document.write( "Question 1187740: It is now between 9 and 10 o’clock.
\n" ); document.write( "a.)At what time after 9’oclock will the minute hand and the hour hand be
\n" ); document.write( "perpendicular for the first time?
\n" ); document.write( "b.)In 4 minutes, the hour hand of the clock will be directly opposite the
\n" ); document.write( "position occupied by the minute hand 3 minutes ago. What time is it?
\n" ); document.write( "c.)In a quarter of an hour the minute hand will be behind the hour hand by only
\n" ); document.write( "half as much as is now behind it. What time is it? \n" ); document.write( "
Algebra.Com's Answer #819182 by Edwin McCravy(20059)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "There are 60 minute graduations around the 360 degree clock face, so the minute\r\n" );
document.write( "graduations are 360/60 = 6 degrees apart.  Therefore the minute hand's angular\r\n" );
document.write( "speed is 6 degrees per minute or 6 dpm.\r\n" );
document.write( "\r\n" );
document.write( "The minute hand rotates 12 times as fast as the hour hand, so the hour hand's\r\n" );
document.write( "angular speed is 1/12 as fast, or 6/12 = 1/2 = 0.5 dpm. \r\n" );
document.write( "\r\n" );
document.write( "So the angle between the hands is increasing at 6.0-0.5 = 5.5 dpm.  Every time\r\n" );
document.write( "the hands are exactly together they are 0 degrees or 360 degrees apart. \r\n" );
document.write( "\r\n" );
document.write( "Notice that we can always add (or subtract) 360 degrees to (or from) any angle\r\n" );
document.write( "between the hands or between the vertical and either hand without incurring any\r\n" );
document.write( "difficulty.\r\n" );
document.write( "\r\n" );
document.write( "We will also use the formulas: \r\n" );
document.write( "\r\n" );
document.write( "Degrees rotated = (degrees per minute)(number of minutes) \r\n" );
document.write( "\r\n" );
document.write( "and\r\n" );
document.write( "\r\n" );
document.write( "Number of minutes = (degrees rotated)/(degrees per minute)

\n" ); document.write( "It is now between 9 and 10 o’clock.
\r
\n" ); document.write( "\n" ); document.write( "a.)At what time after 9’oclock will the minute hand and the hour hand be
\n" ); document.write( "perpendicular for the first time?
\r\n" );
document.write( "At 9 o'clock, the angle between the hands is 90 degrees and the first time they\r\n" );
document.write( "will be perpendicular again is when the angle between them is 270 degrees.\r\n" );
document.write( "So the angle between them must increase by 270-90=180 degrees at the rate of\r\n" );
document.write( "5.5 dpm which will be 180/5.5=1800/55=360/11 = 32 8/11.  \r\n" );
document.write( "\r\n" );
document.write( "Answer:  The time will be 9:32 8/11.

\n" ); document.write( "b.)In 4 minutes, the hour hand of the clock will be directly opposite the
\n" ); document.write( "position occupied by the minute hand 3 minutes ago. What time is it?
\r\n" );
document.write( "Let that time (which we consider as 'now') be x minutes after 9 o'clock.   \r\n" );
document.write( "\r\n" );
document.write( "At 9 o'clock, the minute hand was straight up, or 0 or 360 degrees with the\r\n" );
document.write( "vertical.  So now, x minutes later, the minute hand, which has been rotating at\r\n" );
document.write( "6 dpm, is now at an angle of 6x degrees with the vertical, which, for\r\n" );
document.write( "convenience, we can consider as 6x+360 degrees with the vertical.  \r\n" );
document.write( "\r\n" );
document.write( "At 9 o'clock, the hour hand was \"directly left\" or 270 degrees with the\r\n" );
document.write( "vertical.  So now, x minutes later, the hour hand, which has been rotating at\r\n" );
document.write( "0.5 dpm, is now at an angle of 270+0.5x degrees with the vertical.\r\n" );
document.write( "\r\n" );
document.write( "So right now, x minutes after 9 o'clock, the minute hand makes an angle of\r\n" );
document.write( "6x+360 degrees with the vertical and the hour hand makes an angle of 270+0.5x\r\n" );
document.write( "degrees with the vertical. \r\n" );
document.write( "\r\n" );
document.write( "4 minutes from now, the hour hand, which will be rotating at 0.5 dpm, will have\r\n" );
document.write( "rotated through (0.5)(4)=2 more degrees, which will put it at the position of\r\n" );
document.write( "270+0.5x+3 or 273+0.5x degrees with the vertical, 4 minutes from now.  \r\n" );
document.write( "\r\n" );
document.write( "3 minutes ago, the minute hand, which has been rotating at 6 dpm, has now\r\n" );
document.write( "rotated through (6)(3)=18 more degrees since 3 minutes ago.  That means that 3\r\n" );
document.write( "minutes ago, it was at 6x+360-18 or 6x+342 degrees with the horizontal.\r\n" );
document.write( "\r\n" );
document.write( "The position directly opposite 6x+342 is found by subtracting 180 degrees from\r\n" );
document.write( "it, which is 6x+342-180, or 6x+162.\r\n" );
document.write( "\r\n" );
document.write( "Now the equation is\r\n" );
document.write( "\r\n" );
document.write( "273+0.5x = 6x+162\r\n" );
document.write( "     111 = 5.5x\r\n" );
document.write( " 111/5.5 = x\r\n" );
document.write( "\r\n" );
document.write( "111/5.5 = 1110/55 = 222/11 = 20 2/11 minutes after 9 o'clock.\r\n" );
document.write( "\r\n" );
document.write( "Answer:   The time is 9:20 2/11

