<PRE><B>Question 1172330</B>
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Here is an alternative to the standard algebraic solution shown by tutor @ikleyn.&lt;br&gt;
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.&lt;br&gt;
Consider the least common multiple of the given times.  The LCM of 6, 15, and 20 is 60.&lt;br&gt;
Now see what each worker or combination of workers could do in 60 hours.&lt;br&gt;
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.&lt;br&gt;
That means Joshua alone in 60 hours could paint 10-4-3 = 3 houses.&lt;br&gt;
So Joshua working alone could paint the one house in 60/3 = 20 hours.&lt;br&gt;
ANSWER: 20 hours.&lt;br&gt;
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