# Thread: A new "Work" problem

1. ## A new "Work" problem

$\displaystyle A$ and $\displaystyle B$ can paint a room in $\displaystyle \frac{12}{7}$ hours.
$\displaystyle B$ and $\displaystyle C$ can paint the room in $\displaystyle 6$ hours.
$\displaystyle A$ and $\displaystyle C$ can paint the room in $\displaystyle 12$ hours.

How long would each take to paint the room if working alone?

[Any comments on $\displaystyle C$ ?]

2. Originally Posted by Soroban
$\displaystyle A$ and $\displaystyle B$ can paint a room in $\displaystyle \frac{12}{7}$ hours.
$\displaystyle B$ and $\displaystyle C$ can paint the room in $\displaystyle 6$ hours.
$\displaystyle A$ and $\displaystyle C$ can paint the room in $\displaystyle 12$ hours.

How long would each take to paint the room if working alone?

[Any comments on $\displaystyle C$ ?]

You can use my trick.
If one does work in $\displaystyle x$ the other in $\displaystyle y$. Then together it takes.
$\displaystyle \frac{xy}{x+y}$
Thus,
$\displaystyle \frac{AB}{A+B}=\frac{12}{7}$
$\displaystyle \frac{BC}{C+B}=6$
$\displaystyle \frac{AC}{A+C}=12$

3. ## Anyone? Anyone?

So far, 25 views and only TPHacker has responded.

Hasn't anyone solved it? .The humor is in the solution.
. . (Am I the only one laughing?)

4. Originally Posted by Soroban
So far, 25 views and only TPHacker has responded.

Hasn't anyone solved it? .The humor is in the solution.
. . (Am I the only one laughing?)

I think that B is the most interesting. Gets the room painted 2 hours before he even walks in. (C must be using paint remover....)

-Dan

5. ## Solution

Suppose $\displaystyle A$ can paint the room in $\displaystyle a$ hours,
. . . . . . $\displaystyle B$ can paint the room in $\displaystyle b$ hours,
. . . . . . $\displaystyle C$ can paint the room in $\displaystyle c$ hours.

In one hour, $\displaystyle A$ can paint $\displaystyle \frac{1}{a}$ of the room.
In one hour, $\displaystyle B$ can paint $\displaystyle \frac{1}{b}$ of the room.
In one hour, $\displaystyle C$ can paint $\displaystyle \frac{1}{c}$ of the room.

In one hour, $\displaystyle A$ and $\displaystyle B$ can paint $\displaystyle \frac{1}{a} + \frac{1}{b}$ of the room.
Since it takes both of them $\displaystyle \frac{12}{7}$ hours together,
. . in one hour, they can paint $\displaystyle \frac{7}{12}$ of the room: .$\displaystyle \frac{1}{a} + \frac{1}{b}\:=\:\frac{7}{12}$ . [1]

In one hour, $\displaystyle B$ and $\displaystyle C$ can paint $\displaystyle \frac{1}{b} + \frac{1}{c}$ of the room.
Since it takes both of them $\displaystyle 6$ hours together,
. . in one hour, they can paint $\displaystyle \frac{1}{6}$ of the room: .$\displaystyle \frac{1}{b} + \frac{1}{c} \:=\:\frac{1}{6}$ . [2]

In one hour, $\displaystyle A$ and $\displaystyle C$ can paint $\displaystyle \frac{1}{a} + \frac{1}{c}$ of the room.
Since it take both of them $\displaystyle 12$ hours,
. . in one hour, they can paint $\displaystyle \frac{1}{12}$ of the room: .$\displaystyle \frac{1}{a} + \frac{1}{c} \:=\:\frac{1}{12}$ . [3]

Subtract [2] from [1]: .$\displaystyle \frac{1}{a} - \frac{1}{c} \:=\:\frac{5}{12}$
. . . . . . . . . .Add [3]: .$\displaystyle \frac{1}{a} + \frac{1}{c}\:=\:\frac{1}{12}$

And we have: .$\displaystyle \frac{2}{a} \,= \,\frac{1}{2}\quad\Rightarrow\quad \boxed{a\,=\,4}$

Substitute into [1]: .$\displaystyle \frac{1}{4} + \frac{1}{b} \:=\:\frac{7}{12}\quad\Rightarrow\quad \boxed{b\,=\,3}$

Substitute into [3]: . $\displaystyle \frac{1}{4} + \frac{1}{c}\:=\:\frac{1}{12}\quad\Rightarrow\quad \boxed{c\,=\,-6}$

Punchline: $\displaystyle C$ does negative work.

As Dan pointed out, $\displaystyle C$ must be using paint remover.

(Their names must be: Curly, Larry and Moe.)

6. C would be a very annoying partner

7. I guess that makes since; A and B by themselves have the room done in less than two hours but C makes them take many hours longer.

I'm probably person C XP.