Help with these Math questions

Test Preparation SAT Preparation
1

<li> The least and greatest numbers in a list of 7 real numbers are 2 and 20, respectively. THe median of the list is 6 and the number 3 occurs most often in the list. Which of the following could be the average of the numbers in the list?</li>
</ol>

<p>I. 7
II. 8.5
III. 10</p>

<p>A. I Only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III </p>

<hr>

<p>2.</p>

<p>The number n is a 2-digit number. When n is divided by 10, the remainder is 9, and when n is divided by 9, the remainder is 8. What is the value of n?</p>

<hr>

<li> If j is chosen at random from the set {4, 5, 6} and k is chosen at random from the set {10, 11, 12}, what is the probability that the product of j and k is divisible by 5?</li>
</ol>

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<li> <a href=“http://img.■■■■■■■■■■■■■■■/albums/v289/Amirsan/problem15.jpg[/url]”>http://img.■■■■■■■■■■■■■■■/albums/v289/Amirsan/problem15.jpg</a></li>
</ol>

<hr>

<p>5.</p>

<p>If y=2x+3 and x < 2, which of the following represents all the possible values for y? </p>

<p>A. y < 7
B. y > 7
C. y < 5
D. y > 5
E. 5 < y < 7</p>

<p>(Answer is apparently A…)</p>

<hr>

<li></li>
</ol>

<p>If a does not equal 0 and 5/x = (5+a)/(x+a) -without parentheses-, what is the value of x?</p>

<p>A. -5
B. -1
C. 1
D. 2
E. 5</p>

<hr>

<li>If x^2 = x + 6, which of the following must be true?</li>
</ol>

<p>A. x = 6
B. x < 3
C. x > 0
D. x^2 < x
E. x^2 > x</p>

<p>THANKS guys! If you can only answer one, please do. Please explain it, and also if you can explain the fastest way to solve it.</p>

2

<p>Ohm, I am suck in math, but relatively sensitive to Geometry, so here is my working out in problem 4. I am not sure it is correct or not, other folks shall check it I guess. Meanwhile I hope you can follow :)</p>

<p>Since they say the total area of the entire shape = 9/4, I divide the picture up to 2 parts: a square with all sides = 1 and a rectangle with width = 1 and length = x </p>

<p>

<a href="http://i191.photobucket.com/albums/z119/Boleslaw_I/24.jpg%5B/IMG%5D"&gt;http://i191.photobucket.com/albums/z119/Boleslaw_I/24.jpg

</a></p>

<p>Area of the entity = Area of Square + Area of Rectangle</p>

<p>Area of square = 1^2 = 1</p>

<p>Area of rectangle = 9/4 - 1 = 5/4</p>

<p>Now the area of the rectangle = Width * Length
Thus 5/4 = 1 * x
that is x = 5/4 = 1.25</p>

<p>Now you can wrap things up: The perimeter of the entity should be:</p>

<p>2 + 1 + 1 + 1 + 1.25 = 6.25</p>

<p>But there is also a part small part that I marked red, this part = 0.25</p>

<p>So my answer is 6.5</p>

<p>Please check, I have no idea.</p>

3

<p>It seems that you get stuck in very similar sort of problems that I get stuck too.</p>

4

<ol>
<li><p>E, 2 3 3 6 _ _ 20 satisfies conditions (only way for 3 to be mode)... least in the two blanks are 7 and 8, summing gets you 49, mean is 49/7 = 7, most in blanks could be 18 19, summing gets 71, 71/7 > 10, so all three</p></li>
<li><p>two ways to do this, first is to realize that n must end in 9 and go quickly divide each number with units digit 9 under 100 by 9 to see which gives you a remainder of 8, 89 works (9*9 = 81, so 89/9 has a remainder of 8)</p></li>
</ol>

<p>second is more clever and quickest, add 1 to n and it will be divisible by 9 and 10, the lowest common multiple of 9 and 10 is 90 since they share no common factors. subtract 1 from n to get 89</p>

<ol>
<li><p>9 possibilities, not hard by checking each case, 4<em>10 works, 4</em>11 does not, 4<em>12 does not. 5 is going to work obviously, 6</em>10 works, 6<em>11 does not, 6</em>12 does not => 5/9</p></li>
<li><p>the bottom left side that's not given is 1 (Why?). split the figure right down the middle from top to bottom to get a square and rectangle. the area of the rectangle is 9/4 - 1 = 5/4. Since one side of this rectangle is 1, the only unknown side that's left (to the right) is 5/4. Thus the perimeter is 5/4+5 = 25/4</p></li>
</ol>

5

<p>Snipez90, I think we have the same method, but what about that tiny 0.25, I get like 25/4 but then rethink about it.</p>

