1. A car travels from P to Q at a constant speed. If its speed were increased by 10 km/h, it would have been taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/h. The distance between the two cities is
Write Comment
Comments
By: anil on 05 May 2019 01.53 am
Let the distance between the two cities = $$d$$ km and the speed be $$s$$ km/hr
=> Time taken to travel = $$frac{d}{s}$$ hr Now, in condition 1, the car takes 1 hr lesser when speed is increased by 10 km/h => $$frac{d}{s+10} = frac{d}{s} - 1$$ => $$frac{d}{s} - frac{d}{s+10} = 1$$ => $$10d = s(s+10)$$ ---------Eqn(1) and according to condition 2, => $$frac{d}{s+20} = frac{d}{s} - 1 - frac{45}{60}$$ => $$frac{d}{s} - frac{d}{s+20} = frac{7}{4}$$ => $$20d = frac{7}{4} s(s+20)$$ ---------------Eqn(2) Solving eqns (1) & (2), we get $$s$$ = 60 km/hr and $$d$$ = 420 km
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use
=> Time taken to travel = $$frac{d}{s}$$ hr Now, in condition 1, the car takes 1 hr lesser when speed is increased by 10 km/h => $$frac{d}{s+10} = frac{d}{s} - 1$$ => $$frac{d}{s} - frac{d}{s+10} = 1$$ => $$10d = s(s+10)$$ ---------Eqn(1) and according to condition 2, => $$frac{d}{s+20} = frac{d}{s} - 1 - frac{45}{60}$$ => $$frac{d}{s} - frac{d}{s+20} = frac{7}{4}$$ => $$20d = frac{7}{4} s(s+20)$$ ---------------Eqn(2) Solving eqns (1) & (2), we get $$s$$ = 60 km/hr and $$d$$ = 420 km