vinni.k 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.