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.25hours
B.20 hours
C.15 hours
D.20 hours
Answers
Answer:
Ship A: 300 = (x + 10)*t
Ship B: 300 = x * (t - 8)
Where x is the speed of ship B
and t is the time ship A uses on the journey.
(x + 10 )*t = x*(t - 8)
xt + 10t = xt - 8x
10t = 8x
x = 5/4 t
If we now substitute this back in the eq for ship B:
300 = 5/4t * (t - 8)
300 = 5/4t^2 - 10t
If we multiply every part with 4/5:
240 = t^2 - 8t
This can be written as a quadratic equation:
t^2 - 8t - 300 = 0
t = (-8 +/- sqrt(64 + 1200))/2
t = (-8 + sqrt(1264))/2
t = -4 + 17.776 = 13.776 hours
Step-by-step explanation:
Answer:
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
s
___
s+10=t
t+8
___
s
s+10=t
t+8 Or,
st + 8
s = 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.