document.write( "Question 1047312: Tom and Jerry run 6000m from A to B and then back to A. Tom runs the 12 000 m in 50 minutes, wile Jerry takes 60 minutes. Each runs at a constant rate. If x is the time they've each been running for when they meet each other, find x in minutes.\r
\n" ); document.write( "\n" ); document.write( "I'm not sure how to do this, I found out that Tom is 14 4000m/60 mins and Jerry is 12,000m/60 mins and Tom is running from B to A after 25 mins while Jerry has only run 5000m but then I'm not sure what to do. Help! \n" ); document.write( "

Algebra.Com's Answer #662841 by ikleyn(52803) $\"\"$ $\"About$

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\n" ); document.write( "Tom and Jerry run 6000 m from A to B and then back to A. Tom runs the 12 000 m in 50 minutes, wile Jerry takes 60 minutes.
\n" ); document.write( "Each runs at a constant rate. If x is the time they've each been running for when they meet each other, find x in minutes.
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\n" ); document.write( "\n" ); document.write( "Answer. x = 27.(27) minutes.\r
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\n" ); document.write( "\n" ); document.write( "Solution\r
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document.write( "Tom's   rate is  = 240 m/min.\r\n" );
document.write( "Jerry's rate is  = 200 m/min\r\n" );
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document.write( "Tom is running faster, so he will make his 6000 m and will be in his returning way when he will meet Jerry.\r\n" );
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document.write( "So, the equations are \r\n" );
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document.write( "6000 + x = 240t   (1)   for Tom, and\r\n" );
document.write( "6000 - x = 200t   (2)   for Jerry.\r\n" );
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document.write( "\"x\" is \"additional distance for Tom, t is the time before they meet each other.\r\n" );
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document.write( "Add equations (1) and (2). You will get\r\n" );
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document.write( "12000 = 240t + 200t,  or 12000 = 440t,  hence,  t =  = 27.(27) minutes before they meet each other.\r\n" );
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