a train travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were by 5 km/hr more. find the original speed of the train.
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Answers
Answered by
1
45 km/hr
take speed as xkm/h
therfore eq. is 360/x -360/x+5=48/60 (time has to be converted from 48 min to 48/60 hr)
on solving we get, 1800/ Xsq. + 5=4/5 =Xsq. +5X-2250=0
so x=45 or x=-50(discarded)
hence speed of train=45km/hr
take speed as xkm/h
therfore eq. is 360/x -360/x+5=48/60 (time has to be converted from 48 min to 48/60 hr)
on solving we get, 1800/ Xsq. + 5=4/5 =Xsq. +5X-2250=0
so x=45 or x=-50(discarded)
hence speed of train=45km/hr
Answered by
3
Hey DeepaliJindal,
Here ur answer.
Let original speed of the train be x km/h.
Time taken at original speed = 360/x hours
Time taken at increased speed = 360/(x+5) hours
Now, 360/x - 360/(x+5) = 48/60
360[1/x - 1/(x+5)] = 4/5
x2 + 5x – 2250 = 0
x = 45 or –50
As speed cannot be negative
x = 45 km/h
Hope it will be Help uh.
Thank-you.
Here ur answer.
Let original speed of the train be x km/h.
Time taken at original speed = 360/x hours
Time taken at increased speed = 360/(x+5) hours
Now, 360/x - 360/(x+5) = 48/60
360[1/x - 1/(x+5)] = 4/5
x2 + 5x – 2250 = 0
x = 45 or –50
As speed cannot be negative
x = 45 km/h
Hope it will be Help uh.
Thank-you.
Deepalijindal:
Thanks for helping me
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