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:
Answer: Option
Explanation:
Let speed of the car be x kmph.
Then, speed of the train = | 150 | x | = | ![]() |
3 | x | ![]() |
100 | 2 |
![]() |
75 | - | 75 | = | 125 |
x | (3/2)x | 10 x 60 |
![]() |
75 | - | 50 | = | 5 |
x | x | 24 |
![]() |
![]() |
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.
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.
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.
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.
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.
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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