document.write( "Question 697172: I have been struggling with this for the last hour...\r
\n" ); document.write( "\n" ); document.write( "Three machines are twisting pretzels. Two are old and one is new. The new machine can twist pretzels twice as fast as an old machine can. An old machine can twist the daily quota of pretzels in 20 hours. How long will it take to make the quota if the three machines are operated at the same time?
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Algebra.Com's Answer #429724 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
You must add the rates of twisting for each
\n" ); document.write( "of the machines to get their rate working together.
\n" ); document.write( "Let \"+t+\" = the time to fill daily quota with all
\n" ); document.write( "three machines running
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\n" ); document.write( "The old machine's rate is ( 1 daily quota ) / ( 20 hrs ) = \"+1%2F20+\"
\n" ); document.write( "If the New machine's rate is twice the old one's, then it can do
\n" ); document.write( "( 1 daily quota ) / ( 10 hrs ) = \"+1%2F10+\"
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\n" ); document.write( "With all 3 machines running:
\n" ); document.write( "\"+1%2F20+%2B+1%2F20+%2B+1%2F10+=+1%2Ft+\"
\n" ); document.write( "\"+2%2F20+%2B+1%2F10+=+1%2Ft+\"
\n" ); document.write( "\"+2%2F20+%2B+2%2F20+=+1%2Ft+\"
\n" ); document.write( "\"+1%2Ft+=+4%2F20+\"
\n" ); document.write( "\"+1%2Ft+=+1%2F5+\"
\n" ); document.write( "\"+t+=+5+\"
\n" ); document.write( "It will take 5 hrs to fill daily quota with all
\n" ); document.write( "three machines running
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