Aptitude - Time and Distance - Discussion

Let the speed of the train be x.

So, train speed 50% faster than car means x + x/2 = 3x/2, speed = 75km/hr,
Time delayed = 12.5 mins = 12.5/60 sec.
Now;
D/s = T; (here speed = speed of the car -speed of the train),
So,
75/x -75/3x/2 = 12.5/60.
75/x - 50/x = 5/24.
x = 25 x 24/5.
x = 120 km/hr.

Here, How can we say given time in the question is equal to the value obtained after subtraction? Please explain me.

Thanks for explaining the answer @VIJETH BK.

@All.

So many of us are confused in where the hell this (150x/100) came from right.

So, my simple explanation is that see the train is 50% faster than car.

Now, consider car speed is X so the train speed will be (X+ 50% of X) Now calculate the train speed.

And you will get (X + 50X/100) = 150X/100.

Hope this helps.

Someone help me in this by explaining this.

Most of people saying 1/2 instead of 3/2 is wrong.

The train is 50% more means 1/2 and also 1 for the car speed is added, So it's 3/2.

Can anyone explain to me, how 12.5/60 9is = 5/24?

I didn't understand this, anyone please explain to me?

Speed = Distance/Time.

Step 1 :

If x be the Speed of Car, then Speed of Train is x+50/100 = 3x/2.
According to this problem, Formula can be adjusted as Time = Distance/Speed.

Step 2 :
Time taken by Car => T1 = 75/x.
Time taken by Train => T2 = 75/ 3x/2 => T2 = 50/x.

Step 3 :
However, the train lost about 12.5 minutes while stopping at the stations.
Train's Total Time = Run time + Stopping time.

Convert 12.5 minutes in Hours as follows by 12.5 = 12.5/60
=>125/600 = 25/120 = 5/24.

Train's Total Time = 50/x + 5/24.

Step 4 :
Both reach at the same time, so T1 = T2.
Total time of Car = Total time of Train.
=> 75/x = 50/x + 5/24.
=> 75x - 50/x = 5/24.
=> 25/x = 5/24.
=> 120 = x.
Hence x = 120.

75/x * 1.5 = 75/x-12.5 (speed of car * 1.5 = speed pf train).
X= 37.5 time taken (min).
Car speed 75/(37.5/60) as time was in mins.