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Alan and Peter are cycling at different constant rates on a

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Alan and Peter are cycling at different constant rates on a  [#permalink]

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Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

(1) One hour ago, Alan was 7 miles ahead of Peter.
(2) Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.

Originally posted by summer101 on 05 Aug 2013, 03:54.
Last edited by Bunuel on 11 May 2015, 02:34, edited 2 times in total.
Edited the question.
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Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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New post 05 Aug 2013, 03:58
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
Sufficient

II.Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
\(Speed*time=space\).
\(P*t=A*t+4\) (4=3 miles behind + 1 mile ahead) where \(P=18\) and \(A=14\).
\(18*t=14*t+4\), \(4*t=4\) and \(t=1\) hour or 60 minutes.
Sufficient

For the theory refer here: distance-speed-time-word-problems-made-easy-87481.html
or here: time-speed-and-distance-simplified-150163.html
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Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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New post 05 Aug 2013, 13:09
1
Zarrolou wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
Sufficient

II.Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
\(Speed*time=space\).
\(P*t=A*t+4\) (4=3 miles behind + 1 mile ahead) where \(P=18\) and \(A=14\).
\(18*t=14*t+4\), \(4*t=4\) and \(t=1\) hour or 60 minutes.
Sufficient

For the theory refer here: distance-speed-time-word-problems-made-easy-87481.html
or here: time-speed-and-distance-simplified-150163.html


zarrolou :
One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
i cant understand the reasoning.
peter gained just four miles in an hour, how can we be sure about it?
its only possible if Alan was in a dormant state i mean not moving at all.
but both were moving.
suppose , Alan was 7 miles ahead of peter so peter was 7 miles behind of peter.
in one hour Alan went 6 miles and peter went 2 miles, still they are 3 miles apart.
we dont know still their velocity....
can you explain the subtle fact?
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Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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New post 06 Aug 2013, 13:10
Asifpirlo wrote:
Zarrolou wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
Sufficient

II.Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
\(Speed*time=space\).
\(P*t=A*t+4\) (4=3 miles behind + 1 mile ahead) where \(P=18\) and \(A=14\).
\(18*t=14*t+4\), \(4*t=4\) and \(t=1\) hour or 60 minutes.
Sufficient

For the theory refer here: distance-speed-time-word-problems-made-easy-87481.html
or here: time-speed-and-distance-simplified-150163.html


zarrolou :
One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
i cant understand the reasoning.
peter gained just four miles in an hour, how can we be sure about it?
its only possible if Alan was in a dormant state i mean not moving at all.
but both were moving.
suppose , Alan was 7 miles ahead of peter so peter was 7 miles behind of peter.
in one hour Alan went 6 miles and peter went 2 miles, still they are 3 miles apart.
we dont know still their velocity....
can you explain the subtle fact?



I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour,
Sufficient
so he will take another hour (60 minutes) to be 1 mile ahead of Alan.

he gained 4 miles means? Was Alan stationary ? please explain..
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Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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New post 07 Aug 2013, 08:02
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Since both are travelling at constant speed, peter would gain another 4 miles in next 1 hr since he gained 4 miles in last hour.
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Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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New post 07 Aug 2013, 08:06
1
Asifpirlo wrote:
Asifpirlo wrote:
Zarrolou wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
Sufficient

II.Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
\(Speed*time=space\).
\(P*t=A*t+4\) (4=3 miles behind + 1 mile ahead) where \(P=18\) and \(A=14\).
\(18*t=14*t+4\), \(4*t=4\) and \(t=1\) hour or 60 minutes.
Sufficient

For the theory refer here: distance-speed-time-word-problems-made-easy-87481.html
or here: time-speed-and-distance-simplified-150163.html


zarrolou :
One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
i cant understand the reasoning.
peter gained just four miles in an hour, how can we be sure about it?
its only possible if Alan was in a dormant state i mean not moving at all.
but both were moving.
suppose , Alan was 7 miles ahead of peter so peter was 7 miles behind of peter.
in one hour Alan went 6 miles and peter went 2 miles, still they are 3 miles apart.
we dont know still their velocity....
can you explain the subtle fact?



