document.write( "Question 47643: 4. Two cars started at the same time from towns 50 km apart and travelled toward each
\n" ); document.write( "other at uniform speed. They meet in 20 minutes and immediately started off again,
\n" ); document.write( "but this time, in the same direction, travelling at the same rates as before. Soon after,
\n" ); document.write( "the slower car stopped for 30 minutes and then started as usual. At what rate did each
\n" ); document.write( "car travel if, two hours after they started for the third town, they were 55 km apart?
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Algebra.Com's Answer #31475 by venugopalramana(3286)\"\" \"About 
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Two cars started at the same time from towns 50 km apart and travelled toward each
\n" ); document.write( "other at uniform speed. They meet in 20 minutes and immediately started off again,
\n" ); document.write( "but this time, in the same direction, travelling at the same rates as before. Soon after,
\n" ); document.write( "the slower car stopped for 30 minutes and then started as usual. At what rate did each
\n" ); document.write( "car travel if, two hours after they started for the third town, they were 55 km apart?
\n" ); document.write( "LET THE SPEED OF FASTER CAR =F KMPH
\n" ); document.write( "AND SPEED OF SLOWEWR CAR =S KMPH
\n" ); document.write( "I LEG
\n" ); document.write( "DISTANCE OF SEPERATION AT START= 50 KM
\n" ); document.write( "RELATIVE VELOCITY=F+S
\n" ); document.write( "TIME REQUIRED TO MEET =50/(F+S) HRS=20/60=1/3 HR
\n" ); document.write( "F+S=150............I
\n" ); document.write( "II LEG
\n" ); document.write( "DISTANCE OF SEPERATION AT START =0
\n" ); document.write( "TIME OF TRAVEL FOR SLOWER CAR = 2-30/60=2-0.5=1.5 HRS
\n" ); document.write( "DISTANCE TRAVELLED BY SLOWEWR CAR =1.5S KM
\n" ); document.write( "TIME OF TRAVEL FOR FASTER CAR = 2 HRS
\n" ); document.write( "DISTANCE OF TRAVEL BY FASTER CAR =2F KM
\n" ); document.write( "DISTANCE OF SEPERATION AT THE END =2F-1.5S=55...................II
\n" ); document.write( "EQN.I*2-EQN.II
\n" ); document.write( "2F+2S-2F+1.5S=2*150-55=300-55=245
\n" ); document.write( "3.5S=245
\n" ); document.write( "S=245/3.5=70 KMPH
\n" ); document.write( "F=150-70=80 KMPH
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