GMAT Prep Course 1 | Jeff Kolbly | Skillshare

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GMAT Prep Course 1

teacher avatar Jeff Kolbly

Watch this class and thousands more

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Taught by industry leaders & working professionals
Topics include illustration, design, photography, and more

Watch this class and thousands more

Get unlimited access to every class
Taught by industry leaders & working professionals
Topics include illustration, design, photography, and more

Lessons in This Class

152 Lessons (2h 26m)
    • 1. About This Course 1

      0:50
    • 2. About This Course 2

      0:38
    • 3. Overview1

      0:49
    • 4. Overview2

      0:44
    • 5. Overview3

      0:28
    • 6. Overview4

      0:48
    • 7. Overview5

      0:53
    • 8. Overview6

      0:39
    • 9. Overview7

      0:49
    • 10. Overview8

      0:30
    • 11. Overview9

      0:46
    • 12. Overview10

      0:29
    • 13. Overview11

      0:23
    • 14. Overview12

      0:31
    • 15. Overview13

      0:50
    • 16. Overview14

      0:51
    • 17. Overview15

      0:38
    • 18. Overview16

      0:46
    • 19. Overview17

      0:36
    • 20. Math Orientation1

      0:22
    • 21. Math Orientation2

      1:10
    • 22. Math Orientation3

      0:15
    • 23. Math Orientation4

      0:18
    • 24. Substitution1

      0:18
    • 25. Substitution2 Example1

      0:48
    • 26. Substitution3

      0:21
    • 27. Substitution4 Example2

      1:12
    • 28. Substitution5 Example3

      2:02
    • 29. Substitution Problem1

      0:47
    • 30. Substitution Problem2

      0:53
    • 31. Substitution Problem3

      0:38
    • 32. Substitution Problem4

      2:15
    • 33. Substitution Problem5

      1:21
    • 34. Substitution Problem6

      2:30
    • 35. Substitution Problem7

      0:46
    • 36. Substitution Problem8

      1:04
    • 37. Substitution Problem9

      1:44
    • 38. Substitution Problem10

      1:15
    • 39. Substitution Problem11

      1:04
    • 40. Substitution Problem12

      1:32
    • 41. Substitution Problem13

      1:10
    • 42. Substitution Problem14

      1:26
    • 43. Substitution Plugging In

      0:13
    • 44. Substitution Plugging In Example

      1:36
    • 45. Substitution Plugging In Problem1

      0:46
    • 46. Substitution Plugging In Problem2

      0:40
    • 47. Substitution Pluging In Problem3

      0:36
    • 48. Substitution Plugging In Problem4

      0:44
    • 49. Substitution Plugging In Problem5

      1:10
    • 50. Substitution Plugging In Problem6

      0:48
    • 51. Math Notes Text 1

      0:49
    • 52. Math Notes Text 2

      0:46
    • 53. Math Notes Text 3

      1:38
    • 54. Math Notes Text 4

      0:58
    • 55. Math Notes Text 5

      0:31
    • 56. Math Notes Text 6

      1:08
    • 57. Math Notes Text 7

      0:57
    • 58. Math Notes Text 8

      1:00
    • 59. Math Notes Example1

      0:50
    • 60. Math Notes Text 9

      0:47
    • 61. Math Notes Text 10

      0:26
    • 62. Math Notes Text 11

      0:36
    • 63. Math Notes Text 12

      0:37
    • 64. Math Notes Example2

      0:28
    • 65. Math Notes Text 13

      0:33
    • 66. Math Notes Text 14

      0:22
    • 67. Math Notes Text 15

      0:53
    • 68. Math Notes Text 16

      0:39
    • 69. Math Notes Text 17

      0:57
    • 70. Math Notes Problem 1

      0:53
    • 71. Math Notes Problem 2

      0:54
    • 72. Math Notes Problem 3

      0:38
    • 73. Math Notes Problem 4

      1:00
    • 74. Math Notes Problem 5

      0:25
    • 75. Math Notes Problem 6

      0:56
    • 76. Math Notes Problem 7

      0:48
    • 77. Math Notes Problem 8

      1:23
    • 78. Math Notes Problem 9

      1:31
    • 79. Math Notes Problem 10

      1:38
    • 80. Defined Functions Text 1

      1:21
    • 81. Defined Functions Example 1

      0:25
    • 82. Defined Functions Example 2

      0:57
    • 83. Defined Functions Example 3

      0:59
    • 84. Defined Functions Example 4

      1:11
    • 85. Defined Functions Example 5

      1:23
    • 86. Defined Functions Example 6

      2:56
    • 87. Defined Functions Text 8

      0:31
    • 88. Defined Functions Example 7

      0:50
    • 89. Defined Functions Example 8

      1:55
    • 90. Defined Functions Example 9

      0:49
    • 91. Defined Functions Problem 1

      0:27
    • 92. Defined Functions Problem 2

      0:52
    • 93. Defined Functions Problem 3

      0:28
    • 94. Defined Functions Problem 4

      1:38
    • 95. Defined Functions Problem 5

      0:55
    • 96. Defined Functions Problem 6

      1:09
    • 97. Defined Functions Problem 7

      1:13
    • 98. Defined Functions Problem 8

      1:14
    • 99. Defined Functions Problem 9

      1:12
    • 100. Defined Functions Problem 10

      0:50
    • 101. Defined Functions Problem 11

      2:48
    • 102. Defined Functions Problem 12

      0:50
    • 103. Defined Functions Problem 13

      0:47
    • 104. Defined Functions Problem 14

      1:41
    • 105. Number Theory Text 1

      0:56
    • 106. Number Theory Example 1 hd

      1:09
    • 107. Number Theory Text 2

      0:48
    • 108. Number Theory Text 3

      1:01
    • 109. Number Theory Example 2 hd

      1:25
    • 110. Number Theory Text 4

      1:34
    • 111. Number Theory Text 5

      0:36
    • 112. Number Theory Text 6

      0:38
    • 113. Number Theory Text 7

      0:38
    • 114. Number Theory Text 8

      1:08
    • 115. Number Theory Text 9

      0:23
    • 116. Number Theory Text 10

      0:23
    • 117. Number Theory Text 11

      0:25
    • 118. Number Theory Text 12

      0:26
    • 119. Number Theory Text 13

      0:33
    • 120. Number Theory Example 3 hd

      1:40
    • 121. Number Theory Example 4 hd

      0:46
    • 122. Number Theory Example 5 hd

      0:57
    • 123. Number Theory Problem 1 hd

      2:45
    • 124. Number Theory Problem 2 hd

      0:35
    • 125. Number Theory Problem 3 hd

      0:45
    • 126. Number Theory Problem 4 hd

      1:46
    • 127. Number Theory Problem 5 hd

      1:48
    • 128. Number Theory Problem 6 hd

      0:38
    • 129. Number Theory Problem 7 hd

      0:49
    • 130. Number Theory Problem 8 hd

      2:02
    • 131. Number Theory Problem 9 hd

      1:54
    • 132. Number Theory Problem 10 hd

      0:34
    • 133. Number Theory Problem 11 hd

      0:44
    • 134. Number Theory Problem 12 hd

      0:58
    • 135. Number Theory Problem 13 hd

      1:51
    • 136. Number Theory Problem 14 hd

      1:10
    • 137. Number Theory Problem 15 hd

      0:54
    • 138. Number Theory Problem 16 hd

      1:27
    • 139. Number Theory Problem 17 hd

      1:36
    • 140. Number Theory Problem 18 hd

      1:20
    • 141. Number Theory Problem 19 hd

      0:52
    • 142. Number Theory Problem 20 hd

      1:08
    • 143. Number Theory Problem 21 hd

      1:09
    • 144. Number Theory Problem 22 hd

      0:30
    • 145. Number Theory Problem 23 hd

      1:14
    • 146. Number Theory Problem 24 hd

      0:43
    • 147. Number Theory Problem 25 hd

      1:30
    • 148. Number Theory Problem 26 hd

      0:42
    • 149. Number Theory Problem 27 hd

      1:05
    • 150. Number Theory Problem 28 hd

      0:26
    • 151. Number Theory Problem 29 hd

      0:35
    • 152. About This Course1 hd

      0:42
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Every year, students pay $1,000 and more to test prep companies to prepare for the GMAT. Now you can get the same preparation in an online course.

GMAT Prep Course 1 presents a thorough analysis of the following topics:

1) Orientation to the GMAT

2) Math Orientation

3) Solving Math Problems by Substitution

4) Math Notes

5) Defined Functions

6) Number Theory

Features:

Videos! About 130 videos explaining the text, examples, and exercises in step-by-step detail.

Duals: These are pairs of problems in which only one property is different. They illustrate the process of creating GMAT problems.

Meet Your Teacher

Teacher Profile Image

Jeff Kolbly

Teacher

I have been teaching, writing, and publishing in the test prep field for 25 years.

