OA Explanation:This question is based on the concept of
Relative velocity.
(Refer to the figure below)
The speeds are given in Kmph, so the distance must be taken in Km and time in Hr
Let the length of Train A be \(D\) and speed of Train B be \(S\) kmph
For opposite deirection;
Relative Velocity = \((63 + S)\) Kmph
So, \(t = \frac{27}{3600} = \frac{(D + 0.5)}{(63 + S)}\)
In same deirection;
Relative Velocity = \((63 - S)\) Kmph
Note, since Train A is chasing Train B here, speed of Train A has to be greater than speed of Train B
So, \(t = \frac{162}{3600} = \frac{(D + 0.5)}{(63 - S)}\)
Equate the distance;
\(\frac{27}{3600}(63 + S) = \frac{162}{3600}(63 - S)\)
\(27(63 + S) = 162(63 - S)\)
\((63 + S) = 6(63 - S)\)
\(7S = 5(63)\)
\(S = 45\) Kmph
Put it in any equation to get D;
\(\frac{27}{3600} = \frac{(D + 0.5)}{(63 + S)}\)
\(\frac{27}{3600} = \frac{(D + 0.5)}{(63 + 45)}\)
\(\frac{27}{3600}(108) = (D + 0.5)\)
\(0.81 = (D + 0.5)\)
\(D = 0.31\) Km i.e. 310m
Hence, option A
Attachments
Train A travelling at 63 kmph takes 27 to sec to cross Train B.png [ 4.65 KiB | Viewed 2115 times ]