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# One hour after Adrienne started walking the 60 miles from X

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Joined: 26 Jul 2010
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One hour after Adrienne started walking the 60 miles from X [#permalink]

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Updated on: 31 Oct 2013, 00:54
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Difficulty:

15% (low)

Question Stats:

81% (01:52) correct 19% (01:44) wrong based on 351 sessions

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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.

Originally posted by pgmat on 26 Dec 2010, 08:54.
Last edited by Bunuel on 31 Oct 2013, 00:54, edited 1 time in total.
Renamed the topic and edited the question.
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26 Dec 2010, 09:19
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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.

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26 Dec 2010, 09:27
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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26 Dec 2010, 09:31
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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26 Dec 2010, 09:33
Thanks for your quick response and an easy explanation to the problem.
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22 Feb 2011, 14:01
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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30 Oct 2013, 14:10
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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31 Oct 2013, 06:28
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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Re: One hour after Adrienne started walking the 60 miles from X [#permalink]

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31 Oct 2013, 20:14
1
Kindly refer screenshot below:

Attachment:

s.JPG [ 40.56 KiB | Viewed 4874 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]

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07 Nov 2013, 09:37
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) = r*(t+1)
4t = 3t+3
t=3

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

E) 12 miles
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Re: One hour after Adrienne started walking the 60 miles from X [#permalink]

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14 Aug 2017, 13:23
pgmat wrote:
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.

3(x +1) =4x
3x +3 =4x
x=3

3(4) =12

E
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One hour after Adrienne started walking the 60 miles from X [#permalink]

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14 Aug 2017, 15:40
pgmat wrote:
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.

This way looks time consuming. It isn't; took under a minute.

Travelers in same direction? B-->A-->
SUBTRACT rates. Use $$\frac{D}{r}=t$$ to find time taken to catch her. Finally, use time taken to find distance.

1. Find distance between them

Adrienne walked for one hour before James started. At 3 miles/hr * 1 hr, she walked 3 miles.

Three miles is the distance between them, the "gap."

2. Find combined or relative rate. This is a chase problem. They're walking in the same direction, so

subtract their rates: (4-3) = 1 mi/hr

4. Find time for J to catch A from the distance and relative rate. D/r = t

At 1 mi/hr, for a distance of 3 miles it will take 3 mi/1 mph = 3 hours for him to catch her.

5. How far did James walk? rt= D

4 miles/hr * 3 hrs = 12 miles

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Re: One hour after Adrienne started walking the 60 miles from X [#permalink]

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14 Aug 2017, 15:54
or we can also use the "gap " approach
by setting up an equation

in 1 hour Adrienne is going to cover 3 miles
gap between Adrienne and James 4 - 3 = 1 mph

then 3 = t*1
t = 3 hours

as James' speed is equal to 4 then 4 * 3 = 12 miles
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Re: One hour after Adrienne started walking the 60 miles from X [#permalink]

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18 Aug 2017, 09:19
pgmat wrote:
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

We can let James’ time = t and Adrienne’s time = t + 1. Thus, James’ distance is 4t and Adrienne’s distance is 3(t + 1) = 3t + 3. Because they have each traveled the same distance when James catches up to Adrienne, we equate their two distances and solve for t:

4t = 3t + 3

t = 3

Therefore, when James catches up to Adrienne, he is 4 x 3 = 12 miles from X.

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Re: One hour after Adrienne started walking the 60 miles from X   [#permalink] 18 Aug 2017, 09:19
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