document.write( "Question 717612: A can do a piece of work in 2 days as B can do in 3 days. B can do as much work in 4 days as C can do in 5 days. If A,B& C finish the work together in 10 days. In how many days B can finish the work alone ? \n" ); document.write( "
Algebra.Com's Answer #440484 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
A can do a piece of work in 2 days as B can do in 3 days.
\n" ); document.write( "\"2%2FA\" = \"3%2FB\"
\n" ); document.write( "Cross multiply
\n" ); document.write( "3A = 2B
\n" ); document.write( "A = \"2%2F3\"B\r
\n" ); document.write( "\n" ); document.write( ":
\n" ); document.write( " B can do as much work in 4 days as C can do in 5 days.
\n" ); document.write( "\"4%2FB\" = \"5%2FC\"
\n" ); document.write( "4C = 5B
\n" ); document.write( "C = \"5%2F4\"B\r
\n" ); document.write( "\n" ); document.write( ":
\n" ); document.write( " If A, B & C finish the work together in 10 days.
\n" ); document.write( "Let the completed job = 1
\n" ); document.write( "\"10%2FA\" + \"10%2FB\" + \"10%2FC\" = 1
\n" ); document.write( "Replace A and C
\n" ); document.write( "\"10%2F%28%282%2F3%29B%29\" + \"10%2FB\" + \"10%2F%28%285%2F4%29B%29\" = 1
\n" ); document.write( "Multiply by \"10%2F12\"B, to clear the denominators
\n" ); document.write( "\"5%2F4\"(10) + \"10%2F12\"(10) + \"2%2F3\"(10) = \"10%2F12\"B
\n" ); document.write( "\"50%2F4\" + \"100%2F12\" + \"20%2F3\" = \"10%2F12\"B
\n" ); document.write( "multiply by 12
\n" ); document.write( "3(50) + 100 + 4(20) = 10B
\n" ); document.write( "150 + 100 + 80
\n" ); document.write( "330 = 10B
\n" ); document.write( "B = 330/10
\n" ); document.write( "B = 33 days B can finish the work alone?
\n" ); document.write( "
\n" );