<p>Not so much of a trap though
First part: t=210/50=4.2 hrs
-----> (210+V2)/(1+4.2)=52.5
----->V2=63.</p>
<p>It's here:
<a href="http://talk.collegeconfidential.com/showthread.php?p=997233#post997233%5B/url%5D">http://talk.collegeconfidential.com/showthread.php?p=997233#post997233</a></p>
<p>Parallel thread (btw, anybody has statistics on how many times on average a question is being asked repeatedly within one year?):
<a href="http://talk.collegeconfidential.com/showthread.php?t=74410%5B/url%5D">http://talk.collegeconfidential.com/showthread.php?t=74410</a></p>
<p>twitb, sorry,
you avoided a trap, but got into a pothole.</p>
<p>A student drove the first part of a 210mi trip home at 50mi/h.
Find the speed at which she/he drove the remaining part, if the average speed on this trip was 52.5mi/h, and it took the student 1 hour to finish the second part.</p>
<p>Why don't you just take the information you know and build an equation from there? Ok so you have d=rt, r1=50, d/t=52.5, t2=1 and d=210. It is also assumed that d=d1+d2 and t=t1+t2. Lets see what we can get from these equations:</p>
<p>d=210, d/t=52.5 => t=210/52.5=4hr.</p>
<p>t=t1+t2 =>t1=t-t2=3hr.</p>
<p>r1=50, t1=3, => d1=r1<em>t1=50</em>3=150mi.</p>
<p>d2=d-d1=210-150=60mi.</p>
<p>r2=d2/t2=60/1=60mi/hr.</p>
<p>I know this solution is kind of long and bulky but its the most organized way I know to do it--taking what you have and deducing what you're trying to find. You probably shouldn't use this method on the SAT, however, as it takes too long unless you've run out of options/tricks.</p>