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A helicopter goes 270 miles with the wind in the same time it can go 180 miles against the wind. The wind speed is 6 miles/hour. Find the speed with no wind.

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3 Answers

Hey Tisha -- here's a verbal approach ... +6mph and -6mph creates a 90-mile spread ...
 
12mph means 90mi ... 180mi needs 24mph ... 270mi would need 36mph
 
=> windless is in between at 30mph ... Best wishes, ma'am :)
This is a classic "rate times time equals distance" word problem. If I write that in letters: rt=d, where rate=r, time=t, and distance=d.
 
In this word problem, the rate is made up of two things: the speed of the helicopter (lets call that r) and the speed of the wind (which is 6mph). When the helicopter is traveling with the wind, it goes (r+6)mph and when it is traveling against the wind, the helicopter goes (r-6)mph.
 
OK, we know the helicopter goes 270miles with the wind, so let's set up this "rt=d" equation filling in the numbers we know:
   (r+6)t=270
OK, we also know the helicopter goes 180miles against the wind, so that "rt=d" equation looks like this:
  (r-6)t=180
 
In both of those equations, we know that t=t since in this word problem the times are the same, but we are not given the time. No worries, let's rewrite each of those equations equal to t:
  t=270/(r+6)
  t=180/(r-6)
(I just divided both sides by the (r +/- 6))
 
Since t=t, then:
  270 = 180
 (r+6)   (r-6)
 
Cross-multiply to get rid of the fractions:
270(r-6)=180(r+6)
 
Distribute:
270r-1620=180r+1080
 
Gather your "r" terms on one side and the numbers on the other side:
90r=2700
 
Divide both sides by 90:
r=30.
 
Therefore, the speed of the helicopter is 30mph.
We need to use the formula d=rt to solve this problem: To solve for x we rearrange the formula:
 
t=d/r
 
Let x be the speed with no wind.
 
Let time with the wind be tw and time against be ta
 
tw=270/x+6
 
ta=180/x-6
 
To solve for X;
 
270/x+6=180/x-6  cross multiply
 
270x -1620 = 180x +1080
 
90x = 2700
 
x = 30mph with no wind