Question : A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?
Doubt by Gauri
Solution :
Case I
Let the first speed of the train be x km/h
Distance covered = 54 km
Speed = Distance / Time
Time = Distance/Speed
t1=54/x — (1)
Case II
New Speed = (x+6) km/h
Distance Covered = 63 km
t2=63/(x+6) — (2)
ATQ
t1+t2=3
39x+108=x²+6x
x²+6x-39x-108=0
x²-33x-108=0
x²-(36-3)x-108=0
x²-36x+3x-108=0
x(x-36)+3(x-36)=0
(x-36)(x+3)=0
x-36=0 OR x+3=0
x=36 OR x=-3
But x≠ -3
[As speed can't be negative]
so x=36
First Speed must be 36 km/h.
x²+6x-39x-108=0
x²-33x-108=0
x²-(36-3)x-108=0
x²-36x+3x-108=0
x(x-36)+3(x-36)=0
(x-36)(x+3)=0
x-36=0 OR x+3=0
x=36 OR x=-3
But x≠ -3
[As speed can't be negative]
so x=36
First Speed must be 36 km/h.
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