A man rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is:

Options

A. a) 1 km/hr
B. b) 1.5 km/hr
C. c) 2 km/hr
D. d) 2.5 km/hr

Answer

To solve the problem step by step, we will break it down into manageable parts.

### Step 1: Understand the problem
A man rows to a place 48 km away and returns in 14 hours. He also finds that he can row 4 km downstream in the same time it takes to row 3 km upstream.

### Step 2: Define variables
Let:
- \( b \) = speed of the boat in still water (in km/h)
- \( s \) = speed of the stream (in km/h)

### Step 3: Set up equations based on the given information
1. The total time taken for the round trip is given as 14 hours.
- Time taken to go downstream (to the place) = \( t_1 = \frac{48}{b+s} \)
- Time taken to come back upstream = \( t_2 = \frac{48}{b-s} \)
- Therefore, we have:
\[ t_1 + t_2 = 14 \implies \frac{48}{b+s} + \frac{48}{b-s} = 14 \]

2. From the second piece of information, we know:
- Time taken to row 4 km downstream = \( \frac{4}{b+s} \)
- Time taken to row 3 km upstream = \( \frac{3}{b-s} \)
- Since these times are equal, we have:
\[ \frac{4}{b+s} = \frac{3}{b-s} \]

### Step 4: Solve the second equation
Cross-multiplying gives:
\[ 4(b - s) = 3(b + s) \]
Expanding this:
\[ 4b - 4s = 3b + 3s \]
Rearranging terms:
\[ 4b - 3b = 4s + 3s \implies b = 7s \]

### Step 5: Substitute \( b \) in the first equation
Substituting \( b = 7s \) into the first equation:
\[ \frac{48}{7s + s} + \frac{48}{7s - s} = 14 \]
This simplifies to:
\[ \frac{48}{8s} + \frac{48}{6s} = 14 \]
Finding a common denominator (24s):
\[ \frac{48 \cdot 3}{24s} + \frac{48 \cdot 4}{24s} = 14 \]
This simplifies to:
\[ \frac{144 + 192}{24s} = 14 \implies \frac{336}{24s} = 14 \]
Cross-multiplying gives:
\[ 336 = 14 \cdot 24s \implies 336 = 336s \implies s = 1 \]

### Step 6: Conclusion
The speed of the stream \( s \) is **1 km/h**.