Aptitude - Time and Distance - Discussion

Dist = speed * time.

Let, the speed of car=100x.
Speed of train=150x (as 50% faster).
So, Time taken by car= dist/speed=75/100x.
And by train=75/150x.
And train lost 12.5 min=12.5/60 hr =125/600 hr.
This time is equal to the time taken b/w car and train i.e;
75/100x - 75/150x =125/600.
Solving this eq we get x=6/5.
So, the speed of car is=100*(6/5)= 120km/hr.

Actually, it is time of car to cover the required distance and train take same time but including 12.5 minutes.

So both take same time you can rearrange the formula to understand easily.
(75/x) = (12.5/60)+-(75/(3/2)x).

@Trishita.

Can you please explain, how do you put 75/100x-75/150x = 125/600?

Dist=speed*time.
Let, speed of car=100x.
Speed of train=150x (as 50% faster).
So..Time taken by car= dist/speed=75/100x.
And by train=75/150x.
And train lost 12.5 min=12.5/60 hr =125/600 hr.
This time is equal to the time taken b/w car and train i.e;
75/100x - 75/150x =125/600.
Solving this eq we get x=6/5.
So, the speed of car is=100*(6/5)= 120km/hr.

Let the speed of the car be 100 then speed of the train be 150.
The ratio of the speed of car and train be = 2/3.
When the distance is constant the time is inversly proprotional,
So ratio of time = 3:2,
Diffrence in time =12.5min=5/24hours,
Total time is taken by car=3*5/24=15/24,
Speed = d/t => dis=75km,
Speed of car=75/(15/24)=120km/h.

@All. The time lost by the train (12.5min) was already given in the question. We can't calculate the lost time without knowing other quantities( speed of car, etc).

The train lost about 12.5 minutes while stopping at the stations. How the time loss is calculated? Please anyone tell me.

Both reach same time but suppose TI but train losses 12.5 so train time is TI-12.5 and car time is TI.

Now WKT time=distance/speed, x be the speed of car and 3x/2 speed of the train, both distance is same.

Substitue in formula for train time=TI-12.5 dis=75 speed 3x/2 we get (TI-12.5)=75/3x/2
for car time=TI speed=x dis=75 we get TI=75/x now substitue TI In above trian equation we get;

75/x-12.5 = 75 * 2/3x convert time into 12.5/60 and the result:120km/hr

Here we take 150/100 because it is given 50% more and always we have to take 100 as a base value here it is 50%more so 100+50/100 = 150/100 = 3/2.

(If it is given less than 50% then 100-50/100 = 50/100 = 1/2.

I hope you all understand.

Can anybody tell me why we need to subtract 75/x with 75÷3/2x instead we can add it also right?