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yashikaaggarwal
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25 Jun 2020, 10:47
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
67% (03:13) correct
33%
(02:53)
wrong
based on 221
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A train's journey is disrupted due to an accident on its track after it has traveled 30 km. Its speed then come down to four-fifth of its original, and consequently, it run 45 min late. Had the accident taken place 18 km farther away, it would have been 36 min late. Find the original speed of the train.
A. 24
B. 35
C. 30
D. 20
E. 25
A. 24
B. 35
C. 30
D. 20
E. 25
25 Jun 2020, 10:54
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Bookmarks
Distance between A and B= D
Original speed = v
Because train travelled (30-18=) 18Km at the 4/5th of the original speed, it reached destination (45-36=) 9mins late.
\(\frac{18}{(0.8v)} - \frac{18}{v} = \frac{9}{60}\)
\(\frac{18}{4v} = \frac{9}{60}\)
v= 30Km/h
Original speed = v
Because train travelled (30-18=) 18Km at the 4/5th of the original speed, it reached destination (45-36=) 9mins late.
\(\frac{18}{(0.8v)} - \frac{18}{v} = \frac{9}{60}\)
\(\frac{18}{4v} = \frac{9}{60}\)
v= 30Km/h
yashikaaggarwal
General Discussion
25 Jun 2020, 10:57
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IMO - C
Let the speed be X
18/4/5x - 18/x = (45-36)/60
18(5-4/4x) = 9/60
18/4x = 9/60
X = 30
Posted from my mobile device
Let the speed be X
18/4/5x - 18/x = (45-36)/60
18(5-4/4x) = 9/60
18/4x = 9/60
X = 30
Posted from my mobile device
