Aptitude - Time and Distance - Discussion

How 125/ (10*60) came?. Anyone please explain this point.

Can you please tell me how that 125/(10x60) came?

Here 150/100 how it come means.

Actually they said train 50%more than speed of car.

So train actually speed was 100% means 100/100 and he said 50 more so 150/100 that's it.

please explain:

(x+(50/100)*x)= 3x/2

Where 3x/2 & how 150/100?

Please any one can explain the equation [75/x]-[75/1.5x] = [12.5/60].

Let the speed of the car be 10km/h and so as the speed of the train is 50% more ie 50% of 10 is 5 and so the speed of the train can be assumed to be 15km/h. Based on this assumption also this problem can be solved.

Let as take car speed as =100 than train speed is 50% faster than car so 100+50=150.

@Neo, your method is good. I just want to know how you got that 32.5? please elaborate?

Assume the car speed as 120 km/hr.
If cars speed is 120 km/hr then it reaches 75 km in 37.5 min(explained below).

Since 120 km/hr is 120 km/60 min, which is 2 km/1 min.
Therefore for 75 km it takes 37.5 min.


Train speed is 50% higher so it will be 180km/hr.
180 km/hr is 180 km/60 min, which is 3km/min.

Therefore for 75 km it takes 25 min.
It lost 12.5 min in stopping so total time taken by train is (25+12.5) = 37.5 min.

Therefore our assumption is correct. Answer is 120 km/hr.

Let speed of car = c kmph.

So speed of train be c + (50/100)c = (3/2)c.

Time taken by car to cover 75 km distance t1 = 75/c.

Time taken by train to cover 75 km distance t2 = (75*2/3c).

But it is given train lost 12.5 m while stopping at the stations.

So,

Actual time taken by train when stopping at stations to cover 75 km = (75*2/3c) + (12.5/60).

It is given that.

Both start from point A at the same time and reach point B 75 kms away from A at the same time.

75/c = (75*2/3c) + (12.5/60).

c = 120 kmph.