A motorboat can travel at 10 km/h in still water. It travelled 91km downstream in a river and then re- turned to the same place, taking altogether 20 h. The rate of flow of river is

Options

A. 3 km/h
B. 4 km/h
C. 2 km/h
D. 5 km/h

Answer

To solve the problem step by step, we will use the information given in the question.

### Step 1: Identify the given data
- Speed of the boat in still water (B) = 10 km/h
- Distance traveled downstream (D) = 91 km
- Total time taken for the round trip (T) = 20 hours

### Step 2: Define the speed of the river
Let the speed of the river be \( R \) km/h.

### Step 3: Calculate the downstream and upstream speeds
- Downstream speed = Speed of boat + Speed of river = \( B + R = 10 + R \) km/h
- Upstream speed = Speed of boat - Speed of river = \( B - R = 10 - R \) km/h

### Step 4: Write the time equations for downstream and upstream
Using the formula for time, which is \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \):
- Time taken to travel downstream = \( \frac{D}{\text{Downstream speed}} = \frac{91}{10 + R} \)
- Time taken to travel upstream = \( \frac{D}{\text{Upstream speed}} = \frac{91}{10 - R} \)

### Step 5: Set up the equation for total time
According to the problem, the total time for the round trip is 20 hours:
\[ \frac{91}{10 + R} + \frac{91}{10 - R} = 20 \]

### Step 6: Solve the equation
Multiply through by the common denominator \( (10 + R)(10 - R) \):
\[ 91(10 - R) + 91(10 + R) = 20(10 + R)(10 - R) \]
This simplifies to:
\[ 910 - 91R + 910 + 91R = 20(100 - R^2) \]
\[ 1820 = 2000 - 20R^2 \]
Rearranging gives:
\[ 20R^2 = 2000 - 1820 \]
\[ 20R^2 = 180 \]
\[ R^2 = 9 \]
\[ R = 3 \text{ km/h} \]

### Step 7: Conclusion
The rate of flow of the river is **3 km/h**.

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