350 MCGRAW-HILL’S SAT Using Venn Diagrams to Keep Track of Sets Some counting problems involve “overlapping sets,” that is, sets that contain elements that also belong in other sets. In these situations, Venn diagrams are very helpful for keeping track of things. Example: A class of 29 students sponsored two field trips: one to a zoo and one to a museum. Every student at- tended at least one of the field trips, and 10 stu- dents attended both. If twice as many students went to the zoo as went to the museum, how many students went to the zoo? Set up a Venn diagram of the situation. We repre- sent the two sets—those who went to the museum and those who went to the zoo—as two overlapping circles, because some students went to both. Notice that there are three regions to consider. We know that ten students are in the overlapping region, but we don’t know how many are in the other two regions, so let’s use algebra. Let’s say that x students are in the first region, represent- ing those who went to the museum but not to the zoo. This means that x + 10 students must have gone to the museum altogether. Since twice as many students went to the zoo, the total number in the zoo circle must be 2(x + 10) = 2x + 20. Since 10 of these are already accounted for in the overlapping region, there must be 2x + 20 − 10 = 2x + 10 in the third region. So now the diagram should look like this: The total number of students is 29, so (x) + (10) + (2x + 10) = 29 Simplify: 3x + 20 = 29 Solve: x = 3 So the number of students who went to the zoo is 2(3) + 20 = 26. The Fundamental Counting Principle Some SAT questions ask you to count things. Some- times it’s easy enough to just write out the things in a list and count them by hand. Other times, though, there will be too many, and it will help to use the Fun- damental Counting Principle. Lesson 5: Counting Problems To use the Fundamental Counting Principle (FCP), you have to think of the things you’re counting as coming from a sequence of choices. The Fundamental Counting Principle says that the number of ways an event can hap- pen is equal to the product of the choices that must be made to “build” the event. Example: How many ways can five people be arranged in a line? You might consider calling the five people A, B, C, D, and E, and listing the number of arrangements. After a while, though, you’ll see that this is going to take a lot of time, because there are a lot of possibilities. (Not to mention that it’s really easy to miss some of them.) In- stead, think of “building” the line with a sequence of choices: first pick the first person, then pick the second person, etc. There are five choices to make, so we’ll have to multiply five numbers. Clearly, there are five people to choose from for the first person in line. Once you do this, though, there are only four people left for the second spot, then three for the third spot, etc. By the Fundamental Counting Principle, then, the number of possible arrangements is 5 × 4 × 3 × 2 × 1 = 120. Example: How many odd integers greater than 500 and less than 1,000 have an even digit in the tens place? This seems a lot harder than it is. Again, think of “building” the numbers in question. All integers between 500 and 1,000 have three digits, so building the number involves choosing three digits, so we will multiply three numbers to get our answer. If each number is between 500 and 1,000, then there are only five choices for the first digit: 5, 6, 7, 8, or 9. If the tens digit must be even, we have five choices again: 2, 4, 6, 8, or 0. If the entire number is odd, then we have five choices for the last digit as well: 1, 3, 5, 7, or 9. Therefore, the total number of such integers is 5 × 5 × 5 = 125. To the Museum To the Zoo Museum (but not Zoo) Zoo (but not Museum) Both 10x 2x + 10 MZ CHAPTER 9 / SPECIAL MATH PROBLEMS 351 1. What is the fundamental counting principle? 2. How many different four-letter arrangements of the letters LMNO can be made if no letter can be repeated? Answer this first by listing all of the possible arrangements, then by using the Fundamental Counting Prin- ciple, and check that the two answers agree. 3. If the first digit of a 3-digit area code cannot be 0 and the second digit is either 0 or 1, then how many dif- ferent area codes are possible? 4. A baseball team has six players, each of whom can play in any of the three outfield positions: left field, cen- ter field, and right field. How many possible different arrangements of these players can the team place in the outfield? (This one is a bit harder to do by listing!) 