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A train's journey is disrupted due to an accident on its track 25 Jun 2020, 10:47
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C
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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
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C
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A train's journey is disrupted due to an accident on its track 25 Jun 2020, 10:54
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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




yashikaaggarwal
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
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A train's journey is disrupted due to an accident on its track 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

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A train's journey is disrupted due to an accident on its track Updated on: 25 Jun 2020, 22:58
Originally posted by MayankSingh on 25 Jun 2020, 13:37.
Last edited by MayankSingh on 25 Jun 2020, 22:58, edited 1 time in total.
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Consider the Path to be:

Case 1: A....30Km........C (First breakdown point)...............................B
Usual Sped = x Km/Hr
Speed between AC= x & CB=4/5x
Time:45 min delay

Case 2: A.....30Km.........C..18Km.....D..............................................B
Speed between AC= x & CD=x & DB= 4/5x
Time:36 min delay

Comparing both scenario, difference is only between C & D (accounting for 9 min)
For CD, t1/t2= S2/S1= 5/4 (actual diff= 9min)
So, t1=45 min & t2=36 min

Req speed = 18x60/36= 30Km/hr
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A train's journey is disrupted due to an accident on its track 25 Jun 2020, 15:37
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Given

    • A train meets an accident after traveling 30 km and it is 45 minutes late.
    • If the train meets an accident after traveling another 18 km, then it would be 36 minutes late.

To Find

    • Find the original speed

Approach and Working Out


    • For the first 30 km in both scenarios the speed is the same, i.e, the original speed.
    • After 48 km (30 + 18) in both scenarios the speed is the same, i.e, the reduced speed.
    • The difference of 9 minutes (45 – 36) comes from the 18 km journey.

Speed ratio –
    • Original: Reduced = 5: 4

Time ratio –

    • Original : reduced = 4 : 5
    • Let us take the time as 4x and 5x then,
      o 5x – 4x = 9 minutes
      o x = 9 minutes.
      o Original time, 4x = 36 minutes

Hence, with the original speed it takes 36 minutes to travel 18 km.
    • In 1 hour of 60 minutes it will travel = (18/36) x 60 = 30 km
    • Speed = 30 kmph


Correct Answer: Option C
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A train's journey is disrupted due to an accident on its track 29 Jun 2020, 09:26
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yashikaaggarwal
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
Show more

Solution:

If we let r = the original speed of the train and t = original time reaching the destination, we can create the equation:

30 + (4r/5)(t - 30/r + 45/60) = 48 + (4r/5)(t - 48/r + 36/60)

30 + 4rt/5 - 24 + 3r/5 = 48 + 4rt/5 - 192/5 + 12r/25

3r/5 - 12r/25 = 42 - 192/5

Multiplying the equation by 25, we have:

15r - 12r = 1050 - 960

3r = 90

r = 30

Answer: C
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A train's journey is disrupted due to an accident on its track 08 Mar 2025, 08:22
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