There are 5 houses on a street: house A, B, C, D and E.
The distance between any two adjacent houses is 100 feet.
There are 2 children living in house A, 3 children living in
house B, 4 children living in house C, 5 children living in
house D and 6 children living in house E. If the school
bus can only make one stop on that street, in front of
which house should the bus stop so that the sum of
walking distance among all children will be the least?
(Source: the Noetic Learning Math Contest 2010 Fall, Grade 5)
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If we move the stop from house A to house B, 2 children (who live in house A)
will have to walk 100 feet more each, but 18 children (who live in house B & C &
D & E) will walk 100 feet less each. Therefore, it will be 1800 – 200 =1600 feet
less walk if we move the stop from house A to house B.
If we move the stop from house B to house C, 5 children (who live in house A &
B) will walk 100 feet more each, but 15 children (who live in house C & D & E)
will walk 100 feet less each. Therefore, it will save a total of 1500 – 500 =1000
feet if we move the stop from house B to house C.
If we move the stop from house C to house D, 9 children (who live in house A & B
& C) will walk 100 feet more each, but 11 children (who live in house D & E) will
walk 100 ft. less each. Therefore, it will be 1100 – 900 =200 feet less walk if we
move the stop from house C to house D.
If we move the stop from house D to house E, 14 children (who live in house A &
B & C & D) will walk 100 feet more each, but 6 children (who live in house E )
will walk 100 feet less each. Therefore, it will be 1400 – 600 = 800 feet more walk
if we move the stop from house D to house E.
Therefore, children will walk the least if the bus stop is at House D.
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