A train 100 metres long meets a man going in opposite direction at 5km/hr and passes him in `7(1)/(5)` seconds. What is the speed of the train (in km/hr)?

Options

A. 45km/hr
B. 60km/hr
C. 55km/hr
D. 50km/hr

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the problem
We have a train that is 100 meters long, moving in one direction, and a man walking in the opposite direction at a speed of 5 km/hr. The train passes the man in \(7 \frac{1}{5}\) seconds. We need to find the speed of the train in km/hr.

### Step 2: Convert the mixed fraction time to an improper fraction
The time taken to pass the man is given as \(7 \frac{1}{5}\) seconds. We convert this to an improper fraction:
\[ 7 \frac{1}{5} = \frac{7 \times 5 + 1}{5} = \frac{35 + 1}{5} = \frac{36}{5} \text{ seconds} \]

### Step 3: Set up the equation using the formula for speed
The formula for speed is:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]
In this case, the distance is the length of the train (100 meters), and the time is \(\frac{36}{5}\) seconds. However, since the train and the man are moving towards each other, we need to consider their relative speed.

Let the speed of the train be \(S\) km/hr. The relative speed when two objects move towards each other is the sum of their speeds:
\[ \text{Relative Speed} = S + 5 \text{ km/hr} \]

### Step 4: Convert speeds to meters per second
To convert the speed from km/hr to meters per second, we use the conversion factor:
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
Thus, the relative speed in meters per second is:
\[ (S + 5) \times \frac{5}{18} \text{ m/s} \]

### Step 5: Set up the equation
Using the formula for speed:
\[ \frac{100 \text{ meters}}{\frac{36}{5} \text{ seconds}} = (S + 5) \times \frac{5}{18} \]

### Step 6: Solve for \(S\)
First, simplify the left side:
\[ \frac{100}{\frac{36}{5}} = 100 \times \frac{5}{36} = \frac{500}{36} = \frac{125}{9} \text{ m/s} \]

Now we equate both sides:
\[ \frac{125}{9} = (S + 5) \times \frac{5}{18} \]

To eliminate the fraction, multiply both sides by 18:
\[ 18 \times \frac{125}{9} = (S + 5) \times 5 \]
\[ \frac{2250}{9} = 5(S + 5) \]
\[ 250 = 5(S + 5) \]

Now divide both sides by 5:
\[ 50 = S + 5 \]

Finally, solve for \(S\):
\[ S = 50 - 5 = 45 \text{ m/s} \]

### Step 7: Convert back to km/hr
Now convert \(S\) from m/s back to km/hr:
\[ S = 45 \times \frac{18}{5} = 162 \text{ km/hr} \]

### Final Answer
The speed of the train is \(162 \text{ km/hr}\).

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