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Suppose D is the distance traveled downhill from A to B, L is the distance traveled on level ground from A to B, and U is the distance traveled uphill from A to B. Then the total time to travel from A to B is (D/72)+(L/63)+(U/56)=4. We can multiply through by 504 to get 7D+8L+9U=2016 (call this equation X). On the way back from B to A, D and U switch roles, with D being uphill and U being downhill. So, for the return trip, we get (D/56)+(L/63)+(u/72)=5. Multiply both sides by 504 to get 9D+8L+7U=2520. Adding this to equation X gives 16D+16L+16U=4536. Simplify this to D+L+U=283.5. Since D+U+L is the distance from A to B, we have determined that distance to be 283.5 without specifying what D, L, and U are.

Note that this problem only has a unique solution for some sets of speeds. {D=72,L=63,U=56} happens to be one such set.

Spoiler for a little bit of explanation

Suppose D is the distance traveled downhill from A to B, L is the distance traveled on level ground from A to B, and U is the distance traveled uphill from A to B. Then the total time to travel from A to B is (D/72)+(L/63)+(U/56)=4. We can multiply through by 504 to get 7D+8L+9U=2016 (call this equation X). On the way back from B to A, D and U switch roles, with D being uphill and U being downhill. So, for the return trip, we get (D/56)+(L/63)+(u/72)=5. Multiply both sides by 504 to get 9D+8L+7U=2520. Adding this to equation X gives 16D+16L+16U=4536. Simplify this to D+L+U=283.5. Since D+U+L is the distance from A to B, we have determined that distance to be 283.5 without specifying what D, L, and U are.

Note that this problem only has a unique solution for some sets of speeds. {D=72,L=63,U=56} happens to be one such set.

Well said....! Another unique sets of speeds is D=56, U=42, L=48. In this case (If time of travell remains same); Two equations will be: 6D+7L+8U=6*7*8*4 8D+7L+6U=6*7*8*5 Adding both equations we get: 14(D+L+U)= 6*7*8*9 Therefore D+L+U=216