P, Q and R are the three towns on a river which flows uniformly, Q is equidistant from P and R. Sandeep can row from to P to Q and back in 10 hours and Sandeep can row from P to R in 4 hours compare the speed of his boat in still water with that of the river?

Options

A. 4 : 3
B. 5 : 3
C. 6 : 5
D. 7 : 3

Answer

To solve the problem, we need to compare the speed of Sandeep's boat in still water (let's denote it as \( x \)) with the speed of the river current (let's denote it as \( y \)).

### Step-by-Step Solution:

1. **Understanding the Distances and Times**:
- Let the distance from P to Q (and also from Q to R) be \( d \).
- Sandeep rows from P to Q and back to P in 10 hours.
- He rows from P to R in 4 hours.

2. **Finding the Time for P to Q**:
- The time taken to go from P to Q and back is 10 hours.
- Therefore, the time taken to go from P to Q is \( \frac{10}{2} = 5 \) hours.

3. **Finding the Time for P to R**:
- The time taken to go from P to R is 4 hours.
- Since Q is equidistant from P and R, the distance from P to Q is half of the distance from P to R.
- Thus, the time taken to go from P to Q is \( \frac{4}{2} = 2 \) hours.

4. **Determining the Speeds**:
- The speed of the boat downstream (from P to Q) is \( x + y \) and the speed upstream (from Q to P) is \( x - y \).
- For the journey from P to Q (downstream), the speed is \( x + y \), and for the return journey (upstream), the speed is \( x - y \).

5. **Setting Up the Equations**:
- From P to Q:
\[ \text{Distance} = \text{Speed} \times \text{Time} \implies d = (x + y) \cdot 2 \quad \text{(downstream)} \]
- From Q to P:
\[ d = (x - y) \cdot 5 \quad \text{(upstream)} \]

6. **Equating the Distances**:
- Since both expressions equal \( d \):
\[ (x + y) \cdot 2 = (x - y) \cdot 5 \]

7. **Expanding and Rearranging**:
- Expanding both sides:
\[ 2x + 2y = 5x - 5y \]
- Rearranging gives:
\[ 2x + 2y + 5y = 5x \implies 2x + 7y = 5x \]
- Moving \( 2x \) to the right:
\[ 7y = 5x - 2x \implies 7y = 3x \]

8. **Finding the Ratio**:
- Dividing both sides by \( y \) and \( 3 \):
\[ \frac{x}{y} = \frac{7}{3} \]

9. **Final Ratio**:
- Therefore, the ratio of the speed of the boat in still water to the speed of the river current is:
\[ x : y = 7 : 3 \]