document.write( "Question 631327: A bicycle rider left at 6 a.m. city A to city B at the speed of 15 km an hour. At 8 a.m. a car left city A for city B at the speed of 75 km an hour.
\n" );
document.write( "The car arrived to B, stayed there 2 hours and started driving back to A at the same speed. After a 1/4 of the way back the car met the bicycle.
\n" );
document.write( "What is the distance between A to B?
\n" );
document.write( " \n" );
document.write( "
Algebra.Com's Answer #397687 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A bicycle rider left at 6 a.m. city A to city B at the speed of 15 km an hour. \n" ); document.write( " At 8 a.m. a car left city A for city B at the speed of 75 km an hour. \n" ); document.write( "The car arrived to B, stayed there 2 hours and started driving back to A at the \n" ); document.write( " same speed. \n" ); document.write( "After a 1/4 of the way back the car met the bicycle. \n" ); document.write( "What is the distance between A to B? \n" ); document.write( ": \n" ); document.write( "Let d = distance from A to B \n" ); document.write( "then when they met: \n" ); document.write( ".75d = distance traveled by the bike \n" ); document.write( "1.25d = distance traveled by the car \n" ); document.write( ": \n" ); document.write( "From the information given, we know the bike travel time was 4 hrs more than the car. \n" ); document.write( "(The car left two hrs after the bike and stayed at B for 2 hrs} \n" ); document.write( ": \n" ); document.write( "Write a time equation: time = dist/speed \n" ); document.write( ": \n" ); document.write( "Bike time - car time = 4 hrs \n" ); document.write( " \n" ); document.write( "multiply by 75 \n" ); document.write( "5(.75d) - 1.25d = 75(4) \n" ); document.write( "3.75d - 1.25d = 300 \n" ); document.write( "2.50d = 300 \n" ); document.write( "d = 300/2.50 \n" ); document.write( "d = 120 km from A to B \n" ); document.write( " \n" ); document.write( " |