document.write( "Question 388893: Ahmad and Ushna start off simultaneously from two towns to meet one another. If Ahmad travels 2 km/h faster than Ushna; they would meet in 3 hours. If Ushna travels 1 km/h slower and Ahmad's speed is two-thirds of his previous speed, they would meet in 4 hours. How far apart are the two towns? \n" ); document.write( "
Algebra.Com's Answer #275300 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Ahmad travels 2 km/h faster than Ushna; they would meet in 3 hours. If Ushna travels 1 km/h slower and Ahmad's speed is two-thirds of his previous speed, they would meet in 4 hours. How far apart are the two towns? \n" ); document.write( "... \n" ); document.write( "Ushna --- speed = x km/h \n" ); document.write( "Ahmad x+2 km/h \n" ); document.write( ".. \n" ); document.write( "combined speed = x+x+2 = 2x+2 km/h \n" ); document.write( ".. \n" ); document.write( "D= speed * time \n" ); document.write( "D= 3*(2x+2) \n" ); document.write( "... \n" ); document.write( "Ushna (x-1) \n" ); document.write( "Ahmad 2/3(x+2) \n" ); document.write( "combined speed = 2/3(x+2)+(x-1) \n" ); document.write( "combined speed = 2x/3 + 4/3 +x-1 \n" ); document.write( "combined speed = 5x/3 +1/3 \n" ); document.write( "... \n" ); document.write( "D= 4*(5x/3+1/3) \n" ); document.write( "... \n" ); document.write( "4*(5x/3+1/3)=3(2x+2) \n" ); document.write( "20x/3+4/3 = 6x+6 \n" ); document.write( "20x/3-6x=6-4/3 \n" ); document.write( "(20x-18x)/3=14/3 \n" ); document.write( "2x=14 \n" ); document.write( "2x=14 \n" ); document.write( "x= 7 km/h Ushna's speed. \n" ); document.write( "... \n" ); document.write( "D= 3(2x+2) \n" ); document.write( "D=3(14+2) \n" ); document.write( "D= 48 km. Distance between two towns \n" ); document.write( "... \n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " |