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Retired Moderator
Joined: 16 Apr 2020
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Train A travelling at 63 kmph takes 27 to sec to cross Train B when
[#permalink]
12 Mar 2022, 22:34
12 Mar 2022, 22:34
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Question Stats:
50% (04:04) correct
50% (04:38) wrong based on 4 sessions
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Train A travelling at 63 kmph takes 27 to sec to cross Train B when travelling in opposite direction whereas it takes 162 seconds to overtake it when travelling in the same direction. If the length of train B is 500 meters, find the length of Train A.
A. 310m
B. 400m
C. 500m
D. 650m
E. 810m
A. 310m
B. 400m
C. 500m
D. 650m
E. 810m
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Train A travelling at 63 kmph takes 27 to sec to cross Train B when
[#permalink]
25 Mar 2022, 00:38
25 Mar 2022, 00:38
1
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
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
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gmatclubot
Train A travelling at 63 kmph takes 27 to sec to cross Train B when [#permalink]
25 Mar 2022, 00:38
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