# Work Problem

• Dec 3rd 2009, 10:24 AM
anotherMelissa
Work Problem
I need help with this problem. I have no idea how to go about it! Why can't math just use numbers and not words?!

"Fern can fly her plane 240 miles against the wind in the same time it takes her to fly 360 miles with the wind. The speed of the plane in still air is 30 mph more than four times the speed of the wind. Find the speed of the plane in still air."

Fff, I meant "word problem," not "work."
• Dec 3rd 2009, 11:23 AM
masters
Quote:

Originally Posted by anotherMelissa
I need help with this problem. I have no idea how to go about it! Why can't math just use numbers and not words?!

"Fern can fly her plane 240 miles against the wind in the same time it takes her to fly 360 miles with the wind. The speed of the plane in still air is 30 mph more than four times the speed of the wind. Find the speed of the plane in still air."

Fff, I meant "word problem," not "work."

Hi anotherMelissa,

Seems that lots of students don't like word problems. I believe it has to do with their lack of experience with them. You have to parse the words, extract the data, make some sense about how to arrange the data, create an equation or equations of some kind, and then solve the darn thing. Seems like a "work" problem to me.

Ok, so here goes.

Let w = speed of the wind

Let 4w + 30 = speed of the plane in still air

Distance = rate X time

Time is constant. Time = Distance / rate

Let D1 = 240 (distance traveled against wind)
Let D2 = 360 (distance traveled with wind)

Let r1 = (4w+30)-w (rate against the wind)
Let r2 = (4w+30)+w (rate with the wind)

$\frac{D_1}{r_1}=\frac{D_2}{r_2}$

$\frac{240}{4w+30-w}=\frac{360}{4w+30+w}$

Solve for w (speed of wind)

Solve for 4w + 30 (speed of plane in still air)