<PRE><B>Question 314932</B>
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/(.5x)}}} + {{{40/(.5x)}}} + {{{40/x}}} = 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
{{{40/100}}} + {{{40/100}}} + {{{40/200}}} =
.4 + .4 + .2 = 1 </PRE>