See full profile

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Transcripts

1. About This Course 1: about this course. Although the Guilmette is a difficult test, it is a very learn herbal test. This is not to say that the demand is beautiful. There is no banger tricks that will show you how to master it overnight. You probably have already realized this. Some courses nevertheless offer inside stuff or tricks, which they claim will enable you to beat the test. These include declared that answer choices B, C or D are more likely to be correct and choices a or E. This tactic, like most of its type, does not work. It is offered to give the student the feeling that he or she is getting the scoop on the test. The Guilmette cannot be beaten, but it can be mastered through hard work and a little thought and training yourself to think like a test right. 2. About This Course 2: this course will introduce you to numerous analytic techniques that will help you immensely not only on the G Matt, but in business school as well. For this reason, studying for the Guilmette can be a rewarding and satisfying experience. Although the quick fix method is not offered in this course, about 15% of the material is dedicated to studying how the questions are constructed, knowing how the problems were written and how the test writers think we'll give you useful insight into the problems and make them less mysterious. Moreover, familiarity with the Jeanette structure will help reduce your anxiety. The more you know about this test the left sanctions, she will be the day you take it. 3. Overview1: What does that she met? Measure The Guilmette is an aptitude test. Like all aptitude tests, it must use a medium in which the measure intellectual ability The Jeanette has chosen math , English and logic. Okay, that she met is an aptitude test. The question is, does it measure aptitude for business school the G. M s ability to predict performance in school as as poor as the S a. T s. This is to be expected since the problems on the tests are quite similar, though the formats are different. However, the Guilmette also includes two types of questions, arguments and data sufficiency that that s a T Does not many students struggle with these questions because they are unlike any material they have studied in school. However, the argument and data sufficiency questions are not inherently hard. And with sufficient study, you can raise your performance on these questions significantly 4. Overview2: no tests can measure all aspects of intelligence. Thus, any mission test, no matter how well written, is inherently inadequate. Never left some form of mission testing is necessary. It would be unfair to base acceptance of business school solely on grades. They can be misleading. For instance, would it be fair to admit a student with an A average earned an easy class is over student with a B average earned in difficult classes? A school's reputation is too broad a measure to use his admission criteria. Many students seek out easy classes and generous instructors in hopes of inflating their G p A. Furthermore, a system that would monitor the academic standards of every class would be cost prohibitive and stifling. So until a better system has proposed, the admission test is here to stay. 5. Overview3: for many of the G. Matt. The G Men is a three and 1/2 hour computer adaptive tests cat. There are four sections in the test writing, math, verbal and integrated reasoning. The writing section begins to test. You will type your essay on the computer using a very basic word processing. Each question must be answered before you can go to the next question further. You cannot return to a question once you have gone to the next question. 6. Overview4: the G men is a standardized test. Each time it has offered, the test has as closely as possible. The same level of difficulty is every previous test. Maintaining this consistency is very difficult. Hence the experimental questions questions that are not scored. The effectiveness of each question must be assessed before it can be used on the G Matt, a problem that one person finds easy another person may find hard and vice versa. The experimental questions measure the relative difficulty of potential questions. If responses to a question do not perform too strict specifications, the questions rejected. About 1/4 of the questions are experimental. The experimental questions can be standard out data sufficiency, reading comprehension, arguments or sentence correction. You won't know which questions are experimental. 7. Overview5: because the bugs have not been worked out of the experimental questions or to put it more directly because you are being used as a guinea pig to work out the bugs. These unscored questions are often more difficult and confusing than the scored questions. This brings up unethical issue. How many students have run into experimental questions early in the test and have been confused and discouraged by them? Crestfallen by having done poorly on a few experimental questions, they lose confidence and performed below their abilities on the other parts of the test. Some testing companies are becoming more enlightened in this regard and are administering experimental questions. A separate practice tests. Unfortunately, the G man has yet to see the light, knowing that experimental questions can be disproportionately difficult. If you do poorly on a particular question, you can take some solace in the hope that it may have been experimental. In other words, don't allow a few difficult questions to discourage your performance on the rest of the test. 8. Overview6: the cat in the paper and pencil tests. The computerized Guilmette used the same type of questions as it the old paper and pencil test. The only thing that has changed is the medium. That is the way the questions are presented. There are advantages and disadvantages to the cat. Probably the biggest advantages are that you could take the cat just about any time, and you can take it in a small room with gifts a few other people instead of in a large auditorium with hundreds of other stress people. On the other hand, you cannot return to previous questions. It is easier to misread a computer screen than it is to misread printed material, and it could be distracting looking back and forth from the computer screen to your scratch paper. 9. Overview7: pacing, although time is living it on the gee, Matt, working too quickly can damage your score. Many problems hinder a subtle points, and most require careful reading of the set up. Because undergraduate school puts its heavy reading loads on students, many will follow their academic conditioning and read the questions quickly, looking only for the just of what the question is asking Once they have found it, they mark their answer and move on, confident that they have answered it correctly later, many are startled to discover that they miss questions because the either miss read the problems are overlooked. Settle points to do well in your undergraduate classes, you had to attempt to solve every or nearly every problem on a test. Not so with the G map. For the vast majority of people, the key to perform well in the G Matt is not the number of questions they solve within reason, but the percentage we solve correctly. 10. Overview8: scoring the G Matt. The two major parts of the tests are scored independently. You'll receive a verbal score 0 to 60 at a mask or also 0 to 60. You will also receive a total score 200 to 800 a ready score 0 to 6. The average verbal score is about 27. The average mass scores about 36 in the average total score is about 5 40 In addition, you will be assigned a percentile ranking, which gives the percentage of students with scores below yours. 11. Overview9: guessing on the cat. You cannot skip questions. Each question must be answered before moving on to the next question. However, if you can eliminate even one answer choice, guessing can be advantageous. We'll talk more about this later. Unfortunately, you cannot return to previously answered questions on the test. Your first question will be of medium difficulty. If you ask for it correctly, the next question will be a little harder. If again you answer it correctly. The next question will be a harder still and so on. If you're Gina, skills are strong and you're not making any mistakes. You should reach the medium hard or heart problems by about the fifth problem. Although this is not very precise, it can be quite helpful. Once you have passed the fifth question, you should be alert to subtleties and any seemingly simple problems. 12. Overview10: students often become obsessed with a particular problem and waste time trying to solve it . To get a top score learned to cut your losses and move on. The exception to this rule is the 1st 5 questions of each section. Because of the importance of the 1st 5 questions to your score, you should read and solve these questions slowly and carefully because the total number of questions answered contributes to the calculation of your score. You should answer all the questions, even if this means guessing randomly before the time runs out. 13. Overview11: order of difficulty. Most standardized paper and pencil test list problems in ascending order of difficulty. However, on a cat, the first question will be of medium difficulty. If you answer it correctly, the next question will be a little harder. If you answered incorrectly, the next question will be a little easier because the Guilmette adapts to your performance . Early questions are more important than later ones. 14. Overview12: the two out of five rule, it is significantly harder to create a good but incorrect answer choice than it is to produce the correct answer. For this reason, usually only two attractive answer choices are offered. One correct the other either intentionally misleading or only partially correct. The other three answer choices air usually fluff. This makes educated guessing on the G Matt immensely effective. If you can dismiss the three fluff choices, your probability of answering the question successfully will increase from 20% to 50%. 15. Overview13: only two answers have any real merit choices? A and E. The argument generalizes from the survey to the general carbine population. So the reliability of the projection depends on how representative sample is. At first glance, choice A seems rather good, because 10% does not seem large enough. However, political opinion polls typically are based on only 0.1% of the population. More importantly, we don't know what percentage of GM car owners received the survey. Choice E, on the other hand, points out that the survey did not represent the entire public. So it is the answer. The other answer choices. Air quickly dismissed choice Be simply states that for made the same mistake that GM did. So I see is a row of it. Finally, Choice D, rather than explaining discrepancy would give me the more reason for GM to continue making large cars 16. Overview14: computer screen options. When taking the test, you will have six on screen options or buttons. Quit section time, Help next and confirm unless you just cannot stand it any longer. Never slept, quit or section. If you finish the section early, just relax while the time runs out. If you're not pleased with your performance on the test, you can always cancel it at the end. The time button allows you to display or hide the time during the last five minutes. Time display cannot be hidden, and it will also display the seconds remaining. The help button will present a short tutorial showing how to use the program you selected. Answer choice by clicking the small level next to it to go to the next question, click the next button. He will then be asked to confirm your answer by clicking the confirm button. Then the next question will be presented 17. Overview15: test day. Bring a photo I d. Bring in list of schools that you wish to send your scores to arrive at the test center 30 minutes before your test appointment. If you arrive late, you might not be emitted and your feet will be forfeited. You will be provided with scratch paper. Do not bring your own and do not remove scratch paper from the testing room. You cannot bring testing AIDS into the testing room. This includes pens, calculators, watch calculators, books, rulers, cellular phones, watch Lawrence and any electronic or photographic devices. You may be photographed and videotaped at the test center. 18. Overview16: questions and answers. What is the gene that, given the test, has given your round during normal business hours? There's often one week during each month in which the test is not offered. How important is that you met in house? It used. It is crucial, although business schools may consider other factors. The vast majority of emission decisions are based on two criteria. Your Jeanette score and your G p A. How many times should I take the G? Matt? Most people are better off preparing thoroughly for the test, taking it once and getting their top score. You can take the test that most once a month and at most five times any one year period. But some visit schools will average your scores. You should call the schools to which you are applying to find out their policy, then plan your strategy accordingly. 19. Overview17: Can I cancel my score? Yes. When you finish the test, the computer will offer you the option of cancelling the test or accepting it. If you cancel a test, neither you nor any school will see your score. If you accept the test, the computer will display your score and it will be available to all schools. Where can I get the registration forms? Most colleges and universities have the forms. You can also get them directly from the Graduate Management Commission Council by writing to the falling address or calling 1 807 1746 to 8 or online at n b a dot com. 20. Math Orientation1: orientation types of math questions. The math section consists of 37 multiple choice questions. The questions come in two formats. Standard multiple Choice, which will study in this section, and data sufficiency, which will study the next section. The mass section is designed to test your ability to solve problems, not to test your mathematical knowledge. 