A boat has a speed of 30 km/hr in still water. In flowing water, it covers a distance of 50 km in the direction of flow and comes back in the opposite direction. If it covers this total of 100 km in 10 hours, then what is the speed of flow of the water (in km/hr)?

Let the speed of the boat in still water be \( 30 \, \text{km/hr} \), and let the speed of the flow of the water be \( x \, \text{km/hr} \).

### Step 1: Calculate time for downstream travel
When the boat is moving downstream (in the direction of flow), its effective speed is the sum of the boat's speed and the water's speed, i.e., \( (30 + x) \, \text{km/hr} \).

The time taken to travel 50 km downstream is given by:
\[
t_{\text{down}} = \frac{50}{30 + x}
\]

### Step 2: Calculate time for upstream travel
When the boat is moving upstream (against the flow), its effective speed is the difference between the boat's speed and the water's speed, i.e., \( (30 - x) \, \text{km/hr} \).

The time taken to travel 50 km upstream is given by:
\[
t_{\text{up}} = \frac{50}{30 - x}
\]

### Step 3: Total time for the journey
The total time for both downstream and upstream travel is 10 hours. Therefore, we have the equation:
\[
t_{\text{down}} + t_{\text{up}} = 10
\]
Substituting the expressions for \( t_{\text{down}} \) and \( t_{\text{up}} \):
\[
\frac{50}{30 + x} + \frac{50}{30 - x} = 10
\]

### Step 4: Solve the equation
Multiply both sides of the equation by \( (30 + x)(30 - x) \) to eliminate the denominators:
\[
50(30 - x) + 50(30 + x) = 10(30 + x)(30 - x)
\]

Simplifying the left side:
\[
50(30 - x) + 50(30 + x) = 50 \times 30 - 50 \times x + 50 \times 30 + 50 \times x = 3000
\]

The right side becomes:
\[
10(30^2 - x^2) = 10(900 - x^2) = 9000 - 10x^2
\]

Now, equate both sides:
\[
3000 = 9000 - 10x^2
\]

### Step 5: Isolate \( x^2 \)
Rearranging the equation:
\[
10x^2 = 9000 - 3000
\]
\[
10x^2 = 6000
\]
\[
x^2 = 600
\]

### Step 6: Solve for \( x \)
\[
x = \sqrt{600} \approx 24.49 \, \text{km/hr}
\]

Therefore, the speed of the flow of the water is approximately \( 24.49 \, \text{km/hr} \).

---

Would you like more details or have any questions? Here are some related problems to explore:

1. What would be the total time if the speed of the flow increased to 25 km/hr?
2. How does the time change if the boat's speed in still water decreases to 28 km/hr?
3. What happens to the travel time if the distance is increased to 60 km downstream and upstream?
4. If the boat travels only downstream for 50 km, what is the total time taken?
5. What would the water speed be if the total journey time was reduced to 9 hours?

**Tip:** When solving these types of problems, setting up expressions for downstream and upstream times separately can help avoid errors.

Find the speed of water flow when a boat travels 50 km downstream and back upstream in 10 hours. This problem involves algebraic concepts and quadratic equations, suitable for students in Grades 9-11.

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