SOLUTION: Carlos leaves LA on a cross country car trip at 8 AM, he is going 50 mph; Juanita plans to do the same but leaves at 9 AM and travels 60 mph, develop a diagram or table to determin

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Carlos leaves LA on a cross country car trip at 8 AM, he is going 50 mph; Juanita plans to do the same but leaves at 9 AM and travels 60 mph, develop a diagram or table to determin      Log On

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Question 173469: Carlos leaves LA on a cross country car trip at 8 AM, he is going 50 mph; Juanita plans to do the same but leaves at 9 AM and travels 60 mph, develop a diagram or table to determine at what time Juanita will pass Carlos
Found 2 solutions by checkley77, josmiceli:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
D=RT
50x=60(x-1)
50x=60x-60
50x-60x=-60
-10x=-60
x=-60/-10
x=6 hours after Carlos leaves Juanita will pass him. (@ 3:00 PM.)

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
For both of them, d+=+r%2At, where
d= distance
r= rate
t= time
So, for Carlos,
d%5Bc%5D+=+r%5Bc%5D%2At%5Bc%5D
And for Juanita,
d%5Bj%5D+=+r%5Bj%5D%2At%5Bj%5D
----------------------
The chart would look like this:
-------------------------------
Distances -- rates -- times
-------------------------------
-- d[c] -- -- r[c] -- -- t[c]
-- d[j] -- -- r[j] -- -- t[j]
--------------------------------
Now I can repace some of these variables
with things that I know about them
For instance, d%5Bc%5D+=+d%5Bj%5D because
the distance is from LA to the place
where they meet. I'll call them both d
-------------------------------
Distances -- rates -- times
-------------------------------
-- d -- -- r[c] -- -- t[c]
-- d -- -- r[j] -- -- t[j]
--------------------------------
Carlos travels 50 mi/hr and Juanita travels 60 mi/hr, so
-------------------------------
Distances -- rates -- times
-------------------------------
-- d -- -- 50 -- -- t[c]
-- d -- -- 60 -- -- t[j]
--------------------------------
Juanita leaves LA an hour later, so she has to cover the same
distance in 1 hour less time, so
-------------------------------
Distances -- rates -- times
-------------------------------
-- d -- -- 50 -- --t[c]
-- d -- -- 60 -- --t[c] - 1
--------------------------------
Now I can use this to find out when Juanita will pass Carlos.
The problem doesn't ask for it, but I'll do the algebra
d+=+r%2At for both of them, so
(1) d+=+50t%5Bc%5D
d+=+60%2A%28t%5Bc%5D+-+1%29
(2) d+=+60t%5Bc%5D+-+60
The ds are the same in (1) and (2),
so I'll set them equal to eachother
(3) 50t%5Bc%5D+=+60t%5Bc%5D+-+60
10t%5Bc%5D+=+60
t%5Bc%5D+=+6
And, since t%5Bj%5D+=+t%5Bc%5D+-+1,
t%5Bj%5D+=+6+-+1
t%5Bj%5D+=+5
Carlos left LA at 8AM, so 6 hours later,
it would have been 2PM
Juanita left LA at 9AM, so 5 hours later,
It would have been 2PM, which is when she passed him
check answer:
The distances must be the same, so
(1) d+=+50t%5Bc%5D
d+=+50%2A6
d+=+300 mi
(2) d+=+60t%5Bc%5D+-+60
d+=+60%2A6+-+60
d+=+360+-+60
d+=+300 mi
OK