Comments on: Math word problem… Need help urgently?

By: whabtbob

whabtbob — Thu, 28 Jan 2010 12:16:59 +0000

Since they already give you this info:
"the speed of the plane in still air is 520 mph"
then it is unnecessary to set:
"x = speed of plane in still air" (because you already know it is 520 mph).

You have to use the formula
rate * time = distance
to solve the problem. Notice they give you that both trips are completed in the same amount of time. Solving the formula for time give us:

time = distance/rate

Let’s set w= speed of the wind. Then the speed of the plane with the wind is 520+w, and the speed of the plane against the wind is 520-w. Plug this in the formula:

Let w = the speed of the wind. Then 520 + w = plane’s speed with wind and 520 – w = plane’s speed against the wind.

1725 / (520 + w) = 1395 / (520 – w)
Let w = the speed of the wind. Then 520 + w = plane’s speed with wind and 520 – w = plane’s speed against the wind.

Since the times are the same,
1725 / (520 + w) = 1395 / (520 – w)
Let w = the speed of the wind. Then 520 + w = plane’s speed with wind and 520 – w = plane’s speed against the wind.

Since the times are the same,
1725 / (520 + w) = 1395 / (520 – w)
1725 * (520 – w) = 1395 * (520 + w)
897000 – 1725w = 725400 + 1395w
171600 = 3120w
55 = w

This is much more logical an answer, and that is not rounded; it is exactly 55, which makes me suspect it is the correct answer.

Hope this helps.
References :

By: Brock Sampson

Brock Sampson — Thu, 28 Jan 2010 11:55:59 +0000

Assuming that the wind is pushing the airplane at the same speed the wind is blowing, than
1725=speed with wind
13395=speed against wind
520=speed of plane in still air
x=speed of wind
y=amount of time plane is flying
(520+x)y=1725
(520-x)y=1395
1040y=3120
y=3
(520-x)3=1395
520-x=465
x=55 mph

Remeber this is only right if the plane is being pushed by the wind exactly as fast as the wind is blowing.
References :

By: Ivans

Ivans — Thu, 28 Jan 2010 11:22:59 +0000

You are mostly right, however what it means with 1725 miles with the wind and 1395 miles against the wind in the same amount of time. is that after time t we have these equations:

t(520+y) = 1725 (time * rate = distance)
t(520-y) = 1395
now t=1725/(520+y) from the first equation
and t=1395/(520-y) from the second, I can equate these to get:
1725/(520+y)=1395/(520-y)

Now we can cross multiply to get
1725(520-y)=1395(520+y)
897000 – 1725y = 725400 + 1395y
171600=3120y
y=55

hope that’s clear.
References :

By: hayharbr

hayharbr — Thu, 28 Jan 2010 10:34:59 +0000

x and y are not miles; they are speeds, so you can’t add them up to get 1725 miles.

They give you the speed in still air so don’t use a variable for that. The formula is distance = speed times time, or
time = distance divided by speed

Let w = the speed of the wind. Then 520 + w = plane’s speed with wind and 520 – w = plane’s speed against the wind.

Since the times are the same,
1725 / (520 + w) = 1395 / (520 – w)

So cross multiply and solve for w.
References :