Aptitude Discussion

**Common Information**

Two men are walking towards each other alongside a railway track. A freight train overtakes one of them in 20 seconds and exactly 10 minutes later meets the other man coming from the opposite direction. The train passes this man is 18 seconds. Assume the velocities are constant throughout.

Q. |
The ratio of the velocities of the first man to the second man is: |

✖ A. |
89.7 minutes |

✖ B. |
90 minutes |

✖ C. |
90.3 minutes |

✔ D. |
Cannot be determined |

**Solution:**

Option(**D**) is correct

Let ‘$L$’ be the length of train, ‘$x$’ be the speed of the first man,

‘$y$’ be the speed of the second man and

‘$z$’ be the speed of the train.

\(20=\dfrac{1}{z-x}\text{ and }18=\dfrac{1}{z+x}\)

$⇒ z=10x+9y$

Distance between the two men = $600(z+y)$ mt

Time

= \(\dfrac{600(z+y)âˆ’600(x+y)}{(x+y)}\)

\(=\dfrac{600(9x+9y)}{(x+y)}\)

= 90 minutes

Let ‘$L$’ be the length of train, ‘$x$’ be the speed of the first man, ‘$y$’

$⇒ x/y$ cannot be determined.