how do you do this word math problem?

Question

suppose an airplane travels a distance of 3,000 miles going with the wind, but would have only gone 1,960 miles flying against the wind for the same amount of time. If the speed of the plane in still air is 620 miles per hour, what is the speed of the wind in miles per hour?

Answer 1

First, you cancel out the effect of the wind by taking average of the two distances (with and against the wind):

(3000 + 1960) / 2 = is 2,480 miles. That's how far the plane would fly in the same amount of time with no wind at all.

2,480 mi / (620 mph) = 4 hours.

Thus, the plane flies for 4 hours.

With the wind, the plane travels (3000 - 2480) = 520 miles further than the plane's own speed can account for.

With the wind, the plane went 520 miles further in 4 hours, so the wind speed is 520/4 = 130 miles per hour.

Answer 2

Sometimes it helps to introduce temporary variables to help understand the problem, then remove these variables later.

"suppose an airplane travels a distance of 3,000 miles going with the wind..."

Let a be the airplane speed in still air, also known as the airspeed, and
let w be the wind speed.

Because the plane is travelling with the wind, it is going at a speed relative to the ground of a + w. To go 3,000 miles, it must travel a time t, another temporary variable we will introduce.

This sentence therefore results in the equation
(a + w)t = 3000

"...but would have only gone 1,960 miles flying against the wind for the same amount of time."

Same time as before, but the distance has changed and now the plane is (hypothetically) flying against the wind. The wind speed therefore subtracts from the airspeed to get the ground speed. From this sentence, we get the second equation

(a - w)t = 1960

There is no further mention of the ground speed, so let's get rid of it by solving one of these equations for t and substituting into the other. Let's decide to solve the first equation for t, giving

t = 3000 / (a + w)

and substitute this new expression for t in the second equation, to give for this equation

(a - w) [3000 / (a + w)] = 1960

At this point, we've got that ugly a + w in the denominator, so let's move it to the top by multiplying both sides of the equation by a + w. We now have the equation

3000 (a - w) = 1960 (a + w)

"If the speed of the plane in still air is 620 miles per hour..."

Aha! We now have a value for a and can substitute 620 in our equation everywhere we had an a, giving

3000 (620 - w) = 1960 (620 + w)

"...what is the speed of the wind in miles per hour?"

In other words, solve the above equation for w, which several answerers have already done, so I won't repeat that part yet again.

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At least, that's one way of setting up a problem like this. A more advanced approach would have been to read through the problem once, realizing that airspeed was given and that the ground speed was to be solved for, then a second time to set up the equations. This has the advantage that we could have avoided the temporary variable a.

It is also clear that this is a rate problem, where we can expect some information on rate, time, and distance and relations between them. There is no specific time mentioned except that the plane used the same amount of it to fly two different distances, albeit one flight was only hypothetical, at two different rates of speed. The time variable was therefore a link between two different expressions. Because time = distance/rate, we can go back through the problem a second time to set up the equality of the two expressions of distance/rate, remembering that the airspeed is 620 miles per hour. This gives more directly the equation

3000 / (620 + w) = 1960 / (620 - w)

which we may render more tractable by cross-multiplication to give

3000 * (620 - w) = 1960 * (620 + w)

which is the same equation we got by the first method.

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A third way is to make up a table, something like

| label | r (mi/hr) | time (hr) | d (mi) |
| with wind | 620 + w | 3000/(620+w) | 3000 |
| against | 620 - w | 1960/(620-w) | 1960 |

filling in the table as the information becomes available in the problem. The stuff in the time column comes from using t = d / r. Equating the two time expressions as the "same amount of time" phrase implies gives the basic equation

3000 / (620 + w) = 1960 / (620 - w)

which is the same as in the previous method, and the steps that follow are also exactly the same as in the previous method.

Answer 3

Distance = Rate * Time.

Since the time is the same for the two trips, we have:

D1 / R1 = D2 / R2 = T

where D1 and R1 are the distance and speed of travelling with the wind, and D2 and R2 the distance and speed of travelling against the wind. T is the time for each trip (which is the same of course).

D1 = 3000
D2 = 1960
R1 = 620 + v (since the wind is helping you)
R2 = 620 - v (since the wind is not helping you)

v is the speed of the wind.

Thus, we try to solve for v:

3000 / (620 + v) = 1960 / (620 - v)

=> cross multiply

3000 * (620 - v) = 1960 * (620 + v)

=>

1,860,000 - 3000v = 1,215,200 + 1960v

=>

644,800 = 4960v

=> v = 130 mph.

Answer 4

The speed of the plane is x
The amount of time the plane traveled is t
The speed of the wind is y
x = 620
(x + y)*t= 3000
(x-y)*t = 1960


Now add the two equations
(x+y)t + (x-y)t = 3000 +1960

Simplify
2xt=4960

plug in and simplify
t=4 hours

Plug in x and t to get y
y = 130

Answer 5

let t be the time & w be the wind speed
t=3000/(620+w)
t=1960/(620-w) so
3000/(620+w)=1960/(620-w) take recipricals of both sides
(620+w)/3000=(620-w)/1960 multiply both sides by 147,000
(620+w)49=(620-w)75
30380+49w=46500-75w add 75w to each side
124w+30380=46500 subtract 30380 from each side
124w=16120 divide both sides by 124
w=130 mph wind speed.

Answer 6

i dunno