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Math Help - Rate Problem

  1. #1
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    Question Rate Problem

    Rate Problem:
    "In a bike race, Jake and Kevin started at the same time and rode with constant speeds. Jake rode at 30miles per hour, and Kevin rode at 35miles per hour. If Jake crossed the finish line 10 minutes after Kevin, how many miles long was the race."

    You can successfully solve the problem if you set Kevin's time as T and Jake's as T+1/6. However, I had a student ask if we could set Jake as T and Kevin as T-1/6. We went through the problem that way, but it does not work. Ideas? Anyone know the math principle behind this?
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  2. #2
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    Hi kaitlincpearce

    Both methods definitely will work. Of course you will get different values of T for the methods but the distance will be the same.
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  3. #3
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    Hello, kaitlincpearce!

    In a bike race, Jake and Kevin started at the same time and rode with constant speeds.
    Jake rode at 30miles per hour, and Kevin rode at 35miles per hour.
    If Jake crossed the finish line 10 minutes after Kevin, how long was the race?

    You can successfully solve the problem if you set Kevin's time as T and Jake's as T+1/6.
    However, I had a student ask if we could set Jake as T and Kevin as T-1/6.
    We went through the problem that way, but it does not work. . Really?

    Let T = Kevin's time.
    Then T + \tfrac{1}{6} = Jake's time.

    Kevin rode T hours at 35 mph . . . His distance: . 35T miles.

    Jake rode T + \tfrac{1}{6} hours at 30 mph . . . His distance is: . 30\left(T + \tfrac{1}{6}\right) miles,

    Their distances are equal: . 35T \:=\:30\left(T + \tfrac{1}{6}\right) \quad\Rightarrow\quad T \:=\:1

    Therefore, Kevin rode for 1 hour at 35 mph . . . The distance is 35 miles.



    Let T = Jake's time.
    Then T - \tfrac{1}{6} = Kevin's time.

    Jake rode T hours at 30 mph . . . His distance is: . 30T miles.

    Kevin rode T - \tfrac{1}{6} hours at 35 mph . . . His distance is: . 35\left(T - \tfrac{1}{6}\right) miles.

    Their distances are equal: . 30T \:=\:35\left(t - \tfrac{1}{6}\right) \quad\Rightarrow\quad T \:=\:\tfrac{7}{6}

    Therefore, Jake rode 1\tfrac{1}{6} hours at 30 mph . . . The distance is 35 miles.


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    I would have solved it like this . . .

    Let D = the length of the race (in miles).

    Kevin rode D miles at 35 mph . . . His time is: . \frac{D}{35} hours.

    Jake rode D miles at 30 mph . . . His time is: . \frac{D}{30} hours.


    Jake's time is 10 minutes \left(\tfrac{1}{6}\text{ hour}\right) more than Kevin's time.

    We have: . \frac{D}{30} \;=\;\frac{D}{35} + \frac{1}{6}

    Multiply by 210: . 7D \:=\:6D + 35 \quad\Rightarrow\quad \boxed{D \:=\:35\text{ miles}}

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