Aptitude Discussion

Q. |
A man jogging inside a railway tunnel at a constant speed hears a train approaching the tunnel from behind at a speed of 30 km/h, when he is one third of the way inside the tunnel. Whether he keeps running forward or turns back, he will reach the end of the tunnel at the same time the train reaches that end. The speed at which the man is running is: |

✖ A. |
6 km/hr |

✖ B. |
8 km/hr |

✖ C. |
12 km/hr |

✔ D. |
10 km/hr |

**Solution:**

Option(**D**) is correct

Let the train is at distance $y$ km from the tunnel and the length of the tunnel is $x$ km. Man is at point $C$ which is $x/3$ km away from $B$.

$A$ ----------- $B$ -------------- $C$ ----------------$D$

<---- y------><------\(\dfrac{x}{3}\)-------><------\(\dfrac{2x}{3}\)------->

$AB = y$, $BC$ \(=\dfrac{x}{3}\) and $CD$ \(=\dfrac{2x}{3}\)

Let $M$ km/h be the speed of man.

Now, train is at $A$ and man is at $C$ and both will take same time for reaching from $B$.

\(\dfrac{y}{30}=\dfrac{x}{3M}\)

\(M=\dfrac{10x}{y}\)-------(i)

Also, train and man will take some time for reaching at $D$.

\(\dfrac{x+y}{30}=\dfrac{2x}{M}\)

\(\Rightarrow M=\dfrac{20x}{x+y}\)-------(ii)

From both the equations we get: $x=y$

And on putting value in any equation we get: $M$ = **10 km/hr**

**Darsha**

*()
*

the answer has an error

the correction is:

2x/3M