document.write( "Question 65980: Martina leaves home at 9 A.M., bicycling at a rate of
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document.write( "24 mi/h. Two hours later, John leaves, driving at the rate of 48 mi/h. At what time will John catch up with Martina?
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Algebra.Com's Answer #46728 by uma(370)![]() ![]() ![]() You can put this solution on YOUR website! Speed of Martina = 24 mph \n" ); document.write( "So distance covered by her in 2 hrs = 24*2 \n" ); document.write( " = 48 miles \n" ); document.write( "==> at 11 A.M., Martina is 48 miles away from John.\r \n" ); document.write( "\n" ); document.write( "Speed of John = 48 mph\r \n" ); document.write( "\n" ); document.write( "Let us say the they both meet after x hrs after John's departure.\r \n" ); document.write( "\n" ); document.write( "Distance travelled by Martina in x hrs = 24x + 48\r \n" ); document.write( "\n" ); document.write( "Distance travelled by John in x hrs = 48x\r \n" ); document.write( "\n" ); document.write( "As they meet , they would have travelled the same distance.\r \n" ); document.write( "\n" ); document.write( "==> 48x = 24x + 48 \n" ); document.write( "==> 48x - 24x = 48 \n" ); document.write( "==> 24x = 48 \n" ); document.write( "==> 24x/24 = 48/24 \n" ); document.write( "==> x = 2\r \n" ); document.write( "\n" ); document.write( "==> they meet after 2hrs after John's departure. \n" ); document.write( "==> they meet at 11 A.M., + 2hrs \n" ); document.write( " at 1 P.M\r \n" ); document.write( "\n" ); document.write( "Thus John catches up with Martina at 1 P.M\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Good Luck!!! \n" ); document.write( " |