Aptitude - Time and Distance - Discussion

The speed of train and car = 150 : 100 = 3 : 2.
Speed = 1/time,
Time = 2 : 3.

Delay = 12.5 for train,
So, car time = train delay × 3 = 3 × 12.5 = 37.5min = 37.5/60 = 5/8h.
Finally speed = d/t =75/5/8 =120km/h.

Let the speed of the car be x km/hr.

The train travels 50% faster than the car, so its speed is;
1.5x km/hr.

Travel time of car = 75/x hours.

Travel time of train = 75/1.5 x hours = 50/x hours (time=distance/speed).

But the total travel time taken by train is;

The train lost 12.5 minutes (which is Equal to 12.5/60 hour=5/24 hour) due to stops.
Hence, the total time for the train is the Total time taken by the train = 50/x + 5/24 hour.
In a question given that;
Both the car and the train reach point B at the same time.

Therefore:
75/x = 50/x+50/24.
75/x-50/x = 50/24.
25/x = 50/24.
Solve for x by cross-multiplying.
X = (25*24)/5.
X = 120km/hr.

T:C ( speed) = 3:2.
T:C ( time) = 2:3.
Time taken by car = 12.5× 3.
Speed of the car = 120 km/h.

Speed of car=Sc ; speed of train =St
Sc:St= 100:150 ( A train can travel 50% faster than a car)
= 2:3 -----> (1)
and therefore Tc:Tt= 3:2 -----> (2)

Delay given= 12.5 min = 12.5/60hrs.
Therefore time for car, Tc= 3*12.5/60 = 12.5/20 (refer (2)Tc:Tt= 3:2 )
Distance given= 75km.
We know, Speed= Distance/Time.

Therefore, the speed of the car,
Sc= 75/(12.5/20),
= 120 km/hr.

Speed of car = Sc ; speed of train = St.
Sc : St = 100:150 ( A train can travel 50% faster than a car)
= 2:3 -----> (1)
And therefore Tc:Tt= 3:2 -----> (2)

Delay given= 12.5 min = 12.5/60hrs.
Therefore time for car, Tc= 3 * 12.5/60 = 12.5/20 (refer (2)Tc : Tt = 3:2 )
The distance given 75km.
We know, Speed= Distance/Time.

Therefore, the speed of the car,
Sc = 75/(12.5/20),
= 120 km/hr.

By ratio of Speed.

T :C = 150: 100,
= 3:2.
Since, Distance is constant.
Time ratio = 2:3.
Diff between time ratios 2 and 3 is 1, which means 1 = 12.5 mins.

1 -----> 12.5 min,
3------> 37.5min,
2 ------> 25 min,
,
Speed of Car = Distance/Time.
= (75 /37.5) * 60( For min to hr),
= 120km/hr.

Train 50% faster so,
Tr:ca = 150:100(or) 3 : 2,speed
Train 12.5 min.
So ,3 * (12.5/60) = 12.5/20,time
Distance = 75
Speed = distance ÷ time.
Speed = 75 ÷ (12.5/20)
Ans : 120 km/hr.

150/100 came because train is 50% faster
So, first find the 50% of 100 = that is 50.
Now add the 50 to the total 100% of car = 150.

Anyone, please explain how can be the term (150/100) mean?

Let time is t.
Suppose Car is going at speed x,
Train speed will be x+(x/2) = 3x/2 -------> 1
Speed of Train(dist/time) = 75/(t-12.5) ------> 2
1 =2.
3x/2 = 75/t ----->3
Speed of car = 75/t.
x = 75/t ----> 4.
substitute 4 in 3.

By solving you get;
time of car(t) = 37.5(in min)
By converting into hr.
t = 0.625 hr.
Speed of car = 75/0.625.
Ans : 120kmph