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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 = LK +400K = 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
