<p>There's 3 SAT math problems I can't figure out:</p>
<p>1) Of the 26 cookies in a tin, the most common type is oatmeal. What is the probability that a cookie randomly selected from the tin is NOT oatmeal?
A) 1/26
B) 6/23
C) 1/2
D) 25/26</p>
<p>2) If x and y are integers and xy + x^2 is odd, which of hte following statements must be true?
I. x is odd
II. y is odd
III. x+y is odd</p>
<p>A) I only
B) III only
c) i and II
d) I and III
E) II and III</p>
<p>3) For how many integers n is (2n+1)(3n-1) a negative number?
A) none
B) One
c) two
d) three
E) four</p>
<p>Can you explain how you got your answer? Thank you very much!</p>
<p>I’m in the process of looking at these. The first problem is missing some information, or perhaps there’s a typo in the book from where you obtained it. It can’t be “solved” in its current form.</p>
<p>Rewrite 2 as x(x+y). Neither x nor x+y can be even for if they were then the product is even. So conclude x must be odd and x+y must be odd. The sum of two odd numbers is even, so since we already concluded that (x+y) is odd and x is odd, y must be even.</p>
<p>For 3: For how many integers n is (2n+1)(3n-1) a negative number? Either 2n+1 is negative and 3n-1 is positive or conversely. Start with 2n+1 negative – 2n+1 is negative when n is -1, -2, -3 etc. and 3n-1 is positive when n is 1, 2, 3 etc. There’s no intersection here. So try 2n+1 positive – 2n+1 is positive when n is 0, 1, 2 etc. and 3n-1 is negative when n is 0, -1, -2 etc. Only n=0 works. You can also solve this problem by trial and error. Start with n=0, 1, 2, … Then test out n=-1, -2, etc.</p>
<p>Well, for the first one, it can’t be B, because you couldn’t have a 6/23 probability. It couldn’t be D because if there’s only one oatmeal, it can’t be most common. Further info is required to determine if A or C is correct. (if there are only 2 types, then it can’t be C)</p>
<p>There’s no info missing, for number 1, you can solve it as it is.
First we know that the number of oatmeal cookies is more than half (definition of majority)
If the amount of oatmeal cookies is more than half, then the number of NON-oatmeal cookies has to be less than half.
We can eliminate all choices that are greater or equal to 1/2. [eliminate C and D]</p>
<p>B can be eliminated because there are 26 cookies, it would be impossible for the ratio to be 6/23.</p>
<p>@HCOdude - It never says that oatmeal is the majority of the cookies, merely that it is the most common. Therefore it is entirely possible that there are 2 oatmeal cookies and 1 cookie of 24 other flavors. If the original problem did state oatmeal was the majority, you would be right.</p>
<p>I see fog has already explained the latter two, so I’m just going to try to help with the first:</p>
<p>1) If this problem is authentic, then unless there’s a typo, you would have to divide the contents of the tin into 2 distinct subsets: the oatmeal cookies and the ‘not-oatmeal’ cookies. If interpreted this way, there are only 2 ‘types’ of cookies in the tin. Choices C and D are automatically disqualified, because if either was the probability of picking a ‘non-oatmeal’ cookie, than the oatmeal cookies wouldn’t constitute the majority (aka be the most common ‘type’). Now, I don’t see any way in which 6/23 could be correct, so we’ll have to tentatively say that there are 25 oatmeal cookies in the tin, and 1 cookie that wasn’t made with oatmeal, making the answer A.</p>
<p>actually, C could be correct. The ‘most common’ type just means that there more oatmeal than any other type of cookie. Thus, the tin could cotain 13 oatmeal, and 13 non-oatmeal. Of the 13 nons, it could be any combination of chocolate chip, peanut butter, raisin, snicker doodle, etc. In total, the the non-otmeals add up to 13, but individually, each is “less common” than oatmeal. In this example the probability of pulling out a non-oatmeal is 13/26, or 1/2.</p>
<h1>1 really seems like a flawed problem. I think bluebayou nailed the crux of this problem. If we had 13 oatmeal, 12 chocolate chip, and 1 sugar, oatmeal would still be the most common. I was thinking (C) all along because of that.</h1>
<p>Interestingly enough, #1 is missing a choice E…</p>
<p>When will people learn not to waste their time with garbage like this? If you cannot trust the problem, you cannot trust the answer! What do you think you will learn from working with such material? </p>
<p>By the way, there are dedicated forums for the SAT and the ACT.</p>