Aptitude - Time and Distance - Discussion

Let the speed of the car be x.
Now, the train is 50% faster than a car,
i.e. speed of the train = x + (50/100)x.
= 150x/100.

150/100 came because;
Car speed is 100% and train speed is 50% faster than a car so,
Car speed (100/100)+(50/100),
=(1/1)+(50/100),
Cross multiply then (100+50)/100.
= 150/100.

T:C ( speed) = 3:2.
T:C ( time) = 2:3.
Time is taken by car = 12.5× 3.
Speed of the car = 120 km/h.

The ratio of speeds of train and car is 1.5 : 1

Therefore the ratio of time taken by them to cover the same distance is 1 : 1.5

The train lost 12.5 mins while stopping at stations. This means if the train didn't stop at stations it would reach the destination early by 12.5 mins.

That means the difference between the time taken by train and car is 12.5 mins.

Let consider x as the time taken by train.
Then as per the time taken ratio 1 : 1.5, the difference between 1.5x and x is 12.5 mins,
Then x = 25 mins,
Then 1.5 x = 37.5 mins.

Now the car traveled 75 kms in 37.5 mins,
The speed of car is 75/37.5 kms per min.
Speed of car = 75000/2250 m/s
i.e., 100/3 m/s.
Converting m/s into km/hr-
i.e., (100/3) * (18/5) = 120 kmph.

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.

Let's start by assuming that the speed of the car is x km/h.

Since the train can travel 50% faster than the car, its speed would be 1.5x km/h.
Now, let's calculate the time taken by both the train and the car to travel from point A to point B.
Time is taken by the car = Distance/Speed
= 75/x

Time taken by the train = Distance/Speed.
= 75/(1.5x).

However, we know that the train lost 12.5 minutes while stopping at the stations. So, we need to add this time to the total time taken by the train.
Total time taken by the train = 75/(1.5x) + 12.5/60.

We know that both the train and the car reach point B at the same time. Therefore, we can equate the time taken by the car and the total time taken by the train (including the stoppage time).

75/x = 75/(1.5x) + 12.5/60.

How 150 came? Please explain to me.

Let the speed of the car = x.
So, the speed of the train will be 50% faster than the car i.e;
x + 50% of x = 1.5 x.

The Time taken by the car and train is same and the train lost 12.5 min or 12.5/60 hours on stops.
So,
Time taken by train = Time taken by car.
Time taken by train to cover 75km = (75/1.5x) - (12.5/60)Time lost.
Time taken by car to cover 75km = (75/x).
equate these equations and find x.

Train : car = 150:100 or 3:2.
Difference = 12.5 so 1 part is 12.5.
Speed and time is vice versa.
Time = train : car is 25 min : 37.5 min.
37.5 min = 75 kmph.
60 min = 120 kmph.

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.