<p>Please click the link in my previous post and tell me what do you think, is it right, or it is just my paranoia</p>

<p>Dang it, SAT ==================================> KMA</p>

6

<ol>
<li><p>since x<2, y must be less than 2(2)+3 = 7, so it's A... don't over think this one. you could try some cases but if you think about it, x could be 1.99999 and y still has to be less than 7, it's the only condition</p></li>
<li><p>plugging in leads you to E, but this equation is not hard to solve. cross multiply to get 5(x+a) = x(5+a) => 5x + 5a = 5x +xa -> 5a = xa => x=5</p></li>
<li><p>x^2-x-6 = 0 so (x-3)(x+2) = 0, this tells us that x = 3, or x = -2, so A B and C can be eliminated immediately. checking D shows that it is false as well, E on the other hand works and is the answer</p></li>
</ol>

7

<p>wow thanks for checking that, so the answer is 26/4 = 13/2</p>

<p>i'm glad i didn't make any mistakes like that on the 2 math sections i took :P</p>

8

<p>No problem, I am thinking those questions above. Hey, are you going to take SAT this Dec?</p>

<p>Can you explain for me question 1</p>

9

<p>No I am done taking it. Got a 2210 the first time and a 2230 the second time and perfect in math both times (only my CR went up the second time :-\, nothing else changed).</p>

<p>Well the way I figured number one was by drawing 7 blanks. First we have
2 _ _ 6 _ _ 20 since 6 is right in the middle (median) and we're given 2 is min and 20 is max. We're also told that 3 is the mode, so it must appear twice (or else there is no mode) but since 3 can only go into the two blanks between 2 and 6, it does so. So now we have 2 3 3 6 _ _ 20, now we know that the remaining blanks can't a 6 or 20 (since then that would also be a mode) nor can it have two of the same number (such as 7, 7) for the same reason. </p>

<p>So the minimum integers have to be 7 and 8 and the max 18 and 19. Average is sum divided by number of terms and indeed it can take on any value between (2+3+3+6+7+8+20)/7 = 7 and (2+3+3+6+18+19+20)/7 ~ 10.142 inclusive. So I II and III should work.</p>

10

<p>Thanks a lot. I shall try to imbibe all these things. You must take Cal AB last year and right now Algebra Analysis?</p>

11

<p>No, I am currently taking Calc BC as a senior. If you solve a lot of problems in technical subjects this SAT stuff is pretty routine.</p>

12

<p>Oh I see, that's why I am solving. Hey, did you take Statistics?
My most favourite Math Subject is Logic (Modal Logic)</p>

13

<p>Yes I took AP Statistics last year. It got much more interesting later on in the year as we covered probability (very basic) and inferential statistics. Unfortunately we were basically given a list of 20-30 formulas that were all very symmetrical but we were told to use them without being shown how they were formulated.</p>

14

<p>I got a lot of troubles for Statistics in the beginning. My grade is quite terrible. But I love probability and the inference part. I am taking it this year.</p>

<p>I think if I can get to university I am going to take Topology. If you are still free, drop by my post since I am going to post some more questions. It is interesting to work out SAT online with so many brilliant people. :)</p>

15

<p>You guys are awesome, thank you! I understand every question now, makes sense.</p>

<p>However... it still seems like questions like #3 still irk me... everything about combinations, permutations and probabilities bother me... would you know any good guides that can help me from now until test day??</p>

16

<p>I am not good at permutation and combination, but I study a lot from
Permutation</a> - Wikipedia, the free encyclopedia</p>

<p>Must say that it provides like a bedtime reading more than actually prepare for SAT. But it helps me to think beyond my mind.</p>

17

<p>Last time I checked, SAT did not do permutation, just combination. If this is not true, I am in for a HELL of a night.</p>

<p>EDIT: for that rectangle problem, I am getting 6.5. The area is 9/4. Let's assume the BIG rectangle has an area of (8/4). The small rectangle, therefore, has to have an area of 1/4. Since one side is given as 1, that small little side HAS to be (1/4). Now I added 2+1+1+1+1(1/4)+(1/4)=6.5?</p>

18

<p>Permutation is on there I believe. But if you draw a few blanks for whatever situation they give you, fill them in according to the problem's conditions, and multiply them in the end, you can arrive at the solution most of the time.</p>

<p>And yes the answer is 6.5</p>

19

<p>Correct, hey did you check your mailbox, I sent you already.</p>

20

<p>Yep, I love the grammatix. It is helping me out a lot even this early. What is your plan on going about with SAT studying. Since we both have only a week (6days /omg), you think it'll be easier to just practice,practice,practice or mix in some notes/helpful sections/etc?</p>