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HARISH KUMAR

· started a discussion

· 1 Months ago

HARISH (Question: 111014.0 )

SIR FUNDA IS NOT CLEAR BY THIS METHOD

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

SWADESH SINGH

· commented

· 1 Months ago

ATQ CASE-1 IF ACCIEDENT OCCURS AFTER 100 KM THN IT IS LATE BY 45 MINS.
CASE-2 IF ACCIDENT OCCURS AFTER TRAVELLING FURTHER 50 KM THEN IT IS LATE BY ONLY 30 MINS.
SO USING UNITARY METHOD
15 MINS IS EQUIVALENT TO 50 KM
HENCE,
45 MINS WOULD BE EQUIVALENT TO 150 KM
TOTAL DISTANCE WILL BE 150+100=250KM
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