Q 121 - Intermediate Algebra

Question

A small jet can fly $1435$ miles in $5$ hours with a tailwind but only $1215$ miles in $5$ hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Step-by-Step Solution

Answer

The speed of the jet in still air is $265$ mph.

The speed of the wind is $22$ mph.

Step 1. Given information.

Consider the given question,

A jet can fly a distance of $1435$ miles in $5$ hours with a tailwind.

It can fly $1215$ miles in $5$ hours with a headwind.

Assume $x$ to be the speed of the jet in still air and $y$ to be the speed of the wind.

Step 2. Form the equations.

Consider the given question,

When the jet flies $1435$ miles in $5$ hours with a tailwind in which the wind helps the jet.

So, $$\begin{gathered} 5\left(x+y\right)=1435 \\ x+y=287 ...... \left(i\right) \end{gathered}$$

When the jet flies $1215$ miles in $5$ hours with a headwind in which the wind slows the jet.

So, $$\begin{gathered} 5\left(x-y\right)=1215 \\ x-y=243 ........ \left(ii\right) \end{gathered}$$

Step 3. Solve the equations.

Add equations (i) and (ii),

$$\begin{gathered} \left(x+y\right)+\left(x-y\right)=287+243 \\ 2x=530 \\ x=265 \end{gathered}$$

Substituting $x=265$ in equation (i),

$$\begin{gathered} 265+y=287 \\ y=22 \end{gathered}$$

$265$ mph.

$22$ mph.

Step 4. Check the answers.

Substitute the values in equation (i),

$$\begin{gathered} 265+22=287 \\ 287=287 \end{gathered}$$

This is true.

Substitute the values in equation (ii),

$$\begin{gathered} 265-22=243 \\ 243=243 \end{gathered}$$

This is true.

The speed of the jet in still air is $265$ mph.

The speed of the wind is $22$ mph.