\n" ); document.write( "c.)In a quarter of an hour the minute hand will be behind the hour hand by only
\n" ); document.write( "half as much as is now behind it. What time is it?
\r\n" );
document.write( "Let that time (which we consider as 'now') be x minutes after 9 o'clock.  \r\n" );
document.write( "\r\n" );
document.write( "As explained in part b), the minute hand now makes an angle of 6x, or 6x+360\r\n" );
document.write( "degrees with the vertical, and the hour hand now makes an angle of 270+0.5x\r\n" );
document.write( "degrees with the vertical.  \r\n" );
document.write( "\r\n" );
document.write( "So the number of degrees the minute hand is now behind the hour hand is their\r\n" );
document.write( "difference (270+0.5x)-(6x+360)=270+0.5x-6x-360=-5.5x-90 degrees and we can\r\n" );
document.write( "always add 360 degrees, so we can consider the number of degrees the minute hand\r\n" );
document.write( "is now behind the hour hand as -5.5x-90+360 or 270-5.5x.\r\n" );
document.write( "   \r\n" );
document.write( "In a quarter of an hour (15 minutes), the minute hand, rotating at 6 dpm, will\r\n" );
document.write( "have rotated through (6)(15)=90 degrees. So the position of the minute hand,\r\n" );
document.write( "rotating at 6 dpm, will then be 6x+360+90 or 6x+450 degrees.\r\n" );
document.write( "\r\n" );
document.write( "In a quarter of an hour (15 minutes), the hour hand, rotating at 0.5 dpm, will\r\n" );
document.write( "have rotated through (0.5)(15)=7.5 degrees. So the position of the hour hand,\r\n" );
document.write( "rotating at 6 dpm, will then be 270+0.5x+7.5 or 277.5+0.5x degrees with the\r\n" );
document.write( "vertical.\r\n" );
document.write( "\r\n" );
document.write( "The number of degrees the minute hand will then be behind the hour hand is\r\n" );
document.write( "their difference, or (277.5+0.5x)-(6x+450) or 277.5+0.5x-6x-450=-172.5-5.5x, and\r\n" );
document.write( "to keep that from being negative we can add 360 degrees, making it\r\n" );
document.write( "-172.5-5.5x+360 or 187.5-5.5x.\r\n" );
document.write( "\r\n" );
document.write( "So the equation is\r\n" );
document.write( "\r\n" );
document.write( "187.5-5.5x = (1/2)(270-5.5x)\r\n" );
document.write( "\r\n" );
document.write( "Multiply through by 2\r\n" );
document.write( "\r\n" );
document.write( "375-11x = 270-5.5x\r\n" );
document.write( "    105 = 5.5x\r\n" );
document.write( "105/5.5 = x\r\n" );
document.write( "1050/55 = x\r\n" );
document.write( " 210/11 = x\r\n" );
document.write( "19 1/11 = x\r\n" );
document.write( "\r\n" );
document.write( "Answer:   The time is 9:19 1/11 \r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );