<p>Fabio(i made this name up) walked to school at a rate of 3mph. He ran back at 5mph. If it took him an hour to go to school and back home, how far did he travel?</p>
<p>i'm not posting the answer choices because PR is stupid and doesn't have any explanations for it except to use POE.</p>
<p>On his way there: D = ?, R = 3, T = t
On his way back: D = ?, R = 5, T = 1-t</p>
<p>The reasoning is that if t is the time it takes for him to get there, it takes him 1-t hours to get back as it toally takes him 1 hour.
The distance to school and from school is the same D. Because of this, substitute in:</p>
<p>3<em>t = D = 5</em>(1-t)
3t = 5-5t
8t = 5
t = 5/8
It takes him 5/8ths of an hour to get to school. Plug this back in to find D: </p>
<p>D = 3(5/8) = 15/8 miles</p>
<p>However, this is the ONE WAY distance. To find his total distance, double it = 30/8 = 15/4 = 3.75 miles.</p>
<p>i came across something similar to this ques last night; surprisingly,it was on #7 T_T??
lol took me heck a while,i blanked out for 20 secs. too many methods thats why.</p>
<p>the time spent was like 3 hrs.so xiggi's dont work, the Q is cb's too.
so i have to come up with the old trick spd(3-x)=spd(x)
also ( rate1 + rate 2 ) Distance = Time
so time div by sum of 2 rate, is the distance of one way.
test it out,u'll know and learn.</p>
<p>Well, Xiggi's method solves to get the hour, so wouldn't you just multiply the answer by 3?...</p>
<p>@stl2cali2k1, PR's strategies really, truly suck for the SAT 1 tests.... They always want you to narrow it down to 2 or 3 and then guess. I especially hate their ballparking methods...</p>
<p>(2 * speed1 * speed2)/(speed1 + speed2) [miles/hr]
is the formula for average speed on two-leg trip with equal legs (round-way trip, for example).</p>
<p>If a round-way trip took 1 hr, then total distance is
(2 * speed1 * speed2)/(speed1 + speed2) * 1hr,
and one-way distance is twice less:
<a href="speed1%20*%20speed2">b</a>/(speed1 + speed2)**
- THIS is {xiggi's formula}.</p>
<p>If a round-way trip took 3 hr, then total distance is
(2 * speed1 * speed2)/(speed1 + speed2) * 3hr=
(6 * speed1 * speed2)/(speed1 + speed2),
and one-way distance is twice less:
(3 * speed1 * speed2)/(speed1 + speed2)</p>
<p>i think the real formula should be (2*t* * speed1 * speed 2)/(speed1+speed2).. where t is the time.. usually it's 1 hour so ppl dont care for it.. but if it's 3, just place a 3 in there.</p>
<p>well, i've read over princeton review carefully and i like it. In their examples, they show you how to narrow it down and all (which i dont really care for) but they are trying to make a point. Not necessarily walk you through a problem. Their teaching is sort of like "we ll take you half-way, you are the one that has to figure out on your own and create ure own way for the other half". I like it. Their PITA is great!</p>
<p>although i get annoyed by the Joe Blogg shiit... it's like, "you can eliminate that cuz it's too obvious and joe blogg would choose it!".. i mean i understand what they are saying about it being too obvious, but in my mind, i keep thinking, what if that's the answer?</p>