I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour,
Sufficient
so he will take another hour (60 minutes) to be 1 mile ahead of Alan.

he gained 4 miles means? Was Alan stationary ? please explain..


Hi Asifpirlo,

sorry for the late reply. Here is my reasoning:

Alan and Peter are cycling at different constant rates on a straight track.Now Alan is now 3 miles ahead of Peter, and one hour ago, Alan was 7 miles ahead of Peter.

So I think that you would agree to the fact that in an hour Peter gained 4 miles on Alan: before was 7 miles behind, now is only 3.
Since A and B are are cycling at different constant rates, this pattern : in 1 hour Peter gains 4 miles on Alan will repeat.

So, as I said above, Peter will take another hour (60 minutes) to be 1 mile ahead of Alan.

Hope I've explained myself well
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Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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New post 07 Aug 2013, 08:26
summer101 wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

I.One hour ago, Alan was 7 miles ahead of Peter.
II.Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.


P.S Can some please provide a link or explain how to solve these difference in speed problem?



Zarrolou, yes now what are you saying is true indeed.
1st you said, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour
now you told, in an hour Peter gained 4 miles on Alan .

this two sentences confer two completely disparate meaning, i mean gaining 4 miles and gaining 4 miles on someone......

thats the confusion i had.....
however nice work done again by you.... thanks for replying.....
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Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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New post 23 Nov 2018, 08:30
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summer101 wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

(1) One hour ago, Alan was 7 miles ahead of Peter.
(2) Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.


Given: Alan is NOW 3 miles ahead of Peter

Target question: How many minutes from now will Peter be 1 mile ahead of Alan?

Statement 1: ONE HOUR AGO, Alan was 7 miles ahead of Peter.
Alan is NOW 3 miles ahead of Peter
So, in one hour, the GAP between Alan and Peter decreased by 4 miles
Another way to put it: Peter's speed is 4 mph greater than Alan's speed.
Another way to put it: in one hour, Peter traveled 4 miles more than Alan
So, in ONE HOUR FROM NOW, Peter will travel another 4 miles more than Alan. This means the Peter will not only close the 3-mile gap BUT ALSO travel 1 mile further than Alan travels.
So, the answer to the target question is in 60 MINUTES, Peter will be 1 mile ahead of Alan
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
Another way to put it: Peter's speed is 4 mph greater than Alan's speed.
Another way to put it: in one hour, Peter travels 4 miles more than Alan
At this point, we can apply the same logic we applied to statement 1 to conclude that statement 2 is SUFFICIENT

Answer: D

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Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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New post 23 Nov 2018, 15:21
summer101 wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

(1) One hour ago, Alan was 7 miles ahead of Peter.
(2) Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.

Excellent opportunity for the relative velocity (speed) technique:

\(?\,\,\,:\,\,\,\min \,\,{\rm{for}}\,\,4\,\,{\rm{miles - }}\underline {{\rm{gaining}}} \,\,\,{\rm{Peter}}/{\rm{Alan}}\,\,\)


\(\left( 1 \right)\,\,{\text{RelativeSpee}}{{\text{d}}_{{\text{Peter/Alan}}}}\,\,\, = \,\,\,\,\,\frac{{4\,\,{\text{miles}}}}{{\boxed{1\,\,{\text{hour}}}}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = \boxed{1\,{\text{h}}}\,\,\left( { = 60\min } \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\)


\(\left( 2 \right)\,{\text{RelativeSpee}}{{\text{d}}_{{\text{Peter/Alan}}}}\,\,\,\, = \,\,\,\,\frac{{4\,\,{\text{miles}}}}{{\boxed{1\,\,{\text{hour}}}}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = \boxed{1\,{\text{h}}}\,\,\left( { = 60\min } \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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Re: Alan and Peter are cycling at different constant rates on a   [#permalink] 01 Dec 2019, 08:47
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