SOLUTION: If Craig can paint a house in 18 hours and when working with John can complete the job in 10 hours. How long should it take John to complete the job working alone?

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Question 1196717: If Craig can paint a house in 18 hours and when working with John can complete the job in 10 hours. How long should it take John to complete the job working alone?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52817) About Me  (Show Source):
You can put this solution on YOUR website!
.
If Craig can paint a house in 18 hours and when working with John
can complete the job in 10 hours.
How long should it take John to complete the job working alone?
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The combined rate of work of two workers is  1%2F10  of the job per hour.


The rate of work of one of them (Craig) is  1%2F18  of the job per hour.


HENCE, the rate of work of the other worker (of John) is the difference

    1%2F10 - 1%2F18 = 9%2F90 - 5%2F90 = 4%2F90 = 2%2F45.


It means that John will complete the job in  45%2F2 = 22.5 hours, working alone.     ANSWER

Solved.

On the way, from my solution you learned about rate of work.

You also learned, that when combined rate of work is given,
and the rate of work of one participant is known,
then the rate of work of the other participant is simply the difference of two rates.

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It is a standard and typical joint work problem.

There is a wide variety of similar solved joint-work problems with detailed explanations in this site.  See the lessons
    - Using Fractions to solve word problems on joint work
    - Solving more complicated word problems on joint work
    - Selected joint-work word problems from the archive


Read them and get be trained in solving joint-work problems.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic
"Rate of work and joint work problems"  of the section  "Word problems".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.



Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The response from the other tutor shows a standard formal algebraic method for solving this kind of "working together" problem.

Here is an alternative method that many students like.

(1) Consider the least common multiple of the two given times: LCM(10,18) = 90.
(2) Determine the number of houses that could be painted in 90 hours:
Craig alone: 90/18 = 5 houses
Craig and John together: 90/10 = 9 houses
(3) The difference of 4 houses in 90 hours is the amount of work John could do alone.

John could paint 4 houses in 90 hours, so the number of hours it would take him to paint the one house alone is 90/4 = 22.5.

ANSWER: 22.5 hours