Question

Train Green Arrow leaves station A for B everyday at 7am.On a certainday it was delayed by 2 hours.To coverup the time it increases its speed by 20% bt still arrived at B station 1hours later than the scheduled time.What is the usual duration of train's journey from station A to B??

a)6hrs
b)7hrs
C)8hrs
d)9hrs
e)none of these

Answer 1

The usual duration of the train journey from station A to B is
`6` hours.

Explanation:

Lets try to understand with a real life example of the question then we will frame it in terms of variables.

As given the train starts from
`7` am and reached the destination at
`1`pm.

When it is late by two hours it started its journey at
`9`am so will reach at
`3` pm as it has already covered up with an hour due to increased in its speed then actually it reached at
`2` pm.One hour late from its scheduled time.

Lets frame an equation.

Knowing the fact distances are same for both of the station.

And
`Distance = Speed* time`

Now we take
`s` as our normal day speed and
`s+0.2s=1.2s` speed when the train is late,
`(20)/(100)=0.2` is added with the previous speed.

Similarly with time
`t` as our normal time and
`(t-1)` when the train is running late from its scheduled time.

Now

`d_(1)=s* t` on normal days.

`d_(2)=1.2s(t-1)= 1.2st-1.2s` when the train is late.

Equating both the equations as distances are same.

We have

`st = 1.2st-1.2s`

Subtracting
`st` from both sides.

`0.2st-1.2s=0`

Taking
`s` as common terms.

The remaining equation is

`0.2t-1.2=0`

Adding
`1.2` both sides

`0.2t=1.2`

`t=(1.2)/(0.2) =6`

So the journey fro station A to b is of
`6` hours.