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A train approaches a tunnel AB. Inside the tunnel a cat located at a point i.e. $$\frac{5}{{12}}$$ of the distance AB measured from the entrance A. When the train whistles the Cat runs. If the cat moves to the exit B, the train catches the cat exactly the exit. The speed of the train is greater than the speed of the cat by what order ?

Answer & Solution
Correct Answer: Option C
Train(T)__________ A_____5k____CAT__________B

T<-------x--------------><----12k-------------------------->

Let the speed of train be u and the speed of Cat be v and train whistles at a point T, X km away from A. Let AB = 12k and Cat was 5k distance away from A. Time was constant for both, then
$$\eqalign{ & \Rightarrow \frac{v}{u} = \frac{x}{{5k}} = \frac{{ {x + 12k} }}{{7k}} \cr & \Rightarrow 7x = 5\left( {x + 12k} \right) \cr & \Rightarrow \frac{x}{k} = \frac{{30}}{1} \cr & {\text{Thus}}, \cr & \Rightarrow \frac{u}{v} = \frac{{30}}{5} = \frac{6}{1} \cr} $$
  $${\text{or,}}\,\,6:1$$
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3 Comments
MD Rakib
MD Rakib 8 years ago
intranse=5parts
exits=12-5=7 parts
so df=7-5=2 for cat &12 For train
so ratio of T:C=12:2=6:1...(ANS)
Babjan Bittu
Babjan Bittu 10 years ago
Even though u explained i can't able to understand ur explanation. Can u please explain in brief ??
Pls mail me babjanbitti@gmail.com
Shashank
Shashank 11 years ago
6:1