If a car starts from A towards B with some velocity. Due to some problem in the engine after traveling 30km, the car goes with 4/5 th of its actual velocity The car reaches B 45 min later to the actual time. If the car engine fails often traveling 45km, the car reaches the destination B 36min late to the actual time?

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Jun 21st, 2005
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swati

Mar 31st, 2006

let xkm/s be the velocity of a car,ykm be the distance
(30/x)+(y-30)*5/(4*x)=y/x+45/60
(45/x)+(y-45)*5/(4*x)=y/x+36/60
solve it

kak

Jun 10th, 2006

velocity=20/3;
distance b/w A & B= 110

itclanster
Profile Answers by itclanster

Dec 12th, 2009

Velocity will be 125km/hr.
Equation will be (3/20hr)*(4/5v)=(15km)

mrprajesh
Profile Answers by mrprajesh

Dec 21st, 2009

Total distance x;

|_____|________| |________|____|
30 x-30 45 x-45

4/5 of speed traveled (rest of part) then 1/5 of speed was making the delay time
so
(x-30) / (s/5)= 45
(x-45) / (s/5)= 30

s is speed
solve for x;
x=105

s=20/3 T=63/4
Actual time ~16 hr (slightly above 15)

seth

Nov 3rd, 2011

let velocity as V, distance as D.

equation is: D/V + 3/4 = 30/V + 5(D-30)/4V ...1
D/V + 3/5 = 45/V + 5(D-45)/4V ...2

from 1,2. initial velocity of car is 350/3km/h
distance is 320km