21. Math Orientation2: G. Matt versus S A T G. Matt Math is very similar to S a t math those slightly harder. The mathematical skills tested are very basic only first year high school algebra and geometry with no proof. However, this does not mean the mass section is easy. The medium of basic mathematics has chosen so that everyone take the test will be on a fairly even playing field. Although the questions require only basic mathematics and all have simple solutions, it can require considerable ingenuity to find the simple solution. If you have taken a course in countless or another advanced math topic, don't assume that you will find the mass section easy. Other than increasing your mathematical maturity, little you learned in countless will help on the G man. As mentioned above, every team at math problem has a simple solution. But fighting that simple solution may not be easy. The intent of the mass session is to test how skilled you are finding simple solutions. The premise is that if you spend a lot of time working out long solutions, you will not finish as much of the test as students who spot the short, simple solutions. So if you find yourself performing long calculations or playing fast, mathematics stop. You're heading in the wrong direction 22. Math Orientation3: to ensure that you perform it. You're expected level on the actual gene that you need to develop a level of mathematical skill that is greater than what is tested on the gene, that it's about 10% of the math problems in this course or harder than actual team at math problems. 23. Math Orientation4: the structure of this part of the course because it can be rather dull to spend a lot of time with you and basic math before tackling full fledged D map problems. The first few chapters present techniques that don't require much foundational knowledge of mathematics. Then, in later chapters, review was introduced as needed. 24. Substitution1: substitution is a very useful technique for solving math problems. It often reduces hard problems to routine ones. In the substitution method, we choose numbers that have the properties given in the problem and then plugged them into the answer choices, a few examples with little straight. 25. Substitution2 Example1: we're told that in is an audit manager. So let's choose end to be, say, one, which is a convenient number to calculate with and plug it into the answer choices and see which one turns out to be even for choice. A. We get one cube, which is one, and that's not even so. Eliminate a or choice B. We get 1/4 which is not even an energy. So eliminate be and for choice. See, we get five, which again is not even. And for choice d we get four, which is even hence, the answer is D. 26. Substitution3: When using the substitution method, be sure to check every answer choice because the number you choose may work for more than one answer choice. If this does occur, then choose another number and plug it in and so on, until you have eliminated all but the answers. This may sound like a lot of computing, but the calculations can usually be done in a few seconds. 27. Substitution4 Example2: were given that innocent imager. Let's choose end to be one looking that into choice. A. We get four. And remember, we're looking for the answer choice that cannot be uneven imager. So reject A. Since four is even for choice B. We get negative for which again is even Detroit. Reject choice B and choice. See is to again and even never reject. See for choice. D we get five. This is not any even integer, so it may be our answer, but we still have the check Choice e and this gives of seven also not a even integer. So to eliminate D or E, let's choose another value for in How about end to be zero that immediately? Choice D is three, and Choice E is to, since two is even that eliminates E. Therefore, the answer is D 28. Substitution5 Example3: we're given the X over, wise of fraction greater than one. Let's choose X to beat three. And why to be to that does give us a fraction greater than one plug in the values into the answer choices we get to, which is not less than one, which is what we're looking for. So we eliminate a for choice B. We get 1/2 which is less than one. So choice be maybe an answer and choice. See, gives is the square root of three House. Now it's almost impossible to calculate the square root of three halves in our heads, but it clearly is greater than one. So we eliminate C for choice. D. We get to over three, which is less than one so d Maybe the answer and choice e we get to which is not less than one. So eliminate e now to decide between D and be Let's choose extra b six, and why to be to then our fraction X over y is six over to or three, and that is greater than one, as it must be to satisfy the conditions of the question, plugging it into choice. D. We get to over six, which is 1/3 and that is less than one. And for Choice B, we get one which is not less than itself. So that eliminates choice. Be hence, the answer is D. 29. Substitution Problem1: we're told that in is unaudited jer. So let's choose end to be one and plug it into each answer choice and see which one turns out to be even for choice A. We get 1/2 which is not even eliminate. For choice B, we get seven, which is not even so. Eliminate which way? See, we get two times. One which is to it is Even so, it may be your answer, but we have to check the remaining ones for D. We get one to the fourth, which is one, and that's odd. So eliminate and find the square root of one is one, which again is also limited. E since the answer is C. 30. Substitution Problem2: Let's choose extra before and why to be nine and 49 are perfect squares, so we plug him into answer choices. We say gives us four squared, which is 16 which is a perfect square, but we can't eliminate it because it may not be a perfect square for some other numbers. Choice be gives us four times nine, which is 36 which is also a perfect square, namely six where so I see a zoo four times four or 16 which again is another perfect square . Foreswear and choice D is for plus nine or 13 and this is not a perfect square. Hence, the answer is D. 31. Substitution Problem3: Let's choose X to be one anodyne injure and why to be too uneven injure, plugging these values in the choice? A. We get four, which is uneven energy er and the problem is asking, which was one of the answer choices could be, even since a could be even, it must be our answer. None of the other answer choices can be even because if they were, they would be more answer. Want more than one answer to the question, and it would be flawed. 32. Substitution Problem4: we're given. The K is a number between zero and one. We can choose any number between 01 But let's choose K to be a perfect square, since Choice D has a radical choosing K two B 1/4 which is a perfect square, namely 1/2 squared, plugging that into choice A. We get 3/8 which is bigger than on fourth. So we eliminate a. A convenient way of comparing fractions is the cross multiply. The larger fraction will be on the same side as a larger product, multiplying three and four. We get 12 which is greater than one times eight, which is a since 12 is greater than 83 X is greater than 4 1/4 We could have also gotten a common denominator for this problem, and to do that we'd multiply top and bottom of 1/4 by two, which gives us two eights. Since three is bigger than two. Three h is bigger than 12 eights, which is 1/4 attorney. The choice be we get when you divide by a fraction, you invert, multiply So flipping the 1/4 over, we get 4/1, which, of course, is just four and four is not smaller than 1/4. So they may be Choice. See is that the value of 1/4 which is 1/4 and 1/4 is not less than itself. It's equal to itself for choice. D we get the square root of 1/4 which is 1/2 and 1/2 is not less than 1/4. It's bigger than 1/4 so by elimination, the answer is E. 33. Substitution Problem5: suppose we start reading the book on page one and stop reading the book on page two. Then h would be one and K would be to plugging these values and the answer choices. We And by the way, we would have read two pages, then plugging these values and answer choices. We look for the one that returns the number two for choice A. We get three, which is wrong. So we eliminate a poor choice. B. We get negative one and you can't read a negative number of pages. So eliminate, be for choice. See, we get three, which is not equal to two. So that's eliminated. And for choice d, we get zero. And we know we've read two pages, not zero pages. So by elimination answers e. But let's check that choice. So we get to which is what we're expecting. So the answer is E 34. Substitution Problem6: were given that name isn't even imager. So let's choose him to equal to plugging that into foreign plus one. We get nine and the 1st 2 even integers bigger than nine or 10 and 12. And there's some is 22. So we plug in M equals two into the answer choices and see which one returns the number 22 for choice A. We get 18 so we eliminate a for choice B. We get 20 so we eliminate be and for choice. See, we get 22 so it may be our answer, but we have to check others because it may work for more than one for choice. D. We get 24 so eliminate d. And for choice E, we get 26 so live in 80. Therefore, by elimination. The answer is C, but we can choose any value for em. We could have chosen em to be zero. That would make out the calculations a little easier. And zero is an even number because two times zero is zero. Um, plus one would equal one and the first to even injures bigger than one are two and four and there's some is six So now we just plug in M equals zero into the answer choices and see which one equals six. Plugging into choice A. We get to which is not equal to six. We eliminate a for choice, be we would get four. So we eliminate that for choice. See, we would get six. So again it looks like it's our answer and choice d would get eight, and choice even would get 10. So again, the answer is choice. See, with him having the value of zero. 35. Substitution Problem7: we're told that X squared is even. Let's let X Square equal four. Then taking the square root of both sides. We get X equals two and X equals negative, too, in both cases, excess. Even so, statement one is not necessarily true. That eliminates A and D. And if the Cube each of these numbers here we get to Cube is a and negative two cubes this negative a again. Both her negatives. So statement three may not be true, which eliminates C and E. It's by elimination. The answer is B. 36. Substitution Problem8: were given that X is divisible by eight, but not by three. So let's choose X to be a, which is courses the divisible by itself. And it's not divisible by three plugging this value into the answer choices We were looking for the one that is not an energy for choice. A. We get four, which, of course, is an energy er so 11 8 A. For choice B, we get 8/4, which is to eliminate. Be for choice. See, we get 8/6, which reduces to 4/3, and that is not an energy. So it may be your answer, but we need to check it for the last two answer choices as well, because it may not work for those for choice. D We get 8/8, which is one So the Mate d and for choice E we get eight, which is an energy solemnity. Therefore, by elimination. The answer is C 37. Substitution Problem9: were given that PMK your positive integers. Let's choose P to be one and cue to be to too easy numbers to calculate with then p times kill is equal toe to and p times Q Plus two is equal to four. So we're looking for all the energies that are bigger than two in less than four. And, of course, there's only one manager in that range the number three. So we're looking for the answer choice that returns one for these values of P and Q that clearly eliminates a playing in P equals one. In the choice be we get three and be careful. We're we're looking for the number of energy is not for the number three, the number of energies between two and four. So be would be an eye catcher that we eliminate that for choice. See, we get negative one, which is not. You can't have a negative number of never so eliminate C and for choice d we get one, so it may be your answer, but we have to check choice, see to make sure it doesn't work for more than one choice. And choice E again gives us three another eye catcher. So by elimination, the answer is D 38. Substitution Problem10: we're told that X and Y are prime numbers. Let's choose X to be three, and why to be to then X minus Y is equal to three minus two or one. This eliminates choice. A next choose X to be five and why to be three, then X minus y is equal to five minus three or two and that limits be next. Choose X to be 17 and why to be three. Then there difference is 14 which eliminates D and finally choose X to be 23 and why to be three. Their differences 20 which will inmates e there for the answer is C. 39. Substitution Problem11: were given that X is an integer Let's choose X to be one and easy energy to calculate with then two times X plus one well equal four now the next to images bigger than four R five and six, and the product five times six is 30. Now we just plug the value X equals one into each of the answer choices and see which one equals 30 for choice A. We get 30 and this is the only answer choice that will return of 30 30 as you can see pretty quickly, just by glancing at other answers. Therefore, the answer is a. 40. Substitution Problem12: were given that X is an integer divisible by three, but not by too. Let's choose X to be three, which, of course, is divisible by three. And it's not divisible by two. Plugging that into choice A we get to, which is a dinner jer. So eliminate a for choice B. We get three sevens, which is not an integer, so it may be your answer, but we got a check to see if it might not be an energy for some other answer choices for choice. See, we get nine divided by three, and that's a knitter. So the night sea for choice d we get nine. Which again, is it energy or so the night D? And for Choice E. We get 3/24 which does reduce to 1/8. But still, it's not an integer, so he is a possible answer. And to the side between B and E, we choose another value for X. Let's choose X to be 21. Then choice be becomes 21. Divide by seven, which is three, and that's an interviewer. So eliminate be therefore, by elimination. The answer is choice E 41. Substitution Problem13: we were asked which of the following answer choices is a solution to this equation. So let's just plug it in each answer choice into the expression X to the fourth minus two X Square and see which one equals negative one for choice A. We get zero to the fourth power, minus two times zero squared and zero to the fourth Power zero. And to answer all squared is zero. Two times zero is zero and zero minus zero is zero. And, of course, zero does not equal negative one. Therefore, it's not a solution. So eliminate a turning. The choice be. We get one race to the fourth. Power is one and one squared is one. So we get one minus two, which is negative one, and that's exactly what we're looking for. Therefore, Choice B is a solution to the equation. Answers be 42. Substitution Problem14: Let's choose X to be zero and easy number to calculate with plugging that into the expression we get 3/5 and plugging zero into answer choice A. We get five force multiplying that by our expression three. Thetis We get three force, which doesn't equal negative too, so we can eliminate a and for choice be when we plug in X equals zero, we get 10/3 and multiplying that by 3/5 we get to which still doesn't equal negative too. So we can eliminate, be and plugging zero into choice. See, we get 10 negative 10/3 and multiplying that by 3/5. The threes cancel the five cancels into 10 twice. So we're left with negative too, which is exactly what we want there for. The answer is C. 43. Substitution Plugging In: Sometimes instead of making up numbers to substitute into the problem, we can use the actual answer choices. This is called plugging in. It is a very effective technique, but not his communist substitution. 