document.write( "Question 515481: BOATING TRIP: Cesar starts out on a boating trip at 8 am. Cesar's boat can go 20 miles per hour in still water. How far downsream can Cesar go if the current is 5 miles per hour and he wishes to go down and back in 4 hours? \n" ); document.write( "
Algebra.Com's Answer #344000 by oberobic(2304) ![]() You can put this solution on YOUR website! d = r*t is the basic distance equation. \n" ); document.write( ". \n" ); document.write( "r = 20 mph in still water \n" ); document.write( ". \n" ); document.write( "c = 5 mph is the current in the river \n" ); document.write( ". \n" ); document.write( "going upstream, the speed = r-c = 20-5 = 15 \n" ); document.write( ". \n" ); document.write( "going downstream, the speed = r+c = 25 \n" ); document.write( ". \n" ); document.write( "d = d each way (that is, he returns to the same point he started). \n" ); document.write( "The roundtrip = 2d. \n" ); document.write( ". \n" ); document.write( "d/r = t \n" ); document.write( ". \n" ); document.write( "he wants to make the whole trip in 4 hr \n" ); document.write( ". \n" ); document.write( "d/15 + d/25 = 4 hr \n" ); document.write( ". \n" ); document.write( "(25d + 15d) / (15*25) = 4 \n" ); document.write( ". \n" ); document.write( "cross multiply \n" ); document.write( ". \n" ); document.write( "40d = 4*15*25 \n" ); document.write( ". \n" ); document.write( "d = 4*15*25/40 \n" ); document.write( ". \n" ); document.write( "d = 75/2 = 37 1/2 miles \n" ); document.write( ". \n" ); document.write( "Always check your answer. \n" ); document.write( ". \n" ); document.write( "In this case, what can we check? \n" ); document.write( "Time. \n" ); document.write( ". \n" ); document.write( "37.5 miles/ 15 mph = 2.5 hr going upstream (against the current) \n" ); document.write( "37.5 miles/ 25 mph = 1.5 hr going downstream (with the current) \n" ); document.write( "2.5+1.5= 4 hr \n" ); document.write( "Correct. \n" ); document.write( ". \n" ); document.write( "Done. \n" ); document.write( " |