Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 1)
1.
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
Answer: Option
Explanation:
Speed = | ![]() |
60 x | 5 | ![]() |
= | ![]() |
50 | ![]() |
18 | 3 |
Length of the train = (Speed x Time).
![]() |
![]() |
50 | x 9 | ![]() |
3 |
Discussion:
584 comments Page 58 of 59.
BIJAY said:
4 years ago
@All.
We know that;
Velocity = Distance/Time.
so we can write,
Distance = Velocity*Time.
= (16.66)*(9) meters [=> 60km/hr convert into meters = 60*5/18. = 16.66 meter/sec.].
= 150 meters. [ANS].
We know that;
Velocity = Distance/Time.
so we can write,
Distance = Velocity*Time.
= (16.66)*(9) meters [=> 60km/hr convert into meters = 60*5/18. = 16.66 meter/sec.].
= 150 meters. [ANS].
Harshil Nakrani said:
4 years ago
Great, Thanks all for explaining the answer.
Raju said:
4 years ago
Super, thank you for your detailed explanation.
Ansu said:
3 years ago
Thank you so much for all your explanation.
Hariharan. said:
3 years ago
Thank you everyone for explaining the answer.
Vignesh ernesto said:
3 years ago
Thanks.
Shradha said:
3 years ago
Thank you all for explaining very well.
RISHI NIGAM said:
3 years ago
According to antenna, which we design we say that frequency is between 1 to 5 Mhz or other think we can define values between 1 to 2 Mhz and can use them for infinite frequencies because if we consider frequency between 1 to 2 MHz then if we see then the output can be
1.1,1.111,1.115677,1.9876,1.11153565.....infinity now. Then why we are using this is because if you increase the value before a point it will have a large change as compared to adding values after the point.
And this is because even if there is a small change in frequency then both signals are uncorrelated to large extent.
1.1,1.111,1.115677,1.9876,1.11153565.....infinity now. Then why we are using this is because if you increase the value before a point it will have a large change as compared to adding values after the point.
And this is because even if there is a small change in frequency then both signals are uncorrelated to large extent.
S. Gayathri said:
3 years ago
Speed = distance/time taken.
60 *1000 = length/ 9/3600.
1 hr = 3600 seconds.
9/3600*60*1000 = 150 m.
60 *1000 = length/ 9/3600.
1 hr = 3600 seconds.
9/3600*60*1000 = 150 m.
Adaeze Thompson said:
3 years ago
Thanks for the explanations.
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