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) You can put this solution on YOUR website! When will the two hands of a clock between 10 O'clock and 11 O'clock come in the same line? \n" ); document.write( ": \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 = \n" ); document.write( "m = \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 \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "////////////////////////////////////////////////////////////////////////////// \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 = \n" ); document.write( "subtract 30 from both sides \n" ); document.write( "m = \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 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |