SOLUTION: how to solve Marcus and Will are painting a barn. Marcus paints twice as fast as Will. On the first day, they have worked for 6 h and completed 1/3 of the job when Will gets inj

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Question 747992: how to solve Marcus and Will are painting a barn. Marcus paints twice as fast as Will. On the first day, they have worked for 6 h and completed 1/3 of the job when Will gets injured. If Marcus has to complete the rest of the job by himself, how many additional hours will it take him? answer
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
This one makes more sense than your almost identical posting. Completed 1/3 of the job and Will gets injured.

Marcus rate, 2/x jobs per hour
Will rate, 1/x jobs per hour.
time working together was 6 hours, and 1/3 of the job was completed.

highlight%28%282%2Fx%2B1%2Fx%29%2A6=1%2F3%29
When we find "x" here, we will know how to compute the rate for each of Marcus and Will.

Finding x:
%28%282%2B1%29%2Fx%29%2A6=1%2F3
3%2Fx=1%2F%2818%29
x%2F3=18
x=18%2F3=6

RATES OF THE TWO PAINTERS:
Marcus, 2%2Fx=2%2F6=1%2F3, THREE hours to do ONE job, alone.
Will, 1%2F6, six hours to do one job, if alone.

If Marcus does the rest of the job himself, how long will this take him? Remember, 1/3 of the job was done, so Marcus will do 2/3 of the job.
The situation is a uniform rates situation and so we have r%2At=j, where r is rate in jobs per hour, t is time in hours, and j is how many jobs.

His rate is (1/3), so highlight%28%281%2F3%29y=%282%2F3%29%29 where y is the number of hours to do this 2/3 of the job.
y=%283%2F1%29%282%2F3%29
y=%283%2F3%29%282%2F1%29
highlight%28y=2%29, TWO HOURS MORE THAN when they worked together.