5. Among a set of 40 sophomores, 20 students take French and 27 students take Spanish. If all of the students take either French or Spanish, how many students take both French and Spanish? 6. A box contains buttons, each of which is either blue or green and has either two or four holes. If there are four times as many blue buttons as green buttons and six times as many four-holed buttons as two-holed buttons, what is the least number of buttons that could be in the box? Concept Review 5: Counting Problems 352 MCGRAW-HILL’S SAT SAT Practice 5: Counting Problems 1. A pizzeria offers three different sizes of pizza, two different kinds of crust, and eight different choices for toppings. How many different one- topping pizzas are there to choose from? (A) 13 (B) 16 (C) 24 (D) 48 (E) 60 0, 2, 4, 6, 8 2. How many different integers between 30 and 70 contain only digits from the list above? (A) 7 (B) 10 (C) 15 (D) 20 (E) 25 3. In how many ways can you arrange four different paintings in a line on a wall? (A) 12 (B) 24 (C) 36 (D) 48 (E) 64 4. At Lincoln County High School, 36 students are taking either calculus or physics or both, and 10 students are taking both calculus and physics. If there are 31 students in the calculus class, how many students are there in the physics class? (A) 5 (B) 8 (C) 11 (D) 15 (E) 21 5. Dave’s stickball team has six players. How many different six-player batting lineups can they make if Dave must bat second and either Zack or Paul must bat first? (A) 48 (B) 96 (C) 192 (D) 256 (E) 720 6. Maria gave David x cards, gave Tina two more cards than she gave David, and gave Samuel five fewer cards than she gave Tina. In terms of x, how many cards did Maria give Tina, David, and Samuel all together? (A) 3x + 9 (B) 3x − 1 (C) 3x − 3 (D) x − 3 (E) x − 1 7. From a collection of six paintings, four are to be chosen to hang on a wall. How many different arrangements are possible if every painting is different? (A) 24 (B) 120 (C) 360 (D) 720 (E) 1,296 8. Every marble in a jar has either a dot, a stripe, or both. The ratio of striped marbles to non- striped marbles is 3:1, and the ratio of dotted marbles to nondotted marbles is 2:3. If six mar- bles have both a dot and a stripe, how many marbles are there all together? (A) 16 (B) 18 (C) 20 (D) 36 (E) 40 9. An ant must walk from one vertex of a cube to the “opposite” vertex (that is, the vertex that is farthest from the starting vertex) and back again to its starting position. It may only walk along the edges of the cube. For the entire trip, its path must tra- verse exactly six edges, and it may travel on the same edge twice. How many dif- ferent six-edge paths can the ant choose from? 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 SAT Practice 5 1. D Use the FCP: 3 × 2 × 8 = 48. 2. B “Build” the number: If it’s between 30 and 70, it must be a two-digit number that begins with 4 or 6. That’s two choices. The second digit can be anything in the list, so that’s 5 choices. 2 × 5 = 10 3. B Since there are four spaces, there are four deci- sions to make, so four numbers to multiply. You can choose from four paintings for the first spot, then three paintings for the second spot, etc. 4 × 3 × 2 × 1 = 24 4. D If there are 31 students in calculus but 10 of these are also taking physics, then 31 − 10 = 21 stu- dents are taking only calculus. If there are 36 stu- dents taking either physics or calculus, but only 31 are taking calculus, then 36 − 31 = 5 students are taking only physics. Therefore, the Venn diagram should look like this: As you can see, 5 + 10 = 15 students are taking physics. 5. A “Build” the lineup. You have six spots to fill, and thus six decisions to make and six numbers to multiply. You only have two choices for the first spot (Zack or Paul) and one choice for the second spot (Dave), then you have four players left to fill the other slots, so you have four choices for the third spot, then three for the fourth spot, etc. 2 × 1 × 4 × 3 × 2 × 1 = 48 6. B David gets x cards. Tina gets two more cards than David, which is x + 2. Samuel gets five fewer cards than Tina, which is x + 2 − 5 = x − 3. So all together, x + x + 2 + x − 3 = 3x − 1. 7. C You have six choices for the first spot, then five for the second, then four for the third and three for the fourth. 6 × 5 × 4 × 3 = 360 8. E Set up the Venn diagram: Since the ratio of striped marbles to nonstriped marbles is 3:1, x + 6 = 3y, and since the ratio of dotted marbles to nondotted marbles is 2:3, y + 6 = 2/3x and therefore y = 2/3x − 6. Substituting, we get x + 6 = 3(2/3x − 6) or x + 6 = 2x − 18, so x = 24. Plug this back in to get y = 2/3(24) − 6 = 10. Total = 24 + 6 + 10 = 40. 