A motorboat goes from A to B and comes back located at bank of river. If speed of motor boat becomes double in still water, it take 20% of its original time to travel from A to B and B to A. Find speed of motor boat is how many times than speed of flow of river ?

Options

A. `sqrt((3)/(2))`
B. `sqrt((2)/(3))`
C. `(2)/(3)`
D. `(3)/(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define Variables
Let:
- Speed of the motorboat in still water = \( x \) km/h
- Speed of the river flow = \( y \) km/h

### Step 2: Determine Speeds Upstream and Downstream
- Upstream speed (against the flow) = \( x - y \) km/h
- Downstream speed (with the flow) = \( x + y \) km/h

### Step 3: Establish Time for Original Speed
Let the distance from A to B be \( D \) km. The time taken to travel from A to B and back from B to A is given by:
\[ \text{Total Time} = \frac{D}{x - y} + \frac{D}{x + y} \]

### Step 4: Establish Time for Doubled Speed
When the speed of the motorboat is doubled, the new speed becomes \( 2x \) km/h. Therefore, the new upstream and downstream speeds are:
- Upstream speed = \( 2x - y \) km/h
- Downstream speed = \( 2x + y \) km/h

The time taken for this scenario is:
\[ \text{Total Time} = \frac{D}{2x - y} + \frac{D}{2x + y} \]

### Step 5: Relate the Two Times
According to the problem, the time taken with the doubled speed is 20% of the original time:
\[ \frac{D}{2x - y} + \frac{D}{2x + y} = \frac{1}{5} \left( \frac{D}{x - y} + \frac{D}{x + y} \right) \]

### Step 6: Simplify the Equation
We can cancel \( D \) from all terms (assuming \( D \neq 0 \)):
\[ \frac{1}{2x - y} + \frac{1}{2x + y} = \frac{1}{5} \left( \frac{1}{x - y} + \frac{1}{x + y} \right) \]

### Step 7: Cross Multiply and Rearrange
Cross multiplying gives us:
\[ 5 \left( \frac{1}{2x - y} + \frac{1}{2x + y} \right) = \frac{1}{x - y} + \frac{1}{x + y} \]

### Step 8: Find a Common Denominator
Finding a common denominator for both sides and simplifying will lead to:
\[ \frac{(2x + y) + (2x - y)}{(2x - y)(2x + y)} = \frac{(x + y) + (x - y)}{(x - y)(x + y)} \]
This simplifies to:
\[ \frac{4x}{(2x - y)(2x + y)} = \frac{2x}{(x - y)(x + y)} \]

### Step 9: Cross Multiply Again
Cross multiplying gives:
\[ 4x(x - y)(x + y) = 2x(2x - y)(2x + y) \]

### Step 10: Simplify and Solve for x and y
After simplifying and rearranging, we will arrive at:
\[ 12x^2 = 18y^2 \]
This simplifies to:
\[ \frac{x^2}{y^2} = \frac{18}{12} = \frac{3}{2} \]
Taking the square root gives:
\[ \frac{x}{y} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} \]

### Conclusion
Thus, the speed of the motorboat is \( \sqrt{\frac{3}{2}} \) times the speed of the flow of the river.