Please help me with Math Questions. Details inside

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<p>Hi. I have a few questions. I could really use the help.</p>

<p>Detailed explanations would really help so I can understand the concept. THanks</p>

<p>1)A boostore has 100 books on sale. If 80 of them are paperbacks and 44 of them are mysteries, what is the least possible number of paperback mysteries on sale?</p>

<p>Detailed Explanations would really help. THanks</p>

<p>2)A school supply cabinet stocks numerals to use for room numbers. If the stock consists of 10 each of the numeras 0,1,2,3,4,5,6,7,8,9, how many consecutive room numbers, beginning with room number 1, can be formed?</p>

<p>Detailed Explanations would really help. THanks</p>

<p>3) Of 650 cities surveyed, each city had an art museum, or a nature museum, or both. Of the 320 cities that had art museums, 1/4 also had nature museums. What is the total number of cities surveyred that had nature museums?</p>

<p>Detailed Explanations would really help. THanks</p>

<p>4) Six points on a circle. What is teh greatest number of dfferent triangles that can be drawn so that each triangle consists 3 of these points as vertices.</p>

<p>Detailed Explanations would really help. THanks</p>

<p>5) If three different circles are drawn on a piece of paer, at most how many points can be common to all three?</p>

<p>Detailed Explanations would really help. THanks</p>

<p>Thank you so much for all your help in advance. If anyone has AIM, PM me your username so I can get help if you are willing to help.</p>

<p>Thanks!</p>

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<p>1) There are several different ways to solve this, but the one I like is:
There are 100 books on sale, and 80 of them are paperbacks.<br>
Therefore, 20 of them are not paperbacks.<br>
There are 44 mysteries.<br>
How many of them might not be paperbacks?<br>
Well, at most 20, because there are only 20 non-paperbacks.<br>
Therefore, at least the remaining 24 must be paperbacks, and they are mysteries.
(Note that it is possible that all 44 mysteries are paperbacks, but the question asks for the least possible number of paperback mysteries.)</p>

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<p>2) The key here is to notice that the numbers have to be consecutive (very important restriction) and they have to start with 1. Also, there are only 10 of each of the numerals. So, when you start numbering with 1, everything's great as you're going along, you're always using a different numeral as you number the rooms, until you get to 10. Now you are going to be rapidly using up your stock of 1's. How fast? Well, at 10, you've used up 2 1's (1 for 1 and 1 for 10). At 11, you've already used up 4 numeral 1's. So you have only 6 left. That will get you up to 17, and now you're out of 1's. The next number is 18, and you can't make it. So 17 rooms is the answer.</p>

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<p>3) 650 cities.
Each has to have an art museum, a nature museum, or both.
320 cities have art museums.
Therefore 330 cities do not have art museums. Since the cities all have some museum, these must have nature museums.
But some cities have both art and nature museums.
Of the 320 with art museums, 1/4 have nature museums.
That's 80.
Therefore, the total number of cities with nature museums is 330 + 80 = 410.</p>

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<p>4) You have 6 points on a circle, and you want to know how many different triangles you can make with these points as vertices. To make a triangle, you need to select 3 points out of 6. So, you need the number of combinations of 6 items taken 3 at a time. This is 6!/(3! times 3!). </p>

<p>Are you familiar with the ! symbol for factorials? If you have not studied combinatorics yet, the best book I know of is Algebra and Trigonometry (Functions and Applications) by Paul A. Foerster, published by Addison-Wesley, Chapter 12. You should be able to get it through your local library, on an inter-library loan, if they don't have their own copy. (I'm not Foerster, and I don't work for Addison-Wesley.) Other posters can probably give you other recommendations.</p>

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<p>5) To start this, think about 2 circles. At how many points can they intersect? The options are 0, 1 (circles just tangent to each other), 2 (circles intersecting), or infinity (circles completely overlap). But the circles are different, so the number of intersections of 2 circles is 0, 1, or 2.</p>

<p>So let's suppose the circles intersect in 2 points (we might need to backtrack, but we're trying to find the greatest number of common points of three circles, so it makes sense to start with the greatest number of common points of two circles, and work from there). Could the third circle go through both of those points, without being identical to either of the other 2? Yes. So the three circles could possibly have 2 points in common.</p>

<p>It's useful to think about the following question in connection with questions like this, for the future: How many points on the circumference of a circle do you need to have, in order to define the circle (assuming that you have no other information about the points)?</p>

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<p>Hope this helps. Let me know if anything is not clear.
Other posters--let me know if you disagree with anything.</p>

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<p>Great explanations quant.
number 1 is quite simple. Since 80 and 44 exceed the total number of books, we know that 80 and 44 include both genres. Thus the least possible number of books containing both would be 24. Why cant it be 25? well, subtract 25 from 80 and 44 to find the exact number of books that are ONLY mysteries and ONLY paperbacks, and then add it to 25, the supposed # of books that are ONLY paperback and mystery, and it be 99. Which isnt 100.</p>

<p>The wording of Q3 is somewhat obscure. It states that of the 320 cities that had art museums, 1/4 had natural museums. How do we know if the other 330 had art museums as well? it doesn't state it. it just says that of the 320 that had museums, it also had other museums. It doesn't say only 320 of the cities had art museums, and 1/4 had naturals. So it can be perceived differently. </p>

<p>For 4, is it 36? Im not sure what your final answer is for 4, you seem to have wandered off into a topic about books after giving 6!/3!3!, which idont think would work. Every point enables you to create 6 triangles using the other vertices on the circle 6X6 = 36.</p>

