Old School Math Problem...forgot how to do it.

[Xenon14]

Juan and José are expert house painters. Once, they each timed how long it took the other to paint a tool shed. Juan completed his shed in 3 hours, 20 minutes. José, on the other hand, took an extra 25 minutes to paint an identical shed.
There is one more shed to paint today, identical to the other two they timed themselves on. They've decided to work together on this one. How long will it take if Juan and José work at the same rates that they painted earlier?

 

[nd]

Simple.

Call Juan's performance to be 1 shed / 3:20, or 0.3 sheds per hour.
Jose's performance is 1 shed / 3:45, or 4/15 (~0.2666) sheds per hour.

Let f(t) indicate how many sheds would be complete after 't' hours.

f(t) = 0.3t + (4/15)t

For 1 shed, we do:

1 = 0.3t + (4/15)t

Solve for t and get ~1.76 hours, or about 1 hour and 45 minutes.

 

[Viper GTS]

Easy.

Look at it this way:

In one minute, Juan will paint 1/200th of the shed, & Jose will paint 1/225 of the shed.

Set up an equation:

1 = t(1/200 + 1/225)

Solve for t.

Viper GTS

DISCLAIMER: The above post was produced under the influence of work. It's accuracy is not guaranteed.

 

[Gustavus]

Juan paints a shed in 3 1/3 hours, Jose in 3 3/4 hours. Therefore Juan paints 3/10 of a shed in one hour and Jose paints 4/15 of a shed in one hour. It will therefore take them 1 13/17 hours to paint a shed together, in which time Juan will have painted 9/17 of the shed and Jose will have painted 8/17 of the shed.

 

[nd]

BTW, Both mine and Viper GTS's answer will come out to be the same.. he's just using minutes instead of hours... so, choose whichever way you prefer.