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Bunuel
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A train is involved in an accident 3 hours after starting and is delay 07 Dec 2021, 08:00
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A train is involved in an accident 3 hours after starting and is delayed for 1 hour. After the delay, the train proceeds at 75% of its original speed. It arrives 4 hours late at its destination. Had the accident taken place 150 km further along the tracks, the train would have arrived only 3.5 hours late. What is the total length (distance) of the train's journey?

(A) 850 km

(B) 1000 km

(C) 1200 km

(D) 1750 km

(E) 3000 km


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A train is involved in an accident 3 hours after starting and is delay 13 Dec 2021, 09:36
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Bunuel
A train is involved in an accident 3 hours after starting and is delayed for 1 hour. After the delay, the train proceeds at 75% of its original speed. It arrives 4 hours late at its destination. Had the accident taken place 150 km further along the tracks, the train would have arrived only 3.5 hours late. What is the total length (distance) of the train's journey?

(A) 850 km

(B) 1000 km

(C) 1200 km

(D) 1750 km

(E) 3000 km


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T1 - T2 = .5
150/(.75x) - 150/x = .5,
X =100 Km/hr

Extra Time (4-1 = 3 hrs) or 4/3 of Initial extra time

Distance
=300 + 75X12 = 1200
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A train is involved in an accident 3 hours after starting and is delay 16 Nov 2025, 00:56
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Can you give us complete explanation if possible?
Bunuel
A train is involved in an accident 3 hours after starting and is delayed for 1 hour. After the delay, the train proceeds at 75% of its original speed. It arrives 4 hours late at its destination. Had the accident taken place 150 km further along the tracks, the train would have arrived only 3.5 hours late. What is the total length (distance) of the train's journey?

(A) 850 km

(B) 1000 km

(C) 1200 km

(D) 1750 km

(E) 3000 km


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A train is involved in an accident 3 hours after starting and is delay 21 Nov 2025, 11:31
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delay of 4-1=3 hrs(excluding one hour of breakdown) in first condition is due to the train covering the distance after the breakdown delay of 1 hour at the 3/4th of the original speed------1
however this delay of 3.5-1(excluding one hour of breakdown)=2.5 or 5/2 hrs is due to the train covering 150 kms less distance as mentioned in equation 1 above, at 3/4th of the original speed---2
In 2 , delay is 1/2 an hr less which is due to 150 kms less distance travelled at reduced speed than in 1.
so 150/0.75*s - 150/s=1/2
solving s=100 km/hr ,original speed of train.
now using this value of s in any of the above two equations, d=total distance of journey can be found, for eg. using 1
(d-300)/0.75s - (d-300)/s =3 --------3
{here we are formulating the time of delay 3 hrs, as a result of the difference in time, when the distance travelled after breakdown in the first condition is travelled at the reduced speed to when the distance would have been travelled at the original speed s=100 km/hr if fault hadn't occurred.}
solving 3 will give d=1200 kms
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A train is involved in an accident 3 hours after starting and is delay 21 Mar 2026, 02:15
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Delay = 1/2 hrs less
Why does this happen?
Because 150km distance more is travelled at usual x speed as compared to 3x/4 speed.
Now you can simply subtract and find x

OR

Notice that 3/4 speed happens then time increase to 4/3
Because product remains the same 3/4x*4/3x = x the distance remains the same.
Now 4/3x is time then 1/3x is the increase
Hmm if 1/3x is increase that is equal to our 30 mins!

Hence, 1/3x = 1/2
x=1.5
Speed = 150/1.5 = 100kmph

Now total delay excluding stoppage is 4-1 = 3 hrs.
Now again 1/3x(Increase in time due to accident) = 3 hrs!
now this x = 9 hrs.
So 9 hrs of original travel after accident = 100* 9 = 900
And an additional 300 at the start and thus 300 + 900 = 1200

Answer: Option C
Bunuel
A train is involved in an accident 3 hours after starting and is delayed for 1 hour. After the delay, the train proceeds at 75% of its original speed. It arrives 4 hours late at its destination. Had the accident taken place 150 km further along the tracks, the train would have arrived only 3.5 hours late. What is the total length (distance) of the train's journey?

(A) 850 km

(B) 1000 km

(C) 1200 km

(D) 1750 km

(E) 3000 km


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