A man rows down a river 15 km in 3 hrs. with the stream and returns in `7 1/2` hrs, The rate at which he rows in still water is

Options

A. 2.5 km/hr
B. 1.5 km/hr
C. 3.5 km/hr
D. 4.5 km/hr

Answer

To solve the problem of finding the rate at which a man rows in still water, we can break down the solution into clear steps.

### Step-by-Step Solution:

1. **Define Variables:**
Let:
- \( x \) = speed of the boat in still water (in km/hr)
- \( y \) = speed of the stream (in km/hr)

2. **Determine Speeds:**
- When rowing downstream (with the stream), the effective speed is \( x + y \).
- When rowing upstream (against the stream), the effective speed is \( x - y \).

3. **Use Given Information:**
- The man rows 15 km downstream in 3 hours. Therefore:
\[ \text{Speed downstream} = \frac{\text{Distance}}{\text{Time}} = \frac{15 \text{ km}}{3 \text{ hr}} = 5 \text{ km/hr} \]
This gives us the equation:
\[ x + y = 5 \quad \text{(1)} \]

- The man returns 15 km upstream in \( 7 \frac{1}{2} \) hours. Convert \( 7 \frac{1}{2} \) hours to an improper fraction:
\[ 7 \frac{1}{2} = \frac{15}{2} \text{ hours} \]
Thus, the speed upstream is:
\[ \text{Speed upstream} = \frac{15 \text{ km}}{\frac{15}{2} \text{ hr}} = 2 \text{ km/hr} \]
This gives us the equation:
\[ x - y = 2 \quad \text{(2)} \]

4. **Solve the System of Equations:**
Now we have a system of two equations:
\[ \begin{align*} x + y &= 5 \quad \text{(1)} \\ x - y &= 2 \quad \text{(2)} \end{align*} \]
To eliminate \( y \), we can add both equations:
\[ (x + y) + (x - y) = 5 + 2 \]
This simplifies to:
\[ 2x = 7 \implies x = \frac{7}{2} = 3.5 \text{ km/hr} \]

5. **Find \( y \):**
Substitute \( x = 3.5 \) back into equation (1):
\[ 3.5 + y = 5 \implies y = 5 - 3.5 = 1.5 \text{ km/hr} \]

6. **Conclusion:**
The speed at which the man rows in still water is:
\[ \boxed{3.5 \text{ km/hr}} \]