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Math Help - distance speed problem

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    distance speed problem

    1. Pat biked 1 mile in 3 minutes with the wind at her back, and then she returned in 4 minutes riding against the wind. What was the speed of the wind?

    2. A fraction has a value or 3/4. If 7 is added to the numerator, the resulting fraction is equal to the reciprocal of the original fraction. Find the other fraction.
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    Senior Member Peritus's Avatar
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    \left\{ \begin{gathered}<br />
  \left( {v_{Pat}  + v_{wind} } \right) \cdot t_1  = d  \Rightarrow 3\left( {v_{Pat}  + v_{wind} } \right) = 1 \hfill \\<br />
  \left( {v_{Pat}  - v_{wind} } \right)t_2  = d \Rightarrow 4\left( {v_{Pat}  - v_{wind} } \right) = 1 \hfill \\ <br />
\end{gathered}  \right.<br />
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    Quote Originally Posted by sleigh View Post
    1. Pat biked 1 mile in 3 minutes with the wind at her back, and then she returned in 4 minutes riding against the wind. What was the speed of the wind?
    Let us say that Pat's natural speed (in the absence of wind) is s, and call the speed of the wind v. With the wind at her back, Pat would be traveling at a total speed of s+v, while she would be traveling at a speed of s-v when against the wind.

    We are told that she initially bikes 1 mile in 3 minutes. So:

    \text{total speed}=\frac{\text{distance}}{\text{time}}

    s+v = \frac{1\text{ mi}}{3\text{ min.}}

    But on the trip back we have

    s-v = \frac{1\text{ mi}}{4\text{ min.}}

    Now, you solve two equations in two unknowns. Easy, right?

    Quote Originally Posted by sleigh View Post
    2. A fraction has a value or 3/4. If 7 is added to the numerator, the resulting fraction is equal to the reciprocal of the original fraction. Find the other fraction.
    Assign variables to represent the numerator and denominator, say m and n, and note that the reciprocal of a fraction \frac{a}{b} just means that the fraction should be "flipped" to be \frac{b}{a}.

    Set up your equations and show us your work if you get stuck; you should be able to do this one.
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