document.write( "Question 999958: Jon and Steve had to paint Steve's Basement. John can paint the basement in 8 hours and Steve can paint it in 14 hours. The two agree that they can stop for lunch when 3/4 of the basement is painted. If Steve starts at 7:00 AM and John starts an hour after at 8:00 AM, what time can they stop for lunch at? \n" ); document.write( "
Algebra.Com's Answer #617498 by fractalier(6550)\"\" \"About 
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The setup for this is normally
\n" ); document.write( "t/A + t/B = 1
\n" ); document.write( "Here we are looking for t, A=8, B =14, and it isn't 1 job but 3/4, so our setup is
\n" ); document.write( "t/8 + (t+1)/14 = 3/4
\n" ); document.write( "Multiply by the LCD (which is 56 here)
\n" ); document.write( "56[t/8 + (t+1)/14 = 3/4]
\n" ); document.write( "7t + 4(t+1) = 42
\n" ); document.write( "7t + 4t + 4 = 42
\n" ); document.write( "11t + 4 = 42
\n" ); document.write( "11t = 38
\n" ); document.write( "t = 38/11 hours (that's an ugly number)
\n" ); document.write( "t = 3 5/11 hours or 3 hours 27.3 minutes
\n" ); document.write( "so that lunch break starts at
\n" ); document.write( "11:27.3 am
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