James, by himself, can paint 4 rooms in 10 hours. If he hires June to help, they can do the same job together in 6 hours. If he lets June work alone, how long will it take her to paint 4 rooms?

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- Jan 11th 2007, 12:59 PMsymmetryWorking Alone
James, by himself, can paint 4 rooms in 10 hours. If he hires June to help, they can do the same job together in 6 hours. If he lets June work alone, how long will it take her to paint 4 rooms?

- Jan 11th 2007, 01:05 PMThePerfectHacker
The general formula is,

$\displaystyle \frac{ab}{a+b}$

Where $\displaystyle a,b$ is the amount of time it takes alone and the fraction is the time together they work.

Thus, the Man can do it in $\displaystyle a=10$. The Woman can do it in $\displaystyle b$ hours. Together they spend,

$\displaystyle \frac{10b}{10+b}=6$

Multiply by denominator,

$\displaystyle 10b=6(10+b)$

$\displaystyle 10b=60+6b$

$\displaystyle 10b-6b=60$

$\displaystyle 4b=60$

$\displaystyle b=15$ hours. - Jan 11th 2007, 01:08 PMsymmetryok
I NEVER knew there was such an equation for solving "working alone" problems.

How about that?

I learned something new today.

Thanks a million! - Jan 11th 2007, 01:11 PMThePerfectHacker
- Jan 11th 2007, 09:48 PMAfterShock
Another way to look at this problem:

We know that James can paint 4 rooms in 10 hours; thus, in 30 hours he can paint 12 rooms.

We know that together they can paint 4 rooms in 6 hours. Thus, in 30 hours, they can point 20 rooms.

From this, we can see that June can paint 8 rooms in 30 hours. Thus, in order to paint just 4 rooms, it would take 15 hours. - Jan 11th 2007, 10:35 PMCaptainBlack
James paints 4/10=0.4 rooms per hour

suppose June paints x rooms per hour.

Together they paint x+0.4 rooms per hour.

They paint 4 rooms in 6 hours so their combined rate is 4/6=2/3 rooms per hour

So x+0.4=0.66.., so x=2/3-2/5=4/15 rooms per hour.

Time for June to paint 4 rooms = 4/x =4*15/4=15 hours.

RonL