9. 36 Draw the cube. To get from any vertex to its opposite vertex, the ant has 3 × 2 × 1 possible paths. To see why, trace a path and notice it has three choices for the first edge, then two for the second, then only one option for the third. Since it must return to the opposite vertex, it has 3 × 2 × 1 differ- ent paths it can take back. 3 × 2 × 1 × 3 × 2 × 1 = 36 CHAPTER 9 / SPECIAL MATH PROBLEMS 353 Concept Review 5 1. The number of ways an event can happen is equal to the product of the choices that must be made to “build” the event. 2. Try listing all the “words” that start with L, then all that start with M, and so on: LMNO MLNO NLMO OLMN LMON MLON NLOM OLNM LNMO MNLO NMLO OMLN LNOM MNOL NMOL OMNL LOMN MOLN NOLM ONML LONM MONL NOML ONLM Whew! 24 in total. Annoying, but not impossible. Using the FCP makes it a lot easier: 4 × 3 × 2 × 1 = 24. That’s it! 3. There are too many possibilities to list, but the FCP makes it easy: We have 9 choices for the first digit, 2 choices for the second digit, and 10 choices for the last digit, and 9 × 2 × 10 = 180. 4. “Build” the outfield from left to right. You have 6 players to choose from for left field, but then just 5 for center field and 4 for right field. 6 × 5 ×4 =120. 5. Since 20 + 27 − 40 = 7, there must be 7 students who take both. This Venn diagram shows how it works out: 6. This one’s tough. Say there are g green buttons. If there are four times as many blue buttons as green buttons, then there are 4g blue buttons, so g + 4g = 5g buttons in total. So the total number of buttons must be a multiple of 5. Similarly, if there are n two-holed buttons, there must be 6n four- holed buttons, so the total number of buttons is n + 6n = 7n, so the total number of buttons is also a multiple of 7. The least common multiple of 5 and 7 is 35, so there are 35 buttons: 5 two-holed and 30 four-holed, and 7 green and 28 blue. 713 20 FS Answer Key 5: Counting Problems 10521 PC 6xy DS 354 MCGRAW-HILL’S SAT Probability Geometrical Probability Lesson 6: Probability Problems A probability is a number between 0 and 1 that represents the likelihood of an event. An event with a probability of 0 is impossible, and an event with a probability of 1 is certain. Most probabilities, of course, are somewhere in be- tween 0 and 1. For instance, the probability of rolling a 5 on a fair die is 1 ⁄6. It’s best to think of a probability as a part-to-whole ratio. There are six possible outcomes when you roll a die (the whole), but only one of them is 5 (the part). Thus, the probability of rolling a 5 is 1 ⁄6. Example: What is the probability of rolling a sum of 5 on two dice? Here is a table showing all the possible sums on a roll of two dice: Clearly, there are four ways of getting a sum of 5 out of a possible 36, so the probability is 4 ⁄36, or 1 ⁄9. Die 1 Die 2 1 4 2 3 6 5 123456 234567 345678 456789 5678910 67891011 7 8 9 10 11 12 An SAT question may ask you to find the prob- ability that something hits a certain region, like a dart hitting a dartboard. In these situa- tions, the probability is just the ratio of the par- ticular area to the entire area. Example: A landing target for skydivers consists of two con- centric circles. The smaller circle has a radius of 3 meters, and the larger one has a radius of 6 meters. If a skydiver hits the target, what is the probability that she hits the smaller circle? It might help to sketch the target: If she hits the target, then she hits an area that is π(6) 2 = 36π square meters in area. The smaller circle, though, is only π(3) 2 = 9π square meters in area, so the probability that she lands within the smaller region should be just 9π/36π=1/4. 3 6 CHAPTER 9 / SPECIAL MATH PROBLEMS 355 1. The probability of an impossible event is __________. 2. The probability of an event that is certain is __________. 3. If a jar contains 3 red marbles, 4 white marbles, and 5 blue marbles, then what is the probability of ran- domly choosing a red marble? __________ a white marble? __________ a blue marble? __________ 4. A jar contains 5 red marbles and 10 white marbles. What is the probability of drawing a red marble? __________ If 3 more red marbles are added, then what is the probability of drawing a red marble? __________ 5. A jar contains 24 red and blue marbles. If the probability of selecting a red marble at random is 3 ⁄8, then how many red marbles must be added so that the probability of randomly selecting a red marble becomes 1 ⁄2? 