<PRE><B>Question 946685</B>
It took a plane 50 minutes to fly the 330 kilometers from Washington, D.C. to New York against the wind.
 The return trip with the wind took 5 minutes less.
 Find the rate of the plane in still air and the wind speed in kilometers per hour.
:
From the information given we know, it took 45 min to return with the wind.
50 min = 5/6 hr and 45 min = 3/4 hr
:
let s = the plane speed in still air
let w = speed of the wind
then
(s - w) = ground speed to NY
and
(s + w) = ground speed back to DC
:
write distance equation for each way; dist = speed * time
{{{5/6}}}(s - w) = 330
{{{3/4}}}(s + w) = 330
Get rid of those fractions, multiply the 1st equation by 6/5, the 2nd by 4/3
s - w = 396
s + w = 440
--------------Addition eliminates w, find s
2s = 836
s = 836/2
s = 418 km/h in still air
Find w
415 + w = 440
w = 440 - 418
w = 22 km/h is the speed of the wind
:
;
Use the first original equation to check this out
{{{5/6}}}(418 - 22) = 330
{{{5/6}}}(396) = 330
You can confirm this for yourself in the 2nd original equation
{{{3/4}}}(418 + 22) = 330</PRE>