document.write( "Question 1028963: A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast as the Snail, and they met at 3:40 p.m., find the speed of each. \n" ); document.write( "
Algebra.Com's Answer #644026 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
You can think of this as one of them standing still
\n" ); document.write( "and the other moving at the sum of their speeds
\n" ); document.write( "3:40 minus 3:00 is 40 min
\n" ); document.write( "Convert to hrs:
\n" ); document.write( "\"+%2840%2F60%29+=+2%2F3+\" hrs
\n" ); document.write( "Let \"+s+\" = the speed of the snail
\n" ); document.write( "\"+11s+\" = the speed of the turtle
\n" ); document.write( "------------------------------
\n" ); document.write( "\"+360+=+%28+s+%2B+11s+%29%2A%282%2F3%29+\"
\n" ); document.write( "\"+1080+=+24s+\"
\n" ); document.write( "\"+s+=+45+\"
\n" ); document.write( "and
\n" ); document.write( "\"+11s+=+11%2A45+\"
\n" ); document.write( "\"+11s+=+495+\"
\n" ); document.write( "----------------
\n" ); document.write( "The speed of the snail is 45 m/hr
\n" ); document.write( "The speed of the turtle is 495 m/hr
\n" ); document.write( "-----------------
\n" ); document.write( "check answer:
\n" ); document.write( "Equation for the snail:
\n" ); document.write( "\"+d%5B1%5D+=+45%2A%282%2F3%29+\"
\n" ); document.write( "\"+d%5B1%5D+=+30+\"
\n" ); document.write( "Equation for the turtle:
\n" ); document.write( "\"+d%5B2%5D+=+495%2A%282%2F3%29+\"
\n" ); document.write( "\"+d%5B2%5D+=+330+\"
\n" ); document.write( "----------------
\n" ); document.write( "\"+d%5B1%5D+%2B+d%5B2%5D+=+30+%2B+330+\"
\n" ); document.write( "\"++d%5B1%5D+%2B+d%5B2%5D+=+360+\" m
\n" ); document.write( "OK
\n" ); document.write( "
\n" );