<p>If V = 12R / (r + R) , then R =</p>
<p>A. Vr / (12 - V)
B. Vr + V /12
C. Vr - 12
D. V / r - 12
E. V (r + 1) /12
Correct Answer: A</p>
<p>Does anyone else find this question somewhat... pathetic? my final answer was </p>
<p>V(r+R)=12R ->>Vr+VR=12R Vr+V/11, but that was clearly not the answer. They want R, but the R that answer A gives is the r, not R. So wth? Somebody please help</p>
<p>V = 12R/(r+R)
V(r+R) = 12R
Vr + VR = 12R
Vr = 12R - VR
Vr = (12-V)R
Vr/(12-V) = R</p>
<p>A.</p>
<p>i got the same as daimond's .
thankfully you didn't put If V = 12R / (r + R) as V= 12R / r + R this time.
yes parenthesis matters. =]</p>
<p>Er, I got answer E by plugging in:
V=2 / r=5 / R=1</p>
<p>I'm interested to know how they reached answer A as the conclusion. Is this a BB question?</p>
<p>EDIT: I redid this question by plugging the same values into answer A. It works as well!! This is one of the rarities when plugging in doesn't always work I guess... So, I concur with what diamondbacker said.</p>
<p>Lolz, I got:</p>
<p>v = (-r*v)/(v-12) </p>
<p>when I solved it, which is the same as A, but in a differen format. What number was this... like 12?</p>
<p>mulberrypie: Yes, sometimes that happens. That's one of the problems with the plug-in method. Now, you would have to try another set of values and try them on all of them and keep doing it until you get it right. That's why I usually don't use the plug-in method.</p>
<p>mulberrypie: problems like these do not need to be pluggered. Pluggering is only ever effective for some problems, but I find that (for myself) it wastes too much time and makes me panicky and unsure..</p>
<p>wow this is so easy........... plugging in is your best friend here... say r = 3, R =6, and lets see..V = 12R / (r + R) ... so V = 12(6)/(3+6) = 8... V = 8... now we Know from our created info that R = 6.... so with the other created number r = 3... and solved number v=8, we plug into the expressions and see which equals 6 since R is 6.</p>
<p>A. Vr / (12 - V) so (8)(3)/(12-8) = 24/4 = 6! bingo. Answer is A.</p>
<p>and @ mulberrypie... you ought to avoid 1 as a plug in choice when plugging in.</p>
<p>Nearly ALL of SAT I questions are easy if you plug in. I am simply trying to learn alternatives to plugging just so i have more tools to work with. Thanks everyone.</p>
<p>^ Well then, diamondbacker definitely did it correctly. I had trouble when I first did these types of problems, because I had trouble getting the R's to get to one side (that was my plan--when you're doing these problems it helps to have a plan). Then my thought process went, "Well, there's nothing I can really do but multiply r+R on both sides, so maybe I'll try that and see what happens. Then after that I thought, "Well, the only thing I can think of now is to expand the expression and get rid of the parentheses. Hey! Now I can factor out an R and then divide, and I get the right answer!"</p>
<p>I think examining and correcting thought processes is very important for learning math, because it's so much more applicable than learning specific math concepts. Often just using a correct thinking process can help you on math problems, even ones you've never seen before.</p>
<p>Yeah, I see. And thanks for the advice KeepRolling, I'll keep that in mind.</p>
<p>also,when u plug in, make sure u use diff numbers. try 2 and 4 instead of 2 and 2, just an example =p</p>
<p>Thanks for the tip, Ren!</p>
<p>well, honestly speaking... i don't know "how" to do like half the problems on Math SAT........ i just use plugging in or random methods to do them. I suck at math.</p>
<p>Plugging in is a highly effective method ;)</p>