A car made a run of 390km in 'x' hours. If the speed had been 4kmph more, it would have taken 2 hours less for the journey. Find 'x'

Let the speed be v km/hour
speed of run,v= distance/time = 390/x km/hour
Given that if this speed is increased by 4 km/hour, time will be 2 hours less
so the equation will be
(390/v) – [390/(v+4)] = 2 hours
⇒390[(v+4-v)/v(v+4)]= 2
⇒780 = v²+4v
⇒v²+4v – 780 =0
On factorising
⇒v²+30v -26v -780=0
⇒v(v+30) -26(v+30)=0
⇒(v+30)(v-26)=0
v= 26 km/hours and v = -30 km/hour
So speed can not be negative hence neglecting negative value
v= 26km/hour
Time, x = 390/26 = 15 hours

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