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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