document.write( "Question 624670: Wilbur left the hardware store and traveled toward the recycling plant at an average speed of 33km/h. Mary left 2 hours later and traveled in the same direction but with an avg. spd. of 55km/h. How long did Wilbuer travel before Mary caught up?\r
\n" );
document.write( "\n" );
document.write( "When I solved this, I got the time that Wilbur traveled before Mary caught up to him. But I am not sure if the phrase \"How long did Wilbuer travel before Mary caught up?\" is asking for the distance or the time. Any answers?
\n" );
document.write( " \n" );
document.write( "
Algebra.Com's Answer #393024 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Wilbur left the hardware store and traveled toward the recycling plant at an average speed of 33km/h. \n" ); document.write( " Mary left 2 hours later and traveled in the same direction but with an avg. speed of 55km/h. \n" ); document.write( " How long did Wilbur travel before Mary caught up? \n" ); document.write( ": \n" ); document.write( "Let t = W's travel time \n" ); document.write( "then \n" ); document.write( "(t-2) = M's travel time \n" ); document.write( ": \n" ); document.write( "When M catches W they will have traveled the same distance from the store. \n" ); document.write( "Write a distance equation \n" ); document.write( ": \n" ); document.write( "M's dist = W's dist \n" ); document.write( "55(t-2) = 33t \n" ); document.write( "55t - 110 = 33t \n" ); document.write( "55t - 33t = 110 \n" ); document.write( "22t = 110 \n" ); document.write( "t = 110/22 \n" ); document.write( "t = 5 hrs W's travel time when M caught up; which is what they are asking for \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "We can check this by finding the distance each traveled, they should be the same \n" ); document.write( "55(5-2) = 165 km \n" ); document.write( "33(5) = 165 km \n" ); document.write( " |