A train travels 288km at a uniform speed. If the speed has been 4km/hr, it would have taken 1hr less for the same journey. Find the speed of the train.
Answers
Answered by
3
Let speed of train be x
. distance = 288 km
time = distance / speed
288/x= 288 / (x+4) +1
on solving this equation we will get
(x= 32 , x =-36)
speed can never be negative.
so the usual speed of train is 32km/hr
. distance = 288 km
time = distance / speed
288/x= 288 / (x+4) +1
on solving this equation we will get
(x= 32 , x =-36)
speed can never be negative.
so the usual speed of train is 32km/hr
shankarpop51:
I want a simpler method so that the time will be much less to solve the problem
but there is no method other than if to solve
don't worry if will come in 3 to 4 mark question
Thanks
Answered by
1
Time= Distance/Speed
T=288/4= 72 hour
Now T" = 73
Speed= Distance/Time
=====288/73 = 3.94km/hr
+++-------------------------------------------------------+++++
But I think Question should
4km/h more speed
1 hour less time
------------------------------------------------------------------
Let Speed of train is X
Time = D/S
T=288/X
again
T" = 288/(X+4)
ATQ
T" =T-1
288/(X+4) =( 288/X ) -1
288/X - 288/X+4) = 1
288X+ 1152 -288X= X(X+4)
X^2+4X-1152=0
(X+36)(X-32)=0
X=32 because speed can not be negative.
So the speed of train is 32km/h
Ans- 32 km/h
T=288/4= 72 hour
Now T" = 73
Speed= Distance/Time
=====288/73 = 3.94km/hr
+++-------------------------------------------------------+++++
But I think Question should
4km/h more speed
1 hour less time
------------------------------------------------------------------
Let Speed of train is X
Time = D/S
T=288/X
again
T" = 288/(X+4)
ATQ
T" =T-1
288/(X+4) =( 288/X ) -1
288/X - 288/X+4) = 1
288X+ 1152 -288X= X(X+4)
X^2+4X-1152=0
(X+36)(X-32)=0
X=32 because speed can not be negative.
So the speed of train is 32km/h
Ans- 32 km/h
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