document.write( "Question 1118790: Town X and Town Y are 560 KM apart. At 11 A.M., a van leaves Town X for Town Y, Travelling at a constant rate of speed. At the same time, a bus leaves for Town Y from Town X, also travelling at a constant rate of speed. The 2 vehicle pass each other at 1.20 P.M. The average speed of the van is 60 km per hr faster than the bus.\r
\n" ); document.write( "\n" ); document.write( "a) Find the van's average rate of speed?\r
\n" ); document.write( "\n" ); document.write( "b) How far is the bus from Town X when it passes the van?
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Algebra.Com's Answer #734220 by josgarithmetic(39625)\"\" \"About 
You can put this solution on YOUR website!
11AM to 1:20PM is 2 hours 20 minutes.\r
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\n" ); document.write( "\n" ); document.write( "r, bus speed
\n" ); document.write( "r+60, van speed\r
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\n" ); document.write( "\n" ); document.write( "Distance used between the two was 560 kilometers.\r
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\n" ); document.write( "\n" ); document.write( "\"%28r%2Br%2B60%29%2A%282%261%2F3%29=560\"\r
\n" ); document.write( "\n" ); document.write( "Solve.\r
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\n" ); document.write( "\n" ); document.write( "-\r
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\n" ); document.write( "\n" ); document.write( "\"%282r%2B60%29%287%2F3%29=560\"
\n" ); document.write( "\"2r%2B60=%283%2F7%29%2A560\"
\n" ); document.write( "\"2r%2B60=3%2A80=240\"
\n" ); document.write( "\"2r=180\"
\n" ); document.write( "\"highlight%28r=90%29\"------bus's average speed
\n" ); document.write( "-
\n" ); document.write( "\"90%2B60=highlight%28150%29\"-----van's average speed
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\n" ); document.write( "\"90%2A%287%2F3%29=30%2A7=210\"--------how far the bus traveled from town Y.
\n" ); document.write( "\"560-210=highlight%28350%29\"---------how far was bus from town X after 2 hour 20 minutes
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