A train covered. certain distance at a uniform speed if the train would have been 6km/hr faster it would have taken 8 hrs less then the sheduled time and if the train was slower 6km/hr it would have taken 12 hrs more than the sheduled time find the length of the journey
Answers
Let x km/hr be the speed of train
Let y be the time taken by the train
Distance = speed x time
= x y
To form the first equation let us consider the given information from the question that is If the train had been 6 km/hr faster,it would have taken 4 hours less than the scheduled time.
it clearly says that speed is increased by 6 and time is reduced by 4
So , (x + 6) (y - 4) = x y
x y - 4 x + 6 y - 24 = x y
x y - x y - 4 x + 6 y = 24
- 4 x + 6 y = 24
divided by (-2) => 2 x - 3 y = -12 ----- (1)
To form the second equation let us consider the given information from the question that is If the train were slower by 6km/hr, then it would have taken 6 hours more than the scheduled time.
it clearly says that speed is reduced by 6 and time is increased by 6
So , (x - 6) (y + 6) = x y
x y + 6 x - 6 y - 36 = x y
x y - x y + 6 x - 6 y = 36
6 x - 6 y = 36
divided by (6) => x - y = 6 ----- (2)
2 x - 3 y = -12
(2) x 2 => 2 x - 2 y = 12
(-) (+) (-)
---------------
- y = -24
y = 24
Now we have to apply the value of y in the first equation to get value of x
Substitute y = 24 in the first equation we get
2 x - 3 (24) = -12
2 x - 72 = -12
2 x = -12 + 72
2 x = 60
x = 60/2
x = 30
Speed of the train = 30 km/hr
Time taken by the train = 24 hours
Distance covered by the train = x y = 30 x 24 = 720 km
Verification:
2 x – 3 y = -12
2(30) - 3(24) = -12
60 - 72 = -12
-12 = -12
Hope this helps you
Let t be the usual time taken by the train to cover the distance
Let d be the distance, s be the usual speed
Usual time taken => d/s = t => d=ts
d/(s+6) = t-4
ts/(s+6) = t-4
ts = ts+6t-4s-24
6t - 4s - 24 = 0 --> (1)
d/(s-6) = t+6
ts = ts-6t+6s-36
-6t + 6s - 36=0 --->(2)
Solving (1) nd (2), v get
s = 30 km/h
t = 24 hrs
d = t * s
d = 30 * 24 = 720 km
Ans : 720 km