math
posted by Misty .
They start riding bicycles towards each other in a plan to meet at the midway point. Each is able to ride at 7 MPH. They live 42 miles apart. One of them has a pet pigeon and it starts flying the instant the boys start traveling. The pigeon flies back and forth at 21MPH between the 2 boys until they meet. How many miles does the pigeon travel?

let's look at the pattern
stage 1
they are 42 miles apart
let the time they meet be after t hrs.
distance travelled by boy 2 = 7t
distance travelled by pigeon = 21t
21t + 7t = 42
28t = 42,
t = 1.5
so the pigeon has travelled 21t or 21(1.5) miles
Stage 2
they are 21 miles apart, ( each boy has gone 7(1.5) miles)
define t the same way as before
distance travelled by boy 1 = 7t
distance travelled by pigeon = 21t
21t+7t=21
t = .75
so the pigeon has travelled 21(1.5) + 21(.75) miles
stage 3
they are 10.5 miles apart (each boy has gone 7(.75) miles, leaving 21  2(7)(.75) = 10.5)
distance travelled by boy 2 was 7t
distance travelled by pidgeon was 21t
21t+7t=10.5
t = .375
so the pigeon so far has flown 21(1.5) + 21(.75) + 21(.375)
etc ....
so the total distance covered by the pigeon is
21(1.5) + 21(.75) + 21(.375) + ... to infinity
= 21(1.5 + .75 + .375) + ..
the part inside the bracket is an infinite series with
a = 21.5 , and r = .5
sum = 21 ( 1.5/(1.5) ) = 21(1.5/.5) = 21(3) = 63 miles 
Or, since the boys live 42 miles apart, and they each ride at 7mph, they ride 21 miles each, in 3 hours.
The pigeon, flying at 21 mph for those 3 hours, covers 63 miles. 
Steve, Of course!!!!
How very clever of you 
aw, shucks. > scuff scuff <
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