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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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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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.

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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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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