Two ports A and B are 300 km apart. Two ships leave A for B such that the second leaves 8 hours after the first. The ships arrive at B simultaneously. Find the time the slower ship spent on the trip if the speed of one of them is 10 km/h higher than that of the other.
A. 25 hours
B. 20 hours
C. 15 hours
D. 20 hours
E. 24 hours
Answer: Option D
Solution (By Examveda Team)
A__________300 km_______B Let speed of the 1st ship is s kmph and speed of 2nd ship is (s + 10) kmph. Let 2nd ship takes time t to cover the distance 300 km then 1st ship takes (t + 8) hours. As distance is constant, we have, $$\frac{{\text{s}}}{{{\text{s}} + 10}} = \frac{{\text{t}}}{{{\text{t}} + 8}}$$ Or, st + 8s = st + 10t Or, 8s = 10t If the slower ship took 20 hours (option D) the faster ship would take 12 hours and their respective speeds would be 15 and 25 kmph.This Question Belongs to Arithmetic Ability >> Speed Time And Distance
20 hours
slower=s
faster=s+10
atq, 300/s-300/s+10=8
s=15
so slower time taken=300/15=20 hours...(ans)
(300/s)-(300/s+10)=8...solving this u get s^2+10=375 ...s=25 satisfies this equation.
(300/s)-(300/s+10)=8...solving u get s=15 , so 300/15=20...answer: 20 hours