Question 1025411: Flying against the wind, an airplane travels 3800km in 5 hours. Flying with the wind, the same plane travels 3660km in 3 hours. What is the rate of the plane in still air and what is the rate of the wind?
Found 4 solutions by mananth, ikleyn, n2, josgarithmetic:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! Plane speed =x km/h
wind speed =y km/h
against wind 5 hours
with wind 3 hours
Distance against 3800 km distance with 3660 km
t=d/r against wind -
3800 / ( x - y )= 5
5 ( x - y ) = 3800
5 x - 5 y = 3800 ....................1
3660 / ( x + y )= 3.00
3.00 ( x + y ) = 3660
3.00 x + 3.00 y = 3660 ...............2
Multiply (1) by 4
Multiply (2) by 6
we get
20.00 x + -20 y = 15200
18.00 x + 18 y = 21960
38 x = 37160
/ 38
x = 978 km/h
plug value of x in (1)
5 x -5 y = 3800
4889 -5 y = 3800
-5 y = 3800 -4889
-5 y = -1089 mph
y = 218
Plane speed 978 km/h
wind speed 218 km/h
m.ananth@hotmail.ca
Answer by ikleyn(53742)  (Show Source):
You can put this solution on YOUR website! .
Flying against the wind, an airplane travels 3800km in 5 hours.
Flying with the wind, the same plane travels 3660km in 3 hours.
What is the rate of the plane in still air and what is the rate of the wind?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The solution in the post by @mananth is incorrect.
I came to bring a correct solution.
Let u be the rate of the plane at no wind (in kilometers per hour)
and v be the rate of the wind (in the same units).
Then the effective rate of the plane with the wind is u + v
and the effective rate of the plane against the wind is u - v.
From the problem, the effective rate of the plane against the wind is the distance of 3800 kilometers
divided by the time of 5 hours {{3800/5}}} = 760 km/h.
The effective rate of the plane with the wind is the distance of 3660 kilometers
divided by the time of 3 hours  = 1220 mph.
So, we have two equations to find 'u' and 'v'
u + v = 1220, (1)
u - v = 760. (2)
To solve, add equations (1) and (2). The terms 'v' and '-v' will cancel each other, and you will get
2u = 1220 + 760 = 1980 ---> u = 1980/2 = 990.
Now from equation (1)
v = 1220 - 990 = 280 - 220 = 230.
ANSWER. The rate of the plane in still air is 990 km/h. The rate of the wind is 230 km/h.
Solved.
The solution in the post by @mananth is performed in the style of complete loosing of logic,
so you do not try to find a rational idea in his solution - there is no a mathematical accuracy there.
Simply ignore his post.
Answer by n2(78)  (Show Source):
You can put this solution on YOUR website! .
Flying against the wind, an airplane travels 3800km in 5 hours.
Flying with the wind, the same plane travels 3660km in 3 hours.
What is the rate of the plane in still air and what is the rate of the wind?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let u be the rate of the plane at no wind (in kilometers per hour)
and v be the rate of the wind (in the same units).
Then the effective rate of the plane with the wind is u + v
and the effective rate of the plane against the wind is u - v.
From the problem, the effective rate of the plane against the wind is the distance of 3800 kilometers
divided by the time of 5 hours {{3800/5}}} = 760 km/h.
The effective rate of the plane with the wind is the distance of 3660 kilometers
divided by the time of 3 hours = 1220 mph.
So, we have two equations to find 'u' and 'v'
u + v = 1220, (1)
u - v = 760. (2)
To solve, add equations (1) and (2). The terms 'v' and '-v' will cancel each other, and you will get
2u = 1220 + 760 = 1980 ---> u = 1980/2 = 990.
Now from equation (1)
v = 1220 - 990 = 280 - 220 = 230.
ANSWER. The rate of the plane in still air is 990 km/h. The rate of the wind is 230 km/h.
Solved.
Answer by josgarithmetic(39790) (Show Source):
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