Boat Time puzzle: When will the boats meet first time and second time? There are 2 points A and B in a river which is flowing at 5kmph. Distance between A and B is 300 kms. Two boats with speed 10 kmph and 15 kmph (in still water) start at A and B respectively. When will the boats meet first time and second time?
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To answer the question, first I have to assume the following:
a: river flows from A to B
b: boat-1 starts from point A toward point B and boat-2 starts from point B toward point A
c: second meeting occurs after boat-1 arrive at point B and boat-2 arrive at point A and then both boats return to point of origin
First meeting: 12 hours
boat-1 and boat-2 meet at point C which is L1 km from A or (300-L1) km from B in T1 hours
boat-1 relative speed = (10+5) = 15 kph and boat-2 relative speed = (15-5) = 10 kph
boat-1 travels 15 x T1 = L1 and boat-2 travels 10 x T1 = 300 - L1 -----> L1 = 180 km
First meeting is 180 km : 15 kph = 12 hours
Second Meeting: 40 hours
boat-1 arrives at B in (300 km : 15 kph) = 20 hours and boat-2 arrives at A in (300 km : 10 kph) = 30 hours
when boat-2 returns to origin, boat-1 has traveled to origin (30 - 20) = 10 hours
boat-1 returns with relative speed (10 - 5)= 5 kph and boat-2 (15 + 5)= 20kph
boat-1 and boat-2 meet at point D which is L2 km from A in T2 hours or (300-L2) km from B in (T2+10) hours
boat-2 relative speed = (15+5) = 20 kph and boat-1 relative speed = (10-5) = 5 kph
boat-2 travels 20 x T2 = L2 and boat- travels 5 x (T2+10) = 300 - L2 -----> L2 = 200 km
Second meeting is (200 km : 20 kph) + 30 hours = 40 hours since time of start
a: river flows from A to B
b: boat-1 starts from point A toward point B and boat-2 starts from point B toward point A
c: second meeting occurs after boat-1 arrive at point B and boat-2 arrive at point A and then both boats return to point of origin
First meeting: 12 hours
boat-1 and boat-2 meet at point C which is L1 km from A or (300-L1) km from B in T1 hours
boat-1 relative speed = (10+5) = 15 kph and boat-2 relative speed = (15-5) = 10 kph
boat-1 travels 15 x T1 = L1 and boat-2 travels 10 x T1 = 300 - L1 -----> L1 = 180 km
First meeting is 180 km : 15 kph = 12 hours
Second Meeting: 40 hours
boat-1 arrives at B in (300 km : 15 kph) = 20 hours and boat-2 arrives at A in (300 km : 10 kph) = 30 hours
when boat-2 returns to origin, boat-1 has traveled to origin (30 - 20) = 10 hours
boat-1 returns with relative speed (10 - 5)= 5 kph and boat-2 (15 + 5)= 20kph
boat-1 and boat-2 meet at point D which is L2 km from A in T2 hours or (300-L2) km from B in (T2+10) hours
boat-2 relative speed = (15+5) = 20 kph and boat-1 relative speed = (10-5) = 5 kph
boat-2 travels 20 x T2 = L2 and boat- travels 5 x (T2+10) = 300 - L2 -----> L2 = 200 km
Second meeting is (200 km : 20 kph) + 30 hours = 40 hours since time of start
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