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.

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 t_w and time against be t_a

 

t_w=270/x+6

 

t_a=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