Franklin can do a job in 6h and Mike can do it in 4 h.What part of the job can they do by working together 2h?

Question

Couldn't fit up there. and for y hrs?
Again the Algebra structure and method book. And i'm in section 7-8. Work Problems.

Answer 1

Just figure out the rate at which each goes and add them together.

Franklin does 1 job in 6 hours, so 1 j = 6 h and 1/6 j = h. In one hour he does a sixth of a job.

Mike does 1 job in 4 hours, so 1 j = 4 h and 1/4 j = h. In one hour he does a fourth of a job.

Together, then, they do 1/4 j + 1/6 j every hour. Find common denominators and you get 3/12 j + 2/12 j = 5/12 j. Every hour together they do 5/12 of a job.

Of course, you have two hours this time. So in that time they'll do 2 x 5/12 j = 10/12 j = 5/6 j. Five-sixths of a job.

(Maybe they can talk Franklin into working for one more hour to finish it up!)

Answer 2

Franklin can do the job in 6 hours
In 1 hour,he can do 1/6 part of the job.
Mike does the job in 4 hours
In 1 hour,he does 1/4 part of the job.
Therefore,working together,in 1 hr.they do
1/6 +1/4
=5/12 part of the job
Hence,in 2 hours working together.
they do (5/12)*2 or 5/6 part of the job

Answer 3

(time)/Franklin alone + (time)/Mike alone

2/6 + 2/4 = 4/12 + 6/12 = 10/12 = 5/6

y/6 + y/4 = 2y/12 + 3y/12 = 5y/12