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Combined rate problem

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Combined rate problem

by blackjack » Thu Jul 07, 2011 7:05 am
One hour after Andrew started walking the 60 miles from X to Y, James started walking from X to y as well. Andrew walks 3 mph and James walks 1 mph faster than Andrew. How far from X will James be when he catches up with Andrew.
A. 8 miles
B. 9
C. 10
D. 11
E. 12 miles

For some reason, I am lost. Help!
OA is E
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by Frankenstein » Thu Jul 07, 2011 7:13 am
Hi,
By the time James starts, Andrew has walked for 1 hour. So, distance he traveled is 3*1 = 3miles
Given that James covers a mile more than Andrew does, for every hour
So, the gap of 3 miles will be covered in 3hours by James.
So, in this three hours, James walking at (3+1) = 4 mph covers 4*3 = 12 miles from X.

Hence, E
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by blackjack » Thu Jul 07, 2011 7:25 am
How about the fol strategy?
Dist traveled by Andrew in say t time is 3(t + 1)
Dist traveled by James in t time is 4t

Since dist traveled is same,
3(t + 1) = 4t
3t + 3 = 4t
t = 3
hence dist traveled by james is 12
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by winniethepooh » Thu Jul 07, 2011 8:21 am
James = 3+1 = 4 miles per hr
Andrew = 3 miles per hr

AT the
end of: 1 hr , 2 hr , 3 hr , 4 hr
James =0 miles , 4 miles, 8 miles, 12 miles
Andrew =3 miles , 6 miles, 9 miles, 12 miles
Hence, at 12 miles.E
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by blackjack » Thu Jul 07, 2011 8:54 am
@winniethepooh
great simplistic approach!
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by tryreallyhard » Thu Jul 07, 2011 7:18 pm
I think there are many variations to the rate problems and you sometimes could solve the problem if you happen to see the trick right through it. IMO, I want a consistent and structured method to solve a problem. And for a rate problem like this one, I always build a "rate" table. It's a method I learn from https://www.oojih.com/show/wordproblem/distancerate/

Thus far, this approach has not failed me yet:

Andrew:
Rate = 3mph, time = t+1, Distance= 3(t+1)

James:
Rate = 4 mph, time=t, Distance = 4t

When they meet, the distance is equal, so 3t+ 3 = 4t and you get t =3
So, they meet after Andrew has walked for 4 hours. So, they meet at 12 miles from X.
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by amit2k9 » Thu Jul 07, 2011 11:13 pm
3miles (initial separation) + 3t = 4t

thus t=3

hence distance covered by A = 3+3*3=12.
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