Aptitude - Time and Distance - Discussion

If you take speed of train 50/ more than car, let us assume car speed 100 the train speed is 150. Hello Shobha, minutes convert in seconds multiply with 60 SND.

@Abhishek Banik.

Can you please explain from the T of car - T of train?

You have lost me there.

Can you explain about 12.5*10/60*10?

Can you explain from second step?

Can you explain please?

A to B=75 km.

Distance T = Time.

S = Speed, D = Distance.

Let S of car = x kmph.

S of train = 50% of car speed = x+{(50/100)*x}.

= x+{(1/2)*x}.
= x+x/2 = 3x/2.

T = D/S.

T of car - T of train = 12.5 mins as car speed is low that of train speed.

= 75/x-75/(3x/2) = 12.5/60 (converting 12.5 mins to 12.5/60 hr).
= 75/x-75*2/3x = 12.5/60.

= 75/x-50/x = 12.5/60.
= 25/x = 12.5/60.

x = 25*60/12.5 = 25*60*10/125 = 120.
x = 120 kmph.

Here is simple way of solving this problem.

We have Velocity = Distance/Time.

For car let x be the speed, then speed of the train is 3/2x as per the problem.

Now x = 75/t for car.

1.5x = 75/(t-(5/24)) as train stops for a duration of 5/24 hours.

5/24 obtained by converting 12.5 minute to hour.

Now divide two equations resulting in:

1/1.5 = (24t-5)/24t which upon solving gives t = 0.625.

Put t value in v = d/t where d = 75 km and t = 0.625. Thus you get v = 120 kmph.

Hope this was useful.

Can anyone explain in detail?

Hello friends don't confuse 50% more means x+50% = x+50/100, = x+1/2.

= LCM are 2 and whole equation are 3x/2.

Its not in same time it is at same time if we consider car starts at 10 to reach at 12 then train also have at same time but it is only possible with different speed if have time lag.