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April 2nd, 2017, 10:30 PM   #1
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Algebra, word problem

Looking for the algebraic expression(s) to use for the following problem:

Carrie and Manuel are side by side when they begin to run at the same time in the same direction around a track. Carrie runs at a constant rate of 30 seconds per lap, while Manuel runs at a constant rate of 50 seconds per lap. How many seconds after beginning to run will Carrie have run exactly 1 more lap than Manuel?

(Answer is: 75)
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April 2nd, 2017, 11:12 PM   #2
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the key to this problem is to convert time per lap to lap per time

Carrie runs at 30s/lap, or 1/30 lap per second.

Manuel runs at 1/50 lap per second.

so let $t$ be the number of seconds when Carrie has 1 lap on Manuel

$\dfrac {1} {30} t = \dfrac{1}{50}t + 1$

$\dfrac{t}{30} = \dfrac{t + 50}{50}$

$50t = 30t + 1500$

$20 t = 1500$

$t = 75~s$
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April 3rd, 2017, 01:50 PM   #3
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Pictorially speaking (m = Manual's speed, c = Carrie's speed, u = track length):

@m........u..........>50s........v.........>[50+v/m]s

@c.........u..........>30s..................u+v... ..................>[30+(u+v)/c]s

A way to use ye olde speed = distance / time

Probably not as simple as Romsek's lazy approach
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Last edited by Denis; April 3rd, 2017 at 02:10 PM.
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April 3rd, 2017, 10:23 PM   #4
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Thank you, that helps a lot.
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April 4th, 2017, 09:02 AM   #5
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Carrie runs at (1/30 - 1/50) laps per second relative to Manuel, so Carrie is 1 lap ahead of Manuel after 1/(1/30 - 1/50) seconds, which is 75 seconds.
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