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### Exam krackers mcat_verbal_and_math_book

1. 1. E XAMKRACKERS MeATVERBAL REASONING & MATHEMATICAL TECHNIQUES 7 TH EDITION O St}TE PUBLISHING
2. 2. Acknowledgements Although I am the author, the hard work and expertise of many individuals contributed to this book. The idea of writing in two voices, a science voice and an MeAT voice, was the creative brainchild of my imaginative friend Jordan Zaretsky. I would like to thank David Orsay for his help with the verbal pas- sages. I wish to thank my wife, Silvia, for her support during the difficult times in the past and those that lie ahead. Finally, I wish to thank my daughter Julianna Orsay for helping out whenever possible. Copyright © 2007 Examkrackers. Inc
4. 4. LECTURE 4: How TO STUDY FOR THE VERBAL REASONING SECTION .................................. 5330-MINUTE IN-CLASS EXAMS ........................................................................................55 In-Class Exam for Lecture 1 ......... ........ ........ ......... ....... ......... ....... ..•...... " ..... .................. ......... 55 In-Class Exam for Lecture 2 ......... ........ ......... ........ ................ ................. ...... .. ........ ......... ........ 63 In-Class Exam for Lecture 3 .........•................•...... ................. ........ ................. ........ ................. 71ANSWERS & EXPLANATIONS TO IN-CLASS EXAMS ............................................................ 79 Answers and Scaled Score Conversion for In-Class Exams .................................................. 80 Explanations to In-Class Exam for Lecture 1 ....................................... ........................... ........ 81 Explanations to In-Class Exam for Lecture 2 ................ ......................... ................................. 90 Explanations to In-Class Exam for Lecture 3 ................ ..•....... .• ..... .••......•.......•. ........ .............. 99 Copyright © 2007 Exarn krackers, Inc.
6. 6. 2 VERBAL REASONING & MATHEMATICAL TECHNIQUES The MCAT consists of four sections: 1. Physical Sciences 2. Verbal Reasoning 3. Writing Sample 4. Biological Sciences Physica l Sciences This section covers topics from undergraduate physics and inorganic chemistry. Passages average approximately 200 words in length and are often accompanied by one or more charts, diagrams, or tables. Generally there are 6-10 questions follow- ing each passage, as well as 3 sets of stand-alone multiple-choice questions, for a total of 52 questions. The top score on the Physical Sciences Section is a 15. Verbal Reasoning The Verbal Reasoning Section has been shortened (to everyones pleasure) from 85 to 60 minutes. It now consists of only 40 (previously 60) multiple-choice questions with answer choices A through D. There are 9 passages followed by 4 to 10 ques- tions each. Passages average approximately 600 words in length. There is a wide variety of passage topics ranging from economics and an thropology to poetic analy- sis, and most intentionally soporific. The top score on the Verbal Reas.oning Section is also 15. Writing Sample The Writing Sample Section consists of two 30 minute periods, without any break in between. For each essay, the test-taker is given a general statement to analyze in a standard cookie-cutter fashion. This section is scored on an alphabetic scale from J to T, with T being the highest score. This scale translates to a score of 1-6 on each essay resulting in a combined score of a 2-12 represented by J through T. Bio logical Sciences The Biological Sciences Section covers science topics from a wide range of under- graduate biology topics, organic chemistry and genetics. The set up of this section is exactly the same as the Physical Sciences section, as is the scori ng. ""~, Test Section Questions Time Allotted Time/Question Tutorial ) (Optional) 10 minutes I Physical Sciences 52 70 minutes - 1.35 minutes (81s) Break (Optional) 10 minutes Verbal Reasoning 40 60 minutes 1.5 minutes (90s) Break (Optional) 10 minutes Writing Sample 2 60 minutes 30 minutes/essay Break (Optional) 10 minutes Biological Sciences 52 70 minutes -1.35 minutes (81s) Survey 10 minutes Total Content Time 4 hours, 20 minutes Total Test Time 4 hours, 50 minutes Total Appointment Time 5 hours, 10 m inutes Copyright © 2007 Examkrackers, Inc.