6. A jar contains only black, white, and red marbles. The probability of choosing a white marble is 2 ⁄3. If there are 4 times as many red marbles as black marbles, what is the least possible number of marbles in the jar? Concept Review 6: Probability Problems 356 MCGRAW-HILL’S SAT SAT Practice 6: Probability Problems 1. The figure above shows a spinner in the mid- dle of a disc divided into six equal parts, each labeled with a number. What is the probabil- ity that the spinner will land on a number that is either even or greater than 5? (A) (B) (C) (D) (E) 2. A jar contains 10 blue marbles, 8 green marbles, and 14 red marbles. How many green marbles must be added so that the probability of choosing a green marble at random is ? (A) 16 (B) 32 (C) 40 (D) 64 (E) 72 3. A fair six-sided die has faces bearing the numbers 1, 2, 3, 4, 5, and 6. When the die is thrown, the numbers on the five visible faces are added. What is the probability that this sum is greater than 18? (A) (B) (C) (D) (E) 4. A target consists of three concentric circles, with radii of 1 meter, 2 meters, and 3 meters. If an arrow that hits the target hits any point on the target with equal probability, what is the probability that an arrow that hits the tar- get falls in the outermost region (between the second and third circles)? (A) (B) (C) (D) (E) 5 9 4 9 π 9 1 3 1 9 5 6 2 3 1 2 1 3 1 6 3 4 5 6 2 3 1 2 1 3 1 6 5. The probability of a meteor shower occurring in the skies above a particular island on any given night is . Independently, the proba- bility that any given night will be cloudless is . What is the probability that, on any given night, there will be a meteor shower and it will be cloudless? (A) (B) (C) (D) (E) 6. A basket contains red, green, and yellow balls, all of equal size. The probability of choosing a green ball at random is . If there are 3 times as many red balls as yellow balls, what is the probability of choosing a yellow ball at random? 7. A certain disease occurs in 1 person out of every 101 people. A test for the disease is 100% accurate for patients with the disease and 99% accurate for patients without it. That is, it gives a “false positive” 1% of the time even if the person tested doesn’t have the disease. If you take this test and it returns a positive re- sult, what is the probability that you have the disease? (A) 1 (B) .99 (C) .95 (D) .50 (E) .01 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 4 7 8 25 4 25 17 200 3 100 1 50 1 4 2 25 5 9 2 8 3 6 SAT Practice 6 1. D Put an “x” through any number that is either even or greater than 5, or both. This gives us 8, 9, 2, and 6, which is 4 out of the 6 spaces, giving a probability of 4 ⁄6, or 2 ⁄3. 2. D If the probability of choosing a green marble is to be 3 ⁄4, then 3 ⁄4 of the marbles should be green and 1 ⁄ 4 not green. There are 10 blue and 14 red, for a total of 24 “not green” marbles, and this will not change, since you are adding only green marbles. If this is 1 ⁄4 of the total, then there must be 4(24) = 96 marbles total after you add the extra green marbles. The jar now contains 10 + 8 + 14 = 32 marbles, so you must add 96 − 32 = 64 green marbles. 3. B The six sides of a die add up to 1 + 2 + 3 + 4 + 5 +6 =21. The sum of any five faces can be greater than 18 only if the “down” face is 1 or 2 (so that the sum of the other faces is either 21 − 1 = 20 or 21 −2 = 19. This is 2 possibilities out of 6 for a probability of 2 ⁄6, or 1 ⁄3. 4. E Sketch the target: You want to know the proba- bility of the arrow hitting the outermost ring, which is the ratio of the area of the ring to the entire area of the target. The area of the whole target is π(3) 2 = 9π. The area of the outermost ring is 9π−π(2) 2 (subtract the area of the middle circle from the area of the big circle) = 9π−4π=5π. So the probability is 5π/9π=5/9. 5. A Consider a stretch of 100 consecutive nights. If the probability of a meteor shower is 2 ⁄25, then we should expect a meteor shower on ( 2 ⁄25)(100) = 8 of those nights. If only 1 ⁄4 of the nights are cloudless, though, then ( 1 ⁄4)(8) = 2 of the nights with a meteor shower, on average, should be cloudless. This gives a probability of 2 ⁄100, or 1 ⁄50. Mathematically, we can just multiply the two probabilities (as long as they are independent) to get the joint probabil- ity: ( 2 ⁄25)( 1 ⁄4) = 1 ⁄50. 6. 