document.write( "Question 852091: When will the two hands of a clock between 10 O'clock and 11 O'clock come in the same line? \n" ); document.write( "
Algebra.Com's Answer #513136 by ankor@dixie-net.com(22740)\"\" \"About 
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When will the two hands of a clock between 10 O'clock and 11 O'clock come in the same line?
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\n" ); document.write( "Let m = position of the min had based on a 60 min circle
\n" ); document.write( "The hr hand will be between the 50 and 55th min also, determined by the fraction m/60 times 5 minutes.
\n" ); document.write( "m = \"m%2F60\"5 + 50
\n" ); document.write( "m = \"m%2F12\" + 50
\n" ); document.write( "mult equation by 12 to get rid of the fraction
\n" ); document.write( "12m = m + 600
\n" ); document.write( "11m = 600
\n" ); document.write( "m = 600/11
\n" ); document.write( "m = 54.545 min which is 54 + .545(60) = 54 min 38 seconds
\n" ); document.write( "At 10:54:38 sec the hands will be in line
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\n" ); document.write( "Well that is different, see if we can find that.
\n" ); document.write( "There will be exactly 30 min between the hands
\n" ); document.write( "m + 30 = \"m%2F60\"5 + 50
\n" ); document.write( "subtract 30 from both sides
\n" ); document.write( "m = \"m%2F12\" + 20
\n" ); document.write( "mult equation by 12 to get rid of the fraction
\n" ); document.write( "12m = m + 240
\n" ); document.write( "11m = 240
\n" ); document.write( "m = 240/11
\n" ); document.write( "m = 21.82 min which is 21 + .82(60) = 21 min 49 seconds
\n" ); document.write( "At 10:21:49 sec the hands will be in line\r
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