Aptitude - Time and Distance - Discussion

Distance between A and B = 75kms.
Let 'x' be the speed of the car.

Given that the train can travel 50% faster than the car i.e.,
Car speed = x.
Train speed = x+x*50/100 = 3/2*x.
Distance = speed*time.

Tc(Time of the car) = 75/x.
Tt(Time of the train) = 75/(3/2*x)+12.5 min(while stopping at the station).
=>75/(3/2*x)+12.5/60(into hours).
=>75/x=75/(3/2*x)+12.5/60.

By solving x = 120kmph.

Both trains start at the same time and reach the destination at the same time. So the time taken by both trains should be the same right? So the difference between the times should be zero. How come it is 125/10*60?

The train speed is 50% faster than the car. If the car is moving in 100 then the train moves 50% fast i.e. (100car speed+ 50faster speed) so they taken 150/100.

If the speed of car is 100 then 50% of 100 is 50.

And the speed of the train is 50% more than of car speed.
so 150*x/100.

How did 75/x -75/(3/2x) = 125/(10*60) came? Please explain this.

x+50% of x's speed.
x+50x/100 = 150x/100.

Let the speed of the car=x kmph.

Then, speed of the train is 50% more than car = 50x/100 + x = 3x/2 kmph. (50x/100+x/1=50x+100x/100 =150x/100= 3x/2).

Time is taken by the car to travel from A to B= 75x hours.

Time is taken by the train to travel from A to B=75/ (3x/2) +12.5/60 hours. 1 min = 1/60 hrs. So 12.5/60 hrs.

Since both start from A at the same time and reach point B at the same time,

75/x =75/ (3x/2) +12.5/60.
75/x = 150/3x + 12.5/60.
75/x= 50/x + 12.5/60.
75/x-50/x = 12.5/60.
25/x =12.5/60.
x=25*60/12.5...... 1/x=12. 5/60*25 so, x = 60*25/12.5.
X=2*60.
x = 120.

If it's 50x.

It says that train travels in 50% less speed than the car.

How can we assume that car speed is 100 not other than that speed?

Let Car speed is x.
The train speed is 50% faster than x.
Then the train speed is x+50%of x.
x+50x/100 =150x/100.