# With a head wind, a plane traveled 1000 miles in 4 hours. With the same wind as a tail wind, the return trip took 3 hours and 20 minutes. How do you find the speed of the plane and the wind?

Nov 11, 2017

Speed of the plane $275 \text{ m/h}$ and that of the wind, $25 \text{ m/h.}$

#### Explanation:

Suppose that the speed of the plane is $p \text{ miles/hour (m/h)}$

and that of the wind, $w$.

During the trip of $1000 \text{ miles}$ of the plane with a head wind,

as the wind opposes the motion of the plane, and as such, the

effective speed of the plane becomes $\left(p - w\right) \text{ m/h.}$

Now, $\text{speed"xx"time"="distance,}$ for the above trip, we get,

$\left(p - w\right) \times 4 = 1000 , \mathmr{and} , \left(p - w\right) = 250. \ldots \ldots \ldots \ldots \left(1\right) .$

On the similar lines, we get,

(p+w)xx(3" hour "20" minutes)"=1000......(2).

Note that,

(3" hour "20" minutes)"=(3+20/60" hour")=10/3" hour."

$\therefore \left(2\right) \Rightarrow \left(p + w\right) \left(\frac{10}{3}\right) = 1000 , \mathmr{and} , \left(p + w\right) = 300. \ldots \left(2 '\right) .$

$\left(2 '\right) - \left(1\right) \Rightarrow 2 w = 50 \Rightarrow w = 25.$

Then, from $\left(1\right) ,$ we get, $p = w + 250 = 275 ,$ giving, the desired

Speed of the plane $275 \text{ m/h}$ and that of the wind, $25 \text{ m/h.}$

Enjoy Maths.!