document.write( "Question 1082703: three electricians can fix 10 short circuits in 2 hours. how much time will four electricians need to fix 15 short circuits?
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Algebra.Com's Answer #696764 by ikleyn(52781)![]() ![]() You can put this solution on YOUR website! . \n" ); document.write( "3 electricians at 2 hours each is 6 man-hours \n" ); document.write( "10 shorts are fixed in 6*60 = 360 man-minutes \n" ); document.write( "360/10 = 36 man-minutes per short \n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \n" ); document.write( "For 4 to fix 15: \n" ); document.write( "15*36 = 540 man-minutes \n" ); document.write( "540/4 = 135 minutes, or 2 hours and 15 minutes.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "It is typical joint work problem. See the lesson\r \n" ); document.write( "\n" ); document.write( " - Rate of work problems \r \n" ); document.write( "\n" ); document.write( "in this site.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Also, you have this free of charge online textbook in ALGEBRA-I in this site\r \n" ); document.write( "\n" ); document.write( " - ALGEBRA-I - YOUR ONLINE TEXTBOOK.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The referred lesson is the part of this textbook under the topic \"Rate of work and joint work problems\" of the section \"Word problems\".\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |