document.write( "Question 78254: A painter works on a job for 10 days and then is joined by a helper. Together they finish the job in 6 more days. The helper could have done the job alone in 30 days. How long would have taken the painter have taken to do the job alone.\r
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Algebra.Com's Answer #56119 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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A painter works on a job for 10 days and then is joined by a helper. Together they finish the job in 6 more days. The helper could have done the job alone in 30 days. How long would have taken the painter have taken to do the job alone.
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\n" ); document.write( "Let x = number of days that the painter can do the job by himself
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\n" ); document.write( "Let the completed job = 1
\n" ); document.write( "Painter alone + Painter/Helper = 1
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\n" ); document.write( " $\"10%2Fx\"$ + $\"6%2Fx\"$ + $\"6%2F30\"$ = 1
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\n" ); document.write( "Multiply equation by 30x, gets rid of the denominators:
\n" ); document.write( "30(10) + 30(6) + 6x = 30x
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\n" ); document.write( "300 + 180 = 30x - 6x
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\n" ); document.write( "24x = 480
\n" ); document.write( "x = 480/24
\n" ); document.write( "x = 20 days, the painter by himself
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\n" ); document.write( "Check:
\n" ); document.write( "10/20 + 6/20 + 6/30 =
\n" ); document.write( " 5/10 + 3/10 + 2/10 = 1 \n" ); document.write( "

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