Math Questions help

Test Preparation SAT Preparation
1

<p>Hi this type of questions I always have problems sometimes I get confused solving them.</p>

<p>At Nat's nuts a 2(1/4) or 2.25 pound bag of pistachio nuts costs 6.00$. at this rate, what is the cost in cents,of a bag weighing 9 ounces?
(Note 1 pound = 16 ounces).</p>

<p>1)1.5
2)24
C)150
D)1350
E)2400</p>

<p>A bag contains 4 marbles,1 of each color: red,blue yellow,and green. The marbles are removed at random, 1 at a time. If the first marble is red what is the probablity that the yellow marble is removed before the blue marble?</p>

<p>On certain Russian-American committee,2/3 of the members are men, and 3/8 of the men are Americans. If 3/5 of the committee members are Russians, what fraction of the members are American women?</p>

<p>1)3/20
2)11/60
3)1/4
D)2/5
E)5/12</p>

<p>At a certain university , 1/4 of the applicants failed to meet minimum standards and were rejected immediately.
Of those who met the standards, 2/5 were accepted. If 1200 applicants were accepted,how many applied?</p>

<p>Thanks in advance.</p>

2

<p>C
5/12 (Totally dunno)
A
4000</p>

<p>Im not really sure about answers … :D:D:D</p>

3

<p>1) The answer is 150. Here’s how to crack it:</p>

<p><a href=“http://i.min.us/ilKWVg.JPG[/url]”>http://i.min.us/ilKWVg.JPG&lt;/a&gt;&lt;/p&gt;

4

<p>looooooooooool ^^^^</p>

5

<p>1) First convert 2.25 to ounces. 2.25 * 16 = 36.
Make a proportion. 36 / 6 = 9 / x
x = 1.5
Convert into cents: 150.</p>

<p>2) I forgot how to do this mathematically, but here is what I would do.
Here are the possible choices:
RBYG, RBGY, RGBY, RGYB, RYBG, RYGB
So the possibility of Y before B is 1/2.</p>

<p>3) I’m just going to give a number. Say there are 120 people total on the committee. 80 is men and 40 is women. 30 are American men and 50 are Russian men. 72 are Russian. 22 are Russian women. 18 are American women. 18/120 = 3/20.</p>

<p>4) Set up an equation. x is the number of applicants.
(3x/4) * (2/5) = 1200
6x / 20 = 1200
x = 4000</p>

6

<p>2) Consider this:</p>

<p>There are 4 marbles. But, the first one HAS to be RED. So the only thing that changes is the ORDER of the 3 remaining marbles being drawn. </p>

<p>Your goal is to draw a YELLOW marble before a BLUE marble. </p>

<p>3! = 3 * 2 = 6 = number of possible ways you can draw the remaining 3 marbles:</p>

<p>Yellow, Green, Blue <a href=“check”>/COLOR</a>
[COLOR=“Yellow”]Yellow, Blue , Green<a href=“check”>/COLOR</a>
[COLOR=“Green”]Green, Yellow, [COLOR=“Blue”]Blue <a href=“check”>/COLOR</a></p>

<p>Green, Blue, Yellow (nope)
Blue, Yellow … (nope!)
Blue … (nope!)</p>

<p>So only 3/6 of the possible arrangements work. The answer is 1/2.</p>

<hr>

<p>3) Use the least common multiple of the denominators: </p>

<p>3<em>8</em>5 = 120 members</p>

<p>Here’s a visual breakdown:</p>

<p><a href=“http://i.min.us/ijCkdA.JPG[/url]”>http://i.min.us/ijCkdA.JPG&lt;/a&gt;&lt;/p&gt;

7

<p>Thanks Ice and Silver you cleared my confusions about problem 2</p>

8

<p>No problem realityisadream :)!</p>

9

<p>Does anyone have a good actual mathematical strategy for problem 2? Sometimes you don’t have time to list every possible outcome and then find the favorable ones. What if after you picked the red marble, there were 6 colors left. 6! is way more than you would have time for on the SAT.</p>

10

<p>^ But then it would not be an SAT question…</p>

<p>However, here is a quick argument – I haven’t thought about it long enough to be sure I’m right but: I think the answer will be 1/2 no matter how many marbles there were (as long as there is only 1 yellow and 1 blue). Here’s why: take all of the colors and make a list of every possible order. (Not really, just imagine the list.) Every possible ordering can be paired with its mirror image. So for every time blue comes before yellow, there must be exactly one arrangement that has yellow before blue.</p>

<p>Still, that seems so simple that I am afraid I am missing something! I’m sure someone will catch it and let me know…</p>

11

<p>^
Wow… that sounds so simple, yet it makes so much sense! :)</p>

12

<p>That does make sense pckeller I think you’re right. Do you think the SAT would ever ask this question, but complicate it by adding 2 of the same color? or making the condition “x must be before x AND y” or “x and y must be before y”?</p>

13

<p>No, I think this problem is right at the edge of the SAT envelope. If you look through old sats and such, you’ll see that they do not call for an advanced knowledge of either probability or combinatorics. Beyond the basic counting principle (where you multiply the options) the problems can be done by brute force – writing out the options and counting. And one thing that makes real-life probability much harder than SAT probability is that a subtle change can make a question MUCH harder. Seriously, if you just allow two blue and one yellow, out of say 6 marbles…well now I have to go think for a while…but it won’t ever be on the SAT.</p>

14

<p>Hey guys thanks alot for your explanation,but I wanted to ask @SilverAurora
when you did the equation why did you do "(3x/4) * " where did you get the 3x/4 from,since he only gave 2/5 of those who met standards.</p>

15

<p>@IceQube</p>

<p>Thanks alot for your way of that visual way of explaning third problem it made alot of sense.</p>

<p>@jd989898</p>

<p>I think IceCube way is pretty fast because you see we don’t have 4 combinations anymore if we would it would have been
4 * 3 * 2 * 1 = 24; After we have Red set for us then we have 3 * 2 * 1 ways,which is 6
out of these 6 3 are yellow,so 1/2.</p>

16

<p>3/4 came from the number of applicants who were able to meet minimum standards and were not rejected immediately. Then I just multiplied that with x.</p>

17

<p>^Thanks alot.</p>

18

<p>^No Problem :)</p>