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Ramandeep singh

· started a discussion

· 1 Months ago

great question...keep it up.... (Question: 111014.0 )

such questions are needed toprankers

Question:
A train left station A for station B at a certain speed. After travelling for 100 km, the train meets with an accident and could travel at \(\cfrac{4}{5}\) th of the original speed and reaches 45 minutes late at station B. Had the accident taken place 50 km further on, it would have reached 30 minutes late at station B. What is the distance between station A and B?
Options:
A) 125 km
B) 150 km
C) 200 km
D) 250 km
Solution:
Ans: (d) Let initial speed of the trian = 5 km/h 

Then, speed after the accident \(=\cfrac{4}{5}\times5=4\ km/h\)

Time taken to cover 50 km @ 5 km/h = 10 hours

Time taken to cover 50 km @ 4 km/h \(12\cfrac{1}{2}\) hours

= 150 minutes 

But, actual difference = (45 - 30) minutes 

 = 15 minutes

\(=\cfrac{1}{10}\) of 150 minutes

\(\therefore\) Speed is 10 times of assumed speed.

\(\therefore\) Speed before accident = 10×5 = 50 km/h

and, speed after accident = 10×4 = 40 km/h 

\(\therefore\) Distance between place of accident and B

\(=\cfrac{50\times40}{50-40}\times\cfrac{4}{3}=150\ km\)

\(\therefore\) Distance between A and B = 100 + 150 = 250 km

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