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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?

  • math -

    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

  • math -

    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.

  • math -

    Steve, Of course!!!!
    How very clever of you

  • math -

    aw, shucks. > scuff scuff <

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