Q 123 - Intermediate Algebra

Question

A commercial jet can fly $1320$ miles in $3$ hours with a tailwind but only $1170$ miles in $3$ 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 commercial jet in still air is $415$ mph.

The speed of the wind is $25$ mph.

Step 1. Given information.

Consider the given question,

A jet can fly a distance of $1320$ miles in $3$ hours with a tailwind.

It can fly $1170$ miles in $3$ 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 $1320$ miles in $3$ hours with a tailwind in which the wind helps the jet.

So, $$\begin{gathered} 3\left(x+y\right)=1320 \\ x+y=440 ....... \left(i\right) \end{gathered}$$

When the jet flies $1170$ miles in $3$ hours with a headwind in which the wind slows the jet.

So,$$\begin{gathered} 3\left(x-y\right)=1170 \\ x-y=390 ...... \left(ii\right) \end{gathered}$$

Step 3. Solve the equations.

Add equations (i) and (ii),

$$\begin{gathered} \left(x+y\right)+\left(x-y\right)=440+390 \\ 2x=830 \\ x=415 \end{gathered}$$

Substituting $x=415$ in equation (ii),

$$\begin{gathered} 415-y=390 \\ y=25 \end{gathered}$$

Step 4. Check the answers.

Substitute the values in equation (i),

$$\begin{gathered} 415+25=440 \\ 440=440 \end{gathered}$$

This is true.

Substitute the values in equation (ii),

$$\begin{gathered} 415-25=390 \\ 390=390 \end{gathered}$$

This is true.

The speed of the jet in still air is $415$ mph.

The speed of the wind is $25$ mph.