Jane took 20 min to drive her boat upstream to​ water-ski at her favorite spot. Coming back later in the​ day, at the same boat​ speed, took her 10 min. If the current in that part of the river is 9 km per​ hr, what was her boat speed in still​ water?

Let the boat speed in still water be \( v \) km/h, and let the current speed be 9 km/h.

When Jane drives upstream, her effective speed is \( v - 9 \) km/h (since the current is against her). When she drives downstream, her effective speed is \( v + 9 \) km/h (since the current aids her).

### Step 1: Set up the relationship between time, distance, and speed.

We know that:
- Distance = Speed × Time

Let the distance to Jane's favorite spot be \( d \) kilometers.

Upstream:
- Time taken = 20 minutes = \( \frac{20}{60} = \frac{1}{3} \) hours
- Speed upstream = \( v - 9 \) km/h
- Distance \( d \) = Speed × Time = \( (v - 9) \times \frac{1}{3} \)

Downstream:
- Time taken = 10 minutes = \( \frac{10}{60} = \frac{1}{6} \) hours
- Speed downstream = \( v + 9 \) km/h
- Distance \( d \) = Speed × Time = \( (v + 9) \times \frac{1}{6} \)

### Step 2: Set up the equation for distance.

Since the distance upstream and downstream is the same, we can equate the two expressions for distance:

\[
(v - 9) \times \frac{1}{3} = (v + 9) \times \frac{1}{6}
\]

### Step 3: Solve for \( v \).

Multiply both sides by 6 to eliminate the fractions:

\[
2(v - 9) = (v + 9)
\]

Expand both sides:

\[
2v - 18 = v + 9
\]

Move all terms involving \( v \) to one side:

\[
2v - v = 9 + 18
\]

Simplify:

\[
v = 27 \, \text{km/h}
\]

### Conclusion:
Jane's boat speed in still water is \( \boxed{27} \, \text{km/h} \).

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Would you like further details on any step? 

Here are 5 related questions:
1. What would Jane's time be if the current was 5 km/h instead of 9 km/h?
2. How would the equation change if Jane had traveled a different distance upstream?
3. How does the current affect the boat speed in

Find Jane's boat speed in still water as she travels upstream and downstream with a current of 9 km/h. Using algebraic equations, this solution explains the relationship between speed, distance, and time. Ideal for students in Grades 8-10.

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