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summer101
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Alan and Peter are cycling at different constant rates on a Updated on: 11 May 2015, 03:34
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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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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D
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Alan and Peter are cycling at different constant rates on a 05 Aug 2013, 04:58
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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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Alan and Peter are cycling at different constant rates on a 05 Aug 2013, 14:09
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Zarrolou
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
Show more

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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Alan and Peter are cycling at different constant rates on a 06 Aug 2013, 14:10
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Zarrolou
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?
Show more


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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Alan and Peter are cycling at different constant rates on a 07 Aug 2013, 09:02
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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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Alan and Peter are cycling at different constant rates on a 07 Aug 2013, 09:06
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Asifpirlo
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Zarrolou
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..
Show more

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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Alan and Peter are cycling at different constant rates on a 07 Aug 2013, 09:26
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summer101
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?
Show more


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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Alan and Peter are cycling at different constant rates on a Updated on: 01 Apr 2020, 05:32
Originally posted by BrentGMATPrepNow on 23 Nov 2018, 09:30.
Last edited by BrentGMATPrepNow on 01 Apr 2020, 05:32, edited 1 time in total.
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summer101
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.
Show more

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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Alan and Peter are cycling at different constant rates on a 23 Nov 2018, 16:21
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summer101
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.
Show more
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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Alan and Peter are cycling at different constant rates on a 14 Oct 2020, 13:27
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We're told Alan is currently 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.

In the prompt we were told Alan is currently 3 miles ahead of Peter. From this statement we know Peter's speed is 4 mph faster than Alan's speed. In one hour, Peter will be 1 mile ahead of Alan. Sufficient.

(2) Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.

Peter's speed is 4 mph faster than Alan's speed. That means each hour, Peter will travel miles more than Alan. We know Peter is 3 miles behind Alan currently; in one hour, Peter will be 1 mile ahead of Alan. Sufficient.

Answer is D.
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Alan and Peter are cycling at different constant rates on a 13 May 2024, 01:01
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