<p>Lol.. obviously, the question was posted in jest, but jai6638, here's why your method doesnt work:</p>
<p>If it takes 4 men 8 hours to do a job, then the total work done is 32 man-hours and it would take 1 man 32 hours to do the work. In this case, we want to do "half" a job, so the total time is 16 man-hours. If two men were working on the job, then both men would work for 8 hours (for a total of 16 man-hours) to do half the job.</p>
<p>Think about it. If it takes 1 man 2 hours to dig a hole, how can it take 4 men MORE than 2 hours to dig THE SAME hole? Won't 4 men do the job FASTER than 1 man? (Remember, we're talking about 1 hole here...)</p>
<p>Guys - this could be a real SAT question:</p>
<p>Think about it - it's an inverse proportion problem...</p>
<p>It takes 4 men 8 hours to dig 1 hole.</p>
<p>How long does it take 2 men to dig the same hole?</p>
<p>4*8=2x</p>
<p>X = 16 hours</p>
<p>If we want to complete only half of the hole, divide the total time for the entire hole by half; it takes 8 hours to complete half of the hole.</p>
<p>:D :D</p>
<p>I agree that holes can be quantified, but if you really read into the question that deeply, then you'd agree that I said a "half of a hole" which implies a half of any random hole, not necessarily the hole previously stated. And I could've sworn your answer used to say 16/5 of an hour...</p>
<p>It's good to see that everyone is flexing his math muscles, anyway. :)</p>
<p>Oh, and I'm working on the Barron's Math IIC book right now, and the questions on probability are getting my man-panties in a bind. They seem completely random. How many of these questions appear on the test?</p>
<p>It takes (4)(8) = 32 man_hours to dig 1 hole
In general, (#men)(#hours) should = 32, to dig 1 hole.</p>
<p>So we need 32/2 = 16 man_hours for half a hole.
With 2 men, we need (2)(#hours) = 16
i.e. #hours = 16/2 = 8 for 2 men to dig half a hole.</p>
<p>(P.S. Oh, oh- I missed pg. 2, and TanMan's virtually identical post above. Never mind...)</p>
<p>Optimizerdad and Programmer1234 have just confirmed that my prior posted answer was correct. </p>
<p>I rest my case.</p>
<p>Thank you gentleman.</p>
<p>Now, another question.......make it hard this time.........</p>
<p>One of Euler's famous mathamatical relationships is below:</p>
<p>e^(i*pi) + 1 = 0 </p>
<p>***where e = 2.714... </p>
<p>***where "^" means "to the power of"</p>
<p>***where i = imaginary number = sqrt(-1)</p>
<p>***where pi = 3.14159...</p>
<p>Prove Euler's formula and show that it really equals 0.</p>
<p>Whoever can prove this in less than 1 day without having a course on complex numbers is a genius. Furthermore, note that the above formula contains all the famous and remarkable numbers in exactly 1 equation. It contains, "e", "i", "pi", "1", and "0" all in one single equation. That's pretty awesome. I know that this formula is correct but I don't know the proof for this damn thing.</p>
<p>I read the proof at Wikipedia, but I'm not going to say anything since I cheated. It seems kind of random though... You'd really need to be familiar with Euler's concepts to get the correct answer.</p>
<p>The last question was way too hard and advanced for this forum. It's just beyond your level (at least until you take Calculus II in college). </p>
<p>Don't worry about the question. If you're still interested in the answer, someone answered it on another forum. The URL is below:</p>
<p>In anycase, just move on. If anybody else is interested in presenting another question, feel free...</p>
<p>All I can say is Yuck! - bring back bad memories of complex numbers in Calc III ;)</p>
<p>Just opened this thread after posting Re:
<a href="http://talk.collegeconfidential.com/showthread.php?p=1104086#post1104086%5B/url%5D">http://talk.collegeconfidential.com/showthread.php?p=1104086#post1104086</a></p>
<p>Would you take my 2 holes, I mean 2 cents, here?</p>
<p>1.
4 men's rate = 1/8 hole/hr</p>
<p>2.
2 men's rate = (1/8) / 2 = 1/16 hole/hr</p>
<p>3.
