# Thread: distance speed problem

1. ## 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.

2. 1.

$\left\{ \begin{gathered}
\left( {v_{Pat} + v_{wind} } \right) \cdot t_1 = d \Rightarrow 3\left( {v_{Pat} + v_{wind} } \right) = 1 \hfill \\
\left( {v_{Pat} - v_{wind} } \right)t_2 = d \Rightarrow 4\left( {v_{Pat} - v_{wind} } \right) = 1 \hfill \\
\end{gathered} \right.
$

3. Originally Posted by sleigh
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?

Originally Posted by sleigh
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.