document.write( "Question 267422: Here is a confusing question for me: The ratio of Aldo's cycling speed to Jose's cycling speed is 6:5. Jose leaves school at 3 P.M. and Aldo leaves at 3:10 P.M. By 3:30, Aldo is only 2 km behind Jose. How fast is each cycling? \n" ); document.write( "
Algebra.Com's Answer #196322 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! The ratio of Aldo's cycling speed to Jose's cycling speed is 6:5. \n" ); document.write( "Let m = the multiplier \n" ); document.write( "then \n" ); document.write( "6m = A's speed \n" ); document.write( "and \n" ); document.write( "5m = J's speed \n" ); document.write( ": \n" ); document.write( " Jose leaves school at 3 P.M. and Aldo leaves at 3:10 P.M. By 3:30, \n" ); document.write( "J's travel time 30 min; \n" ); document.write( "A's travel time 20 min; \n" ); document.write( ": \n" ); document.write( "By 3:30, Aldo is only 2 km behind Jose. How fast is each cycling? \n" ); document.write( "; \n" ); document.write( "Write a distance equation. Dist = time * speed \n" ); document.write( "J's dist - A's dist = 2 km \n" ); document.write( " \n" ); document.write( "Multiply by 6 to get rid of the denominators, results \n" ); document.write( "3(5m) - 2(6m) = 6(2) \n" ); document.write( "15m - 12m = 12 \n" ); document.write( "3m = 12 \n" ); document.write( "m = 4 is the multiplier \n" ); document.write( "then \n" ); document.write( "4(6) = 24 km/hr is A's speed \n" ); document.write( "and \n" ); document.write( "4(5) = 20 km/hr is J's speed \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution by finding the dist traveled by each \n" ); document.write( "J: .5(20) = 10 km \n" ); document.write( "A: .33(24)= 8 km \n" ); document.write( "--------------------- \n" ); document.write( "difference = 2 km \n" ); document.write( " |