Word problem help!!!

[mathstupid]

5. Person T, Person S, and Person F assembled a very large puzzle together in 40 hours. If Person S worked twice as fast as Person F and Person T worked just as fast as Person F, then how long would it have taken Person F to assemble the puzzle alone?

If you can please help. I came up with the answer 1/8 and I don't think that is right. I need it really soon. Thanks a bunch. I also have to show my work.

 

[Guest]

Let x is part of the work, that F will do for 1 hour.
Then S does 2x, and T will also do x.
Together T, S and F did all work for 40 hours.
Therefore, for 1 hour they together perform 1/40 part of the work.

2x+x+x=1/40
4x=1/40
x=1/160

i.e. F for 1 hour does 1/160 part of the work.
All work he will do after 1/x = 160 hours.

 

[soroban]

Hello, mathstupid!

Here is another approach . . .

5. T, S, and F assembled a very large puzzle together in 40 hours.
If S worked twice as fast as F and T worked just as fast as F,
then how long would it have taken F to assemble the puzzle alone?

Let \(\displaystyle X\) = number of hours for F to assemble the puzzle alone.
\(\displaystyle \;\;\)If one hour, F will assemble \(\displaystyle \frac{1}{X}\) of the puzzle.

S works twice as fast as F.
\(\displaystyle \;\;\)In one hour, S will assemble \(\displaystyle \frac{2}{X}\) of the puzzle.

T works at the same speed as F.
\(\displaystyle \;\;\)In one hour, T will assemble \(\displaystyle \frac{1}{X}\) of the puzzle.

Working together, in one hour they will assemble \(\displaystyle \frac{1}{X}\,+\,\frac{2}{X}\,+\,\frac{1}{X}\:=\:\frac{4}{X}\) of the puzzle.

But we've been told that together it takes them 40 hours to assemble the puzzle.
\(\displaystyle \;\;\)That is, in one hour they will assemble \(\displaystyle \frac{1}{40}\) of the puzzle.

And there is our equation! . . . \(\displaystyle \:\frac{4}{X}\:=\:\frac{1}{40}\)

Therefore: \(\displaystyle \,X\,=\,160\) hours.