A, B and C are situated at the bank of river which is flowing at a constant rate. B is at an equal distance with A and C. A swimmer Avinash takes 10 h to swim from A to B and B to A .Also ,he takes 4 h to swim from A to C. What is the ratio of speed of Avinash in still water and speed of stream?

Options

A. `5 :3`
B. `3 :5`
C. `2 :5`
D. `1 :2`

Answer

To solve the problem, we need to establish the relationship between the speeds of Avinash in still water and the speed of the stream. Let's denote:

- Speed of Avinash in still water = x km/h
- Speed of the stream = y km/h

### Step 1: Understand the distances and times
Avinash takes 10 hours to swim from A to B and back to A. This means the total time for the round trip from A to B is 10 hours.

### Step 2: Calculate the effective speed from A to B
Let the distance from A to B be d km. The effective speed when swimming downstream (from A to B) is (x + y) km/h, and the effective speed when swimming upstream (from B to A) is (x - y) km/h.

The time taken to swim downstream from A to B is:
\[ \text{Time}_{AB} = \frac{d}{x + y} \]

The time taken to swim upstream from B to A is:
\[ \text{Time}_{BA} = \frac{d}{x - y} \]

Since the total time for the round trip is 10 hours:
\[ \frac{d}{x + y} + \frac{d}{x - y} = 10 \]

### Step 3: Simplify the equation
To simplify, we can combine the fractions:
\[ \frac{d(x - y) + d(x + y)}{(x + y)(x - y)} = 10 \]
This simplifies to:
\[ \frac{2dx}{(x + y)(x - y)} = 10 \]

### Step 4: Rearranging the equation
Multiplying both sides by (x + y)(x - y):
\[ 2dx = 10(x^2 - y^2) \]
\[ 2dx = 10x^2 - 10y^2 \]
Dividing everything by 2:
\[ dx = 5x^2 - 5y^2 \]
Rearranging gives us:
\[ 5x^2 - dx - 5y^2 = 0 \] (Equation 1)

### Step 5: Calculate the effective speed from A to C
Avinash takes 4 hours to swim from A to C. Since B is equidistant from A and C, let the distance from A to C also be d km.

The effective speed when swimming downstream from A to C is still (x + y) km/h, and the time taken is:
\[ \frac{d}{x + y} = 4 \]
This implies:
\[ d = 4(x + y) \] (Equation 2)

### Step 6: Substitute Equation 2 into Equation 1
Substituting d from Equation 2 into Equation 1:
\[ 5x^2 - 4(x + y)x - 5y^2 = 0 \]
Expanding gives:
\[ 5x^2 - 4x^2 - 4xy - 5y^2 = 0 \]
This simplifies to:
\[ x^2 - 4xy - 5y^2 = 0 \]

### Step 7: Factor the quadratic equation
Factoring gives:
\[ (x - 5y)(x + y) = 0 \]
This gives us two possible solutions:
1. \( x - 5y = 0 \) → \( x = 5y \)
2. \( x + y = 0 \) → Not possible since speeds cannot be negative.

### Step 8: Find the ratio
From \( x = 5y \), the ratio of the speed of Avinash in still water to the speed of the stream is:
\[ \frac{x}{y} = 5 \]

### Final Answer
The ratio of the speed of Avinash in still water to the speed of the stream is **5:1**.