SOLUTION: Two ferryboats leave opposite sides of a lake at the same time. They pass eachother when they are 800 meters from the nearest shore. When it reaches the opposite side, each boat

Question 31230: Two ferryboats leave opposite sides of a lake at the same time. They pass
eachother when they are 800 meters from the nearest shore. When it reaches
the opposite side, each boat spends 30 minutes at the dock and then starts
back. This time the boats pass each other when they are 400 meters from the
nearest shore. Assuming that each of the boats travels at the same speed in
both directions, how wide is the lake between the two ferry docks?
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
LET US MARK 2 SHORES AS S1 AND S2 AND THE 2 BOATS AS B1 STARTING FROM
S1 AND B2 STARTING FROM S2
LET SPEED OF FASTER BOAT B1 BE KX WHERE K IS GREATER THAN 1 .
LET THE DISTANCE BETWEEN 2 SHORES BE L�.HENCE �.S1S2 = L
NOW B1 BEING FASTER WILL GO FARTHER THAN B2 AND HENCE WILL MEET NEARER TO S2
SO FIRST MEET IS 800 M FROM S2.AND L-800 M FROM S1.
OBVIOUSLY SINCE B1 IS ALREADY NEARER TO S2 IT WILL REACH S2 EARLIER
AND WILL START EARLIER THAN B2 AFTER 30 MTS REST.
HENCE BY THE SAME LOGIC THE SECOND MEET WILL BE NEARER TO S1 AND IT
WILL BE 400M FROM S1.AND L-400 M.FROM S2.
NOW THAT WE GOT THE PICTURE LET US TRACK EACH BOAT SEPERATELY.
MOVEMENT OF B1�.
DISTANCE TO FIRST MEETING POINT = L-800
TIME TAKEN TO TRAVEL THIS DISTANCE = (L-800)/KX��.I
FURTHER TIME TAKEN TO REACH S2 =800/KX��II
BREAK TIME = 30 MTS��..III
FURTHER TIME TAKEN TO REACH SECOND MEETING POINT = (L-400)/KX��...1V
TOTAL TIME TAKEN UPTO SECOND MEET.=(L-800)/KX + 800/KX + 30 +
(L-400)/KX��..V
MOVEMENT OF B2
DISTANCE TO FIRST MEETING POINT = 800
TIME TAKEN TO TRAVEL THIS DISTANCE = 800/X��VI
FURTHER TIME TAKEN TO REACH S1 =(L-800)/X��VII
BREAK TIME = 30 MTS��..VIII
FURTHER TIME TAKEN TO REACH SECOND MEETING POINT = 400/X��.1X
TOTAL TIME TAKEN UPTO SECOND MEET.=800/X +(L- 800)/X + 30 + (400)/X��..X
NOW WE EQUATE THE TIME OF TRAVEL UPTO THE 2 MEETING POINTS BY B1 AND
B2 TO SOLVE FOR L
TIME UP TO MEET 1.=EQN.I = EQN.VI
(L-800)/KX = 800/X��OR�..L-800 = 800K��XI
TIME UPTO MEET 2.=EQN.V=EQN.X.
(L-800)/KX + 800/KX + 30 + (L-400)/KX =800/X +(L- 800)/X + 30 +(400)/X�....��XII
(2L-400)/K = L + 400�OR�.2L-400 = K(L+400)
SUBSTITUTING FOR K FROM EQN.XI..WE GET
2L-400 = (L+400)(L-800)/800
1600L-320000 =L^2-400L-320000
L^2-2000L=0
L(L-2000)=0
L=2000
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