document.write( "Question 107977: ok, here we go.\r
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document.write( "John and Harry can paint a house in 10 days. when working alone John can paint the house in 30 days. How many days will it take Harry to paint the house.\r
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document.write( "I have no idea how to do this probolem. \n" );
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Algebra.Com's Answer #78689 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! John and Harry can paint a house in 10 days. when working alone John can paint the house in 30 days. How many days will it take Harry to paint the house. \n" ); document.write( ": \n" ); document.write( "Here's a method you can use on all these kind of problems \n" ); document.write( ": \n" ); document.write( "Let the completed = 1, each will do a fraction that will add up to 1 \n" ); document.write( "It's obvious that John will complete 1/3 of the job in 10 days (10/30) \n" ); document.write( "Therefore John has to do 2/3 of the job. What denominator of 10 = 2/3? \n" ); document.write( ": \n" ); document.write( "Let x = Harry's time by himself \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "Get rid of the denominators, multiply equation by 30x \n" ); document.write( ": \n" ); document.write( "30x* \n" ); document.write( ": \n" ); document.write( "Cancel out the denominators and you have; \n" ); document.write( "10x + 30(10) = 30x \n" ); document.write( ": \n" ); document.write( "10x - 30x = -300 \n" ); document.write( ": \n" ); document.write( "-20x = - 300 \n" ); document.write( ": \n" ); document.write( "x = -300/-20 \n" ); document.write( ": \n" ); document.write( "x = 15 days required by John by himself \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check the solution by adding up the fractions \n" ); document.write( " \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "Did this make sense to you? Any questions? \n" ); document.write( " |