Question:easy

A train 300 meters long is running at a speed of 25 meters per second, it will cross a bridge 200 meters long in

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The train clears the bridge only when its last coach leaves it, so the distance is train length plus bridge length, that is 300 plus 200. Then divide by the speed.
Updated On: Jul 17, 2026
  • 5 seconds
  • 10 seconds
  • 20 seconds
  • 25 seconds
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The Correct Option is C

Solution and Explanation

Step 1: Track the two ends of the train separately.
Picture the bridge from left to right, 200 m long. Let the position of the engine be measured from the left end of the bridge. The crossing begins when the engine is at 0 m, at the start of the bridge.
The crossing ends when the rear of the train passes 200 m, the far end. At that instant the engine is 300 m further ahead, because the train itself is 300 m long, so the engine stands at $200 + 300 = 500$ m.

Step 2: Read off the distance travelled.
The engine moved from the 0 m mark to the 500 m mark, so the train covered
\[ 500 \text{ m} \]
This is why the two lengths get added rather than subtracted.

Step 3: Convert speed into distance per second.
The train runs at 25 m per second, meaning it covers 25 m in every single second. So counting how many 25 m chunks fit into 500 m gives the number of seconds directly.
\[ \frac{500}{25} = 20 \]
So 20 chunks, so 20 seconds.

Step 4: Verify by multiplying back.
In 20 seconds at 25 m per second the train covers
\[ 25 \times 20 = 500 \text{ m} \]
which is exactly the bridge plus the train. The check holds.

Step 5: Sort out the distractors.
If someone uses only the bridge, they get $\frac{200}{25} = 8$ s, which is not even listed, showing that the paper did not reward that error directly.
The 10 second choice matches $\frac{250}{25}$, an average of the two lengths, which has no physical meaning here.
The 5 second choice would need a distance of only 125 m, less than the train itself, so the train could not even have left the bridge.
The 25 second choice mirrors the speed value and would need 625 m of travel, which is 125 m more than the crossing requires.

Final Answer:
With 500 m to cover at 25 m per second, the crossing takes 20 seconds.
\[ \boxed{20} \]
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