Aptitude Discussion

Q. |
A train covered a certain distance at a uniform speed. If the train had been 6 km/hr faster, it would have taken 4 hour less than the scheduled time. And, if the train were slower by 6 km/hr, the train would have taken 6 hr more than the scheduled time. The length of the journey is: |

✖ A. |
700 km |

✖ B. |
740 km |

✔ C. |
720 km |

✖ D. |
760 km |

**Solution:**

Option(**C**) is correct

Let the length of the journey be d km and the speed of train be $S$ km/hr.

Then,

\(\dfrac{d}{S+6}=t-4\)-----(i)

\(\dfrac{d}{S-6}=t+6\)------(ii)

Subtracting the 1 equation from another we get:

\(\dfrac{d}{S-6}-\dfrac{d}{S+6}=10\)------(iii)

Now \(t=\dfrac{d}{s}\)

Substitute in equation (i) and solve for $d$ and $S$

We get $S = 30$ and $d$ = **720 km**

**Mithun Behera**

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60 came from where?? i can not understand....

**Madhu**

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By solving equation (3) I got d=5/2 anybody help me how d= 720km

simplifying equation 1

d=ts-4s+6t-24

put ts=d

we get eq

-4s+6t=24

-2s+3t=12......eq(3)

similiarly solve for equation 2

we get

s-t=6.........eq(4)

from 3 and 4 we get t=24 and s=30

I couldn't understand but the ans will be correct just see my solve

Here d1=v1*t1

=6*(4*60)

=1440

d2=v2*t2

=6*(6*60)

=2160

Hence D=d2-d1

=2160-1440

=720-(ans)