SOLUTION: An experienced bricklayer can work twice as fast as an apprentice bricklayer. After they worked together on a job for 12 h, the experienced bricklayer quit. The apprentice required

Algebra ->  Rate-of-work-word-problems -> SOLUTION: An experienced bricklayer can work twice as fast as an apprentice bricklayer. After they worked together on a job for 12 h, the experienced bricklayer quit. The apprentice required      Log On


   

Question 794133: An experienced bricklayer can work twice as fast as an apprentice bricklayer. After they worked together on a job for 12 h, the experienced bricklayer quit. The apprentice required 16 more hours to finish the job. How long would it take the experienced bricklayer, working alone, to do the job?
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Let h = number of hours
rate*days=jobs
rate=jobs/days

Note carefully that the units for rate are JOBS per HOUR.

Layer________________rate____________time___________Jobs
Experienced__________(1/h)____________h______________1
Apprentice___________(1/2h)____________2h_____________1
BOTH_______________(1/h+1/(2h))________12_________%281%2Fh%2B1%2F%282h%29%29%2A12

After "BOTH", part of the job was done, and then apprentice worked 6 hours at his rate to finish the whole job. The sum of the parts of the job equal 1 job. Understanding this requires some thought, and should seem logical to reach this conclusion.

highlight%2812%281%2Fh%2B1%2F%282h%29%29%2B%281%2F%282h%29%29%2A16=1%29
Again, this is just sum of the parts of the job equals the whole job.

Solving for h will give you the information about the work rate of each person brick layer.

Simplifying,
12%28%282%2B1%29%2F%282h%29%29%2B8%2Fh=1
12%2A3%2F%282h%29%2B8%2Fh=1
18%2Fh%2B8%2Fh=1
26%2Fh=1
highlight%2826=h%29

MEANING: Looking back at what this variable means, the experienced brick layer can do the job in 26 hours by himself; and the apprentice would need 52 hours if working alone.