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. |
How long after the train has passed the second man will the two men meet? |

✖ A. |
89.7 minutes |

✔ B. |
90 minutes |

✖ C. |
90.3 seconds |

✖ D. |
91 seconds |

**Solution:**

Option(**B**) 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.

\(20=\dfrac{L}{z-x}\) and

\(18=\dfrac{L}{z+x}\)

$⇒ z=10x+9y$

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

Time = \(\dfrac{600(z+y)-600(x+y)}{(x+y)}\)

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

= **90 minutes**

**RAHUL GAUTAM**

*()
*

**Aditya Pant**

*()
*

sister can you please explain it again.

**Nimesh Goyal**

*()
*

Can someone explain to me how was the value '600' arrived at ?

Never mind. Got it........................................................................................

**Sudhir Kumar Roy**

*()
*

Thanks for correcting me

**Zoha Amjad**

*()
*

I don't understand the explanation at all :(

That's bad for you Zoha :P

**Sudhir Kumar Roy**

*()
*

Answer will be 100 minutes not 90 minutes.

$z=10x+9y$

Distance between 2 men $=600(z+y)$

$=600(10x+9y+y)=6000(x+y)$

Time $=6000(x+y)$

------------

$x+y$

$=6000 \tex{ sec}$

$\textbf{=100 minutes}$

It will be 100 min only. You forgot to subtract the distance which men will reduce by walking.

IN SOLUTION PART YOU TAKE X IN BOTH EQUATION...REMOVE X WITH Y IN EQ 2