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, then how long would it have

Algebra ->  Customizable Word Problem Solvers  -> Misc -> 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, then how long would it have      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 314932: 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, then how long would it have taken Fred to assemble the puzzle alone?

Answer by ankor@dixie-net.com(22740) 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,
then how long would it have taken Fred to assemble the puzzle alone?
:
Let x = time required by Fred
then
.5x = time required by Stacy, and time required by Tracy
:
Let the completed puzzle = 1
:
A shared work equation
40%2F%28.5x%29 + 40%2F%28.5x%29 + 40%2Fx = 1
multiply by .5x, results
40 + 40 + .5(40) = .5x
:
40 + 40 + 20 = .5x
:
.5x = 100 hrs,
Mult by 2
x = 200 hrs Fred alone
:
:
Check solution in shared work equation
40%2F100 + 40%2F100 + 40%2F200 =
.4 + .4 + .2 = 1