A particle goes from `A` to `B` with a speed of `40km//h` and `B` to `C` with a speed of `60Km//h`. If `AB = 6BC`, the average speed in `km//h` between `A` and `C` is-
[Hint : Average speed `= ("total distance travelled")/("time taken")]`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the distances
Let the distance from B to C be \( BC = x \). According to the problem, the distance from A to B is \( AB = 6BC = 6x \).

### Step 2: Calculate the total distance
The total distance from A to C is:
\[ AC = AB + BC = 6x + x = 7x \]

### Step 3: Calculate the time taken for each segment
1. **Time taken from A to B**:
- Speed from A to B = \( 40 \, \text{km/h} \)
- Distance from A to B = \( 6x \)
- Time taken \( t_1 \) is given by:
\[ t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{6x}{40} = \frac{3x}{20} \, \text{hours} \]

2. **Time taken from B to C**:
- Speed from B to C = \( 60 \, \text{km/h} \)
- Distance from B to C = \( x \)
- Time taken \( t_2 \) is given by:
\[ t_2 = \frac{\text{Distance}}{\text{Speed}} = \frac{x}{60} \, \text{hours} \]

### Step 4: Calculate the total time taken
The total time \( T \) taken to travel from A to C is:
\[ T = t_1 + t_2 = \frac{3x}{20} + \frac{x}{60} \]

To add these fractions, we need a common denominator. The least common multiple of 20 and 60 is 60. Thus, we convert:
\[ \frac{3x}{20} = \frac{9x}{60} \]
Now we can add:
\[ T = \frac{9x}{60} + \frac{x}{60} = \frac{10x}{60} = \frac{x}{6} \, \text{hours} \]

### Step 5: Calculate the average speed
The average speed \( V_{avg} \) is given by:
\[ V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{7x}{\frac{x}{6}} = 7x \cdot \frac{6}{x} = 42 \, \text{km/h} \]

Thus, the average speed between A and C is:
\[ \boxed{42 \, \text{km/h}} \]