If the speed of a bus is increased by 5 km/h from its normal speed , it takes 2 hour less for a journey of 300 km, then the normal speed of bus (in km/h) is

Options

A. 25
B. 30
C. 45
D. None of these

Answer

To solve the problem step-by-step, we can follow these steps:

### Step 1: Define the Variables
Let the normal speed of the bus be \( s \) km/h.

### Step 2: Write the Time Equations
The time taken to cover 300 km at the normal speed \( s \) is given by:
\[ \text{Time at normal speed} = \frac{300}{s} \]
When the speed is increased by 5 km/h, the new speed becomes \( s + 5 \) km/h. The time taken at this increased speed is:
\[ \text{Time at increased speed} = \frac{300}{s + 5} \]

### Step 3: Set Up the Equation
According to the problem, the time taken at the normal speed is 2 hours more than the time taken at the increased speed. Therefore, we can set up the equation:
\[ \frac{300}{s} - \frac{300}{s + 5} = 2 \]

### Step 4: Solve the Equation
To solve the equation, first, we can multiply through by \( s(s + 5) \) to eliminate the denominators:
\[ 300(s + 5) - 300s = 2s(s + 5) \]
This simplifies to:
\[ 300s + 1500 - 300s = 2s^2 + 10s \]
Thus, we have:
\[ 1500 = 2s^2 + 10s \]

### Step 5: Rearrange to Form a Quadratic Equation
Rearranging gives:
\[ 2s^2 + 10s - 1500 = 0 \]
Dividing the entire equation by 2 simplifies it to:
\[ s^2 + 5s - 750 = 0 \]

### Step 6: Factor the Quadratic Equation
Next, we need to factor the quadratic equation. We look for two numbers that multiply to -750 and add to 5. The numbers are 30 and -25. Thus, we can factor the equation as:
\[ (s + 30)(s - 25) = 0 \]

### Step 7: Solve for \( s \)
Setting each factor to zero gives us:
1. \( s + 30 = 0 \) → \( s = -30 \) (not valid since speed cannot be negative)
2. \( s - 25 = 0 \) → \( s = 25 \)

### Conclusion
The normal speed of the bus is \( \boxed{25} \) km/h.