two guys painting a barn. one of them paints twice as fast as the other. On the first day, they have worked for 6 h and completed 1/3 of the job, when the 2nd guy gets hurt. If the 1st guy has to complete by himself how many additional hours will it take?

To solve this problem, we need to break it down into two parts: the work done by both guys together and the work that needs to be completed by the first guy alone.

Let's assume that the first guy can complete the job in x hours. Since the second guy paints twice as fast, he can complete the same job in half the time, which is x/2 hours.

On the first day, both guys worked for 6 hours and completed 1/3 of the job. This means that in 6 hours, they completed 1/3 of the job together.

So, the work done by both guys in 6 hours is 1/3 of the job. Therefore, the work remaining to be completed by the first guy alone is 1 - 1/3 = 2/3 of the job.

Since the first guy is now working alone, we can set up the following equation:

(x/2) * 6 = (2/3) * 1

Simplifying this equation, we get:

3x = 4

Dividing both sides by 3, we find:

x = 4/3

So, the first guy can complete the job on his own in 4/3 hours. Since he has already worked for 6 hours, he would need to work for an additional (4/3) - 6 = 2/3 hours to complete the job.

Therefore, it would take the first guy an additional 2/3 hours to complete the job by himself.

1/(1/3) * 6h = 18h = Time for 2 to do the full job.

T*2T/(T+2T) = 18.
2T^2/3T = 18,
2T/3 = 18,
2T = 54 Hours.
T = 27 Hours = Time for the faster painter to do the full job alone.
2/3 * 27 = 18 h = Time for faster painter to do the final 2/3 of the job alone.