document.write( "Question 1091689: What time after 4 o'clock will the minute hand be nine minutes behind the hour hand? \n" ); document.write( "

Algebra.Com's Answer #706139 by greenestamps(13200) $\"\"$ $\"About$

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I find it easiest to work problems involving the angle between the hands of a clock by measuring all angles relative to 12 o'clock.

\n" ); document.write( "The arithmetic required to solve this kind of problem is based on the fact that the hour hand moves 1/12 of a revolution, or 30 degrees, in 1 hour (60 minutes), so its rate of rotation is 0.5 degrees per minute; and the minute hand moves one full rotation (360 degrees) in an hour (60 minutes), so its rate of rotation is 6 degrees per minute.

\n" ); document.write( "You want to find the time after 4 o'clock when the minute hand is \"9 minutes\" behind the hour hand. 9 minutes is 3/20 of an hour; so you want the minute hand to be 3/20 of 360 degrees, or 54 degrees, behind the hour hand.

\n" ); document.write( "At x minutes after 4, the angle the hour hand makes with 12 o'clock, in degrees, is
\n" ); document.write( " $\"120%2B0.5x%29\"$ (120 degrees to get to 4; then 0.5 degrees for each minute after 2)

\n" ); document.write( "And x minutes after 4, the angle the minute hand makes with 12 o'clock, in degrees, is
\n" ); document.write( " $\"6x\"$

\n" ); document.write( "You want the minute hand to be \"9 minutes\", or 54 degrees, behind the hour hand, so you want
\n" ); document.write( " $\"%28120%2B0.5x%29-6x+=+54\"$
\n" ); document.write( " $\"-5.5x+=+-66\"$
\n" ); document.write( " $\"x+=+12\"$

\n" ); document.write( "The minute hand will be \"9 minutes\" behind the hour hand at 4:12. \n" ); document.write( "

\n" );