(1/2<hole>) / (1/16<hole hr="">) = 8 hours.</hole></hole></p>
<p>Threading Euler in a half hole reminded me of a very funny and deep book ("The Fateful Adventures of the Good Soldier Svejk...").</p>
<p>Somebody's asking there a question similar to this:
"If my girlfriend lives on the third floor across the street from a pub, how old is my grandmother?"</p>
<p>4 mans rate = 1/8 per hour
2 mans rate = 1/8 per hour
1 mans rate = job aint getting done.
Applied mathematics.</p>
<p>I have a different answer for 18.</p>
<pre><code>4 * 4 * 3 * 1 * 1= 48
</code></pre>
<p>da answer is 72</p>
<p>ok.... heres another one for u math wizrds for those of u hav da CB blue book:
1)pg 489 #8
2)pg 490#11
3)pg 490#12</p>
<p>PS:wud also appreciate it if y'all cud gimme some explanation on graphin quadratic functions..
thanks..
i</p>
<h1>8. This is a simple logic problem that can seem complex if you try to use algebra or geometry to figure it out.</h1>
<p>The logic of the problem is this: Knowing that the center is (4,0) is the key to finding the two different x-coordinates that belong to points with the same y-coordinate on the circle. Since a circle is symmetric, what it really means is you have to move the same number of spaces from the center to the left, as from the center to the right, in order to get two different points whose HEIGHTs (y coordinates) are the same. Once you have that insight, it's clear that the answer is (C) 2 and 6, because 2 is 2 units to the left of 4 and 6 is 2 units to the right of 4. This takes no real mathematical skill, and is all about seeing through the logic of the puzzle. </p>
<h1>11. This question, unlike #8, is a straightforward test of math knowledge: It's basically asking if you know how to graph a quadratic equation. If you don't, you should review algebra 1 because that's one of the main things you should've come away with from that class. Or if you didn't take algebra 1 yet, take it, because it will be invaluable to standardized tests. Anyway, the basic idea is this: In ax^2 + bx + c (the equation of a parabola), "a" determines whether your parabola opens up or down (if a = + or - respectively) and c determines the y-coordinate of your vertex (center) of the parabola. So if c is zero, your parabola is on the x-axis, if c = 2, your parabola is 2 units above the x-axis, if it's -5, it's 5 units below the x-axis, etc. They're telling you a and c are BOTH negative, so this means the parabola opens down and that the parabola is below the x-axis. Choice (A).</h1>
<h1>12. This one is another mainly logic problem. The fastest way to answer this one is to mark the diagram up with all the information that they give you, and then the answer should quickly become apparent.</h1>
<p>Try it and you'll probably figure it out. </p>
<p>If you need still need an explanation, here it is: BC is 4, and line segments PQ and QR are each symmetric about the vertical sides of the rectangle. What this ends up meaning is that the part of PR that lies outside the rectangle is equal to the part of PR that lies inside the rectangle. Since the part that lies inside the rectangle is the same length as BC (4), PR is equal to twice that, or 8 (B). </p>
<p>Hope this helps.</p>
<p>andre.. man.. thanks a lot.... hey dude do u knw any resources/websites that cud help me learn more bout quadratic functions specially graphs..</p>
<p>"Anyway, the basic idea is this: In ax^2 + bx + c (the equation of a parabola), "a" determines whether your parabola opens up or down (if a = + or - respectively) and c determines the y-coordinate of your vertex (center) of the parabola. So if c is zero, your parabola is on the x-axis, if c = 2, your parabola is 2 units above the x-axis, if it's -5, it's 5 units below the x-axis, etc. They're telling you a and c are BOTH negative, so this means the parabola opens down and that the parabola is below the x-axis. Choice (A)."</p>
<p>Actually, in the standard form of a quadratic, y=ax^2+bx+c, c is where the parabola intercepts the y axis, or the y intercept. It is not the y coordinate of the vertex.</p>
<p>Completed square form, y = a(x-h)^2 + k gives the coordinates of the vertex (k is the y coord of the vertex).</p>
<p>skywalker99, this website might be helpful with quadratics:</p>
<p>
</p>
<p>You're right. My bad. My statement is true only of quadratics where b = 0 or completed square form where k subs for c.</p>