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Two stations L and M are 180 km apart from each other. A train leaves from L to M and simultaneously another train leaves from M to L. Both trains meet after 6 hours. If the speed of the first train is 3 km/hr more than the second train, then what is the speed of the slower train?

1. 12 km/hr

2. 13.5 km/hr

3. 12.5 km/hr

4. 13 km/hr

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Correct Answer - Option 2 : 13.5 km/hr

Given:

The distance between L and M = 180 km

Formula used:

\(S\ =\ {D\over T}\) (Where S = The speed, D = The distance, and T = The time)

If A and B travel in opposite directions to each other then relative speed = (Speed of the A) + (Speed of the B)

If A and B travel in the same direction to each other then relative speed = (Speed of the A) - (Speed of the B)

Calculation:

Let us assume the speed of slower train be X

⇒ The speed of the faster train = X + 3

⇒ \(6\ = {180 \over X\ +\ X\ +\ 3}\)

⇒ 12X + 18 = 180

⇒ 12X = 180 - 18

⇒ 12X = 162

⇒ X = 162/12 = 13.5 km/h

∴ The required result will be 13.5 km/h.

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