SOLUTION: A jet plane flying with the wind went 2060 mi in 4 h. Flying against the wind, the plane could only fly 1800 mi in the same amount of time. Find the rate of the plane in calm air

Question 933809: A jet plane flying with the wind went 2060 mi in 4 h. Flying against the wind, the plane could only fly 1800 mi in the same amount of time. Find the rate of the plane in calm air and the rate of the wind.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
Questions 933811, 933810, and 933809 are the same fundamental problem as different examples.

A vehicle traveling with the wind went w miles in t hours. Traveling against the wind, the vehicle could only go c miles in the same amount of time. Find the rate of the vehicle in calm air and the rate of the wind.

The unknowns are r for rate in calm air, and n for the rate of the wind.

____________rate_______time_______distance
WITH______r+n________t___________w
AGNST____r-n__________t___________c

The equations are and .
Solve for r and n.

Add corresponding members to Eliminate(Method) n;

. This is one of the variables solved purely in symbols.

You can use Elimination again to solve for n. Seeing that the system has coefficients of 1 on both variables, just subtract corresponding members.

. The other variable solved completely in symbolic form.

This question example uses values assigned to the given variables as
t=4, w=2060, c=1800.
Substitute them into the formulas and evaluate.
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

A jet plane flying with the wind went 2060 mi in 4 h. Flying against the wind, the plane could only fly 1800 mi in the same amount of time. Find the rate of the plane in calm air and the rate of the wind.

Let speed of plane in calm air be S, and wind speed, W
Average speed, with wind = , or 515 mph
Therefore, we get: S + W = 515 -------- eq (i)
Average speed, against wind = , or 450 mph
Therefore, we get: S - W = 450 -------- eq (ii)
2S = 965 ------ Adding eqs (ii) and (i)
S, or speed of plane in still air = , or mph
482.5 + W = 515 --------- Substituting 482.5 for S in eq (i)
W, or wind speed = 515 � 482.5, or mph
You can do the check!!
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