document.write( "Question 1064251: Machine A can do a certain job in 8 hours. machine B can do the same job in 10 hours. machine C can do the same job in 12 hours. all three machines start the job at 9.00 am, machine B breaks down at 11.00am and the other two machines finish the job. approximately at what time would the job be completed?
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Algebra.Com's Answer #679298 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! in one hour \n" ); document.write( "A does 1/8, B does 1/10, C does 1/12 \n" ); document.write( "x is the number of total hours worked. \n" ); document.write( "B worked 2 hours \n" ); document.write( "2/8+2/10+2/12=30/120+24/120+20/120=74/120 of the job done by all three. \n" ); document.write( "The other two account for 46/120; that comes from the fact that if 74/120 was done by all 3, the remaining 46/120 (74+46=120) were done by the two remaining machines. \n" ); document.write( "x/8+x/12=46/120 \n" ); document.write( "15x/120+10x/120=46/120 \n" ); document.write( "15x+10x=46 \n" ); document.write( "25x=46 \n" ); document.write( "x=1.84 hours or 12:50.24 ANSWER \n" ); document.write( "check \n" ); document.write( "A worked 3.84 hours, B 2 hours and C 3.84 hours \n" ); document.write( "3.84/8+2/10+3.84/12= \n" ); document.write( "common denominator=120 \n" ); document.write( "57.6+24+38.4 is the numerator, which adds to 120. \n" ); document.write( " |