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Alan and Peter are cycling at different constant rates on a
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Updated on: 11 May 2015, 02:34

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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.

Re: Alan and Peter are cycling at different constant rates on a
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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

Re: Alan and Peter are cycling at different constant rates on a
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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

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?

Re: Alan and Peter are cycling at different constant rates on a
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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

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..

Re: Alan and Peter are cycling at different constant rates on a
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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

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.

Re: Alan and Peter are cycling at different constant rates on a
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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.....

Re: Alan and Peter are cycling at different constant rates on a
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23 Nov 2018, 08:30

Top Contributor

1

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

Re: Alan and Peter are cycling at different constant rates on a
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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:

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

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