Aptitude - Time and Distance - Discussion

Why do we have to subtract the speed of a car - the speed of a train?

Please explain to me

If the car speed is 100kmph then the train speed is 150 kmph.
Because train 50% extra speed than a car.

Train in station stopping time is 12.5min.
But a car and train cover the same distance.
In case the train running non-stoping train cover more distance.
car: train 100%:150%.
car : train 100% ====(100% + 12.5min)

That 50% is 12.5min.
100% = 2 * 12.5min.
100% = 25min..
37.5min =75kmph
1min = 75/37.5
1min = 2 kmph
1hour = 60min = 2 * 60 = 120kmph.

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.

n any of the time distance sum out of time, distance, speed one of the quantities common
Here the time is common;

So we;
d = s*t.
Car Time = train Time.
Dist of car/Speed of car = Dist of train/Speed of train.
But here is train takes extra time for pause which must be equal.
Time train = Time of train running + Time train pass.
Time train running + Time train pass = Dist of car/Speed of the car.
Time train running - Dist car/Speed car = Time train pass.
75/1.5(x) -75/x = 12.5/60.
FYI (12.5/60 = 125/10*60),
x = 120Km/hr.

The speed of the train: speed of the car = 150 : 100 = 3 : 2.
From the formula ; ratio of the speed = a:b,
if the same distance then, ratio of the time = b : a
Time ratio = 2:3
1u = 12.5 min
2u = 25 min
3u = 37.5 min.

The time taken by the car in hr = 37.5/60 = 375/600,
= 5/8 hr.
SPEED OF THE CAR IS = 75 divided by 5/8.
= 15 * 8.
= 120 km /hr.

Formula:
Time = distance/speed.
Speed = distance/time.

Let speed of train = x.
Then speed of the car = x + (x(50)/100),
= x + x/2,
= 3x/2,
The time taken by the train = 75/x.
The time taken by the car = 75/(3x/2).

Difference between them = 12.5 minutes,
75/x - 75/(3x/2) = 125/10 minutes,
75/x - 75/(3x/2) = 125/(10*60) hr,
(225-150)/3x=125/600,
75/3x = 125/600,
25/x = 125/600,
x = (25*600)/125,
x = 120kmph.

I did not understand. Please help me to get it.

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

Convert km/h to m/s.

S(c) = 75000/T -->(1)
S(tr) = 75000/(T-12.5x60)-->(2)
S(tr) = 3/2S(c) -->(3)

Sub (3) in (2).
3/2s(c) = 75000/(T-12.5x60).

sub (1) in above;

(3/2).(75000/T) = 75000/(T-12.5x60).
T = 3x12.5x60 ---->(4)

Sub (4 in (1) [to get the speed of the car].
S(c) = (75000/3x12.5x60 ).(18/5).
s(c) = 120km/h.

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.