SOLUTION: Tracy, Stacy and Fred assembled a very large puzzle together in 40 hours. If Stacy worked twice as fast as Fred and Tracy worked just as fast as Stacy, how long would it have taken

Algebra ->  Expressions-with-variables -> SOLUTION: Tracy, Stacy and Fred assembled a very large puzzle together in 40 hours. If Stacy worked twice as fast as Fred and Tracy worked just as fast as Stacy, how long would it have taken      Log On


   

Question 348096: Tracy, Stacy and Fred assembled a very large puzzle together in 40 hours. If Stacy worked twice as fast as Fred and Tracy worked just as fast as Stacy, how long would it have taken Fred to assemble the puzzle alone?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Tracy, Stacy and Fred assembled a very large puzzle together in 40 hours. If Stacy worked twice as fast as Fred and Tracy worked just as fast as Stacy, how long would it have taken Fred to assemble the puzzle alone?
--------------
Together rate: 1/40 job/hr
---
Fred rate = 1/2x job/hr
---
Stacy rate = 1/x job/hr
---
Tracy rate = 1/x job/hr
---------------------------
Equation:
1/x + 1/x + 1/2x = 1/40
---
40 + 40 + 20 = x
x = 100 hrs
---
Fred's time alone = 2x = 200 hrs.
====================================
Cheers,
Stan H.