<p>5.I somewhat diagree with quant on this one. I say its 6 because the max for 2 circles is 2. However, you can create more intersections using a smaller circle that intersects the arcs created by the 2 circles. These new intersections would be common to all 3 because it is on at least 1 point of each circle. Am i doing something wrong?</p>

<p>Please give us the answers freesn3!</p>

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<p>for number 5 nvm, quant is correct.</p>

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<p>Quix, on #4, I'm not sure what you are thinking when you say that every point enables you to create 6 triangles using the other vertices on the circle.</p>

<p>Say you pick one of the points on the circle. Call it *. How many triangles can you make with * and the other 5 points on the circle? Since you already have one vertex, you need to choose 2 more out of 5.<br>
This gives 5!/(2! times 3!) possibilities, which is a total of 10 (rather than 6).
If you label the remaining 5 points A, B, C, D, E, then you can make triangles that contain * and
1) A and B
2) A and C
3) A and D
4) A and E
5) B and C
6) B and D
7) B and E
8) C and D
9) C and E
10) D and E
This exhausts the list of possibilities, so there would be 10 triangles that contain a given point as one of the vertices.</p>

<p>However, the total number of triangles is only 20. [6!/(3! times 3!)]</p>

<p>The reason that it's not 6 times 10: When you're counting the triangles that contain point A, you don't get 10 completely new triangles. A lot of the ones that you would be counting as containing point A also contain * and they were already counted.</p>

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<p>On #3, I agree with Quix that the wording could be clearer. I think the trick here is that the question says "Of the 320 cities that had art museums . . ."
Strangely, the key word here is "the." You probably wouldn't think it's that important. But by including the word "the," <strong>by the usual convention</strong> it means that 320 is actually the total number of cities that had art museums.</p>

<p>This question should definitely be re-worded, because as is, it's not testing what it's trying to test.</p>

<p>This reminds me of a simple question on Stanford's EPGY math placement exams, which described a team that played 24 games and won 2 out of 3 of them. How many games did the team win? </p>

<p>Well, it's trivial if you know that winning "2 out of 3" means that the team won 2/3 of its games. But this is also a conventional way of writing, and you can't use the literal meaning of the words. If you took it completely literally, you'd say that you know the outcome of 3 games: they won 2 and lost (or tied) 1. But what about the other 21 games? </p>

<p>Alternatively, if you tried to say that for every set of 3 games, the team won 2, that's not right (assuming that they actually won 2/3 of their games). In that case, they lost or tied 8. Pick 3 of the games that they lost or tied--they definitely did not win 2 out of those 3.</p>

<p>Not thinking that anyone would have trouble with the game question--just trying to show that there are some conventional ways of writing math questions that you need to be familiar with--otherwise the question will be ambiguous, as far as you're concerned.</p>

<p>CB should try to eliminate the ambiguities (and they are probably working on it), but you can't count on that to happen this year.</p>

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<p>Wow, i never knew that! I would have skipped that Q if it was on the actual SAT. The question, as quant mentioned, should definitely be reworded because it can pose to be a problem for people who has english as a second language. Having questions infused with various subtle grammar rules just defeats the purpose of testing "aptitude". I highly doubt that Q3 is from the bluebook, or any of CBS materials for that matter. Freesn3, can you tell us where is it that you got those questions from? Im not questioning quants methods and answers to the Qs, but it would be quite reassuring if you posted the answers!</p>

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<p>Good explanations both of you. Im still trying to understand 4, but its beyond me. Can anyone help?</p>

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<p>nbafan135,
I'd be surprised if #4 is on the SAT I Math test. It's really an application of a formula from combinatorics, or else a long "listing" problem. It might be on one or both of the SAT II math tests, but--agree with Quix--this one doesn't seem likely to come from the Blue Book, either.</p>

<p>Here's another way to think about it:</p>

<p>Suppose you had only 3 points on the circle. Then you could make 1 triangle.</p>

<p>If you had 4 points, you could make 4 triangles (leave out a different point each time, keeping the other 3 to make the triangle). </p>

<p>If you have 5 points, then call them A, B, C, D, and E as above. To make triangles, you pick three points at a time. This means that you are leaving out two of the points at a time. Make a list of possibilities for the two you are leaving out. There are 10 possibilities--the list of A, B, C, D, and E, taken two at a time, is identical to the list in post #10. But there's one significant difference in the interpretation. There, "A and B" on the list meant that you were making a triangle using *, A, and B. But now, when we're trying to figure out how many triangles you can make with 5 points, the list item "A and B" means that A and B are left out of the set A, B, C, D, E, when you're making the triangle. So "A and B" left out, means that we're making the triangle with C, D, and E . . . and similarly for the other items on the list. Therefore, there are 10 triangles that can be made with 5 points.</p>

<p>When we add a sixth point, we just need to count the new triangles that can be made, now that we have a sixth point. So we don't need to recount the 10 that we could make with just A, B, C, D, and E. The new triangles all contain *. So, as figured out in post #10, there are 10 of them.</p>

<p>Those 10 plus the 10 you can make with just the points A, B, C, D, and E give 20 all together.</p>

<p>This is easier to see with pictures of the labeled points than with words. If it doesn't seem clear, try sketching a picture.</p>

<p>But my general advice, if you are taking a test that has this sort of question, then you really need to read background material on combinatorics and become comfortable with it. The problem is really quick, done combinatorially, but long (and error prone), done with lists.</p>

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<p>oh wow. thanks for the explanation quantmech. i think its interesting how for 5 points, instead of using the 3 points that make triangles, you see what points are left over. thats a cool way to look at it. I get it now. Thanks again. But like you said, this isnt likely in a SAT I, i hope.</p>