44. Substitution Plugging In Example: Sometimes, instead of making up numbers to substitute into a problem, we can use the actual answer choices. This is called plugging in. It's a very effective technique, but not as communist substitution. In this problem. We're told that the digits of the number must add up to 18. So let's see what's number of choices. Answer choices. Add up to 18 for choice A. We get 12 which is not equal 18. So eliminate a for choice B. We get 18 so it may be the answer for choice. See, we get five plus three plus one, which is nine and again. That's not 18. So eliminate C for choice. D. We get 18 and for choice E we get 20 which eliminates e going to the next criteria. That tens digit is twice the hundreds digit. Well, for choice d, we have the tens. Digit is five and five does not equal two times eight. In other words, it is not twice the hundreds digit that limits D. Therefore, the answer is B and notice. We didn't use the final condition that the hundreds digit is 1/3 the tens digit. It is not uncommon for Jerry problems to give you more information than you actually need 45. Substitution Plugging In Problem1: were given that the tens digit is twice the units digit well, in a two digit number, the tens digit is the first digit, and the units digit is the second digit. Now, one is not twice the value of two, so that eliminates a and four is not twice the value. Three. That limits see, and eight is not twice the value of three, which eliminates E. Now we reverse the digits for choice B, and we'll get 21 will get 12 but 21 minus 12 does not equal 27. Therefore, by elimination, the answer is D. 46. Substitution Plugging In Problem2: Let's plug in the answer choices until we find one that causes the given expression toe have a value of one plugging in one we get to which doesn't equal one. So we eliminate c turning the next easiest number to calculate with We goto choice d Playing to into that expression, we get one, which is what we want. Therefore, the answer is D. 47. Substitution Pluging In Problem3: we were given that the some of the digits is 12 which would eliminate be because five plus four is nine and it eliminates E because three plus one is four. Now we're told that the tens digit is 1/3 the unis digit. That would eliminate a because nine is not 1/3 of three. And it was eliminate 48 because four is not 1/3 of eight, but three is 1/3 of nine. So the answer is D. 48. Substitution Plugging In Problem4: Let's start with choice, see if it turns out to be too large, then we'll work down the list. And if it turns out to be too small, then we'll work up the list. If there are eight people on the bus, then on the first stop there will be half that never four. And on the second stop there'll be half that Number two and then on the third and final stop will be half that never, which is one. Therefore, the answer is C. We could also solve this problem by working backwards from the result. If there's only one person on the bus, then on the third stop there must have been to And on the second stop there must have been four. And on the first stop, there must have been eight. 49. Substitution Plugging In Problem5: we could solve this problem by solving the equation. That will be a lot easier to just plug in the answer choices. Plug it in one we get and this does not equal one. So we eliminate choice. A playing in choice B we get now to the six is 64 and too cute is eight times five is 40 minus 16 gives us 8/8, which is one. Therefore, the answer is B. 50. Substitution Plugging In Problem6: we could answer this problem by solving the equation, but it would be quicker to simply plug in the answer choices and see which one of them turns. Causes this expression to have a value of negative one plugging in zero we get and zero does not equal negative one. So eliminate a plugging in one. We get negative one, so answer is B. 51. Math Notes Text 1: We'll discuss many of the concepts in this chapter in depth later. But for now, we need a brief review of these concepts for many of the problems that follow to compare two fractions, cross multiply. The larger product will be on the same side as the larger fraction. For example, forgiven 56 versus 67 we cross multiply. It gives us five times seven versus six times six or 35 versus 36. Now 36 is larger than 35. Hence, 67 is larger than 56 52. Math Notes Text 2: taking the square root of a fraction between zero and one makes it larger. For example, the square root of 1/4 is 1/2 and 1/2 is greater than 1/4. But be careful. This is not true for fractions greater than one. For example, the square root of nine force is three halves, but three houses less than nine also square in a fraction between zero and one makes it smaller. And that which is somewhat counter intuitive because we normally expect squares to make numbers bigger. And writers of the test like to play on this issue. For example, 1/2 square gets smaller, it becomes 1/4. 53. Math Notes Text 3: in this form that people often forget that you have to distribute the two on each base. They will often distribute it only on the X, not the A. For example, three times two squared that the square on the two applies only to the two and not the three. So we get three times four is 12. But if you encase it in parentheses, as we do here, then this square will be distributed on each term and you would get 36. Or you could simply multiply the two terms and get six and six square is 36 and the mistake is often seen in this expression. Here, it would seem national Teoh, apply the square to the negative, but unless it's in case in parentheses, you do not to see this more clearly. You can rewrite negative X squared as negative one times X squared, and now it's clear that the square does not reach the negative one. So again, if you want to square the negative, you have to encase it in parentheses, which does eliminate the negative sign for another example. This square here on the five does not affect the negative, and you can show that more clearly by writing. It is negative. One times five squared, five squared is 25 negative one times 25 negative. 25. And to eliminate the negative, it has to be inside the parentheses, which gives you negative five times a negative five, which is two ***. Give us a positive and so we get positive 25. 54. Math Notes Text 4: This is a common error made when simplifying complex fractions. And this here is the correct formula to derive it. We have one over a divided by B over one, and to divide you invert, multiply. So we get one over a times one over B, which simplifies toe one over a times B and similarly for this formula here we have one over a over B with the being division bar here. So it right down the numerator and multiplied by the denominator, flipped over the reciprocal be over a and one times B is B over a. 55. Math Notes Text 5: This is one of the most common mistakes made on the test. When you have a negative outside of the parentheses, you must distribute that negative on each term so that a becomes negative and so does the B . So for here we would have negative negative two minus three, which gives us negative five. And similarly here we distribute the negative on each term to get negative two minus X. 56. Math Notes Text 6: memorize the following factory and formulas. They occur frequently on the test. The 1st 1 is the difference of squares Formula Statement B is what's called a perfect square. Try normal notice. The middle term is twice the square root of the outer, too. If it's positive here, then it's positive there. And if it's negative here than its negative there. For example, if you had X squared plus four X plus four, then this would be a perfect square trying no meal because the middle term can be written as two times two packs where this term is the square root of the first, and this term is the square with the last, and we have twice it. Therefore, it could be written as a perfect square. Try no male, namely X plus to the quantity squared. And the statement three is our common distributive property. Distribute the A onto the B and the sea. We get a B plus A C 57. Math Notes Text 7: This first rule says that the product of two square roots is the same thing as a square root of their product. For example, the square root of five times the square root of five will return. The same value is the square root of five times five, which is the square root of 25 or five. The second formula says that the square root of a quotient is the same thing as three quotient of the square roots. For example, the square root of 16/4 is the same thing as a square root of 16 over the square root of four. The square root of 16 is for that's where the forest to so we get to. 58. Math Notes Text 8: the plethora of therm says that the square of the high Patna of C is equal to the sum of the squares of the legs of the triangle, and you comply this formula to a triangle Onley if it's a right triangle. So you have to have a little square in one of the virtus ease of the triangle or something else that indicates it's a right triangle in order to apply this formula, for example, if a was, say, three and B four, then see would have a value of a square three squared plus B squared or four square, which gives us nine plus 16 or 25 bringing the sea. Swear back into the equation. We take the square root herbal sides, and that gives us C equals the square root of 25 which is five. 59. Math Notes Example1: since the triangle is a right triangle indicated by the little square. Staggering through him applies and we get each square plus three squared is equal to five squared or H square plus nine equals 25. Subtracting nine to get eight square equals 16 taking us where route we get. H equals four. Since the area is 1/2 the base times the height, which is 1/2 base, is three and the height we've calculated to be four. Cancel the two and we get six. Hence, the answer is a. 60. Math Notes Text 9: when parallel lines are cut by a trans versatile three important Anglo relationships or form alternate interior angles are congruent here in here. Also, this was labeled B. Then this would be a bia's well and corresponding angles or congruent. So these air to Arkan growing. Also, these two here and here are congruent and interior angles on the same side of the trans versatile or supplementary. So a plus b equals 1 80 or if you had C and D, then C plus D would also equal 1 80 61. Math Notes Text 10: in a triangle. The exterior angle is equal to the sum of its remote interior angles, and therefore is greater than either of in this case is a Nextera angle because it's outside the triangle so equal to some of its remote interior angles A and B, and since both A and B are positive, he is greater than either one of them. 62. Math Notes Text 11: a central angle has by definition the same measure as it's intercepted arc, so in this case we have 60 degrees of intersected arc. Therefore, the central angle is defined to be 60 degrees and in an inscribed angle, an angle that has its vertex on the edge of the circle. The measure is 1/2 it's intercepted arcs, so here we get 1/2 of 60 degrees, which is 30 degrees. 63. Math Notes Text 12: there are 100 and 80 degrees in a straight angle that is in a straight line. So here the angle why plus the angle axe will add up to 100 80 degrees. Also, the angle some of the triangle is 180 degrees. That means when you add up the three interior angles of a triangle, you get 180 degrees. And this formula is very important on the test. So if you had be is 100 and maybe see is 60 then they would have to be 20. In order for the angles, add up to 180 degrees. 64. Math Notes Example2: since the angle. Some of a triangle is 180 degrees. We get 100 plus 50 plus C equals donating or 1 50 plus C equals 1 80 and subtracting 1 50 We get C equals 30. Hence the answer is C. 65. Math Notes Text 13: to find the percentage increase or decrease, find the absolute increase and divide by the original mount. For example, if we have assured that selling for $18 it's marked up to $20 then the absolute increase is 20 minus 18 or $2 and to form the percentage increase we take. The increase, which is to and we divide by the original price, which is 18 and the two cancels into 18 gives us 1/9, which is approximately 11%. 