To cover D km downstream, Boat A takes 1 hour 40 min and to cover D km upstream Boat B takes 3 hours 45 min. If the speed of Boat A (in still water) is 9 kmph more than that of Boat B then what is the speed of Boat B in still water (speed of current being 3 kmph constant for both boats)?

Options

A. 24 kmph
B. 27 kmph
C. None of those given'as options
D. 15kmPh

Answer

To solve the problem step by step, we will first define the variables and then use the information provided to set up equations.

### Step 1: Define Variables
Let the speed of Boat B in still water be \( x \) km/h.
Then, the speed of Boat A in still water will be \( x + 9 \) km/h (since Boat A is 9 km/h faster than Boat B).

### Step 2: Convert Time to Hours
Boat A takes 1 hour and 40 minutes to cover \( D \) km downstream.
1 hour and 40 minutes can be converted to hours:
\[ 1 \text{ hour} + \frac{40 \text{ minutes}}{60} = 1 + \frac{2}{3} = \frac{5}{3} \text{ hours} \]

Boat B takes 3 hours and 45 minutes to cover \( D \) km upstream.
3 hours and 45 minutes can be converted to hours:
\[ 3 \text{ hours} + \frac{45 \text{ minutes}}{60} = 3 + \frac{3}{4} = \frac{15}{4} \text{ hours} \]

### Step 3: Set Up Equations for Downstream and Upstream
For Boat A (downstream), the effective speed is:
\[ \text{Speed of Boat A} + \text{Speed of Current} = (x + 9) + 3 = x + 12 \text{ km/h} \]
Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \):
\[ \frac{5}{3} = \frac{D}{x + 12} \]
Cross-multiplying gives us:
\[ 5(x + 12) = 3D \quad \text{(Equation 1)} \]

For Boat B (upstream), the effective speed is:
\[ \text{Speed of Boat B} - \text{Speed of Current} = x - 3 \text{ km/h} \]
Using the same formula:
\[ \frac{15}{4} = \frac{D}{x - 3} \]
Cross-multiplying gives us:
\[ 15(x - 3) = 4D \quad \text{(Equation 2)} \]

### Step 4: Express D from Both Equations
From Equation 1:
\[ D = \frac{5(x + 12)}{3} \]
From Equation 2:
\[ D = \frac{15(x - 3)}{4} \]

### Step 5: Set the Two Expressions for D Equal
Setting the two expressions for \( D \) equal:
\[ \frac{5(x + 12)}{3} = \frac{15(x - 3)}{4} \]

### Step 6: Cross Multiply to Solve for x
Cross-multiplying gives:
\[ 5(x + 12) \cdot 4 = 15(x - 3) \cdot 3 \]
This simplifies to:
\[ 20(x + 12) = 45(x - 3) \]
Expanding both sides:
\[ 20x + 240 = 45x - 135 \]

### Step 7: Rearranging the Equation
Rearranging the equation:
\[ 240 + 135 = 45x - 20x \]
\[ 375 = 25x \]

### Step 8: Solve for x
Dividing both sides by 25:
\[ x = 15 \]

### Conclusion
The speed of Boat B in still water is \( \boxed{15} \) km/h.