A fast train takes 2 h less for a journey of 300 km in comparison to a slow train whose speed is 5 km/h less than that of the fast train. The speed of the fast train is equal to

Options

A. 30 km/h
B. 25 km/h
C. 40 km/h
D. 45 km/h

Answer

To solve the problem, we need to find the speed of the fast train. Let's denote the speed of the fast train as \( x \) km/h. Consequently, the speed of the slow train will be \( x - 5 \) km/h since it is stated that the slow train is 5 km/h slower than the fast train.

### Step 1: Set up the equations based on the information given.

The distance for both trains is the same, which is 300 km. We can express the time taken by each train using the formula:

\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]

For the fast train, the time taken can be expressed as:

\[ \text{Time}_{\text{fast}} = \frac{300}{x} \]

For the slow train, the time taken is:

\[ \text{Time}_{\text{slow}} = \frac{300}{x - 5} \]

According to the problem, the fast train takes 2 hours less than the slow train. Therefore, we can write the equation:

\[ \frac{300}{x} = \frac{300}{x - 5} - 2 \]

### Step 2: Solve the equation.

To eliminate the fractions, we can multiply through by \( x(x - 5) \):

\[ 300(x - 5) = 300x - 2x(x - 5) \]

Expanding both sides gives:

\[ 300x - 1500 = 300x - 2x^2 + 10x \]

Now, simplifying this equation:

\[ 300x - 1500 = 300x - 2x^2 + 10x \]

Subtracting \( 300x \) from both sides:

\[ -1500 = -2x^2 + 10x \]

Rearranging gives:

\[ 2x^2 - 10x - 1500 = 0 \]

### Step 3: Simplify the quadratic equation.

Dividing the entire equation by 2:

\[ x^2 - 5x - 750 = 0 \]

### Step 4: Use the quadratic formula to find \( x \).

The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -5 \), and \( c = -750 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-5)^2 - 4 \cdot 1 \cdot (-750) = 25 + 3000 = 3025 \]

Now substituting into the quadratic formula:

\[ x = \frac{-(-5) \pm \sqrt{3025}}{2 \cdot 1} = \frac{5 \pm 55}{2} \]

Calculating the two possible values:

1. \( x = \frac{60}{2} = 30 \)
2. \( x = \frac{-50}{2} = -25 \) (not valid since speed cannot be negative)

Thus, the speed of the fast train is:

\[ \boxed{30} \text{ km/h} \]