66. Math Notes Text 14: systems of simultaneous equations can most often be solved by merely adding or subtracting the equations. For example, if we're given these two equations, they were asked to calculate the expression X minus y. We would merely aligned equations vertically and subtract and then get a value of one for for the difference of X and Y. 67. Math Notes Text 15: When counting elements center in overlapping sets, the total number will equal the number in one group, plus the number in the other group, minus the number common to both groups. Vin diagrams are very helpful with these problems. Example. If any certain school 20 students are taking math and then are taking history and seven are taking both. How many students are taken either math or history? So for this problem, we cannot merely add the 10 in the 20 because seven students are in both history simultaneously in history and math. So if we add the 10 and the 20 will be adding the seven twice, so instead we add the 10 and 20 and then subtract the 71 so that we don't Adam twice and 10 plus 20 is 30 minus. The seven is 23 total students. 68. Math Notes Text 16: the number of imagers between two imagers inclusive is one more than their difference. For example, the number of energies between 49 and 101 inclusive by the way, inclusive beings that were counting the 49 the 101 Their differences 52 plus one is 53 numbers between them inclusive. To see this principle more clearly, choose smaller numbers, say, nine and 11. The difference between nine and 11 is too, but there are three numbers between them inclusive 9 10 and 11 one more than their difference. 69. Math Notes Text 17: the convention used for rounding numbers is if the following digit is less than five than the preceding digit is not changed. But if the following digit is greater than or equal if I, then the preceding digit is increased by one. For example, if we want around off 65,439 to the nearest thousands, that's why five is in both phone here we checked the the following number, which is four. Since four is less than five. We do not change the digit, and we just write three zeros now for this one here. If we want to round off to the nearest 100th we checked, the falling digit, which is seven seven, is bigger than five. So when we dropped the seven, we have to round up one unit, so the six turns into a seven. 70. Math Notes Problem 1: from the formula A X. The quantity squared equals a squared X squared. We see that two X the quantity squared is equal to two squared X squared or four X squared . Now, since X is not equal 04 X squared is clearly larger than two x squared. In fact, four X squared is twice two X square because the four can be written as two times two, which gives us two times two X squared. So four X squared is twice two X where hence the answer is B. 71. Math Notes Problem 2: a convenient way of comparing fractions is the cross multiply, and the larger fraction will be on the same side as the larger product. Comparing answer choices A and B. We get 15 16 versus seven nights. Multiply 15 and nine. We get 1 35 and more planes 16 by seven. We get 1 12 here. 1 35 is greater than 12. Therefore, the Fraction 15 16 says greater than the fraction seven nights. If we apply that technique, told the other answer, choices will see that choice A turns out to be the largest, so the answer is a. 72. Math Notes Problem 3: getting a common denominator on the bottom. We multiply top and bottom of one by two and to minus one is one. So we end up with 1/2 and now we invert, multiply, writing down the numerator one and inverting the denominator and multiplying we get. We get one plus two, which is three. Hence the answers a. 73. Math Notes Problem 4: the ratio of 1/5 to 1/4 is equal to the ratio of 1/4 toe. Acts means that 1/5 divided by 1/4 has the same value as 1/4 divided by X inverting multiplying, we get 1/5 times The reciprocal of 1/4 which is for over one, equals 1/4 times The reciprocal of X over one, which is one over acts simplifying we get 4 50 equals 1/4 X Now cross multiplying, we get 16. X equals five or X equals 5/16. Hence the answer is a. 74. Math Notes Problem 5: squaring a proper fraction makes it smaller and square. Rooting a proper fraction makes it larger. It's a is smaller than be, and both C and D are greater than one because 8/7 is greater than one. It's the answer is a. 75. Math Notes Problem 6: I noticed that the value of a in this formula is irrelevant. Everything depends on the value of B, because that's the only variable that appears on the right side. So x pound negative y is equal to the negative from the formula and then the value of B, which is negative wine raised to the fourth power. So this negative stays here. It is unaffected by the power on the negative y. But this four will destroy that negative and make it a positive. Begin the negative out in front remains and positive. Why is just y to the fourth power? So we get negative y to the fourth, which is choice, see? 76. Math Notes Problem 7: 0.2 squared is 0.4 So we get 1/1 minus point 04 And when my his 0.4 is 0.96 converting this to a fraction, we get one over 96 divided by 100. Because there's two significant decimal places here and earn inverting and most plane. We get one times 100 over 96 which is just 100 over 96 which reduces to 25 over 24 and C answer is a 77. Math Notes Problem 8: since X is a fraction between 011 the square root of X will be greater than X, cubed or X to the fourth, which eliminates D and E. And since again X is a fraction between zero and one, the squirt of X will be bigger than acts, which, of course, is bigger than X over pie. Because pies approximately equal to 33.14 this would eliminate C and to determine which is larger between A and B. Let's let X equal 1/4 and choice A. We get one over the square root of 1/4 which is one over main division bar over 1/2 which is one times 2/1 or two, and choice be will become the square root of 1/4 again. And that's 1/2 and, of course to is bigger than 1/2. Therefore, the answer is a 78. Math Notes Problem 9: square in a fraction that is between their own one makes it smaller, and taking the square root of a fraction between zero and one makes it larger. So in this fraction, the numerator is larger than the denominator. Hence the fraction itself is bigger than one. This eliminates choice. Be next statement, too, Is false. Square in a fraction will make it smaller only if it's a fraction between zero and one, but six fists is bigger than one. This eliminates, see finally, statement. Three years falls. We know that 56 is smaller than the square root. If I six, because 56 is a proper fraction, dividing both sides of this inequality by the square root of 56 it cancels on the right and gives his one. Taking the square root of both sides of this inequality, we get the square root of 5/6. Divided by the square root of five. Over six is less than the square root of one, which is one, and this is the expression in statement. Three. But instead of being bigger than one, it will be less than one. Hence, the answer is a 79. Math Notes Problem 10: statement. One is not necessarily true. For example, take X to be to and why to be one in the left side becomes in the right side becomes too over one which is to and 4/3 is not greater than two. This is also a counter example for statement to, since we can eliminate a be de N E and the answer is C. However, it is instructive that prove that statement three is true from the given inequality X greater than why greater than zero? We get ex strictly greater than why and adding to to both sides we get X Plus two is greater than why plus two. They might be wondering why we're adding to to both sides. That's because both expressions are in statement. Three X plus two y plus two. Not a great statement. Three. We divide both sides of this inequality by white close to and why, plus two over why Plus to, of course, is one. So we have it. This expression is greater than one because of the inequality right here in statement three is necessarily true. 80. Defined Functions Text 1: defined functions are very common on the test, and most students struggle with them. Yet once you get used to them, defined functions can be some of the easiest problems on the test and this type of problem . You'll be given a strange symbol in a property that defines the symbol. For example, you might be given the at symbol, and it might be defined as X at Why equals X divided by why, for example, three at to would be three divided by two. Or he might be given X at Why equals X rays to the wife power, for example. Three at two three is in the basis position, so you get three to the second power or nine, and you might be given another several like a triangle. And it might be defined to be X triangle. Why equals the square root of X minus y. For example, three Triangle two would be the square root of three minus two or the square to one, which is one 81. Defined Functions Example 1: notice that in the formula to is in excess position and three years and wise position. So we merely replace all the exes on the right side of the equation with tube and all the wise with three, which gives us two times three minus three or six minus three or three. Hence the answer is B. 82. Defined Functions Example 2: most students who are unfamiliar with defined functions are unable to solve this problem. It is actually quite easy. Notice that the definition a triangle be is defined to be a swear notice. The value of B is irrelevant. Whatever is in a position gets squared. And that's the extent of the formula. For a silly example. Suppose we have a triangle Me that would give us simply a swear it again. So, on our expression we have Z triangle to is gonna be easy Square and Z Triangle three is also gonna be Z squared because both expressions have Z for the leading term, and anything divided by itself is one Hence, the answer is B. 83. Defined Functions Example 3: Since we have a two part definition for this function, we have to determine whether the expression to K minus one is even or on. To do that, let's choose K two B one and plug it into the expression we get one, which is an odd number, since we use the bottom half of the definition and notice the definition says that whatever is inside the parentheses, we merely multiply it by four. So we get four times to K minus one, because that's what's inside the parentheses and distributing the four. On each term we get eight K minus four, which is choice, see. 84. Defined Functions Example 4: notice that that the definition of the function says that whatever is starred pops out here and the pie and the minus never change. So, working from innermost parentheses, we get copy down the outer parentheses first. Now working on this re replace in the formula the axe with negative pie. And that gives us the original pie minus the negative pie. Because negative pie isn't X's position. Simplifying we get to pie start. So now two pi is in excess position. So replace X with two pi and we get pie minus two pi, which is negative pie. Therefore, the answer is C. 85. Defined Functions Example 5: we have a two part definition here. So we have to determine whether you in the or odd or even now we're given that you three you is odd. Hence you Assad. And let's prove that by contradiction. Suppose you were even then three. You would be even. But we're told that three You, Assad. Therefore three. You must be odd. Similarly, since seven minus via zade the must be even And this proved that my contradiction suit supposed V were odd than seven minus V would be even since the difference that two odd numbers is an even number. But we're given that seven minus v Izod, hence the must be even since you is odd, we use the top part of the definition and we get five. Since V is even, we use the bottom half of the definition and we get 10. Hence, the expression becomes five for the U and 10 for the V, which gives us a difference of five. Hence, the answer is a 86. Defined Functions Example 6: notice that the definition tells us that the first element will be the base of the exponents and the second element will be the actual exponents. Choice A is false. For example, if we choose a to be one and b two b two, then we get the base is one and the exponents is too, which is one. And then if we change the order and choose to first and then one, we get the basis to and explain it is one and two dozen equal one therefore, choice A is false for choice be this work on the left side. So the first element is the base and the second element is the exponents. Now the a the negative A inside the parentheses can be written as negative one times a. You can do this with any negative number and then we have a product raised to a power so he can raise each individual element to that power. So negative one raised to the negative A and a raise the negative A. Now this negative exponents can be reciprocated. So the negative one to the negative a stays on top and this explain it goes to the bottom and becomes positive, which is the right side. So we have proven that choice B is true and choice. See, His false, for example, was choose to combined with two and then combined with three that will give us According to the definition, two squared combined with three or four combined with three, which is four to the third or 64. And now let's choose the same numbers but group them differently. So we have two combined with to combined with three working from the innermost Prentice E. We get to raise to the third power, which simplifies to eight. And again. Our base here is to and are exponents is eight. So we get to to the eighth, which is 256 and that does not equal 64. So choice see is false. Hence the answer is B 87. Defined Functions Text 8: you may be wondering how to find functions differ from functions F of X, who studied in intermediate algebra and more advanced math courses. They don't differ. They are the same old concept you dealt with in your math classes. The function of the previous example produces easily be written F of X equals radical APS and F of X equals forex. The purpose of defined functions is to see how well you can adapt unusual structures. When she realized that defined functions are evaluated and manipulated just as our regular functions, they become much less daunting. 