Aptitude - Time and Distance - Discussion

Discussion Forum : Time and Distance - General Questions (Q.No. 4)
4.
A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
100 kmph
110 kmph
120 kmph
130 kmph
Answer: Option
Explanation:

Let speed of the car be x kmph.

Then, speed of the train = 150 x = 3 x kmph.
100 2

75 - 75 = 125
x (3/2)x 10 x 60

75 - 50 = 5
x x 24

x = 25 x24 = 120 kmph.
5

Discussion:
342 comments Page 32 of 35.

Jen said:   5 years ago
Thanks all.

Rishika said:   5 years ago
Why are we subtracting 75/x - 75/3/2x? Please explain.

Naga sai said:   5 years ago
@Rishika.

We are subtracting it because the time difference between them is 12.5 min.

Anandhi said:   5 years ago
@Anjali.

12.5 minutes lost in stopping so(time difference).
12.5/60 (dividing by 60 for converting minutes to hours).

To remove decimal dividing by 10.
125/10 * 60.

Piyush said:   5 years ago
Can anyone please explain whole solution again?

Dimple said:   5 years ago
How come 3/2?

Aditya Mohanty said:   5 years ago
How come is the speed 125/10 * 60?

Nahid said:   5 years ago
Let,
Speed of the car= x kph.
So, speed of the train = x + 50% of x = (3x/2) kph.

Priyam said:   5 years ago
@Aditya Mohanty.

For converting the minute into the hour we take 12.5/60 and when we remove the decimal point it became 125/60*10.

Shavra yaqub shah said:   5 years ago
T1 - T2 = 12.5 minutes
speed1/distance1 - speed2/distance2 = 12.5 minutes .

@All.

This is how you form the equation and solve the problem related to the time difference (early/late than usual),or speed difference or distance difference.


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