A motorboat travels 10 km. upstream in the stream which is following at a speed of 2 km./hr. and returns to starting point in 55 min. What is the speed of motorboat in still water ?

Options

A. 24 km./hour
B. 22 km./hour
C. 32 km./hour
D. 28 km./hour

Answer

To find the speed of the motorboat in still water, we can follow these steps:

### Step 1: Understand the Problem
The motorboat travels 10 km upstream against the current and then returns downstream to the starting point. The speed of the current (stream) is given as 2 km/hr, and the total time for the round trip is 55 minutes.

### Step 2: Convert Time to Hours
Since the speed is given in km/hr, we need to convert the total time from minutes to hours.
- Total time = 55 minutes = 55/60 hours = 11/12 hours.

### Step 3: Define Variables
Let the speed of the motorboat in still water be \( b \) km/hr.
- The speed of the boat upstream (against the current) = \( b - 2 \) km/hr.
- The speed of the boat downstream (with the current) = \( b + 2 \) km/hr.

### Step 4: Calculate Time Taken for Each Leg of the Journey
- Time taken to travel upstream (10 km) = Distance / Speed = \( \frac{10}{b - 2} \) hours.
- Time taken to travel downstream (10 km) = Distance / Speed = \( \frac{10}{b + 2} \) hours.

### Step 5: Set Up the Equation
The total time for the journey (upstream + downstream) is equal to the total time given:
\[ \frac{10}{b - 2} + \frac{10}{b + 2} = \frac{11}{12} \]

### Step 6: Solve the Equation
Multiply through by \( 12(b - 2)(b + 2) \) to eliminate the denominators:
\[ 12(b + 2) \cdot 10 + 12(b - 2) \cdot 10 = 11(b - 2)(b + 2) \]
\[ 120(b + 2) + 120(b - 2) = 11(b^2 - 4) \]
\[ 120b + 240 + 120b - 240 = 11b^2 - 44 \]
\[ 240b = 11b^2 - 44 \]
Rearranging gives:
\[ 11b^2 - 240b - 44 = 0 \]

### Step 7: Use the Quadratic Formula
To solve for \( b \), we can use the quadratic formula:
\[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
where \( A = 11, B = -240, C = -44 \).

Calculating the discriminant:
\[ B^2 - 4AC = (-240)^2 - 4 \cdot 11 \cdot (-44) = 57600 + 1936 = 59536 \]

Now substituting back into the formula:
\[ b = \frac{240 \pm \sqrt{59536}}{22} \]
Calculating \( \sqrt{59536} \approx 244 \):
\[ b = \frac{240 \pm 244}{22} \]

Calculating the two possible values:
1. \( b = \frac{484}{22} = 22 \) km/hr
2. \( b = \frac{-4}{22} \) (not possible since speed cannot be negative)

### Conclusion
Thus, the speed of the motorboat in still water is **22 km/hr**.

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