Painting together

"A" paint one house per hour , or he paint each hour a house. And "B" paint one house per 3 hours , so he paint each hour the third of a house. And if they work together they will paint: "1+(1/3)=(4/3)" of the house per hour , so they will paint a house in: "(3/4)" hour , which is 45minute .

It takes A 60 minutes to paint one house, and it takes B 180 minutes.
Let m = number of minutes. Therefore, the number (or portion) of houses painted for A is $\frac{m}{60}$ and for B is $\frac{m}{180}$ . Adding the two together will be the whole house. So we have:

$\frac{m}{60} + \frac{m}{180} = 1$

Solving for m :

m = 45 or 45 minutes

Suppose the surface area of house is $S m^{2}$

Then, rate of painting by A is $­\frac {S}{60} \frac{m^{2}}{min}$

Similarly, rate of painting by B is $­\frac {S}{180} \frac{m^{2}}{min}$

Now, let the time taken for both to paint be 'T'

Then area painted by A in time 'T' is $­\frac {S \times T}{60} m^{2}$

Similarly, area painted by B in time 'T' is $­\frac {S \times T}{180} m^{2}$

Since, sum of area painted by A and B is equal to S

$­\frac {S \times T}{60} m^{2} +­\frac {S \times T}{180} m^{2} =S$

Solving for T gives,

$\boxed{T=45 minutes}$