Q. 4.45 - Intermediate Algebra

Question

Translate to a system of equations and then solve:

A small jet can fly $1,325$miles in $5$hours with a tailwind but only $1,035$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 $236$ mph and the speed of wind is $29$ mph.

Step 1. Given

A jet can fly $1,325$ miles in five hours with a tail and can fly $1,035$ miles in same time into a headwind.

Step 2. Translating into system of equations.

We know that distance is the product of rate and time.

So we will calculate the distance for the jet by assuming the speed of jet and wind as $j\&w$ respectively.

1. When jet flies with a tailwind, the time taken is $5$ hours and the rate will be $(j+w)$. So the distance comes out to be $5(j+w)$.
2. When jet flies into a headwind, the time taken is same and the rate will be $(j-w)$. So the distance comes out to be $5(j-w)$.

The distance travelled when jet has a tailwind is $1,325$ miles and on the other hand, when flies into headwind, the distance is $1,035$ miles.

Step 3. Finding the speed of jet.

We have two equations, i.e.,

$$\begin{gathered} 5(j+w)=1,325 \\ 5(j-w)=1,035 \end{gathered}$$

Taking $5$ common from both the equations give us,

$$\begin{gathered} j+w=265 \\ j-w=207 \end{gathered}$$

Adding both the equations give us

$$\begin{gathered} 2j=472 \\ j=236 \end{gathered}$$

This shows that the speed of the jet is $236$ miles.

Step 4. Finding the speed of the wind.

We know that the speed of jet, i.e., $j=236$.

Putting this value in $j-w=207$gives us

$$\begin{gathered} 236-w=207 \\ w=29 \end{gathered}$$

This shows that the speed of the wind is $29$ mph.

Step 5. Check the solution.

We will calculate the tailwind and headwind rate by adding or subtracting the speed of wind and the speed of jet and then calculate the distance covered. If the equation gives us the given value, our answers are right. 

1. Tailwind rate becomes $236+29=265$ mph.
   In $5$ hours, the distance covered will be $5\times 265=1,325$ miles.
2. Headwind rate becomes $236-29=207$ mph.
   In $5$ hours, the distance covered will be $5\times 207=1,035$ miles.

Thus, we get a given distance which shows our answer is right.