An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. What is the rate of the plane in still air and what is the rate of the wind?
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Let the speed of the air plane in still air be x km/hr
And the speed of the wind be y km/hr.
Then speed of the airplane going with the wind = (x + y) km/hr
and speed of the airplane going against the wind = (x - y) km/hr.
We know that....
Distance = Speed × Time
Speed = Distance
Time
According to the problem,
An airplane travels 4362 km against the wind in 6 hours
x - y = 4362/6
or, x - y = 727 ----------- (1)
Again, the airplane travels 5322 km with the wind in the same amount of time i.e. 6 hours
x + y = 5322/6
or, x + y = 887 ----------- (2)
Now add (1) and (2) we get,
x - y = 727
x + y = 887
2x = 1614
or, 2x/2 = 1614/2, (Divide both sides by 2)
or, x = 807
Now substitute the value of value of x = 807 in equation (2) we get,
807 + y = 887
-807 -807, (Subtract 407 from both sides)
y = 80
Answer - Rate of the plane in still air = 807 km/hr
Rate of the wind = 80km/hr
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