Latest SSC Aptitude Question SOLUTION: A train start running from station A,if made an accident after travelling 150 km.After accident the speed train reduces by 5/6 of the previous speed,so it reach 15 minute late at s

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(#M40034518) SSC Numerical Ability question algebra Keep an EYE

A train start running from station A,if made an accident after travelling 150 km.After accident the speed train reduces by 5/6 of the previous speed,so it reach 15 minute late at station B.Had the accident taken place 30 km further it have been 7 minutes late .find the speed of the train and distance between A & B station.

Asked In SSC abhishek shah (10 years ago)

Solved Ann Theressa Read Solution (3)

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Read Solution (3) :

Ann Theressa  (10 years ago)

1125km/hr and 206.25km

Let the distance be x and speed be s

time t = x/s hours

Since the total distance is x, distance traveled after accident = (x-150)
and Speed at that time is reduced by 5/6, speed = s - 5/6*s = s/6
time taken is increased by 15 min ,
time = t + 15/60 hours
= x/s + 15/60

150/s + (x - 150)/s/6 = x/s + 15/60 ---(1)

similarly the 2nd equation is

180/s + (x-180)/s/6 = x/s + 7/60 ---(2)

Solving (1) and (2)

distance, x = 206.25km
speed, s = 1125km/hr

Is this solution Helpfull? Yes (8) | No (1)  

krishna murari mishra  (10 years ago)

distance=1605km,speed=45km/hr

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Sudhanshu Mishra  (9 years ago)

Let the initial speed= x km/minute & distance= y km.
original time taken= y/x min.
After the accident speed reduces by 5/6 of previous speed,so speed after accident= x - (x(5/6)) = x/6.
1st condition:
(150/x) + ((6(y-150))/x) =(y/x) +15,
or, 150 + 6y -900 = y + 15x,
15x - 5y = -750 ----- (1)
2nd condition,
(180/x) + ((6(y-180))/x) = (y/x) + 7,
or, 180 + 6y - 1080 = y + 7x,
or, 7x - 5y = -900 ----(2)
Solving (1) & (2), we get x= 75/4 & y= 206.25
so, x = (75/4)km/minute = 1125 km/hr
&, y = 206.25km

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