3 ⁄28 Call the probability of choosing a yellow ball x. If there are three times as many red balls as yellow balls, the probability of choosing a red ball must be 3x. The probability of choosing a green ball is 4 ⁄7. These probabilities must have a sum of 1: x + 3x + 4 ⁄7 = 1 Simplify: 4x + 4 ⁄7 = 1 Subtract 4 ⁄7:4x = 3 ⁄7 Divide by 4: x = 3 ⁄28 7. D Most people would say that this probability is quite high, because the test is so reliable. But intu- ition is often wrong. Imagine that you test 101 peo- ple. Of these, on average, one will have the disease, and 100 will not. Since the test is 100% accurate for those who have the disease, that person will test positive. Of the 100 who do not have the disease, 99 will test negative, but one will test positive, be- cause of the 1% “false positive” rate. So of those two who test positive, only one will have the dis- ease; thus, the probability is 1 ⁄2. CHAPTER 9 / SPECIAL MATH PROBLEMS 357 Concept Review 6 1. 0 2. 1 3. red marble: 3 ⁄12, or 1 ⁄4 white marble: 4 ⁄12, or 1 ⁄3 blue marble: 5 ⁄12 4. What is the probability of drawing a red marble? 5 ⁄15, or 1 ⁄3 If 3 more red marbles are added, what is the prob- ability of drawing a red marble? 8 ⁄18, or 4 ⁄9 5. If the jar contains 24 red and blue marbles and the probability of selecting a red marble at random is 3 ⁄8, there must be ( 3 ⁄8)(24) =9 red marbles, and 24 −9 =15 blue marbles. If the probability of drawing a red marble is to be 1 ⁄2, there must be as many red as blue marbles, so you must add 15 − 9 = 6 marbles. 6. Let’s say that the probability of drawing a black marble is x. Since there are 4 times as many red marbles as black marbles, the probability of drawing a red marble must be 4x. The probabil- ity of choosing a white marble is 2 ⁄3. Since we are certain to pick one of these colors, the sum of all of these probabilities must be 1: x + 4x + 2 /3 = 1 Simplify: 5x + 2 ⁄3 = 1 Subtract 2 ⁄3:5x = 1 ⁄3 Divide by 5: x = 1 ⁄15 Therefore, 1 ⁄15 of the marbles are black, 4 ⁄15 of the marbles are red, and 2 ⁄3 of the marbles are white. The least common denominator of these fractions is 15, which means that 15 is the least possible number of marbles. In that case, there are 1 black marble, 4 red marbles, and 10 white marbles. Answer Key 6: Probability Problems 1 2 3 CHAPTER 10 ESSENTIAL GEOMETRY SKILLS 1. Lines and Angles 2. Triangles 3. The Pythagorean Theorem 4. Coordinate Geometry 5. Areas and Perimeters 6. Similar Figures 7. Volumes and 3-D Geometry 8. Circles 358 ✓ Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use. When Two Lines Cross When two lines cross, four angles are formed. “Vertical” angles are equal, and look like this: Don’t be fooled by diagrams that look like ver- tical angles, but aren’t. Vertical angles are formed by two and only two crossed lines. When two pairs of vertical angles are formed, four pairs of adjacent angles (side-by-side) are also formed. Adjacent angles add up to 180°: When a Line Crosses Parallel Lines Imagine taking two crossed lines, making a “copy” of them, and sliding the copy down one of the lines so that together they look like this: This produces a pair of parallel lines crossed by a third line. "slide" v ertical not vertical vertical angles CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 359 When two parallel lines are crossed by another line, all acute angles are equal, and all obtuse angles are equal. Also, every acute angle is sup- plementary to every obtuse angle (that is, they add upto 180Њ). To show that two lines are parallel, use the arrow marks “>” like those in the figure in the previous column. To show that two angles are equal, use the arc marks “)” like those in the figure in the previous column. Don’t be fooled by diagrams that only look as if they have two parallel lines crossed by an- other line. Don’t assume that two lines are par- allel just because they look parallel. It must be given that they are parallel. To help yourself to see the relationships be- tween angles in parallel line systems, you might try looking for these special “letters”: Angles that make Z’s are equal: Angles that make C’s or U’s are supplementary (they have a sum of 180°): Angles that make F’s are equal: x° x° y° y° b° a° a + b = 180° c° d° c + d = 180° a° a° Lesson 1: Lines and Angles adjacent angles . times as many four-holed buttons as two-holed buttons, what is the least number of buttons that could be in the box? Concept Review 5: Counting Problems 352 MCGRAW-HILL’S SAT SAT Practice 5: Counting. Probability Problems 356 MCGRAW-HILL’S SAT SAT Practice 6: Probability Problems 1. The figure above shows a spinner in the mid- dle of a disc divided into six equal parts, each labeled with a. 35, so there are 35 buttons: 5 two-holed and 30 four-holed, and 7 green and 28 blue. 713 20 FS Answer Key 5: Counting Problems 10521 PC 6xy DS 354 MCGRAW-HILL’S SAT Probability Geometrical Probability Lesson