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# Algebraic word problem (d=rt)

Fri 25 Sep, 2009 05:20 pm
Yissania (not even a real name!) and Charlotte walked on treadmills at the gym for the same amount of time. Yissania walkes at 5 miles per hour, and Charlotte walks at 4.2 miles per hour. If Yissania walked o.6 miles farther than Charlotte, how long did they use the treadmills?

I've been trying to figure this out for the better part of an hour with no help from the book or the internet. But I'm fairly positive it's a d=rt problem. I know the answer but I can't get there.
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Type: Question • Score: 0 • Views: 3,146 • Replies: 3

Brandon9000

3
Fri 25 Sep, 2009 05:44 pm
@brokencdplayer,
brokencdplayer wrote:

Yissania (not even a real name!) and Charlotte walked on treadmills at the gym for the same amount of time. Yissania walkes at 5 miles per hour, and Charlotte walks at 4.2 miles per hour. If Yissania walked o.6 miles farther than Charlotte, how long did they use the treadmills?

I've been trying to figure this out for the better part of an hour with no help from the book or the internet. But I'm fairly positive it's a d=rt problem. I know the answer but I can't get there.

5t = 4.2t + .6
(5 - 4.2)t = .6
.8t = .6
t = .6/.8 = 3/4 hour
brokencdplayer

1
Fri 25 Sep, 2009 05:51 pm
@Brandon9000,
wow thank you. you make it look so easy! was there a specific equation that you used? it didn't look like the usual d=rt one.
engineer

1
Fri 25 Sep, 2009 06:24 pm
@brokencdplayer,
Each side uses the d=rt.

You know that
d1 = d2 + .6

d1 = 5t
d2 = 4.2t
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