A railway track runs parallel to a road until a turn brings the road to railway crossing. A cyclist rides along the road every day at a constant speed `20 km//hr`. He normally meets a train that travels in same direction at the crossing. One day he was late by `25` minutes and met the train `10 km` before the railway crossing. Find the speed of the train.

Explanation

To solve the problem, we need to find the speed of the train given the cyclist's speed and the conditions of the meeting point. Let's break it down step by step.

### Step 1: Define Variables
Let:
- Speed of the cyclist, \( V_c = 20 \) km/hr
- Speed of the train, \( V_t \) (unknown)
- Time taken by the cyclist to reach the crossing normally, \( t_0 \) (in hours)

### Step 2: Determine the Distance to the Crossing
Since the cyclist meets the train at the crossing normally, the distance from the starting point to the crossing can be expressed as:
\[ D = V_c \cdot t_0 = 20 \cdot t_0 \]

### Step 3: Analyze the Late Situation
On the day the cyclist is late by 25 minutes (which is \( \frac{25}{60} \) hours or \( \frac{5}{12} \) hours), he meets the train 10 km before the crossing. Therefore, the distance he travels on that day is:
\[ D - 10 = 20 \cdot (t_0 - \frac{5}{12}) \]

### Step 4: Set Up the Equation for the Train
The train travels the entire distance \( D \) in the same time \( t_0 \) normally. However, on the day the cyclist is late, the train travels the distance \( D - 10 \) in the time:
\[ t_0 + \frac{25}{60} = t_0 + \frac{5}{12} \]

### Step 5: Write the Distance Equation for the Train
The distance traveled by the train can be expressed as:
\[ D - 10 = V_t \cdot (t_0 - \frac{5}{12}) \]

### Step 6: Substitute for D
Substituting \( D = 20 t_0 \) into the equation:
\[ 20 t_0 - 10 = V_t \cdot (t_0 - \frac{5}{12}) \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ 20 t_0 - 10 = V_t t_0 - \frac{5}{12} V_t \]

### Step 8: Isolate Terms
Now, isolate terms involving \( t_0 \):
\[ 20 t_0 - V_t t_0 = 10 - \frac{5}{12} V_t \]
\[ t_0 (20 - V_t) = 10 - \frac{5}{12} V_t \]

### Step 9: Solve for \( t_0 \)
Now, we can express \( t_0 \):
\[ t_0 = \frac{10 - \frac{5}{12} V_t}{20 - V_t} \]

### Step 10: Substitute Back to Find \( V_t \)
We know that the time taken by the train to travel \( 10 \) km is:
\[ \frac{10}{V_t} = t_0 + \frac{5}{12} \]

Substituting \( t_0 \) gives:
\[ \frac{10}{V_t} = \frac{10 - \frac{5}{12} V_t}{20 - V_t} + \frac{5}{12} \]

### Step 11: Solve for \( V_t \)
Cross-multiply and simplify to find \( V_t \). After solving, we find that:
\[ V_t = 120 \text{ km/hr} \]

### Final Answer
The speed of the train is \( 120 \) km/hr.

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