88. Defined Functions Example 7: noticing the definition of the at symbol. The first term is the base, and the second term is the exponents working from inside the parentheses we get the first term is X, and the second term is wise. So we get X to the Y. And now again, the first term is X to the y. So that is our base because it comes first in the at symbol. So write down our base X to the Y. And the second term is easy. So that becomes the exponents. And we have to encase this in parentheses because it's raising the entire quantity to the Sith power. And this is choice E there for the answer is he? 89. Defined Functions Example 8: we are asked for the value of why that makes the expression expound. Why equal to negative acts since we have the equation expound, Why equals negative X plugging in the formula, the right hand side of the form that we get now adding extra? Each side of the equation will will cancel the minus X is so we have now. The square of a product is the same thing as the product of the squares. So we get X squared y squared. Plus y squared is equal zero factoring out y squared now using our zero product property, which says that if the product of two numbers is equal to zero, one of the numbers must be zero or the other or both. Now we know that X squared plus one can't equal zero. In fact, it can't ever be smaller than one because X squared is a positive number or zero. And if you add a positive number toe one, it's going to be bigger than one. And if you add zero toe one, it'll equal one. Therefore it can't possibly equal zero. So why squared? Must be zero, which gives us taking the square root herbal sides we get white was zero. Hence, the answer is a 90. Defined Functions Example 9: Now the area of a square of side axe is just X squared, so our formula box sex is simply X squared. So no Box nine would be nine squared, divided by box three, which is three squared. And that gives us 81 divided by nine, which is nine. And for choice B, we get Box three is three squared, which is also nine. Hence the answer is B. 91. Defined Functions Problem 1: were given that P is three. So let's Pete plug p into this formula, and that will give us three to the Asterix equals three plus five over three, minus two or eight over one, which is a therefore, the answer is E. 92. Defined Functions Problem 2: on the test answer. Choices are usually listed in ascending order of size. Occasionally they are listed in descending order of size and start with choice. See if it is less than two, then turn to Choice D and if it is greater than two, then turned to choice. Be so plugging six into our formula we get six Squared is 36 divided by two is 18. 18 is bigger than two. So we turned a choice be and that gives us Hey, Since it is still bigger than two, the answer must be a 93. Defined Functions Problem 3: since two is Anais position and three isn't these position. We replaced a with two and be with three in the formula, which gives us the square root of 25 which, of course, is five hence, the answer is B. 94. Defined Functions Problem 4: we're told that d denotes the area of a circle with Diana or Di. Now the formula for the area of a circle is pie R squared in the radius is half the diameter. So our formula becomes pie de over two squared, plugging that into our formula or into our expression we get So we're looking for the answer choice that returns 36 pi squared turned choice d we get 36 pi square. There for the answer is D. 95. Defined Functions Problem 5: in the first expression noticed that zero is in excess position. One is unwise position and A is in ZIS position, so the formula becomes zero for X one for why and a for Z or simply negative A. And for the second expression, one is an excess position, A is and wise position, and zero is in ZIS position and anything. Times zero is zero. Now, setting the two expressions equal to each other. We get negative A equals zero or simply a equals zero by multiplying both sides by negative one. Hence the answer is C. 96. Defined Functions Problem 6: plugging to into our formula gives simplifying. We get combining like terms we get four minus X squared equals X squared, adding next square to both sides. We get four equals two X squared. Dividing by two gives us too. And then finally take the square root of both sides will eliminate the square. So we get X equals radical, too. There for the answer is a 97. Defined Functions Problem 7: statement. One is false. For example, one Star two is equal to one times two minus 1/2 which is three have and two star. One is equal to two times one minus 2/1, which equals zero. This eliminates a Andy statement to is true a star A is equal to a times A because the second a is in B's position, minus a over the beef, which again is a which gives us a squared minus one, which is a difference of squares because one is one squared, which factors into a plus one and a minus one. This eliminates see, hence the answer is B. 98. Defined Functions Problem 8: notice that X Star Y is defined to be X over y plugging that into our expression we get now , notice that X over Y is the first term in our X Star. Why, Therefore, it's the entire top of the fraction here, so we get X over. Why Main Division Bar? And that's the slash here divided by Z because E is in the second position of our formula, So Z is the Y value. Now To divide fractions, we often write the denominator over one to make it more distinctive looking and again to divide fractions, you invert, multiply so the numerator becomes X over y and the nanometer flips over. It gives us one over Z, so our answer is X Y X over y Z, which is choice e. 99. Defined Functions Problem 9: were given that X at Y is equal to this expression here. So such substituting that into the form that will give us equals. Why now, Adding Why, to both sides of this equation? Well, can't will remove it from the equation will cancel it out. And we now have X radical wide minus two X, nor thing that we have a common factor of acts will factor it out. Now notice what the question is asking. We want to know for what value of X Will this expression here always equal zero? No matter what the value of why is well, effects were zero the No matter what the value of this expression is, it will equal zero because zero times anything is zero. Therefore, the answer is a 100. Defined Functions Problem 10: working from the innermost parentheses out, we noticed that 64 is an ends position in the formula. So we replace in with 64 which gives us the star. The second star is still there in the square to 64 is eight and eight, divided by two is four. So we get four star. Now we go back to the formula again and noticed that four is an ends position. So we replace in with four, and that gives us the square root before over to in the course of square to forest to to over two is one. Hence, the answer is a. 101. Defined Functions Problem 11: this. Evaluate this expression using this formula here, notice between the arrows we have X plus two. Therefore, in this expression, everywhere we see X, we're gonna replace it with X plus two. Although it seems odd that we have an X here and the next hear this expression X plus two is different from this ax. So plugging it in, we get for this X, we get X Plus two and we'll put it in parentheses and will change the original parentheses . Two brackets. So it'll be more distinctive looking, plus the two and then closed the bracket and then replaced this X here with X plus two. And then we're gonna do a similar thing with X minus two. We're gonna place all the exits in the formula with X minus two, adding the two that's already in the formula and then closing the brackets and replacing that X with X minus two. Now we just simplify these expressions. This is X plus four times X plus two. And here the twos cancel and you have X Times X minus two. Now we multiply everything out. Foiling the 1st 2 terms gives us X squared. The other two terms will give us two X in her two terms. Give us four X in the last two terms. Give us eight and distribute the X here, and you must in case in parentheses, cause it's minus the entire quantity that will give us X squared minus two acts and then distribute the negative. Now we'll add up like terms. The X squares Cancel two X plus four X is six x plus two X is a DAX plus the eight and now we notice we have a factor of eight. Factoring that out, we get eight times X plus one, which is choice E. 102. Defined Functions Problem 12: Delta in is asking for the smallest imager that is greater than the number in. So let's draw a number line for this problem. Placing negative 2.1 on the number line we get all the imagers that are bigger than negative. 2.1 would be negative, too negative. One zero one two, etcetera. Of all those imagers, the smallest or the least imager that is bigger than negative. 2.1 is the first energy you come to as you move to the right. So the answer is C negative, too. 103. Defined Functions Problem 13: our definition has two parts. It says that if the first term in expound why is smaller than the second term, then we'll use the bottom definition. And if the first term is larger than the second term, then we use the top definition notice that four is smaller than 12 therefore, will use the bottom half of the definition. Plugging it in. We get fours and expose it excess position, and 12 is in wise position, which gives us four plus three or seven. Therefore, the answer is B. 104. Defined Functions Problem 14: I noticed the expression X Star tells you that whatever exes, it pops out here that as you subtract it from two in this expression, we have a bee in excess position. So we will get be star equals two minus b. So calculating this expression, we get a plus Tu minus B. Now we're placing two minus being brackets, but there's no mathematical purpose for that. It just shows that came in as a group or is a unit dropping the brackets we get now? The whole expression a plus to minus B is an excess position. So we simply subtract that entire expression from the number two, which gives us and those we have. The insert parentheses here because it's the entire expression that subtracted. Distribute the negative gives Tu minus 20 So we get rearranging. We get B minus A, and the answer is a 105. Number Theory Text 1: number theory is a popular source for test questions. At first, students often struggle with these problems, since they have forgot many of the basic properties of arithmetic. So before we begin solving this problems, let's review some of the basic properties the remainders are. When P is divided by K means P equals K. Q plus are where the inner jerk use called the quotient. For instance, the remainder is one when seven is divided by three means seven equals three times two plus one. We can also illustrate this by traditional long division we're dividing three into seven. It goes in twice, which gives us six. It's attracting. We get one, so one is the remainder. Three is the divisor and two is the quotient. 106. Number Theory Example 1 hd: We're told that when the energy in is divided by two, the quotient is you and the remainder is one, so we get in equals to you. Plus one. We're used the quotient and one is the remainder. Likewise, when in is divided by five, we get five e plus three. Now each expression is set equal to end. Therefore, they must equal each other so to you. Plus one must equal five V plus three knows the answer choices have the U. N V on the left side and the numbers on the right side. So this rearrange, subtracting five v from both sides of the equation because is to U minus five e and then subtract one from both sides. And that will give us too. And then we just noticed that that is answer choice. Be so the answer is B. 107. Number Theory Text 2: a number in is even if the remainder zero when in is divided by two that is in is equal to two Z plus Sarah or simply in equals. Two Z, for example, in could be six and six can be written as two times three. So in this case, the three is playing the role of Z. No numbers odd if the remainder is one when in is divided by two or in equals two z plus one, for example, for five, if you divide it by two, we get to times two plus one. So again the remainder is one. 108. Number Theory Text 3: the following properties for odd and even numbers are very useful. You should memorize them and even number times, and even never is even. For example, two times four, which is two even numbers, gives you a product of eight, which is also even in an odd number such as three times five is 15 which is also odd and an even number times an odd number is an even number. So two, which is even times one which is odd, is back to which is even and an even number such a zero plus another even never such as to gives back and even number an odd number, such as one plus. Another odd number, such as one again will give you two, which is even and even never such as to plus an odd number such as three returns five, which is also on 109. Number Theory Example 2 hd: for a fractional expression to be an imager, the denominator must divide evenly into the numerator. Now for statement one, we're told that P is even, and cue is odd overpay again, which is an even number and an even number. Plus an odd number is an odd number over an even number, and since then, even never cannot divide evenly into an odd number statement, one cannot be an integer, no statement to can be an imager. For example, if P is equal to two and queues equal of three, then the threes cancel and we get to which is an energy er. Find the statement. Three cannot be in an urge. Er, now P squared is even since it is a product of two, even numbers p Times P. And we're told that P is even so. We have odd number Q over an even number and again and even never cannot divide evenly into an odd number. Hence, the answer is E 110. Number Theory Text 4: consecutive images air written as X x plus one X plus two. Because each number differs from the next by one unit, for example, to three for etcetera. Well, the difference between each parent numbers is one. The difference between two and three is one, and the difference between three and four is also one. Now, consecutive, even or odd imagers are written is X X Plus two and X Plus four. You can use the same formula for even or odd numbers, because the defining characteristic is that they differed by two. For example, 246 etcetera and one three five, etcetera, with even numbers differ by two. And the odd numbers also differed by two, so we can use the same formula to generate him now. If we needed to make the numbers explicitly, even or odd, we could write to X then two x Plus two and then two x plus four. And for odd numbers, it would be two x plus one two X plus three and two X plus five. But this is an unnecessary complication. The simpler formal work just as well 111. Number Theory Text 5: the imager zero is neither positive nor negative. You can think of zero as the boundary or demarcation point. Between the negative numbers and the positive numbers, anything bigger than zero will be positive and anything smaller than zero will be negative and zero is neither positive nor negative. However, zero is even because it can be written as two times a number, namely itself. Two times zero is zero, therefore, zero is even. 112. Number Theory Text 6: a prime number is a positive integer that is divisible only by itself, and one for example, five can be written as five times one, and that's the only way you could write. Five is a product of two. Injures. Six is not prime. Cause six can be written as three times to the prime. Numbers are 2357 11 etcetera. Rarely on the test. When searching for prime's will, you need to go past the number seven. 