Aptitude - Time and Distance - Discussion

***Assume the speed of car be 'a'.

TIME OF THE CAR = distance of car/speed of car.
= 75/a.


Since train is 50% faster than car, therefore
50% of a=====50/100 X a===1/2 X a====a/2 (this is 50% of the car speed),
a + a/2====3a/2 (train is 50% faster than car).

TIME OF THE TRAIN = distance of train/speed of train.
= 75/(3a/2),
= 150/3a,
= 50/3a.

Time of the car - time of the train = 12.5 min ----------------------> MAIN FORMULA.

75/a - 150/3a = 12.5/60 (since we are calculating in km/hr, so divide it by 60),
75/a - 50/a = 12.5/60,
= 120.

Train and car covers the same distance in same time. But train 50%faster than the car.

The Train stops at stations 12.5.
It means they cover 75km in 25minutes.
So convert minutes into hr (75÷60=12/5).
So that the speed of the train = 75*12÷5=180,
50% of 120 is 60,
180 - 60 = 120.
So Speed of car is 120km/hr.
Because train is faster than car 50% is given above.

Many have asked how 3/2x came?

According to me "train travels 50% faster than car"
Let the speed of car be x.
Then the speed of train is,
The speed of car+ 50%speed of car (as per question)
ie,
x+ 1/2x.
=3/2x.

Let me clear this.

We can assume the speed of car as X
Now the speed of train (as per question) = X+X/2 = 3X/2
Distance (as per question) = 75 km
So time = distance/speed
For car its= 75/x
For train = 75/3x/2 now multiply 2 to 75 = 150/3x = 50/x
Now we have to add waist time to the train speed = 50/x+ 12.5/60
12.5/60 because time is given in minutes and we need it in hour so multiply 12.5 with 1/60 So it's = 12.5/60 or we can write it like 125/600= 5/24

Now the new speed of the train is 50/x+ 5/24.

put car speed = trains speed

75/x = 50/x+5/24.
or
75/x -50/x =5/24.
or
75 - 50
--------- = 5/24.
x
or
25/x = 5/24.
cross multiply.

24*25 = 5x.
600 = 5x.
x= 120 kmph.

Just read the question carefully.

Both car and train starting and finishing at a same time that means they are taking the same time to reach A to B.

now we know that,
time=distance/speed

distance= 75
speed of car = x
speed of train = speed of car + 50% of speed of car = x+[50/100]x or 3x/2

now
travelling time of car = time taken in travelling +no waiting
75/x + no waiting
time of train = time taken in the travelling + waiting
= 75/[3x/2] + 12.5/60 hr

and both times are equal as in question it is given they are starting and reaching at the same time let we compare them.

75/x =75/[3x/2] + 12.5/60
which will give x= 50.

speed : car=x;train =y;
x=100% of speed;y=150% of speed; from this wkt, train covers a 150 km per hour but as we know from the detail.
It loses the 12.5 min due to stoppages.so it covers the 150km in 47.5min.

Speed=distance/time.(150*60/47.5)=180kmph.so train speed is 180kmph.
Hence car speed is 120kmph.

Could we take 75/ (3/2) x before 75/x ?

Time taken by train minus time taken by the car is how come that?

Can you please explain 150/100x=3/2x?

Hi, now see a simple and driveway.
Train speed is 50% more than car speed it"s given

Now,
Lets the speed of car be x and 50% of x be x*50/100=x/2.
The speed of the train is x+50% more than car (x/2) =3x/2

Now,
At the same time, both cover same distance ie. 75km so,

75/x=75/3x/2 .

But it is also given that train stop at any station nd waste time 12.5 minutes (and we have to change it in km/hr so it becomes 125/10 same as 12.5 divides it by 60 to make it in hour=125/10*60).

From question, we understand speed of the train is more than car so it reaches first but both reach at same time due to train stop at any station so,

To calculate speed of car,

75/x- (75/3x/2+125/10*60) =0 (according to question (75/3x/2+125/10*60) is the time taken by train to calculate same distance) now,

75/x-75/3x/2=125/600.
75/x-50/x=25/120.
25/x=25/120.
1/x=1/120.
X=120km/h is the speed of car now,
I think everyone understands and get the answer. thank you.