_{stream} $$

Now, let's rearrange the terms to group $V_{stream}$ on one side and constants on the other:

$$ V_{stream} + 3 V_{stream} = 28 - \frac{28}{3} $$ $$ 4 V_{stream} = \frac{28 \times 3 - 28}{3} $$ $$ 4 V_{stream} = \frac{84 - 28}{3} $$ $$ 4 V_{stream} = \frac{56}{3} $$

Finally, divide by 4 to find $V_{stream}$:

$$ V_{stream} = \frac{56}{3 \times 4} $$ $$ V_{stream} = \frac{56}{12} $$

We can simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 4:

$$ V_{stream} = \frac{56 \div 4}{12 \div 4} = \frac{14}{3} $$

So, the velocity of the stream is $\frac{14}{3}$ km/h.

Checking the Options

Let's compare our calculated stream velocity with the given options:

Our calculated velocity of the stream is $\frac{14}{3}$ km/h, which matches Option 4.

Revision Table: Key Concepts

Additional Information on Boat and Stream Problems

Understanding the relative speeds is crucial in solving boat and stream problems. The water current either helps (downstream) or hinders (upstream) the movement of the boat.

• If the speed of the boat in still water is $V_b$ and the speed of the stream is $V_s$, then:
  • Speed downstream = $V_b + V_s$
  • Speed upstream = $V_b - V_s$
• Often, you might be given the speed downstream and upstream and asked to find the speed of the boat in still water or the speed of the stream.
  • Speed of boat in still water = $\frac{\text{Speed downstream} + \text{Speed upstream}}{2}$
  • Speed of stream = $\frac{\text{Speed downstream} - \text{Speed upstream}}{2}$
• Remember that the speed of the boat in still water must always be greater than the speed of the stream for the boat to be able to move upstream ($V_b > V_s$). In our problem, $\frac{28}{3} \approx 9.33$ and $\frac{14}{3} \approx 4.67$, so $V_{man} > V_{stream}$, which is consistent.