document.write( "Question 151277: Mr Tan et off from Town A to Town B at an average speed of 75km/h at 08 00. Mr Chen set off from Town A after some time and overtook Mr Tan at 10 00. When Mr Chen reached Town B at 12 00, Mr Tan was still 30km away.
\n" ); document.write( "(a)What was Mr Chen's average speed?
\n" ); document.write( "(b)What time did Mr Chen leave Town A?\r
\n" ); document.write( "\n" ); document.write( "pls. try not to use algebra but if there is no choice, then, algebra would be fine.
\n" ); document.write( "

Algebra.Com's Answer #111170 by vleith(2983)\"\" \"About 
You can put this solution on YOUR website!
Not sure how one would solve this without using algebra:) But will try to get it simple.\r
\n" ); document.write( "\n" ); document.write( "Tan left at 8 going 75. Chen caught up to him at 10. So when Chen passes Tan, Tan has been driving for 2 hours. He has gone 75*2 = 150km.\r
\n" ); document.write( "\n" ); document.write( "Chen reaches B in another 2 hours. But Tan is still 30km from town. Tan has driven another 2 hours since being passed by Chen, so Tan has driven another 150km. And he is till 30km from B. So the total distance between A and B is 150+150+30 = 330km.\r
\n" ); document.write( "\n" ); document.write( "1) The distance from when Chen passed Tan to B is 150+30 = 180.
\n" ); document.write( "Chen took 2 hours to cover 180km, so Chen's average speed is 180/2 = 90km.\r
\n" ); document.write( "\n" ); document.write( "2) The distance from A to where Chen catches Tan is 150km. Chen's average speed is 90. 150/90 = 5/3 hours. Chen left 20 minutes after Tan. So Chen left at 8:20\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );