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.

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.

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 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.

Simplifying this equation, we get:
5/x = 4 / (1.5x) + 1/6.

Multiplying both sides by 6x, we get:
30 = 16 + x.

Therefore, the speed of the car (x) is:
x = 14 km/h.

So, the speed of the car is 14 km/h.

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.

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 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 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.

Case1: For car;
D = TV.
75=TV.

Case2:
75={T+(12.5/60)}*1.5V.
75={(60T+12.5)*V}/40.

Now, compare case 1&2.
You get, T = 12.5/20;
As. D = TV.
75 = 12.5/20*V,
V = 120.

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

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.