document.write( "Question 276491: If 5 adults and 2 children work together, a job can be done in a day. If only 2 adults work, then 6 children must work in order to complete the job in a day. The number of days that it takes for a child to do the job alone is:\r
\n" ); document.write( "\n" ); document.write( "a 9 b 25/3 c 26/3 d 8 e none of above
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Algebra.Com's Answer #201625 by ankor@dixie-net.com(22740)\"\" \"About 
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If 5 adults and 2 children work together, a job can be done in a day.
\n" ); document.write( "If only 2 adults work, then 6 children must work in order to complete the job in a day.
\n" ); document.write( " The number of days that it takes for a child to do the job alone is:
\n" ); document.write( ":
\n" ); document.write( "Let a = time required when adult does the job alone
\n" ); document.write( "let c = time required when a child does it
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\n" ); document.write( "Let the completed job = 1
\n" ); document.write( ":
\n" ); document.write( "\"5%2Fa\" + \"2%2Fc\" = 1
\n" ); document.write( "and
\n" ); document.write( "\"2%2Fa\" + \"6%2Fc\" = 1
\n" ); document.write( ":
\n" ); document.write( "Therefore:
\n" ); document.write( "\"2%2Fa\" + \"6%2Fc\" = \"5%2Fa\" + \"2%2Fc\"
\n" ); document.write( "\"6%2Fc\" - \"2%2Fc\" = \"5%2Fa\" - \"2%2Fa\"
\n" ); document.write( "\"4%2Fc\" = \"3%2Fa\"
\n" ); document.write( "Cross multiply
\n" ); document.write( "4a = 3c
\n" ); document.write( "a = \"3%2F4\"c
\n" ); document.write( "a = .75c
\n" ); document.write( ":
\n" ); document.write( "Using the 1st equation replace a with .75c, find c
\n" ); document.write( "\"5%2F%28.75c%29\" + \"2%2Fc\" = 1
\n" ); document.write( "Multiply equation by 3c, results
\n" ); document.write( "4(5) + 3(2) = 3c
\n" ); document.write( "20 + 6 = 3c
\n" ); document.write( "c = \"26%2F3\" days for a child alone\r
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