SOLUTION: 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 begi
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Question 85199: 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 yards from one end and continue on their ways to their own ends of the pool, turn around, and meet the second time 8 yards from the opposite end (from which they first met), how long is the pool? Answer by ankor@dixie-net.com(22740) (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 yards from one end and continue on their ways to their own ends of the pool, turn around, and meet the second time 8 yards from the opposite end (from which they first met), how long is the pool?
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Let x = length of the pool
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(It will be easy to understand if you draw a rough diagram of what's happening here.)
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Distance traveled to the 1st meeting:
Swimmer A = 21 yds
Swimmer B = (x-21) yds
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Distance traveled from the 1st meeting to the 2nd meeting:
Swimmer A: (x-21) + 8 = (x - 13)
Swimmer B: 21 + (x - 8) = (x + 13)
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The ratio of the two swimmers' distances remains the same in both instances:
Write a ratio equation =
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Cross multiply:
(x-21)(x-13) = 21(x+21)
:
x^2 - 34x + 273 = 21x + 273
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x^2 - 34x - 21x + 273 - 273 = 0
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x^2 - 55x = 0
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x(x - 55) = 0
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x = 55 yds is the length of the pool
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Did this make sense to you?
