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 April 2nd, 2017, 09:30 PM #1 Newbie   Joined: Apr 2017 From: Australia Posts: 2 Thanks: 0 Math Focus: The easier stuff 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)
 April 2nd, 2017, 10:12 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 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 $\dfrac {1} {30} t = \dfrac{1}{50}t + 1$ $\dfrac{t}{30} = \dfrac{t + 50}{50}$ $50t = 30t + 1500$ $20 t = 1500$ $t = 75~s$ Thanks from deesuwalka and Brian Leon
 April 3rd, 2017, 12:50 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 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 Thanks from Brian Leon Last edited by Denis; April 3rd, 2017 at 01:10 PM.
 April 3rd, 2017, 09:23 PM #4 Newbie   Joined: Apr 2017 From: Australia Posts: 2 Thanks: 0 Math Focus: The easier stuff Thank you, that helps a lot.
 April 4th, 2017, 08:02 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2202 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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