Algebraic word problem (d=rt)

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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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[+3] - Brandon9000 - 09/25/2009[quote="brokencdplayer"] 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...

@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

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@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.

1 Reply

@brokencdplayer,

Each side uses the d=rt.

You know that
d1 = d2 + .6

d1 = 5t
d2 = 4.2t

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Algebraic word problem (d=rt)

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