<p>The very common question of if I travel to work at 20 mph and return home at 30 mph, what is the average rate. The formula 2xy/x+y instantly gives you the average rate, which is less than the average of the two speeds. It's only good when you have 1 hour I believe. How would you use this formula if it took 3 hours to go to work and back home?</p>
<p>can someone explain to me xiggi's method to solving those type of problems? for some reason i cant say i know it for sure..</p>
<p>This formula does not depend on the travel time.
For distance d, speed one way v1, speed on the way back v2, average speed v
d/v1 + d/v2 = 2d/v
1/v1 + 1/v2 = 2/v
(v1+v2) / (v1 v2) = 2/v
(v1+v2) / (2 v1 v2) = 1/v
v = 2 v1 v2 / (v1 + v2)</p>
<p>If the total travel time is 1 hour, then
the distance one way is
v<em>(1)/2 = v1 v2 / (v1 + v2);
if the travel time is 3 hours, then the distance is v</em>(3)/2 = 3 v1 v2 / (v1 + v2).</p>
<p>I highly doubt you'll ever see on the SAT the travel time other than 1 hr.</p>
<p>Okay, think about the problem like this (hopefully this will make very much sense!):</p>
<p>EASY METHOD.</p>
<p>Let's say the distance from home to work is 60 miles.</p>
<p>It takes 3 hours to get there. It takes 2 hours to get back. The total distance is 120 miles. The total time is 5 hours. Therefore the average rate is 120/5, or 24 miles per hour.</p>
<p>MORE COMPLICATED METHOD.</p>
<p>If x is the distance from home to work, the total distance for the round trip is 2x.</p>
<p>It takes x/a (a is the rate to get to work, or 20mph) + x/b (b is the rate to get home) hours. Total time is x/a + x/b, or (ax + bx)/(ab) hours. </p>
<p>Average rate is 2xab/(ax+bx).</p>