A passenger train departs from Delhi at 6 p.m. for Mumbai.At 9 p.m., an express train,whose average speed exceeds that of the passenger train by 15 kmph leaves Mumbai for Delhi. Two trains meet each other mid-route. At what time do they meet,given that the distance between the cities is 1080 km ?
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540/x - 540/x+15 =3
540(x+15)-540x/x(x+15) = 3
540x+ 540×15- 540x = 3x^2+45x
540×15= 3x^2 +45x
X^2 + 15x- 2700=0
X^2+60x-45x-2700 =0 ( quadratic equation)
X(x+60)-45(x+60)= 0
X-45 =0 , x=45
X+60=0 , x=-60
Speed cannot be negative, that's why we will take positive speed x= 45km/hr.
540/45= 12 hrs
6:00pm+ 12hrs = 6:00 am morning ( answer).
12 midnight
verify the answers by given options.. and 6 a.m. is correct answer ... to save your time only way you have to verify the answer of this type of method.... thank you