Aptitude - Time and Distance - Discussion

train lost 12.5 min , convert into hrs = 12/60 = 1/5.

Remaining time = 4/5.

4/5*75 = 60.

As it is 50% then total speed of car is 120.

75/x - 75/(3/2)x is in KMPH and the term 12.5 mins is converted into secs. how could the answer can be given in KMPH by equating these two terms? please anyone who knows the explanation, explain me?

150/100 come using this strategy.

<-------------simple logic:------------->

Suppose car speed =x km/hr.

Then train speed= x + 0.5x.

1.5x or 150/100 or 3/2.

{0.5x comes by breaking the 1x into half (50%) i.e 1/2 i.e 0.5}.

Since the train is 50% faster than than the car and also delayed for 12.5 minutes,

It implies it would have used the time delayed to cover half of the distance=75/2 km.

Therefore its speed =180 km/hr (75/2*60/12.5).

Since trains' speed assuming x to be the speed of the car,

Then x + 0.5x = 180 km,

Thus x = 120 km/hr.

Actually it is 50% more.

i.e (100% + 50 %) * x.

((100/100) + (50/100)) * x.

(1 + (50/100)) * x.

(150/100) * x.

Can anyone please explain me this solution easily?

Hi, Actually we get speed as 25 only since.

Speed = Distance/(Time x Extra time).

Speed = 25/(125/600) => (25 x 24)/5 = 120.

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.

Time taken for car - time taken for train = 5/24 hr.

This equation is derived because both start at same time and end at same time.

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.