Math Problems Help Needed

Test Preparation SAT Preparation
1

<p>I have a couple of questions that I am either unsure about or need further elaboration upon. </p>

<p>The first few questions come from the January 2011 SAT.</p>

<p>QAS: Section 2, Question 18, Page 12</p>

<p>There is a picture but I can't really post it. </p>

<p>


The sequence of figures above starts with one hexagon. Each successive figure is formed by adding a ring of identical hexagons around the preceding figure. What is the total number of hexagons in the fifth figure of the sequence.

Because there is no illustration, I will post what they look like: the first figure just has one hexagon, the second figure has one hexagon in the middle and six on the outside, the third figure has the preceding figure (one hexagon and six on the outside of the hexagon) but with 13 more on the outside of the six.

The answer is 61.

Section 5, Question 19, Page 29

``` 1/ (1)(2), 1/ (2)(3), 1/ (3)(4)

The first three terms of a sequence are given above. The nth term of the sequence is 1/ (n)(n+1), which is equal to (1/n) – (1/n+1). What is the sum of the first 50 terms of this sequence? ```

The answer is 50/51.

Section 8, Question 13, Page 40

``` Of 6 radios in a store, 2 are defective and 4 are not. A customer chooses 2 of these radios at random. If the first radio chosen is not defective, what is the probability that the second radio chosen is also not defective? ```

The answer is 3/5.

The following are questions from other prep books.

``` An urn above contains 4 white marbles and 5 black marbles, all of equal size. If two marbles are drawn at random with no replacement, what is the probability that two marbles are different colors? ```

The answer is 5/9.

This was the explanation given to me: Because you do (5/9)(4/8) to find the possibility and then multiply it by 2 because you can have it the reverse way. I don't really understand the concept. I thought it was just (5/9)(4/8) because (5/9) represents the first draw and (4/8) represents the second draw. Please try to explain this more thoroughly to me.

``` *Picture not shown The picture shows a circle with a center at (0,0) with a radius of 10.

In the xy-plane above, a circle is at center (0,0). Which of the following points is inside the circle? ( a ) (6,9) ( b ) (7,8) ( c ) (-8,-8) ( d ) (-9,4) ( e ) (5,-9) ```

The answer is (d).

``` How many ways can seven books be arranged on a shelf if two of them are math books and must be kept together? ```

The answer is 1440 because you do 6!*2, I have no idea why though, please try to explain this clearly as well.

*Picture not shown so letters are used instead.

(CARD A) (CARD B) (CARD C) (CARD D) (CARD E)


If the five cards shown above are placed in a row so that (CARD C) is never at either end, how many different arrangements are possible?

</p>

<p>The answer is 3213 because you do 4<em>4</em>3<em>2</em>4. Try to explain this one clearly as well because probability is all about concept.</p>

2

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</p>

<p>The answer cannot be 3213. The answer to this old problem is 72.</p>

<p>Tip: Google search THIS site for discussions on the official questions. You should find plenty of help. For the questions from other sources, best of luck to you!</p>

3

<p>I’ll help with a few…</p>

<p>For the second one, see this thread: </p>

<p><a href=“http://talk.collegeconfidential.com/sat-preparation/1180319-sat-math-help.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/1180319-sat-math-help.html&lt;/a&gt;&lt;/p&gt;

<p>For the marbles: 4 white and 5 black…</p>

<p>It’s 4/9 x 3/8 + 5/9 x 4/8 </p>

<p>The first product is the probablilty of drawing white followed by white. The next product is for drawing black followed by black. I don’t understand/agree with the answer you gave.</p>

<p>For the 7 books…If two of them have to be together, then think of them as one unit. So that gives you 6 “units” to arrange (5 books and one book-book set). That’s where the 6! comes from. But then you have to double the answer because you get a new arrangement if you take the two books that make a unit and reverse them.</p>

<p>As for the 5 cards, this is a retread of an old blue book question. Quickest way is to start with card C, for which there are 3 places. Then, the next card you place has 4 spots available. Then the next one has 3, then 2, and the last card has 1 spot left. So to get the answer, multiply 3x4x3x2x1.</p>

<p>OK, one more: the circle one – just use the distance formula to check how far each point is from the origin. Only one is less than 10 away.</p>

4

<p>Oh yeah, that’s wrong. The whole thing multiplied out isn’t even 3213. I guess it was an oversight on my part. I googled it and the answer seems to be here: <a href=“http://talk.collegeconfidential.com/sat-preparation/880530-sat-math-counting-problem.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/880530-sat-math-counting-problem.html&lt;/a&gt;&lt;/p&gt;

<p>@pckeller The answer is 5/9. This has been agreed on by many people I’ve asked. I just don’t understand the full reasoning. What answer are you thinking about?</p>

5

<p>Woops – I just scrolled to the end of your question and saw that they want the probabiluty that the marbles are different! I answered for the probability that they are the same. And if you calculate what I gave you, it comes out to 4/9. So the answer is 1- 4/9 (b/c if you subtract out the two ways to have it come out the same, you get left with the prob that it comes out different!)</p>

<p>Still, if I had read it all the way through, I would not use the approach I gave you. Instead:</p>

<p>Think of it as probability of white followed by black + probability of black followed by white:</p>

<p>4/9 x 5/8 + 5/9 x 4/8 </p>

<p>or, since those products are the same, just calculate one of them and then double it (which is what your explanation said to do – but if you don’t understand where the products come from, you are not likely to realize that you can double it…)</p>

6

<p>The best way to solve the hexagon problem is to draw the next sequence of hexagons.
You should get:
1 hexagon in the center
6 hexagons around the center
12 hexagons around the 6 previous hexagons
18 hexagons around the 12 previous hexagons
24 hexagons around the 18 previous hexagons
So the total hexagons is 1 + 6 + 12 + 18 + 24 = 61
If you want to see the picture of the 61 hexagons, go to Wikipedia and search for “magic hexagons”</p>

7

<p>You have 6 radios with 2 of them bad.
You draw 1 radio, and it is good.
So now there are 5 radios with 2 of them bad. Therefore, 3 of them are good.
The probability of drawing a good radio next = the number of good/ total radios left
3 good radios/ 5 total radios = 3/5</p>

8

<p>I’ll do question 2:</p>

<p>1/(1)(2), 1/(2)(3), 1/(3)(4)</p>

<p>The first three terms of a sequence are given above. The nth term of the sequence is 1/ (n)(n+1), which is equal to (1/n) – (1/n+1). What is the sum of the first 50 terms of this sequence?</p>

<p>On hard problems - and this one qualifies - the SAT will rarely give you information that’s not useful, so the fact that 1/(n)(n +1) is equal to (1/n) - (1/n + 1) is a good route to try. Also, on almost all sequence questions, writing out the terms let’s you see a pattern. In this case, write out the first three terms and the last term in the form of (1/n) - (1/n + 1):</p>

<p>1st term: (1/1) - (1/2)
2nd term (1/2) - (1/3)
3rd term (1/3) - (1/4)
50th term (1/50) - (1/51) </p>

<p>Now add them together:
1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + . . . + 1/50 - 1/51</p>

<p>All the middle terms are going to cancel out and you’ll be left with the first and last term: </p>

<p>1/1 - 1/51</p>

<p>So the answer is 50/51</p>

9

<p>Ahh I remember. There was controversy over that problem. It is asking for the second drawing’s probability of being not defective only.</p>

<p>I did 4/6* 3/5. The wording is so off. Would this wording likely show up again? (A explanation of how to dissect the word problem would be nice).
Edit: The “is also” in the wording made it seem like the first draw had to be calculated too.</p>