Aptitude - Time and Distance - Discussion

HOW 150 comes in the initial step ?

@maheshwaran
take x as a spd of car
T=75/x -->*
spd of train is x+x/2(train 50% faster than car)
and time reduce by 12.5 min( by qus)
12.5min=12.5/60hr= 2.5/12hr
T-2.5/12=75/(x+x/2)
T=(75/x+0.5x)+ 2.5/12
-->#
by solving * and# we will get ans as 120

If speed is x, how do you'll apply the formula in 75/x i.e distance upon speed gives time isn't it?

Yes, Meetali T=d/s. The difference between the the time of car and train is 12. 5 min. i.e. we are calculating 75/x-75/3/2= 125/10*60.

The train can travel 50% faster than car. So we are assuming car speed as X kmph,then we want to know the train speed so we are adding car assumed speed + train speed (data given 50% faster so it was in percentage we are converting into normal form as 50/100)
Train speed= (X+ 50/100)
= (100X +50)/100
=150X/100
= 3/2 X
=1.5X



2.next step as we see
we want to find car speed only and if u see in the data he didn't ask exact train speed. See both car and train reach at point B 75kms away from point A.

75/x -------> car speed
75/(3/2)X ----->train speed
75/X-75/(3/2)X=12.5min (we have to add the train delayed time to get car speed)
75/X-75/(3/2)X=12.5/60 ------------>1min = 60 secs
Finally car speed X=120 kmph
Train speed = (3/2)X =(3/2)120
=180 kmph.

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.