A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to cover the total distance. Assuming the uniform speed to be 'x' km/h, form an equation and solve it to evaluate 'x'.

Answer

To solve the problem step by step, we will start by defining the variables and then formulating the equations based on the information provided.

### Step 1: Define the variables
Let the uniform speed of the bus be \( x \) km/h.

### Step 2: Calculate the time taken at the original speed
The distance covered by the bus is 240 km. The time taken to cover this distance at speed \( x \) is given by the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{240}{x} \text{ hours} \]

### Step 3: Calculate the time taken at the reduced speed
Due to heavy rain, the speed of the bus is reduced by 10 km/h. Therefore, the new speed is \( x - 10 \) km/h. The time taken to cover the same distance at this reduced speed is:
\[ \text{Time} = \frac{240}{x - 10} \text{ hours} \]

### Step 4: Set up the equation based on the time difference
According to the problem, the time taken at the reduced speed is 2 hours longer than the time taken at the original speed. Therefore, we can write the equation:
\[ \frac{240}{x - 10} = \frac{240}{x} + 2 \]

### Step 5: Clear the fractions
To eliminate the fractions, we can multiply through by \( x(x - 10) \):
\[ 240x = 240(x - 10) + 2x(x - 10) \]

### Step 6: Expand and simplify the equation
Expanding both sides gives:
\[ 240x = 240x - 2400 + 2x^2 - 20x \]
Now, we can simplify this equation:
\[ 0 = 2x^2 - 20x - 2400 \]

### Step 7: Divide the entire equation by 2
To simplify further, divide the entire equation by 2:
\[ 0 = x^2 - 10x - 1200 \]

### Step 8: Factor the quadratic equation
We need to factor the quadratic equation \( x^2 - 10x - 1200 = 0 \). We look for two numbers that multiply to -1200 and add to -10. The numbers are -40 and 30:
\[ (x - 40)(x + 30) = 0 \]

### Step 9: Solve for \( x \)
Setting each factor to zero gives:
1. \( x - 40 = 0 \) → \( x = 40 \)
2. \( x + 30 = 0 \) → \( x = -30 \) (not valid since speed cannot be negative)

### Step 10: Conclusion
Thus, the speed of the bus is:
\[ \boxed{40} \text{ km/h} \]