document.write( "Question 984471: Workman A's rate of doing work is twice that of Workman B's. One day, A & B work together for 8 hrs, then B due to injury stops and A finishes the rest of the job in 2 hrs. How long would it take B to complete the job alone? \n" ); document.write( "
Algebra.Com's Answer #605345 by ankor@dixie-net.com(22740)\"\" \"About 
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Workman A's rate of doing work is twice that of Workman B's.
\n" ); document.write( " One day, A & B work together for 8 hrs, then B due to injury stops and A finishes the rest of the job in 2 hrs.
\n" ); document.write( " How long would it take B to complete the job alone?
\n" ); document.write( ":
\n" ); document.write( "let a = time required by A to do the job
\n" ); document.write( "He works twice a fast as B, therefore
\n" ); document.write( "let 2a = time required by B to do the same job
\n" ); document.write( "let the completed job = 1
\n" ); document.write( ":
\n" ); document.write( "A works a total of 10 hrs, B works 8 hrs
\n" ); document.write( "The shared work equation
\n" ); document.write( "\"10%2Fa\" + \"8%2F%282a%29\" = 1
\n" ); document.write( "Multiply equation by 2a, cancel the denominators and you have:
\n" ); document.write( "2(10) + 8 = 2a
\n" ); document.write( "20 + 8 = 2a
\n" ); document.write( "28 = 2a
\n" ); document.write( " a = 14 for A to do the job alone
\n" ); document.write( "then
\n" ); document.write( "2(14) = 28 hrs for B to do the job alone
\n" ); document.write( ":
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\n" ); document.write( "See if that works out using decimals
\n" ); document.write( "\"10%2F14\" + \"8%2F28\" = 1
\n" ); document.write( ".7143 + .2857 = 1
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