<p>ditto
++++++</p>
<p>1.
x^3 = (x^2)/2
2(x^3) = (x^2)
2x = 1
x = 1/2</p>
<p>y^3 = (y^2)/3
3(y^3) = (y^2)
3y = 1
y = 1/3</p>
<p>x + y = 1/2 + 1/3 = 5/6</p>
<p>2.</p>
<p>There is always a danger of loosing solutions x by dividing both sides of eqations by variables/expressions.
If there were no condition x>0 y>0, you'd loos roots x=0, y=0.
(of course, question would have multiple correct answers then).</p>
<p>b
u
m
p</p>
<p>This thread is crucial.</p>
<p>I think I remember what this wierd question I had on the May SAT was.
Roy, Matt, and Fred want to buy something together. They split the price in the ratio r:m:f. Roy's share is r<em>m</em>f. What expression represents the other 2 boys shares added together?</p>
<p>This quuestion really confused me because there are absolutely no real numbers mentioned, only variables.</p>
<p>From the CB Blue Book, on the second practice test, page number 476. #16.</p>
<p>This question includes i diagram, but instead i will just use letters.</p>
<p>A B C D E</p>
<p>If the five cards (lettters) above are place in a row so that "C" is never at either end, how many different arrangments are possible?</p>
<p>This question, which many of you have probobly already done, gave me a hard time. How did you solve this problem : compinations and permutations??? </p>
<p>the answer is 72.</p>
<p>thank you</p>
<p>There are 3 possibilities for C (because C can't be on the edges)</p>
<p>3</p>
<p>There are 4 possibilites for the next letter.</p>
<p>3x4</p>
<p>There are 3 possibilities for the next letter</p>
<p>3x4x3</p>
<p>There are 2 possibilites for the next letter.</p>
<p>3x4x3x2</p>
<p>There is 1 possibility for the next letter</p>
<p>3x4x3x2x1</p>
<p>========================================</p>
<p>3x4x3x2x1 = 72</p>
<p>"I think I remember what this wierd question I had on the May SAT was.
Roy, Matt, and Fred want to buy something together. They split the price in the ratio r:m:f. Roy's share is r<em>m</em>f. What expression represents the other 2 boys shares added together?</p>
<p>This quuestion really confused me because there are absolutely no real numbers mentioned, only variables."
Assuming that you've copied the problem correctly (which i doubt as there's more syntactic ambiguity in it than they usually allow on the SAT), we know that Roy has r<em>m</em>f shares. </p>
<p>Since the ratio of the number of Roy's shares to the number of Matt's shares is
r:m, we have (# of Matt's shares) = ((# of Roy's shares)/r)<em>m, or ((r</em>m<em>f)/r)</em>m = m^2<em>f. Similarly, (# of Fred's shares) = ((# of Roy's shares)/r)</em>f = f^2*m.</p>
<p>The sum is m<em>m</em>f + m<em>f</em>f or mf(f + m).</p>
<p>how do you do the questions where they say that there was a race of 700 meters that had two laps of equal lengths and going up they person ran 7mps and on the way back her ran 5mps. So, what is the average speed? How do I solve for this?</p>
<p>CMU:</p>
<p>Average speed is equal to total distance covered over total time. The distance is obviously 2x700m = 1400m. The first lap took (700m/(7 m/s)) = 100s and the second took (700m/(5 m/s)) = 140 s, so the total time is 240s. The average speed is 1400m/240s = 5.83 m/s</p>
<p>I'm not as mathematically inclined as some of you are here and in fact, math is my weakest area on the SAT's, so far. So guys, I need a lot of help. The question below, was one I encountered on a practice test. I managed to get it correct by working backwards but I don't know how to solve it w/out working backwards. </p>
<p>At one instant, two meteors are 2,500 kilometers apart and traveling toward each other in straight paths along the imaginary line joining them. One meteor has a velocity of 300 meters per second while the other travels at 700 meters per second. Assuming that their velocities are constant and that they continue along the same paths, how many seconds elapse from the first instant to time of their collision? (1 kilometer = 1000 meters)</p>
<p>I wasn't sure how to solve this. I realized that I had to convert the 2,500 kilometers to meters, so I multiplied (2500)(1000) = 2,500,000 meters. I had absolutely no idea what to do next and panicking, I took one random answer (it happened to be 2,500) and decided to work backwards. If one meteor has a velocity of 300 m/s, then (300 m/s)(2500 s) = 750,000 meters and if the other meteor has a velocity of 700 m/s, then (700 m/s)(2500 s) = 1,750,000 meters. I added 750,000 meters + 1,750,000 meters = 2,500,000 meters. </p>
<p>Now, I was fortunate to pick the choice that was the correct answer but if it hadn't worked out, I would've used the same procedure on all the choices until I arrived at my answer. How do you solve this problem without going through all the work that I did? Most of the time on the Math Section, I know how to get the answer, one way or another, but I always manage to waste a lot of time doing so. My procedure was very time consuming, so I'm looking forward to hearing an alternative method.</p>
<p>Cryptic:
The two meteors are (2500)(1000) = 2,500,000 meters apart, and they are closing the gap at a rate of (700 + 300) meters/second. How long does it take them to close the gap to 0?</p>
<p>Answer: (2500)(1000) / (700 + 300) = 2500 seconds.</p>
<p>If you have trouble visualizing this, ask yourself - what if they started out 1000 meters apart? In 1 second, meteor A would travel 300 meters, and B would travel 700 meters; they would meet. Now, what if they had started 2000 meters apart? 3000? 2,500,000?</p>
<p>(This is an example of the distance = speed * time formula
or equivalently time = distance / speed.</p>
<p>Be sure you use consistent units like distance in meters,
time in seconds and speed in meters/second; don't mix & match).</p>
<p><em>nevermind</em> I had a question, but I figured it out.</p>
<p>Pyroclastic, thanks for the explanation. Could you explain why "(# of Matt's shares) = ((# of Roy's shares)/r)*m"?<br>
Thanks. For some reason I don't get this problem, and I don;t want to mess up on a similar problem when I retake the SAT.</p>