A train coverd certain distance at a uniform speed.If the train would have been 10km/hr faster ,it would have taken 1 hours less than the scheduled time.And if the train were slower by 6km/hr, it would have taken 1 hour more than the scheduled time .Find the length of the journey
Answers
Hey there!
Length of the Journey => Distance covered by the train.
Let Speed of the train = x Kmph
Time taken = y Hr
Distance = Speed × time = xy. -----[1]
If the train would have been 10 kmph faster,
Speed of the train = x+ 10
Time taken = y -1
Distance = Speed × time = (x+10)(y-1)
= xy - x +10y -10 ------[2]
Equating [1] & [2]
=> xy = xy-x +10y -10
=> x -10y +10 = 0. ---------------[A]
If the train would have been 6 kmph slower,
Speed of the train = x-6
Time taken = y + 1
Distance = Speed × time = (x-6) (y+1) -----[3]
= xy+x-6y-6
Equating [1] & [3]
=> xy = xy+x-6y-6
=> x - 6y -6 = 0 -------------[B]
The two equations are :
Take [A] =>
x - 10y +10 = 0 => x = 10y-10
Substitute in [B]=>
x - 6y - 6 = 0
=> (10y-10) -6y-6 = 0
=> 4y-16 = 0
=> 4y = 16
=> y = 4
Substitute in [A]
x = 10y-10
=> x = 10(4)-10
=> x = 40-10
=> x = 30
Thus,
Speed of the train = x = 30 Kmph
Time taken by the train = y = 4 Hrs
Distance covered by the train = Speed × Time
= 30 × 4
= 120 Km
Therefore,
Length of the Journey = 120 Km
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