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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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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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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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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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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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Posts: 85
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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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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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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Senior Manager  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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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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Posts: 4000
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

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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

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

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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.

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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Our high-level "quant" preparation starts here: https://gmath.net Re: Alan and Peter are cycling at different constant rates on a   [#permalink] 23 Nov 2018, 16:21
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