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Pradeep Kumar Teki

· started a discussion

· 1 Months ago

simply solution (Question: 98796.0 )

if the train passed A to B 120 km with x hrs and had accident taken place from B is 300 km. then acdnt place to B is X x 5/6 so that difference time is 2 hrs late then x - 5x/6 = 2hrs then remaining time to reach in 2 hrs 30 kmph but 1 hr remaing 300 km to reach in 60 kmph

Question:
A train met with an accident 120 km from station A. It completed the remaining journey at \(\cfrac{5}{6}\) of its previous speed and reached 2 hours late at station B. If the accident took place 300 km further, it would have been only 1 hour late. What is the speed of the train?
Options:
A) 100 km/h
B) 120 km/h
C) 60 km/h
D) 50 km/h
E) 70 km/h
Solution:
Ans: (c)


P1 → Place of accident

P2 → Imaginary place of accident

For the distance x

Speed                                   Time

\(\cfrac{1}{6}\) ↓                                    \(\cfrac{1}{5}\) ↑ = 1 h

Thus, the usual time required for the distance x km is 5 × 1 = 5 hours.

For the distance (x + 300)

Speed                                   Time

\(\cfrac{1}{6}\) ↓                                    \(\cfrac{1}{5}\) ↑ = 2 h

Thus, the usual time required for the distance (x + 300)km is 5 × 2 = 10 h.

It means he covers 300 km distance in 5 h
∴ Speed = \(\cfrac{300}{5}=60\) km/h
(normal speed)

Dheeraj Kumar

· commented

· 1 Months ago

ye kya samjhaya hai bhai tumne, achhe se samjhao yaar.
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