SOLUTION: Vernon, Joshua, and Dino are painting a house. Working together they can paint the house is 6 hours. Working alone Vernon can paint the house in 15 hours and Dino can paint the hou

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Question 1172330: Vernon, Joshua, and Dino are painting a house. Working together they can paint the house is 6 hours. Working alone Vernon can paint the house in 15 hours and Dino can paint the house in 20 hours. How long would it take Joshua to paint the house working alone?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.

The combined rate of work of the three workers is    of the job per hour.


The individual rates of the two participants are    and    of the job per hour.


Then the rate of work of the remained third participant is  the difference


     -  -  =  =  =   of the job per hour.


Hence, Joshua can complete the job in 20 hours, working alone.    ANSWER

Solved.

---------------

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
    - Rate of work problems
    - Using Fractions to solve word problems on joint work
    - Solving more complicated word problems on joint work
    - Using quadratic equations to solve word problems on joint work
    - Solving rate of work problem by reducing to a system of linear equations
    - Solving joint work problems by reasoning
    - Selected joint-work word problems from the archive
    - Joint-work problems for 3 participants
    - HOW TO algebreze and solve these joint work problems ?
    - Had there were more workers, the job would be completed sooner
    - One unusual joint work problem
    - Special joint work problems that admit and require an alternative solution method
    - Joint work word problems for the day of April, 1
    - OVERVIEW of lessons on rate-of-work problems

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.

- - - - - - - -

The lessons to learn from this solution :

    1)  You can solve this problem/(such problems) even without using equations.

        Freely manipulating fractions is enough.


    2)  Use the notion "rate of work" and remember:

            a)  when two or three persons work together, their combined rate of work is the sum of individual rates;


            b)  when the combined rate of three workers is given along with the individual rates of two of them, then the rate 
                of the third worker is the difference of combined rate and two given individual rates.



Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Here is an alternative to the standard algebraic solution shown by tutor @ikleyn.

Many students prefer this method because it avoids using fractions. Note, however, that the required arithmetic is nearly the same as in the solution shown by @ikleyn.

Consider the least common multiple of the given times. The LCM of 6, 15, and 20 is 60.

Now see what each worker or combination of workers could do in 60 hours.

Together the three of them in 60 hours could paint 60/6=10 houses.
Vernon working alone in 60 hours could paint 60/15 = 4 houses.
Dino working alone in 60 hours could paint 60/20 = 3 houses.

That means Joshua alone in 60 hours could paint 10-4-3 = 3 houses.

So Joshua working alone could paint the one house in 60/3 = 20 hours.

ANSWER: 20 hours.


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