Nick can complete a job in 12 hours. Adam can complete a job in 6 hours. Mike can complete a job in 8 hours. If all three of them start working together, after 2 hours, Adam leaves, and after 4 hours, Mike leaves, how many hours until Nick gives up? :) :)
im too lazy to solve it since i have to do my homework :P
but for the first 2 hours all of them combine efforts so
(1/12+1/6+1/8)*2 that much work done
then you use same logic and stuff for when the kids leave and stuff
glhf
The three men working together can finish the job in the reciprocal of:
1/12 + 1/8 + 1/6=3/8 or in 8/3 =2 2/3 hours. Therefore, when Adam leaves after 2 hours, 2 / (2 2/3) or 75% of the work has already been finished. The remaining 1/4 of the job left, or 2 2/3 / 4=40 minutes, while all 3 men were working together. However, this 40 minutes, or 1/4 of the job remaining can be finished by the two remaining men in: 1/12 + 1/8 =5/24 or (24/5) x 1/4=1.2 hours, or 72 minutes. Therefore, before Mike leaves, the job will have been finished.
+129840 In one hour, Nick does 1/12 of the job, Adam does 1/6 of the job and Mike does 1/8 of the job
So....... 1/12 + 1/6 + 1/8 = 3/8 of the job is done after one hr and 6/8 = 3/4 of the job is done after two hours.....so only 1/4 remains = 6/24
When Adam leaves......1/12 + 1/8 = 5/24 of the job gets done in one hr.....so....it will take the two other guys working together another [ 6/24] / [5/24] = 6/5 of an hour to finish = 72 more minutes
So.....the whole job is completed in 3 hrs and 12 miniutes.........therefore......Nick needn't worry
