A boat covers a round trip journey in a river in a certain time . If its speed in still water is doubled and the speed of the stream is tripled, it would take the same time for the round trip journey. Find the ratio of the speed of boat in still water to the speed of the stream ?

Answer

To solve the problem, we need to establish the relationship between the speed of the boat in still water (let's denote it as \( b \)) and the speed of the stream (denote it as \( s \)).

### Step-by-Step Solution:

1. **Define the Speeds:**
- Let the speed of the boat in still water be \( b \).
- Let the speed of the stream be \( s \).

2. **Calculate Effective Speeds:**
- When the boat is going downstream, its effective speed is \( b + s \).
- When the boat is going upstream, its effective speed is \( b - s \).

3. **Time for Round Trip:**
- Let the distance covered in one direction be \( d \).
- The time taken to go downstream is \( \frac{d}{b+s} \).
- The time taken to go upstream is \( \frac{d}{b-s} \).
- Therefore, the total time for the round trip is:
\[ T = \frac{d}{b+s} + \frac{d}{b-s} \]

4. **Simplify the Total Time:**
- To combine the fractions, find a common denominator:
\[ T = d \left( \frac{1}{b+s} + \frac{1}{b-s} \right) = d \left( \frac{(b-s) + (b+s)}{(b+s)(b-s)} \right) \]
- This simplifies to:
\[ T = d \left( \frac{2b}{b^2 - s^2} \right) = \frac{2bd}{b^2 - s^2} \]

5. **New Speeds After Changes:**
- If the speed of the boat is doubled, it becomes \( 2b \).
- If the speed of the stream is tripled, it becomes \( 3s \).
- The new effective speeds are:
- Downstream: \( 2b + 3s \)
- Upstream: \( 2b - 3s \)

6. **New Total Time for Round Trip:**
- The new time for the round trip is:
\[ T' = \frac{d}{2b + 3s} + \frac{d}{2b - 3s} \]
- Simplifying this gives:
\[ T' = d \left( \frac{1}{2b + 3s} + \frac{1}{2b - 3s} \right) = d \left( \frac{(2b - 3s) + (2b + 3s)}{(2b + 3s)(2b - 3s)} \right) \]
- This simplifies to:
\[ T' = d \left( \frac{4b}{4b^2 - 9s^2} \right) = \frac{4bd}{4b^2 - 9s^2} \]

7. **Equate the Two Times:**
- Since the total time for both trips is the same:
\[ \frac{2bd}{b^2 - s^2} = \frac{4bd}{4b^2 - 9s^2} \]

8. **Cancel \( bd \) (assuming \( b \neq 0 \) and \( d \neq 0 \)):**
- This simplifies to:
\[ \frac{2}{b^2 - s^2} = \frac{4}{4b^2 - 9s^2} \]

9. **Cross Multiply:**
- Cross multiplying gives:
\[ 2(4b^2 - 9s^2) = 4(b^2 - s^2) \]
- Expanding both sides:
\[ 8b^2 - 18s^2 = 4b^2 - 4s^2 \]

10. **Rearranging the Equation:**
- Rearranging gives:
\[ 8b^2 - 4b^2 = 18s^2 - 4s^2 \]
- This simplifies to:
\[ 4b^2 = 14s^2 \]
- Dividing both sides by 2:
\[ 2b^2 = 7s^2 \]

11. **Finding the Ratio:**
- Taking the square root of both sides gives:
\[ \frac{b}{s} = \sqrt{\frac{7}{2}} = \frac{\sqrt{14}}{2} \]

### Final Answer:
The ratio of the speed of the boat in still water to the speed of the stream is:
\[ \frac{b}{s} = \sqrt{\frac{7}{2}} \text{ or approximately } 1.87:1 \]