3+7+11+15+ . . . . . (1)
1+9+17+25+ . . . . . (2)
The condition is satisfied for the 4 seconds in 2^{nd} journey, ant covered 17 m more than 1^{st} one.
The sum of speeds would be 0.8 m/s (relative speed in opposite direction). Also if we go by option (b) the speeds will be 0.03 and 0.05 m/s respectively. At this speed the overlapping would occur in every 60 second.
A _______18miles_____B ___x(let)___C
Let the speed of the ship be 1 mile/h and then speed of the plane becomes 10 miles/h. Let time t is taken to cover the distance x from point B to point C by ship. Hence, in time t plane will cover (18+x) miles.
Now, ^{x}/_{1} = ^{(18+x)}/_{10}
Or, 10x = 18+x
Or, x = 2 miles.
Hence, ship will cover 2 miles in 2 hours and plane will cover 20 miles in same time.
Short method:
Relative speed = (10-1) = 9 miles/h. hence, time taken to catch up the ship is 2 hours so distance covered by the ship = 20 miles.
In the final minute of collision, the two boats travel ^{5}/_{60} km and ^{10}/_{60} km.
As they moves towards each other relative distance covered in 1 minute, = (^{1}/_{12})+(^{1}/_{6})
= ^{1}/_{4} km.
A__________300 km_______B
Let speed of the 1^{st} ship is s kmph and speed of 2^{nd} ship is (s+10) kmph. Let 2^{nd} ship takes time t to cover the distance 300 km then 1^{st} ship takes (t+8) hours.
As distance is constant, we have, ^{s}/_{(s+10)} = ^{t}/_{(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.
Since, the two horses meet after 200 minutes; they cover 0.5% of the distance per minute (combined) or 30% distance per hour. This condition is satisfied only if the slower rider takes 10 hours (thereby covering 10% per hour) and the faster rider takes 5 hours.
Distance travelled by man-train in 11 minute = distance traveled by sound in (11 min 45 sec - 11 min) = 45 sec.
Let the speed of train be x kmph.
Now,
Or, x * ^{11}/_{60} = (^{45}/_{3600}) * 330 * ^{18}/_{5}
Or, x = 81 kmph.
In 6 minutes, the car goes ahead by 0.6 km. Hence, the relative speed of the car with respect to the pedestrian is equal to 6 kmph.
Hence, Net speed of the car is 8 kmph.