What is the usual speed of the train ?
(i) The speed of the train is increased by 25 km/hr to reach the destination 150 km away in time
(ii) The train is late by 30 minutes.

Answer

To solve the problem of finding the usual speed of the train, we can break it down step by step.

### Step 1: Define Variables
Let the usual speed of the train be \( V \) km/hr.

### Step 2: Understand the Situation
The train has to cover a distance of 150 km. If the speed is increased by 25 km/hr, the new speed becomes \( V + 25 \) km/hr.

### Step 3: Set Up the Time Equations
1. The time taken to cover 150 km at the usual speed \( V \) is given by:
\[ \text{Time}_{\text{usual}} = \frac{150}{V} \]

2. The time taken to cover 150 km at the increased speed \( V + 25 \) is:
\[ \text{Time}_{\text{increased}} = \frac{150}{V + 25} \]

### Step 4: Relate the Times
According to the problem, the train is late by 30 minutes (or 0.5 hours). Therefore, we can set up the equation:
\[ \text{Time}_{\text{increased}} = \text{Time}_{\text{usual}} - 0.5 \]

Substituting the time equations:
\[ \frac{150}{V + 25} = \frac{150}{V} - 0.5 \]

### Step 5: Clear the Fractions
To eliminate the fractions, multiply through by \( V(V + 25) \):
\[ 150V = 150(V + 25) - 0.5V(V + 25) \]

### Step 6: Expand and Rearrange
Expanding the right side:
\[ 150V = 150V + 3750 - 0.5V^2 - 12.5V \]

Now, simplify:
\[ 0 = 3750 - 0.5V^2 - 12.5V \]

### Step 7: Rearranging into Standard Form
Rearranging gives:
\[ 0.5V^2 + 12.5V - 3750 = 0 \]

### Step 8: Multiply by 2 to Simplify
To eliminate the decimal, multiply the entire equation by 2:
\[ V^2 + 25V - 7500 = 0 \]

### Step 9: Factor or Use the Quadratic Formula
Now we can solve this quadratic equation using the quadratic formula:
\[ V = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 25, c = -7500 \).

### Step 10: Calculate the Discriminant
Calculate the discriminant:
\[ b^2 - 4ac = 25^2 - 4 \cdot 1 \cdot (-7500) = 625 + 30000 = 30625 \]

### Step 11: Solve for V
Now substitute back into the quadratic formula:
\[ V = \frac{-25 \pm \sqrt{30625}}{2} \]
Calculating \( \sqrt{30625} \) gives approximately 175.

Thus:
\[ V = \frac{-25 + 175}{2} \quad \text{(since speed cannot be negative)} \]
\[ V = \frac{150}{2} = 75 \text{ km/hr} \]

### Final Answer
The usual speed of the train is **75 km/hr**.

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