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May 9th, 2010, 01:49 PM   #1
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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?
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May 9th, 2010, 03:50 PM   #2
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Re: word problem

Around 36-37 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 36-37 . Crappy way to solve, but that's what I got .
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May 9th, 2010, 05:09 PM   #3
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Re: word problem

Hello, alfarabi!

There is a back-door approach to these "catch-up" problems.


Quote:
A girl leaves home on her bicycle and travels at 5 mph.
Four hours later her father leaves home and travels the same direction in his car at 38 mph.
How many hours doess it take the father to catch up to to the daugther?

The girl has a 4-hour 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.

Mind-stretching 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]

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May 11th, 2010, 11:44 AM   #4
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Re: word problem

Thank you all.
Can we translating this problem into an equation as 38(t-4)=5t? I think that is'nt correct.
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May 13th, 2010, 11:11 AM   #5
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Re: word problem

Quote:
Originally Posted by alfarabi
Thank you all.
Can we translating this problem into an equation as 38(t-4)=5t? I think that is'nt correct.
Since you're looking for how long it took for her father, you instead want:
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