SOLUTION: A plane flies 720 miles with the wind in 3 hours. the return trip (against the wind) takes 4 hours. what is the speed of the wind and the speed of the plane in still air

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Question 1112308: A plane flies 720 miles with the wind in 3 hours. the return trip (against the wind) takes 4 hours. what is the speed of the wind and the speed of the plane in still air
Found 3 solutions by josgarithmetic, ikleyn, josmiceli:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
r, plane
w, wind
system%28r%2Bw=240%2Cr-w=180%29

2r=240%2B180
r=120%2B90
r=210------plane

2w=60
w=30------wind

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
720%2F3 = 240 = u + v   is the effective speed with the wind.

720%2F4 = 180 = u - v   is the effective speed against the wind.


So, you have this system of 2 eqs in 2 unknowns


u + v = 240    (1)

u - v = 180    (2)


where "u" is the speed of the plane at no wind;  "v" is the speed of the wind.


To solve the system, add the equations. You will get

2u = 240+180 = 420  ====>  u = 420%2F2 = 210.


So, 210 mph is the speed of the plane at no wind.


Now from eq(1)  v = 210-180 = 30 mph is the speed of the wind.

Solved.

------------------
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.


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.



Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the speed of the plane in still air in mi/hr
Let +w+ = the speed of the wind in mi/hr
--------------------------------------------
Equation for flying with the wind:
(1) +720+=+%28+s+%2B+w+%29%2A3+
Equation for flying against the wind:
(2) +720+=+%28+s+-+w+%29%2A4+
---------------------------------
(1) +3s+%2B+3w+=+720+
Multiply both sides by +4+
(1) +12s+%2B+12w+=+2880+
and
(2) +4s+-+4w+=+720+
Multiply both sides by +3+
(2) +12s+-+12w+=+2160+
-------------------------------
Add (1) and (2)
(1) +12s+%2B+12w+=+2880+
(2) +12s+-+12w+=+2160+
---------------------------
+24s+=+5040+
+s+=+210+
and
(1) +3s+%2B+3w+=+720+
(1) +3%2A210+%2B+3w+=+720+
(1) +630+%2B+3w+=+720+
(1) +3w+=+90+
(1) +w+=+30+
----------------------
The wind speed is 30 mi/hr
The speed of the plane in still air is 210 mi/hr
----------------------------------------------
check answer:
(1) +720+=+%28+s%2Bw+%29%2A3+
(1) +720+=+%28+210+%2B+30+%29%2A3+
(1) +720+=+240%2A3+
(1) +720+=+720+
and
(2) +720+=+%28+s-w+%29%2A4+
(2) +720+=+%28+210+-+30+%29%2A4+
(2) +720+=+180%2A4+
(2) +720+=+720+
OK