Question

Two trains A and B were moving in opposite directions, their speeds being in the ratio 5 : 3. The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other. It took another 69 seconds for the rear ends of the trains to cross each other. The ratio of length of train A to that of train B is

• A. 2 : 3
• B. 2 : 1
• C. 5 : 3
• D. 3:2

Answer 1

The Correct Option is D

Trains A and B are traveling in opposing directions with a speed differential of 5:3. Train A travels at a five-fold pace and Train B travels at a three-fold pace.

Let \( L_a \) and \( L_b \) represent the lengths of Train A and Train B, respectively.

Step 1: Analyzing the relative speeds

When the front ends of the two trains cross, the relative speed is the sum of the speeds of Train A and Train B. Since the speeds have a ratio of 5:3, the relative speed is:

\[ \text{Relative speed} = 5x + 3x = 8x \] where \( x \) is a constant factor.

Step 2: Crossing the back end of Train B

Forty-six seconds after the front ends of the trains crossed, the front end of Train A crosses the back end of Train B. The distance covered during this time is \( L_b \), and it takes 46 seconds to complete this:

\[ \text{Distance covered} = L_b = 46 \times (5x + 3x) = 46 \times 8x = 368x \]

Step 3: Crossing the back ends of the trains

It took an additional 69 seconds for the back ends of the trains to pass one another. The distance covered during this time is \( L_a \), and it takes 69 seconds to complete this:

\[ \text{Distance covered} = L_a = 69 \times (5x + 3x) = 69 \times 8x = 552x \]

Step 4: Relating the lengths of the trains

Now, using the given equation:

\[ \frac{\frac{L_a}{5x + 3x}}{\frac{L_b}{5x + 3x}} = \frac{69}{46} \] Simplifying this, we get: \[ \frac{L_a}{L_b} = \frac{69}{46} = \frac{3}{2} \]

Step 5: Final Conclusion

The ratio of the lengths of Train A to Train B is 3:2.