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One hour after Adrienne started walking the 60 miles from X

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One hour after Adrienne started walking the 60 miles from X [#permalink] New post 26 Dec 2010, 07:54
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One hour after Adrienne started walking the 60 miles from X to Y, James started walking from X to Y as well. Adrienne walks 3 miles per hour and James walks 1 mile per hour faster than Adrienne. How far from X will James be when he catches up to Adrienne?

A) 8 miles
B) 9 miles
C) 10 miles
D) 11 miles
E) 12 miles

Whats the best way to solve this problem? Is the distance from X to Y in the problem matter? Please give me detailed steps. Thank you.
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Oct 2013, 23:54, edited 1 time in total.
Renamed the topic and edited the question.
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Re: RTD problem [#permalink] New post 26 Dec 2010, 08:19
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pgmat wrote:
Whats the best way to solve this problem? Is the distance from X to Y in the problem matter? Please give me detailed steps. Thank you.

One hour after Adrienne started walking the 60 miles from X to Y, James started walking from X to Y as well. Adrienne walks 3 miles per hour and James walks 1 mile per hour faster than Adrienne. How far from X will James be when he catches up to Adrienne?

A) 8 miles
B) 9 miles
C) 10 miles
D) 11 miles
E) 12 miles


James walked for t hours at the rate of 4 miles per hour and Adrienne walked for t+1 hours at the rate of 3 miles per hour;

When James catches up to Adrienne they will walk the same distance, so 3(t+1)=4t --> t=3 --> d=4t=12.

Answer: E.
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Re: RTD problem [#permalink] New post 26 Dec 2010, 08:27
Thanks Bunuel. So, we don't care about the miles apart (60 miles, in this case) to solve the problem? Whatever the distance, we just equate both RT?
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Re: RTD problem [#permalink] New post 26 Dec 2010, 08:31
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pgmat wrote:
Thanks Bunuel. So, we don't care about the miles apart (60 miles, in this case) to solve the problem? Whatever the distance, we just equate both RT?


Yes, this info is just to confuse us.
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Re: RTD problem [#permalink] New post 26 Dec 2010, 08:33
Thanks for your quick response and an easy explanation to the problem.
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Re: RTD problem [#permalink] New post 22 Feb 2011, 13:01
we can solve it by using relative speed method,

time taken for James to catch up to Adrianne = 3/1 =3 hours.

Therefore, James travels 4 *3 =12miles to catch up
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Re: RTD problem [#permalink] New post 30 Oct 2013, 13:10
Bunuel wrote:
pgmat wrote:
Thanks Bunuel. So, we don't care about the miles apart (60 miles, in this case) to solve the problem? Whatever the distance, we just equate both RT?


Yes, this info is just to confuse us.


Bunuel,

Why is it that in the referenced problem we disregard the total distance from x to y and just set RT (Adrienne) = RT (James), but in this problem (a-bullet-train-leaves-kyoto-for-tokyo-traveling-240-miles-30242.html) we calculate the distance traveled of the KT and TK train?

Is it because if you're chasing someone, the distance traveled once you've caught up is the same - whereas if two objects are colliding, and if their rates are different, the distance traveled canot be assumed to be the same?
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Re: RTD problem [#permalink] New post 31 Oct 2013, 05:28
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NvrEvrGvUp wrote:
Bunuel wrote:
pgmat wrote:
Thanks Bunuel. So, we don't care about the miles apart (60 miles, in this case) to solve the problem? Whatever the distance, we just equate both RT?


Yes, this info is just to confuse us.


Bunuel,

Why is it that in the referenced problem we disregard the total distance from x to y and just set RT (Adrienne) = RT (James), but in this problem (a-bullet-train-leaves-kyoto-for-tokyo-traveling-240-miles-30242.html) we calculate the distance traveled of the KT and TK train?

Is it because if you're chasing someone, the distance traveled once you've caught up is the same - whereas if two objects are colliding, and if their rates are different, the distance traveled canot be assumed to be the same?

__
Yes, that's correct.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: One hour after Adrienne started walking the 60 miles from X [#permalink] New post 31 Oct 2013, 19:14
Kindly refer screenshot below:

Attachment:
s.JPG
s.JPG [ 40.56 KiB | Viewed 836 times ]


60 miles distance (between X & Y) is not required for calculation in this problem. However, had the problem been "How far from point Y these two met"? then it was required
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Re: One hour after Adrienne started walking the 60 miles from X [#permalink] New post 07 Nov 2013, 08:37
One hour after Adrienne started walking the 60 miles from X to Y, James started walking from X to Y as well. Adrienne walks 3 miles per hour and James walks 1 mile per hour faster than Adrienne. How far from X will James be when he catches up to Adrienne?

First, determine how far Adrienne has walked in the one hour. She has walked three miles which means she is three miles ahead of James when he sets off. James walks at four miles/hour which means that every hour, james will get one mile closer to Adrienne. If he gets one mile closer every hour, it will take him three hours to catch up to her which means he travels 3hours * 4 miles/hour = 12 miles and she travels 4 hours * 3 miles/hour = 12 miles. He will be 12 miles from X when he catches up to her.

A slightly different way to solve...

We don't know how long they will walk before they catch up to one another but we do know that A walks for one hour more than J. J = T and A = T+1. We are looking for the distance at which they meet up which means the distance will be the same. D=r*t so,

r*t (james) = r*t (adrienne)
r*(t) = r*(t+1)
4t = 3t+3
t=3

d=r*t
d=4*3
d=12

E) 12 miles
Re: One hour after Adrienne started walking the 60 miles from X   [#permalink] 07 Nov 2013, 08:37
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