Question

An airplane travels 640 miles from Topeka to Houston in 3.2 hours, going against the wind. The return trip is withthe wind, and takes only 2 hours. Find the rate of the airplane with no wind. Find the rate of the wind.

Answer 1

Solution:

Let V represent the rate of the airplane, let W represent the rate of the wind.

But

$\text{speed}=\frac{\text{distance covered}}{time\text{ }}$

Thus, against the direction of wind, we have

$\begin{gathered} V-W=\frac{640}{3.2} \\ \Rightarrow V-W=200\text{ ---- equation 1} \end{gathered}$

When the airplane is in the direction of wind, we have

$\begin{gathered} V+W=\frac{640}{2} \\ \Rightarrow V+W=320\text{ ---- equation 2} \end{gathered}$

From equation 1, make V the subject of the formula or equation.

$\begin{gathered} V-W=200 \\ \Rightarrow V=200+W\text{ --- equation 3} \end{gathered}$

Substitute equation 3 into equation 2.

Thus, we have

$\begin{gathered} V+W=320 \\ \text{where }V=200+W \\ \text{thus,} \\ 200+W+W=320 \\ \Rightarrow200+2W=320 \\ \text{subtract 200 from both sides of the equation,} \\ 200+2W-200=320-200 \\ \Rightarrow2W=120 \\ \text{divide both sides by the coeff}icient\text{ of W, which is 2} \\ \frac{2W}{2}=\frac{120}{2} \\ \Rightarrow W=60 \end{gathered}$

Substitute the value of 60 for W into equation 3.

Recall that

$\begin{gathered} V=200+W \\ \text{where W=60} \end{gathered}$

Thus,

$\begin{gathered} V=200+60 \\ \Rightarrow V=260 \end{gathered}$

Hence, the airplane flies at 260 mi/h with no wind. The rate of wind is 60 mi/h.

The first option is the correct answer.