Distance between A and B is 60 kms. When they move in opposite direction they meet in 6 hours. If A moves at 2/3rd of its actual speed and B at 2 times of its speed then they meet in 5 hours what are their respective speed.

Answer

To solve the problem step by step, we will define the speeds of A and B, set up equations based on the information provided, and then solve those equations.

### Step 1: Define the Speeds
Let the speed of A be \( S_A \) km/h and the speed of B be \( S_B \) km/h.

### Step 2: Set Up the First Equation
When A and B move in opposite directions, they meet in 6 hours. The total distance between A and B is 60 km. The relative speed when moving in opposite directions is the sum of their speeds:
\[ S_A + S_B = \frac{\text{Distance}}{\text{Time}} = \frac{60 \text{ km}}{6 \text{ hours}} = 10 \text{ km/h} \]
Thus, we have our first equation:
\[ (1) \quad S_A + S_B = 10 \]

### Step 3: Set Up the Second Equation
In the second scenario, A moves at \( \frac{2}{3} \) of its actual speed and B moves at \( 2 \) times its speed. They meet in 5 hours:
\[ \frac{2}{3} S_A + 2 S_B = \frac{60 \text{ km}}{5 \text{ hours}} = 12 \text{ km/h} \]
This gives us our second equation:
\[ (2) \quad \frac{2}{3} S_A + 2 S_B = 12 \]

### Step 4: Solve the Equations
Now we have a system of equations:
1. \( S_A + S_B = 10 \)
2. \( \frac{2}{3} S_A + 2 S_B = 12 \)

From equation (1), we can express \( S_B \) in terms of \( S_A \):
\[ S_B = 10 - S_A \]

### Step 5: Substitute \( S_B \) in the Second Equation
Substituting \( S_B \) in equation (2):
\[ \frac{2}{3} S_A + 2(10 - S_A) = 12 \]
Expanding this gives:
\[ \frac{2}{3} S_A + 20 - 2 S_A = 12 \]

### Step 6: Combine Like Terms
To combine like terms, convert \( 2 S_A \) into thirds:
\[ \frac{2}{3} S_A - \frac{6}{3} S_A + 20 = 12 \]
This simplifies to:
\[ -\frac{4}{3} S_A + 20 = 12 \]

### Step 7: Isolate \( S_A \)
Subtract 20 from both sides:
\[ -\frac{4}{3} S_A = 12 - 20 \]
\[ -\frac{4}{3} S_A = -8 \]
Multiplying both sides by -1:
\[ \frac{4}{3} S_A = 8 \]
Now multiply both sides by \( \frac{3}{4} \):
\[ S_A = 8 \times \frac{3}{4} = 6 \text{ km/h} \]

### Step 8: Find \( S_B \)
Now substitute \( S_A \) back into equation (1) to find \( S_B \):
\[ S_B = 10 - S_A = 10 - 6 = 4 \text{ km/h} \]

### Final Answer
The speeds of A and B are:
- Speed of A, \( S_A = 6 \) km/h
- Speed of B, \( S_B = 4 \) km/h