A train 'P' crosses a pole in 6.75 sec and a 240-meter long platform in 15.75 sec. If train 'Q' which is 120meter-long running in same direction crosses train 'P' in 45 sec, then find time required by train 'Q' to cross train 'P' running in opposite direction?

Options

A. A)6sec
B. B)7sec
C. C) 5 sec
D. D)9 sec

Answer

To solve the problem step-by-step, let's break it down:

### Step 1: Find the length of Train P
When Train P crosses a pole, it travels a distance equal to its own length in a given time.

Let the length of Train P be \( L_P \) and its speed be \( V_P \).

Using the formula:
\[ L_P = V_P \times \text{time} \]
Given that Train P crosses a pole in 6.75 seconds:
\[ L_P = V_P \times 6.75 \quad \text{(1)} \]

### Step 2: Find the speed of Train P using the platform
When Train P crosses a 240-meter long platform, it travels a distance equal to its own length plus the length of the platform in a given time.

Using the formula:
\[ L_P + 240 = V_P \times 15.75 \]
Substituting \( L_P \) from equation (1):
\[ V_P \times 6.75 + 240 = V_P \times 15.75 \]
Rearranging gives:
\[ 240 = V_P \times (15.75 - 6.75) \]
\[ 240 = V_P \times 9 \]
\[ V_P = \frac{240}{9} = \frac{80}{3} \text{ m/s} \]

### Step 3: Calculate the length of Train P
Substituting \( V_P \) back into equation (1):
\[ L_P = \frac{80}{3} \times 6.75 \]
Calculating gives:
\[ L_P = \frac{80 \times 6.75}{3} = \frac{540}{3} = 180 \text{ meters} \]

### Step 4: Find the speed of Train Q
Train Q, which is 120 meters long, crosses Train P in the same direction in 45 seconds.

Using the formula:
\[ L_P + L_Q = (V_Q - V_P) \times \text{time} \]
Substituting the known values:
\[ 180 + 120 = (V_Q - \frac{80}{3}) \times 45 \]
\[ 300 = (V_Q - \frac{80}{3}) \times 45 \]
Dividing both sides by 45:
\[ \frac{300}{45} = V_Q - \frac{80}{3} \]
\[ \frac{20}{3} = V_Q - \frac{80}{3} \]
Adding \( \frac{80}{3} \) to both sides:
\[ V_Q = \frac{20}{3} + \frac{80}{3} = \frac{100}{3} \text{ m/s} \]

### Step 5: Calculate the time taken for Train Q to cross Train P in opposite directions
When crossing in opposite directions, the speeds add up:
\[ \text{Total distance} = L_P + L_Q = 180 + 120 = 300 \text{ meters} \]
\[ \text{Total speed} = V_P + V_Q = \frac{80}{3} + \frac{100}{3} = \frac{180}{3} = 60 \text{ m/s} \]
Using the formula for time:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{300}{60} = 5 \text{ seconds} \]

### Final Answer
The time required by Train Q to cross Train P when running in opposite directions is **5 seconds**.

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