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Alan and Peter are cycling at different constant rates on a
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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
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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: distancespeedtimewordproblemsmadeeasy87481.htmlor here: timespeedanddistancesimplified150163.html
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Re: Alan and Peter are cycling at different constant rates on a
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05 Aug 2013, 14:09
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: distancespeedtimewordproblemsmadeeasy87481.htmlor here: timespeedanddistancesimplified150163.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
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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: distancespeedtimewordproblemsmadeeasy87481.htmlor here: timespeedanddistancesimplified150163.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
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07 Aug 2013, 09:02
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
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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: distancespeedtimewordproblemsmadeeasy87481.htmlor here: timespeedanddistancesimplified150163.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
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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 hournow 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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