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An airplane travels 5096 kilometers against the wind in 7 hours
and 5796 kilometers with the wind in the same amount of time.
What is the rate of the plane still air and what is the rate of the wind
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Let u be the airplane rate in still air (in kilometers per hour), and
let v be the rate of wind.
Then the airplane's effective speed flying with the wind is u+v kilometers per hour,
while its effective speed flying against the wind is u-v kilometers per hour.
From the condition, we have this system of equations
u + v = 5796/7 = 828 km/h (1)
u - v = 5096/7 = 728 km/h (2)
To solve the system, add the equations. You will get
2u = 828 + 728 = 1556; hence u = 1556/2 = 778 km/h.
Then from equation (1),
v = 828 - u = 828 - 778 = 50 km/h.
ANSWER. The rate of the airplane in still air is 778 kilometers per hour;
the rate of the wind is 50 kilometers per hour.
Solved.
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It is a typical "tailwind and headwind" word problem.
See the lessons
- Wind and Current problems
- Wind and Current problems solvable by quadratic equations
- Selected problems from the archive on a plane flying with and against the wind
in this site, where you will find other similar solved problems with detailed explanations.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the section "Word problems", the topic "Travel and Distance problems".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.