Question 1031238
Assuming the bus can only stop once, if it stops at
the first house (A), then the distance to reach the stop is:
0 (from A) + 500 from B + (500+500 from C) + (500+500+500 from D) + (500+500+500+500 from E) + (500+500+500+500+2000 from F) = 9000
at (B): then (500 from A) + 0 + (500 from C) + (500+500 from D) + (500+500+500 from E) + (500+500+500+2000 from F) = 7000
at (C): then (500+500 from A) + (500 from B) + (0 from C) + (500 from D) + (500+500 from E) + (500+500+2000 from F) = 6000
at (D): then (500+500+500 from A) + (500+500 from B) + (500 from C) + (0 from D) + (500 from E) + (500+2000 from F) = 6000
at (E): then (500+500+500+500 from A) + (500+500+500 from B) + (500+500 from C) + (500 from D) + (0 from E) + (2000 from F) = 7000
at (F): then (500+500+500+500+2000 from A) + (500+500+500+2000 from B) + (500+500+2000 from C) + (500+2000 from D) + (2000 from E) + (0 from F) = 15000
.
The best place for the bus to stop is at either C or D.