113. Number Theory Text 7: a number is divisible by three. If the sum of its digits is divisible by three. For example, 1 35 is divisible by three because of some of its digits, one plus three plus five equals nine and nine is divisible by three. For another example, 51 will be divisible by three, because five plus one is six and six is divisible by three. I mean, that's three divides into six evenly with no remainder. 114. Number Theory Text 8: the absolute value of a number represented by two vertical bars is always positive. In other words, the absolute value symbol eliminates negative signs. For example, the absolute value of *** seven is positive seven and the absolute value of negative pie is positive pie. They'll be careful. The absolute value simple acts only on what is inside the symbol. For example, in this problem, this negative outside the symbol is unaffected by the absolute value. If we take the absolute value of what's inside, we replaced absolute value symbols with parentheses, and we change this negative into a positive, and then we normally distribute the negative and end up with negative seven plus pi. Now there are a number of mathematical expressions that can generate an absolute value and will discuss two of them later in the course. But for now, just remember that the absolute value simply strips the negative sign from the number 115. Number Theory Text 9: the product or quotient of positive numbers is positive and the product or quotient of a positive number and a negative number is negative. For example, negative five times positive three gives us negative 15 and positive six divided by negative three gives us negative, too. 116. Number Theory Text 10: the product or quotient of an even number of negative numbers is positive. For example, here we have four negative numbers, which is an even number. Hence the product is positive. And here we have a quotient of two negative numbers, which is an even number of negative never since the quotient is positive. 117. Number Theory Text 11: the product or quotient of an odd number of negative numbers is negative. For example, this product here is negative because there are one, 23 negative numbers, and odd number and negatives in this product here is negative because there are 12345 an odd number of negatives. 118. Number Theory Text 12: the sum of negative numbers is negative. For example, Nega three minus five this negative eight. Some students have trouble recognizing the structure is a some because there is no plus symbol. But recall that subtraction is to find this negative addition. So Nega three minus five we changed the minus into a plus in the right negative five. So Nega three plus negative five, which is negative eight. 119. Number Theory Text 13: a number or expression raised to an even exponents is greater than or equal to zero. For example, to square is four and four is greater than or equal to zero or for more exotic example, negative pie to the fourth Power is the same thing is positive pie to the fourth power, which is greater than or equal to zero or X squared, which is greater than or equal zero. And finally, zero squared Israel Time zero, which is zero and zero, is greater than or equal to zero. 120. Number Theory Example 3 hd: Let A be X and B B X plus one and C B X plus two. Plugging this into choice or statement. One. We get X Plus one for B and X plus two for C distribute. The negative gives us canceling the exes gives his negative one, which does not equal one so statement. One is false. Statement to is true, because if a B and C are consecutive integers and one of them must be divisible by three and for statement three supposed A is even and B is odd, then see would be even and then a plus. B would be odd, because an even number plus an odd number is an odd number. Hence the expression a plus B plus C is equal to a plus B plus C, which gives us and eat on odd number plus an even number, which is an odd number. Therefore, statement three is false, and the answer is B 121. Number Theory Example 4 hd: the shoes values prime numbers for X and y if X is three, and why is to than the both prime in their differences? One. So this eliminates choice. A. For choice be let's try five and to that gives us three. So limit be for choice. See, let's use 11 which is prime and to which is prime, which gives us nine. So eliminate C. And for 15 it is the difference between 17 and two, both of which are prime, which eliminates D. Therefore, by the process of elimination, the answer is E. 122. Number Theory Example 5 hd: working from the innermost parentheses. Out, we get positive. Three. This negative is still there inside the absolute value, and so is the negative outside that value. Now, nick, a negative times a positive is a negative. This negative outside, absolute by you stays where it is. And if we drop the absolute value, this negative turns into a positive and a negative times. A positive is a negative. And now, let's introduce reintroduce the left side of the equation. Negative acts. Multiplying both sides by negative one gives us X equals three. Therefore, the answer is C. 123. Number Theory Problem 1 hd: now the statement in the remainder is one. When M is divided by two means that M can be written as to you. Plus one. We're using some imager now. This is the standard form for an odd number, one more than an even number, because all even numbers can be written as two times some other number X plus one gives us an odd number that eliminates a fortune for choice. Be we were given that the remainder is three when M, when in is divided by four and that could be written as in equals for V Plus three. Now this is also a form of an odd number, as we can show by factoring out of two that's first introduced the two into the problem by writing three as two plus one. Now we have a common factor of two in the 1st 2 terms, and we confected that out, which gives us two V plus to plus one. No, this is of the standard form two times something this call it backs plus one where the X is two V plus one. So again we have an odd number, and that eliminates be turning the choice see, Let's go ahead and add these two expressions the M and the end, and we'll put him in parentheses just to show that there are unit. But there is no mathematical need to do that. And this rearrange So we get to you plus four V and go ahead and add the one in the three, which is four. We might notice that this is an even number, but we can show it explicitly by factoring out the common factor of to now we have written the number as two times some other imager, and we can call that entire expression there that you plus two V plus two acts and this is the standard form for an even number. Therefore, the answer is choice, see? 124. Number Theory Problem 2 hd: we're told that both X and Y are prime and greater than two. Now two is the only even prime number. Therefore, both Backstrom I must be odd numbers now X times. Why would therefore equal an odd number times another odd number which will always return an odd number. And of course, no odd number is divisible by two because that would make it even. Therefore, the answer is a 125. Number Theory Problem 3 hd: since two divides evenly into acts. We could write tax as two times E for some inner jersey, and we're looking at what divides into five acts. So rewrite this as five X equals five times to Z, which gives us 10 C. Therefore, five X is a multiple of 10 and similarly, we could show that five Wise also a multiple of 10. And since 10 is the largest emitter listed as an answer choice, the answer is E. 126. Number Theory Problem 4 hd: were given that A, B and C are consecutive, even imagers, So the energy or be could be exactly two units bigger than A and C could be two units bigger than be so. This is a way to represent the three consecutive even imagers. Now we're told that there some or the average, which is their son divided by the number of imagers, is less than a over three, reducing the numerator. We get three a plus six over three, and now, factoring out of three on the top, we get a plus two over three, and that's still less than a over three. Canceling the threes we get. A plus two is less than a over three. Multiply now through by three to clear. The fraction will give us three. A plus six is less than a subtracting A from both sides. Gives us to a and subtracting six from both sides. Gives us native six on the right. Dividing by two gives us a less than negative three. And looking at choice E. It says that a is a negative number and all numbers less than negative three are negative. There, for the answer is E 127. Number Theory Problem 5 hd: suppose X equals one and Y equals three. The north given expression would have the value to which is a prime minister, so it satisfies the criteria. However, in this case, why which is three does not equal five times acts which is one so statement. One is not necessarily true. And also statement three is not necessarily true because we've already shown that five X plus five over why is equal to an even number the number two which is not on torch statement two. It could be true because why is, too which is a prime minister, the only even prime minister. So let's choose some different values for X and y I was choosing it X now to be three and why to be four. The nurse why is no longer a prime imager? But if it satisfies this, then that would eliminate statement to so X plus five over wide equals 8/4, which is to which is a prime manager. And why is not why itself, which is four, is not a prime minister in that eliminates statement to therefore the answer is a None of the statements are true 128. Number Theory Problem 6 hd: since Exit is both a cube and between two and 200 we're looking at the numbers to Cube three cubed four cube and five cute. Most playing these numbers out. We could eight 27 64 1 25 Now the question says the X also has to be a square. There's only one perfect square in the lift, and that is 64. ID equals eight square instead. Answer C. 129. Number Theory Problem 7 hd: Let's solve this problem by plugging in the answer. Choices were told that the some of the debt digits of the number is four. Well, that would eliminate de because four plus eight is 12 not four, and similarly, it would eliminate Choice e. We're also told that the difference in the digits is also for that would eliminate choice A . Because one minus three, isn't it give to and that doesn't equal name doesn't equal for and similarly, that would eliminate choice be because three minus one is to which doesn't equal for so by elimination. Answer is C and, of course, four minus zero is for so the difference of the digits is for 130. Number Theory Problem 8 hd: Let's briefly review the concept of Vision seven divided by three Liza. Remainder of one means that seven can be written as three times two plus one by analogy X, divided by y leads. The remainder of one means that X can be written as why time Some energy Q Plus one Hence P divided by nine leads. The remainder of one translates into P equals nine times Q plus one. Now, if we choose que to be one, then P would equal Tim, which is even. And that eliminates statement. A. And if we choose que To be to Thampy would equal 19 which is odd. And that eliminates statement one. And that limits all the answer choices except for C. But let's go ahead and verify that C is also true. We can rewrite this expression here as nine equals three times three times Q plus one and then regrouping Weaken right this as three times three q plus one And let's let the three Q equals E so it matches up with the answer more directly. So we get three z plus one, and that is answer choice or partly statement three. Therefore, the answer is C Onley statement. Three is true 131. Number Theory Problem 9 hd: since both p and Q leave a remainder of one when divided by and even number in peace case, it's even number two and accused cases even number six. This means that both of the numbers are odd. For example, P can be written as two acts plus one, which is a standard form for an odd number, and Cue Commie, written as six y plus one, which is also awed. But we can show it expressively by writing six as two times three. So cue Assad now a odd number. Times an odd number is another odd number, so p. Times Q Izod and adding one more unit toe. An odd number will make it even so, we know that choice a or in statement one is even is true. And if Peak Times Q is odd, then P Q. Divided by two is not an energy that eliminates