document.write( "Question 1110002: At 09 00 car A starts its journey and traveling at 70 km/h at 10 30.Car B started from the same place and traveled steadily on the same road. Car B took 3 1/2 h to catch up with car A. \r
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Algebra.Com's Answer #725005 by ikleyn(52777)\"\" \"About 
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\n" ); document.write( "(1) At 09:00 car A started its journey and traveled at 70 km/h.\r
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document.write( "Catching up moment was at 10:30 + 3 h 30 min = 14:00.\r\n" );
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document.write( "From 9:00 till 14:00 the car A moved 5 hours and covered  5*70 = 350 kilometers.  \r\n" );
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document.write( "From 10:30 till 14:00 car B moved 3.5 hours and covered the same distance.  So, its rate was  \"350%2F3.5\" = 100 km/h.\r\n" );
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document.write( "After catching up moment, car B was on the way from 14:00 till 15:24, i.e 1 hour and 24 minutes = \"1\"\"24%2F60\" hours = 1.4 hours.\r\n" );
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document.write( "and covered 100*1.4 = 140 kilometers.\r\n" );
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document.write( "The car A took \"140%2F70\" = 2 hour to cover the same 140 km.  So, car A arrived to the destination point at 14:00 + 2 hours = 16:00.\r\n" );
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