Aptitude - Time and Distance - Discussion

Hi guys let me explain in this way.

Given that both reached the distance 75 km at the same time.

That indicates the time taken by car is equal to the time taken by train.

However the time taken by train also includes the delay times due to stations.

Thus,

Train time = distance/speed + 12.5 min (total time).

Cars time = distance/speed (total time).

As distance = 75 km it is clear but.

The speed of train is 50% faster, means 50% more.

If car speed = S then 50% more is.

S+(50/100)S.

Substitute the equation and you can see the answer your self.

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.

Let speed of car = c kmph.

So speed of train be c + (50/100)c = (3/2)c.

Time taken by car to cover 75 km distance t1 = 75/c.

Time taken by train to cover 75 km distance t2 = (75*2/3c).

But it is given train lost 12.5 m while stopping at the stations.

So,

Actual time taken by train when stopping at stations to cover 75 km = (75*2/3c) + (12.5/60).

It is given that.

Both start from point A at the same time and reach point B 75 kms away from A at the same time.

75/c = (75*2/3c) + (12.5/60).

c = 120 kmph.

Since the ratio of speed => train : car.
150 : 100 -----> A.

Then the ratio of time will be 100 : 150 ----> B.
(speed = distance/time).

Now given 12.5 min is the difference of time as the train stopped for the same then, from equation B.

=> 50 = 12.5 (ratio difference).
then 1=12.5/5.
So time taken by train = (12.5/50)x100.
=25 min (25/60 hr) ----> C.

Now the speed of train = distance covered by train/time taken.
= (75km) x 25/60 hr (from C).
= 180km/hr.

Hence, the answer will be 180km/hr.

Let speed of car = x (km/h).
So, speed of train = x + 50% of x = x + x/2 = 3x/2 (km/h),
To travel 75 km, car needs time of 75/x hours,
To travel 75 km, train needs time of 75/(3x/2) hours + 12.5/60 hours,
As the two reach destination on same time on the clock, so these times should be equal. That is:
75/x = 75/(3x/2) + 12.5/60,
=> 75/x = (75*2)/3x + 12.5/60,
=> 75/x = 50/x + 12.5/60,
=> 75/x - 50/x = 12.5/60,
=> 25/x = 12.5/60,
=> 12.5x = 25 * 60,
=> x = 1500/12.5.
So, x = 120 (km/h).

Here is simple way of solving this problem.

We have Velocity = Distance/Time.

For car let x be the speed, then speed of the train is 3/2x as per the problem.

Now x = 75/t for car.

1.5x = 75/(t-(5/24)) as train stops for a duration of 5/24 hours.

5/24 obtained by converting 12.5 minute to hour.

Now divide two equations resulting in:

1/1.5 = (24t-5)/24t which upon solving gives t = 0.625.

Put t value in v = d/t where d = 75 km and t = 0.625. Thus you get v = 120 kmph.

Hope this was useful.

Assume the car speed as 120 km/hr.
If cars speed is 120 km/hr then it reaches 75 km in 37.5 min(explained below).

Since 120 km/hr is 120 km/60 min, which is 2 km/1 min.
Therefore for 75 km it takes 37.5 min.


Train speed is 50% higher so it will be 180km/hr.
180 km/hr is 180 km/60 min, which is 3km/min.

Therefore for 75 km it takes 25 min.
It lost 12.5 min in stopping so total time taken by train is (25+12.5) = 37.5 min.

Therefore our assumption is correct. Answer is 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.

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.

Both reach same time but suppose TI but train losses 12.5 so train time is TI-12.5 and car time is TI.

Now WKT time=distance/speed, x be the speed of car and 3x/2 speed of the train, both distance is same.

Substitue in formula for train time=TI-12.5 dis=75 speed 3x/2 we get (TI-12.5)=75/3x/2
for car time=TI speed=x dis=75 we get TI=75/x now substitue TI In above trian equation we get;

75/x-12.5 = 75 * 2/3x convert time into 12.5/60 and the result:120km/hr