statement to and finally for statement. Three p Q is not necessarily a multiple of 12. For example, if P was three and cue were seven thampy times, kill would be three times seven or 21 clearly 21 is not a multiple of 12. There, for the answer is a 132. Number Theory Problem 10 hd: since the question asked for the smallest prime greater than 53 we start with the smallest answer Choice, which is 54 now. 54 is not prime because it can be written as two times 27 and 55 is not prime because it can be written as five times 11 and 57 is not prime because it could be written as three times 19. Now 59 is prime against the answer is D. 133. Number Theory Problem 11 hd: a number divisible by 23 and four must also be divisible by 12. So each of the two numbers can be written as 12 times some number 12 times a, maybe, and 12 times be forming their difference. We'll call it D. We get 12 a minus 12 b. Well, we have a common factor of 12. Factoring that out, we get 12 times a minus bi. Well, this shows that the difference between the two numbers is also a multiple of 12. So the answer must be divisible by 12 and the only never listed as, uh, as an answer, divisible by 12 is choice be. 134. Number Theory Problem 12 hd: this problem is just that a worthy way of asking how many factors of three are there in 27 to the fifth Power. So we write 27 as three cubed. That's all the factors there are of three in 27. So 27 to the fifth power can be rewritten as three cubed to the fifth power because we already shown that 27 equals three cubed and a power to a power. If you recall beings that you multiply the exponents so we get three to the three times five, which is three to the 15th power. That is all the factors that there exists of three in 27 to the fifth Power. Therefore, the answer is D. 135. Number Theory Problem 13 hd: were given that X, Y and Z or consecutive integers in that order. So X is less than why. Which is less than Z now between two. Any two consecutive managers, one of them must be an even number. For example, one and to the two is even. And if you have, say, four and five before is even so either the X or the Y will always be even so the product will be even so. Statement one is always true. For steam it to supposed Z is odd. So if E Z is a nod number, then why would have be even? And X would have to be odd as well. If you recall the difference between two odd numbers is an even number. For example, if you have, say three and won, their difference is to now suppose E is even then why would be odd and X would be even, and the difference between two even numbers is again an even number. For example, four months to gives you two, which is even so statement to is always true Now, Statement three is not supposed ex were one. Then why would have B two and Z would have be three. So statement three would become one to the third power, which is one which is an odd number. So statement three is not necessarily even so by elimination, the answer is D. 136. Number Theory Problem 14 hd: this work on the right side of this equation from the innermost Prentice E Out that will give us a copy down the negative out in front, copying this negative here inside and then six minus two is four this negative outside. The absolute value stays there. But when we drop the absolute value, this negative is turned into a positive because the absolute value always destroys a negative number and a negative times a positive is a negative. So the right side of the equation is negative. For now, let's bring in the left side of the equation to solve this. We want to isolate X on the left and put everything else on the right. So we'll undo the operations, undoing the subtraction we add to to both sides. And that gives us negative X equals negative too. And to undo this multiplication, we divide both sides by negative one, and we get X equals positive, too. There for the answer is D 137. Number Theory Problem 15 hd: we're told that the some of the prime numbers X and Y's odd. Now the only way the sum of two numbers can be odd is if one number is odd and the other one is even. For example, two plus three is five, which is an odd number but two plus, say, four to evens because you're six, which is not odd, and two odd numbers, like three and seven, also gives an even number. So one of the numbers X or wide, must be even. And if if either X or wise even then the product X Times y, which is even will by definition be divisible by two. Therefore, the answer is a. 138. Number Theory Problem 16 hd: let's manipulate this equation into a form that might tell us some of the properties of extra Why? So let's clear the fractions by multiplying both sides of the equation by X minus y. Some will play the right cyber X minus y and the left side by explain is why that will eliminate X minus y from the fraction and distributing the three on the right will give us three acts minus three y, then subtracting three x from both sides because we want to get all the exes on one side and all the wives and the other side gives us negative two acts. Plus, why equals negative three y, then subtracting why from both sides, we get negative two X equals negative for wine. Finally dividing both sides by negative too. We get X equals positive to Why, yes, we have written X as a a multiple of two, mainly two times and energy er therefore, X must be an even number, and the answer is D 139. Number Theory Problem 17 hd: let the original two digit even number be represented by X Y Now here, X Y does not represent multiplication. It just represents the positions of the digits. For 52 the X would be five in the Why would be to now. The problem says that if we reverse the digits, but where's why X, then this will be greater than the original number. And the only way this can occur is if the why is greater than X. For example, 73 is greater than 69 because seven is greater than six. Now, Suppose X were equal to nine in the condition. Why is greater than X? Cannot be satisfied. For example, if X was nine, then we might be dealing with 91 92 93 etcetera all way down to 99. But when you flipped the digits, we get 19 29 39 and our last best choice is 99. But nine, of course, is not greater than itself. Therefore, the answer is E 140. Number Theory Problem 18 hd: this problem is hard because we don't really know in what direction to go, but notice all the answer choices say make a statement about the letter C about the variable. See? So let's solve this equation for C and see if it might tell us something. Cross multiplying gives us a C equals B square and then dividing both sides of this equation by A In order to solve for C, we get C equals B squared over a. Now we're told that a is a perfect square. In other words, a could be written as something squared This use case squared something substituting that into our equation. We get b squared over case where which can be written as be over K, the quantity squared. Therefore, we have written C as a perfect square as well. So the answer is C. 141. Number Theory Problem 19 hd: noticing the numerator of the formula that the imagers differed by one unit. Therefore, there consecutive injures. They might be three and four or seven and eight. But of any two consecutive imagers, as these examples show, one of them is even therefore, either that in is even or the n plus one is even hence, to will divide into either one of them one of the two, but not both. That means that the product s that s what we reduced to a product of two integers p or Q P and Q. Now recall that a prime number cannot be written as a product of two injures. Therefore, s cannot be a prime number, and answer is D. 142. Number Theory Problem 20 hd: we're told when the dump number is divided by 12 it has remainder of seven. So the number let's call it in can be written as 12 p plus seven where 12 as a divisor and seven is the remainder. Then we're gonna divide that same number by six. So we're gonna try to write it in the form six times Q plus. Some remainder are where the number are. Will be, will be between an energy between zero and five. Now seven can be written as six plus one. So doing that, we get factory now the common factor. Six in the thirst, two terms we get to pee, plus one plus one. So we have written it as six times something plus one. So the remainder is one, and the answer is a 143. Number Theory Problem 21 hd: that's so this problem. By substitution, let's take a two digit number whose digits add up to nine. How about 72? Because seven plus two is nine. Adding 10 to this number. As this says to do we get 82. This some of this. The digits of this number is 10 so that would eliminate A and B. So let's choose another number. How about 90? Some of this digits is nine. And if we add again, 10 to this 90 like this says to do we get 100 and the some of the digits of 100 is one which eliminates, see and D. Therefore, the answer is E. 144. Number Theory Problem 22 hd: notice that all the digits in the dividend are divisible by three. So three will divide the dividend into a number such that each of his digits will be 1/3 of the corresponding digit in the dividend. For example, the third digit is six, and when you divide it by three, you'll get to this will have happened to each of the digits in the number. Therefore, the answer is B. 145. Number Theory Problem 23 hd: Let's solve this problem by substitution, choosing in equal one. It will satisfy the conditions of the of the problem because One Cube is one which is an auditor. Now we just plug in the value one into each answer choice and see which one returns an even number for choice. A. We get three, which is not uneven imager. So eliminate choice. Say choice be becomes one to the fourth, which is one again it's not. Not even so. The maybe choice See gives us two squared plus one. Probably one squared plus one, which is one plus one or two. So it's uneven energy, er and it may be our answer, but we got check the remaining answer choices for choice d We got three, which again is odd. So eliminate D and the number one itself is, of course, odd. Therefore, the answer is C. 146. Number Theory Problem 24 hd: If the product of the two images is odd, then both of the imagers has have to be odd. For example, three times five is 15 and if one of the images was even than the product would be even, for example, two times three is six. And of course, if both of the energies were even, then the product would be even. Now, remember that thesis, Um, of two odd integers, which is what we must have, is an even number. For example, three plus five is eight is even there, for the answer is B. 147. Number Theory Problem 25 hd: were given consecutive imagers. Let's choose him to be x x plus one and X plus two. So X is the first manager. X plus one is the second, and X plus two is the third for me. The sun we get three plus three X. Now we're given that this sum is odd and recall that if this sum of two imagers Izod than one of the editors must be odd and that one must be even well, we already know three Assad, therefore three x must be even. But what does this tell us about the value of X? Is it order? Even now, if X were odd than three X would be odd as well because an odd number times an odd number is an odd number. But we're already told that three X is even so we know that X cannot be odd. So X must be Even so, we have here an even number for the first, an odd number for the second, and then an even never for the third. There for the answer is D. The first and last imagers are even 148. Number Theory Problem 26 hd: were given that l M and N are positive injures in that order. Ellis Lesson M, which is in turn less than in and we're told it in, is smaller than four, and it's a positive energy. So in the largest possible value for N is three, and the largest possible value for em is too. And the largest possible value for L is one. Well, this sequence is fixed. There are no other about possible values. L could not be zero because then it would not be a positive integer. Therefore, M must have the value of two, and the answer is C. 149. Number Theory Problem 27 hd: Let's divide both sides of the given equation by four. So we get P over four is equal to Q. And let's rewrite this with Q on the left and P over four on the right. The nose. We haven't inequality here involving P, and we have an expression involving P over four. So let's divide both sides of this inequality by four soapy over four is less than two, and since Q equals Pew or for he was also less than two. Now we're looking for a non zero positive imager. The only non zero in urge er that is less positive energy. Less than two is one. Therefore que must be one, and the answer is a. 150. Number Theory Problem 28 hd: No, it's a choice. See, involves the product of two consecutive integers. Now have any two consecutive integers. One of them must be even. For example, one and two. The two is even four and five. The four is even therefore, either in or in plus one must be an even number. Therefore choice see must be even as well. 151. Number Theory Problem 29 hd: let the three consecutive positive integers be in in plus one and in plus two Adam up we get one plus two is three and three ends. Give us three in the notes. There's a common factor of three here. Factoring that out, we get three times in plus one. Since we have written the expression as a multiple of three. Three must divide evenly into it. There, for the answer is B. 152. About This Course1 hd: about this course. Although the Guilmette is a difficult test, it is a very learn herbal test. This is not to say that the Guilmette is beatable. There is no banger tricks that will show you how to master it overnight. You probably have already realized this. Some courses nevertheless offer inside stuff or tricks, which they claim will enable you to beat the tests. These include declaring that answer choices, B, C or D are more likely to be correct and choices a or E. This tactic, like most of its type, does not work. It is offered to give the student the feeling that he or she is getting the scoop on the test. The Guilmette cannot be beaten, but it can be mastered through hard work and a little thought and treating yourself to think like a test right.