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

Answer:

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

Step-by-step 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 $Distance = Speed\times time$

Now we take $s$ as our normal day speed and $s+0.2s=1.2s$ speed when the train is late, $\frac{20}{100}=0.2$ 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

$d_{1}=s\times t$ 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

$0.2t=1.2$

$t=\frac{1.2}{0.2} =6$

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