A boat goes 4 km upstream and 4 km downstream in 1 hour. The same boat goes 5 km downstream and 3 km upstream in 55 minutes. What is the speed (in km/hr) of boat in still water ?

Options

A. 6.5
B. 7.75
C. 9
D. 8.5

Answer

To solve the problem, we need to find the speed of the boat in still water. Let's denote:

- Speed of the boat in still water = \( x \) km/hr
- Speed of the stream = \( y \) km/hr

### Step 1: Set up the equations based on the given information

1. The first part of the problem states that the boat goes 4 km upstream and 4 km downstream in 1 hour.
- Upstream speed = \( x - y \)
- Downstream speed = \( x + y \)

The time taken for upstream and downstream can be expressed as:
\[ \frac{4}{x - y} + \frac{4}{x + y} = 1 \]
This is our **Equation 1**.

2. The second part states that the boat goes 5 km downstream and 3 km upstream in 55 minutes (which is \( \frac{11}{12} \) hours).
- The time taken can be expressed as:
\[ \frac{5}{x + y} + \frac{3}{x - y} = \frac{11}{12} \]
This is our **Equation 2**.

### Step 2: Solve Equation 1

From **Equation 1**:
\[ \frac{4}{x - y} + \frac{4}{x + y} = 1 \]
Multiplying through by \( (x - y)(x + y) \) to eliminate the denominators:
\[ 4(x + y) + 4(x - y) = (x - y)(x + y) \]
This simplifies to:
\[ 4x + 4y + 4x - 4y = x^2 - y^2 \]
Combining like terms gives:
\[ 8x = x^2 - y^2 \]
Rearranging:
\[ x^2 - 8x - y^2 = 0 \quad \text{(Equation 3)} \]

### Step 3: Solve Equation 2

From **Equation 2**:
\[ \frac{5}{x + y} + \frac{3}{x - y} = \frac{11}{12} \]
Multiplying through by \( 12(x + y)(x - y) \):
\[ 60(x - y) + 36(x + y) = 11(x^2 - y^2) \]
Expanding gives:
\[ 60x - 60y + 36x + 36y = 11x^2 - 11y^2 \]
Combining like terms:
\[ 96x - 24y = 11x^2 - 11y^2 \]
Rearranging:
\[ 11x^2 - 96x - 11y^2 + 24y = 0 \quad \text{(Equation 4)} \]

### Step 4: Solve the system of equations

Now we have two equations (Equation 3 and Equation 4) involving \( x \) and \( y \). We can express \( y^2 \) from Equation 3:
\[ y^2 = x^2 - 8x \]
Substituting this into Equation 4:
\[ 11x^2 - 96x - 11(x^2 - 8x) + 24y = 0 \]
This simplifies to:
\[ 11x^2 - 96x - 11x^2 + 88x + 24y = 0 \]
Thus:
\[ -8x + 24y = 0 \implies y = \frac{8x}{24} = \frac{x}{3} \]

### Step 5: Substitute back to find \( x \)

Substituting \( y = \frac{x}{3} \) back into Equation 3:
\[ x^2 - 8x - \left(\frac{x}{3}\right)^2 = 0 \]
This leads to:
\[ x^2 - 8x - \frac{x^2}{9} = 0 \]
Multiplying through by 9 to eliminate the fraction:
\[ 9x^2 - 72x - x^2 = 0 \]
Combining like terms:
\[ 8x^2 - 72x = 0 \]
Factoring out \( 8x \):
\[ 8x(x - 9) = 0 \]
Thus, \( x = 0 \) or \( x = 9 \). Since speed cannot be zero, we have:
\[ x = 9 \text{ km/hr} \]

### Conclusion

The speed of the boat in still water is \( 9 \) km/hr.