A motor boat went downstream for 120 km and immediately returned it took the boat 15 hours to complete the round trip.if the speed of the river were twice as high the trip downstream and back would take 24 hours.What is the speed of the boat in still water?

Options

A. 20 km/h
B. 18 km/h
C. 15 km/h
D. 16 km/h

Answer

To solve the problem step by step, let's break it down:

### Step 1: Define Variables
Let:
- \( x \) = speed of the boat in still water (in km/h)
- \( y \) = speed of the river current (in km/h)

### Step 2: Write the Time Equation for the First Scenario
The boat travels downstream and upstream a total distance of 120 km each way. The time taken for the downstream trip is given by:
\[ \text{Time downstream} = \frac{120}{x + y} \]
The time taken for the upstream trip is given by:
\[ \text{Time upstream} = \frac{120}{x - y} \]
According to the problem, the total time for the round trip is 15 hours:
\[ \frac{120}{x + y} + \frac{120}{x - y} = 15 \]
This is our **Equation 1**.

### Step 3: Simplify Equation 1
To simplify, we find a common denominator:
\[ \frac{120(x - y) + 120(x + y)}{(x + y)(x - y)} = 15 \]
This simplifies to:
\[ \frac{240x}{x^2 - y^2} = 15 \]
Cross-multiplying gives:
\[ 240x = 15(x^2 - y^2) \]
Dividing both sides by 15:
\[ 16x = x^2 - y^2 \]
Rearranging gives us:
\[ x^2 - 16x - y^2 = 0 \quad \text{(Equation 1)} \]

### Step 4: Write the Time Equation for the Second Scenario
If the speed of the river current doubles, the new speed of the current is \( 2y \). The time for the downstream trip becomes:
\[ \text{Time downstream} = \frac{120}{x + 2y} \]
The time for the upstream trip becomes:
\[ \text{Time upstream} = \frac{120}{x - 2y} \]
According to the problem, the total time for this round trip is 24 hours:
\[ \frac{120}{x + 2y} + \frac{120}{x - 2y} = 24 \]
This is our **Equation 2**.

### Step 5: Simplify Equation 2
Finding a common denominator:
\[ \frac{120(x - 2y) + 120(x + 2y)}{(x + 2y)(x - 2y)} = 24 \]
This simplifies to:
\[ \frac{240x}{x^2 - 4y^2} = 24 \]
Cross-multiplying gives:
\[ 240x = 24(x^2 - 4y^2) \]
Dividing both sides by 24:
\[ 10x = x^2 - 4y^2 \]
Rearranging gives us:
\[ x^2 - 10x - 4y^2 = 0 \quad \text{(Equation 2)} \]

### Step 6: Solve the System of Equations
Now we have two equations:
1. \( x^2 - 16x - y^2 = 0 \)
2. \( x^2 - 10x - 4y^2 = 0 \)

From Equation 1, we can express \( y^2 \):
\[ y^2 = x^2 - 16x \]
Substituting \( y^2 \) into Equation 2:
\[ x^2 - 10x - 4(x^2 - 16x) = 0 \]
This simplifies to:
\[ x^2 - 10x - 4x^2 + 64x = 0 \]
Combining like terms:
\[ -3x^2 + 54x = 0 \]
Factoring out \( x \):
\[ x(54 - 3x) = 0 \]
Thus, \( x = 0 \) or \( x = 18 \).

Since speed cannot be zero, we have:
\[ x = 18 \text{ km/h} \]

### Step 7: Find the Speed of the Current
Using \( y^2 = x^2 - 16x \):
\[ y^2 = 18^2 - 16 \times 18 = 324 - 288 = 36 \]
Thus, \( y = 6 \text{ km/h} \).

### Conclusion
The speed of the boat in still water is:
\[ \boxed{18 \text{ km/h}} \]