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

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Manager
Joined: 06 Jun 2012
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Alan and Peter are cycling at different constant rates on a  [#permalink]

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Updated on: 11 May 2015, 03:34
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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.

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Originally posted by summer101 on 05 Aug 2013, 04:54.
Last edited by Bunuel on 11 May 2015, 03:34, 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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05 Aug 2013, 04:58
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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Joined: 10 Jul 2013
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Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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05 Aug 2013, 14:09
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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06 Aug 2013, 14:10
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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Joined: 06 Jul 2013
Posts: 96
GMAT 1: 620 Q48 V28
GMAT 2: 700 Q50 V33
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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07 Aug 2013, 09:02
1
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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07 Aug 2013, 09:06
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
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
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Senior Manager
Joined: 10 Jul 2013
Posts: 315
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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07 Aug 2013, 09:26
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......

however nice work done again by you.... thanks for replying.....
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Posts: 8419
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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17 Apr 2018, 07:22
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Re: Alan and Peter are cycling at different constant rates on a &nbs [#permalink] 17 Apr 2018, 07:22
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