# distance rate time problem

• Dec 11th 2009, 09:48 AM
esther
distance rate time problem
I can't figure out
plane travels 680 mi one way with the wind at 300 mph plus tailwind
plane traves 520 mi back against the wind at 300 mph less headwind
the time to go and return are equal

I keep getting 300 as wind but that can't be

Thank you
• Dec 11th 2009, 10:35 AM
bigwave
since Distance = Rate x Time
Quote:

Originally Posted by esther
I can't figure out
plane travels 680 mi one way with the wind at 300 mph plus tailwind
plane traves 520 mi back against the wind at 300 mph less headwind
the time to go and return are equal

I keep getting 300 as wind but that can't be

Thank you

since time is = then we have
let w = wind

since Distance = Rate x Time

$\frac{D_1}{R_1}= \frac{D_2}{R_2}$

$
\frac{680m}{300mph - w} = \frac{520m}{300mph + w}
$

$680(300+w) = 520(300 - w)$

$204000 + 680w = 156000 -520w$

$48000 = -1200w$

$|- 40| = w = 40mph$
• Dec 11th 2009, 11:01 AM
Soroban
Hello, esther!

Quote:

A plane travels 680 mi one way with the wind at 300 mph plus tailwind.
The plane travels 520 mi back against the wind at 300 mph less headwind.
The time to go and return are equal
Find the rate of the wind.

Let $x$ = rate of the wind.

With the wind, the plane flies 680 miles at $300+x$ mph.
. . This takes: . $\frac{680}{300+x}$ hours.

Against the wind, the plane flies 520 miles at $300-x$ mph.
. . This takes: . $\frac{520}{300-x}$ hours.

Since the times are equal, we have: . $\frac{680}{300+x} \;=\;\frac {520}{300-x}$

Divide by 40: . $\frac{17}{300+x} \:=\:\frac{13}{300-x}$

Cross-multiply: . $5100 - 17x \:=\:3900 + 13x \quad\Rightarrow\quad 30x \:=\:1200\quad\Rightarrow\quad x \:=\:40$

The rate of the wind is 40 mph.