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Flying against the jetstream, a jet travels 2700 miles in 5 hours. Flying with the jetstream, the same jet travels 8820 miles in 9 hours. What is the rate of the jet in still air and what is the rate of the jetstream?

Question

Flying against the jetstream, a jet travels 2700 miles in 5 hours. Flying with the jetstream, the same jet travels 8820 miles in 9 hours. What is the rate of the jet in still air and what is the rate of the jetstream?

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Solution

Let's denote the rate of the jet in still air as J and the rate of the jetstream as S.

When the jet is flying against the jetstream, the effective speed of the jet is (J - S) because the jetstream is slowing it down. So, we can write the equation for the distance covered as:

2700 = 5(J - S) ...(equation 1)

When the jet is flying with the jetstream, the effective speed of the jet is (J + S) because the jetstream is helping it along. So, we can write the equation for the distance covered as:

8820 = 9(J + S) ...(equation 2)

Now we have a system of two equations, and we can solve it step by step.

First, let's simplify both equations:

From equation 1: 2700 = 5J - 5S => 540 = J - S From equation 2: 8820 = 9J + 9S => 980 = J + S

Now we have a new system of equations:

540 = J - S ...(equation 3) 980 = J + S ...(equation 4)

We can solve this system by adding these two equations together:

540 + 980 = J - S + J + S 1520 = 2J J = 1520 / 2 J = 760 mph

Substitute J = 760 mph into equation 4:

980 = 760 + S S = 980 - 760 S = 220 mph

So, the rate of the jet in still air is 760 mph and the rate of the jetstream is 220 mph.

This problem has been solved

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