Aptitude - Time and Distance - Discussion

Another shortcut yet simple method I would like to share wherein forget the hassle of x and y so the method is as follow:

Time taken by the car and train to reach the destination is the same and if you read the question carefully it is mentioned that the train has halted for 12.5 minutes at the stations that means just because of 12.5 minutes halt the train and the car covered 75 kilometers at the same time.

Just because of 12.5 minutes the train misses to cover 50% of the route that means the total time taken by the train and the car to reach the destination is 37.5 minutes ( working- 75km/2 =37.5, so in 12.5 minutes stop it missed 37.5km, hence the train took total 25 mins to reach the destination + 12.5 mins halt = 37.5 mins. )

Now comes the cross multiplication part. 37.5 mins = 75 kms
60 mins = ? (working- we already know how much kilometers it can cover in minutes but we want to find km/hour, so 1 hour = 60 mins.)
= (60*75)/37.5
=120.

To solve in this way you require a bit logic and good knowledge of cross multiplication. Still if you don't understand this method I suggest you to go for the Admins method.

The train can travel 50% faster than car. So we are assuming car speed as X kmph,then we want to know the train speed so we are adding car assumed speed + train speed (data given 50% faster so it was in percentage we are converting into normal form as 50/100)
Train speed= (X+ 50/100)
= (100X +50)/100
=150X/100
= 3/2 X
=1.5X



2.next step as we see
we want to find car speed only and if u see in the data he didn't ask exact train speed. See both car and train reach at point B 75kms away from point A.

75/x -------> car speed
75/(3/2)X ----->train speed
75/X-75/(3/2)X=12.5min (we have to add the train delayed time to get car speed)
75/X-75/(3/2)X=12.5/60 ------------>1min = 60 secs
Finally car speed X=120 kmph
Train speed = (3/2)X =(3/2)120
=180 kmph.

Yes, Meetali T=d/s. The difference between the the time of car and train is 12. 5 min. i.e. we are calculating 75/x-75/3/2= 125/10*60.

If speed is x, how do you'll apply the formula in 75/x i.e distance upon speed gives time isn't it?

@maheshwaran
take x as a spd of car
T=75/x -->*
spd of train is x+x/2(train 50% faster than car)
and time reduce by 12.5 min( by qus)
12.5min=12.5/60hr= 2.5/12hr
T-2.5/12=75/(x+x/2)
T=(75/x+0.5x)+ 2.5/12
-->#
by solving * and# we will get ans as 120

HOW 150 comes in the initial step ?

It's a very easy question
suppose train running 12.5 min. & lost 12.5 mint in station
s=d/t====>
s=(75/25*60)km/s
s=1/20km/s change in m/sec
s=1000/20 m/sec =50 m/sec
change in km/hr
50*18/5=180km/h train speed.....in option of this question
only 120 car speed 50% is satisfy train speed 120+120/2=180 that's it......

Say, the speed of the car be 1 km/hr (0. 5 +0. 5).

And if there is 50% increase in speed means (1 +0. 5) = 1. 5 km /hr.

That's how 3/2 came.

Let Speed of the car=x.

Train Speed=Speed of car+50% (Speed of car).

=> x + 50/100 (x).

=> x + x/2.

=> 3x/2.

Let the speed of the car be 100% and the train is 50% more
100%+50%=150%
that is 150/100