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Question 34721This question is from textbook Intermediate Algebra
: An experienced bricklayer can work twice as fast as an apprentice bricklayer. After they work together for 6 hours, the experienced bricklayer quits. The apprentice requires 10 more hours to finish the job. How long would it take the experienced bricklayer working alone to do the job?
I have been working on this for two days, and I have set it up 100 different ways and I STILL can't get the correct answer which is: Working alone the experienced bricklayer could do the job in 14 hours. Please help if you can....Thank you so very much!!!
This question is from textbook Intermediate Algebra
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! Let a = number of bricks per hour laid by the apprentice.
So, 2a is the number of bricks laid by the experienced bicklayer.
Working together for 6 hours, they lay (2a+a) bricks in an hour, so after 6 hours... 6(2a+a) --> 6(3a) --> 18a.
Then the apprentice is left to finish the work, taking 10 hours --> 10a bricks in total.
So, the job laid 18a + 10a bricks --> 28a.
Now, the question says how many hours would it take the experienced bricklayer to lay this many bricks?
well, his rate is 2a and the number of hours is the unknown...call it x --> 2ax. And this has to equal the total from the first way: 28a. So,
2ax = 28a
2x = 28
x = 14
jon.
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