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May 9th, 2010, 12:49 PM  #1 
Newbie Joined: Mar 2010 Posts: 8 Thanks: 0  word problem
I hope that you help me to solve this problem: A girl leaves home on her bicycle and travels at 5 mi/h.4 hours later her father leaves home and travels the same direction in his car at 38 mi/h.How many hours doess it take the father to cotch up to to the daugther? 
May 9th, 2010, 02:50 PM  #2 
Newbie Joined: May 2010 Posts: 1 Thanks: 0  Re: word problem
Around 3637 minutes, probably. It's not an exact time but more or less an approximation. I thought It would simply be 5 x 4 =20 miles for the father to catch up, which is approximated to 31.6 minutes. However, I realized that by then, the daughter would be ahead by that many miles. Therefore, I just repeated the numbers several times until it got too small and ended up with 3637 . Crappy way to solve, but that's what I got .

May 9th, 2010, 04:09 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: word problem Hello, alfarabi! There is a backdoor approach to these "catchup" problems. Quote:
The girl has a 4hour headstart at 5 mph. [color=beige]. . [/color]She is already 20 miles ahead. The father follows her at 38 mph. The difference in speeds is:[color=beige] .[/color] mph. Mindstretching thought: It is as if the girl has stopped 20 miles away [color=beige]. . [/color]and her father is approaching her at 33 mph. How long does it take him to cover 20 miles at 33 mph? [color=beige]. . [/color]  
May 11th, 2010, 10:44 AM  #4 
Newbie Joined: Mar 2010 Posts: 8 Thanks: 0  Re: word problem
Thank you all. Can we translating this problem into an equation as 38(t4)=5t? I think that is'nt correct. 
May 13th, 2010, 10:11 AM  #5  
Senior Member Joined: Apr 2010 Posts: 215 Thanks: 0  Re: word problem Quote:
 

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