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Question 23351: Two friends are swimming laps back and forth across the length of a pool. They swim with constant speeds and essentially instantaneous turns at the ends of the pool. If at the beginning they leave from opposite ends of the pool and meet for the first time 21 meters from one end and continue on their ways to their own ends of the pool, trun around and meet the second tome 8 meters from the opppsite end (from which they first met), how long is the pool?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Two friends are swimming laps back and forth across the length of a pool. They swim with constant speeds and essentially instantaneous turns at the ends of the pool. If at the beginning they leave from opposite ends of the pool and meet for the first time 21 meters from one end and continue on their ways to their own ends of the pool, trun around and meet the second tome 8 meters from the opppsite end (from which they first met), how long is the pool?
LET THE LENGTH OF POOL = L......LET THE TWO FRIENDS BE F1 AND F2.
LET THE SPEEDS OF F1 AND F2 BE X AND Y AND LET THEM START FROM ENDS.1 & 2 RESPECTIVELY
I SWIM.....
DISTANCE BETWEEN THE TWO =L...SINCE THEY LEAVE FROM OPPOSITE ENDS.
RELATIVE SPEED =X+Y..SINCE THEY ARE SWIMMING IN OPPOSITE DIRECTIONS.
TIME TAKEN TO MEET IN I SWIM =L/(X+Y)
DISTANCE COVERED BY F1 DURING THIS TIME = (L*X)/(X+Y)= 21 METRES FROM END 1 SAY.
DISTANCE COVERED BY F2 DURING THIS TIME = (L*Y)/(X+Y)= L-21 METRES FROM END 2.
II SWIM........THERE IS AN AMBIGUITY IN WORDING OF THE PROBLEM HERE
QUOTE
......meet for the first time 21 meters from one end and continue on their ways to their own ends of the pool(THEIR OWN ENDS MEAN THEY GO BACK TO THEIR ORIGINAL STARTING ENDS???OR CONTINUE ON THEIR WAYS MEANS THEY GO TO OPPOSITE ENDS FROM WHERE THEY STARTED) , trun around and meet the second tome 8 meters from the opppsite end
UNQUOTE...CAPITAL LETTERS INDICATE THE AMBIGUITY IN WORDING..LET US WORK OUT THE PROBLEM ON THE FOLLOWING ASSUMPTION.
...THEY GO BACK TO THEIR OWN ENDS THAT IS TO REACH THE SAME ENDS AS THEY STARTED FROM....
DISTANCE TRAVELLED BY F1 TWO TO MEET A SECOND TIME
=21 M.TO GO BACK TO END 1 + L-8 M. = 21+L-8=L+13
DISTANCE TRAVELLED BY F2 TWO TO MEET A SECOND TIME
= L-21 M.TO GO TO END 2 + 8 M.= L-21+8=L-13.
ANALYSIS.......
SINCE THEY WERE TOGETHER AT THE SAME TIME / PLACE WHEN THEY FIRST MET IN I SWIM,THE DURATION FOR WHICH THE TWO SWAM TO MEET A SECOND TIME IS SAME SINCE THEY WERE MEETING AT THE SAME TIME THERE DURING II SWIM.
HENCE ..
(L+13)/X=(L-13)/Y...OR....Y/X=(L-13)/(L+13)......EQN.III
BUT WE GOT FROM I SWIM THAT
(L*X)/(X+Y)=21...AND........EQN.I
(L*Y)/(X+Y)=L-21............EQN.II
EQN.II/EQN.I...GIVES
Y/X=(L-21)/L..........EQN.IV..USING EQN.III...WE GET
(L-21)/L=(L-13)/(L+13)...CROSS MULTIPLYING
(L-21)(L+13)=L(L-13)
L^2+13L-21L-21*13=L^2-13L
L^2+13L-21L-L^2+13L=21*13
5L=273
L=273/5=54.6 M ...THIS APPEARS TO BE MORE REASONABLE ASSUMPTION.
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