11. 11. INTRODUCTION TO THE MCAT INCLUDING MCAT MATH . 7The MCAT is likely to give yo u any values that you need for trigonometric func-tions; however, since MCAT typically uses common angles, it is a good idea to befamiliar w ith trigonometric values for common angles. Use the paradigm below toremember the values of common angles. Notice that the sine values are the reverseof the cosine values. Also notice tha t the numbers under the radical are 0, 1, 2, 3 and4 from top to bottom for the sine function and bottom to top for the cosine function,and all are divided by 2. e sine cosine 0° ro 2 14 2 30° .J1 -!3 2 2 45° .J2 .J2 2 2 60° -!3 - - .J1 -- 2 2 90" 14 ro 2 2Less practiced test takers may perceive a rounding strategy as risky. On the con-trary, the test makers actually design their answers with a rounding strategy inmind. Complica ted numbers can be intimidating to anyone not comfortable with arounding strategy.Copyright © 2007 Examkrackers, Inc.
12. 12. 8 . VERBAL REASONING & MATH EMATICAL TECHNIQUESQuestionsSolve the following problems by rounding. Do not lise n pencilor n enlculator. 5.4 x 7.1 x 3.2 1. 4.6 A. 2.2 B. 3.8 C. 5.8 D. 7.9 .,/360 x 9.8 2. 6.2 A. 9.6 B. 13.2 C. 17.3 D. 20.2 (F2) X23 3. 50 A. 0.12 B. 0.49 C. 0.65 D. J.J 4. A. II B. 39 C. 86 D. 450 5. ~2X9.8 49 A. 0.3 B. 0.8 C. 1.2 D. 66,,6r 51 laM5ue pexa a4~ pallOJ SJ a s 2 (;08<;6 511aMSUU pexa al{~ pallOJ sf V r ,86[<; 511aMsue pexa a4~ paUOJ sf :::J 1 SJaMSUf Copyright © 2007 Examkrackers, Inc.
14. 14. 10 V ERBAL REASONI NG & MATHEMATICAL T ECH NIQUES where y = 10. We simply add the coefficients of y. Rounding, we have 3y + 69y = 72y. Thus 72 x 10 , or 7.2 x 10 is the answer. When rearranging 6.91 x 10 to 69.1 x 10, we simply multiply by 10/ 10 (a form of 1). In other words, we divide 72 by 10 and multiply 10 by 10. xlO ~X~=69.1 XI04 .,. 10 A useful mnemonic fo r remembering which way to move the decimal point when we add or subtract from the exponent is to use the acronym LARS, i.7 Multiplication and Division When multiplying similar bases with exponents add t.he exponents; when dividing, subtract the exponents. 10 x 10 = 109 10/10-6 = 10 10 When multiplying or dividing with scientific notation, we deal with the exponen- tial terms and m antissa separately, regardless of the nllmbet of terms. For instance: (3.2 x 10) x (4.9 x 10.... ) (2.8 x 10- 7 ) should be rearranged to: - - x 10 xlO-s 3x5 _ 3 10 giving us an MeAT answer of something greater than 5 x 10 . (The exact answer, 5.6 x 10, is greater than our estimate because we decreased one term in the numer- ator by more than we increased the other, which would result in a low estimate, and because we increased the term in the denominator, w hich also results in a low es- timate.) When taking a term written in scientific notation to some power (such as squaring or cubing it), we also deal with the decimal and exponent separately. The MeAT an- swer to: (3.1 X 10)2 14 is something greater than 9 x 10 • Recall that when taking an exponential term to a power, we multiply the exponents. The first step in taking the square root of a term in scientific notation is to make the exponent even. Then we take the square root of the mantissa and exponential term separa tely. .JS.l x 10 Make the exponent even. .JS1x10 Copyrig ht © 2007 Exam krackers, Inc.
15. 15. INTRODUCTION TO THE MCAT INCLUDING MCAT MATH • 11Take the square root of the mantissa and exponential term separately. .,J8i x M = 9 xl02Notice how much more efficient this method is. What is the square root of 49,000?Most students start thinking about 700, or 70, or something with a 7 in it. By usingthe scientific notation method, we quickly see that there is no 7 involved at all. .J49,000 x.J4.9xl0 4 =2.1xl02Try finding the square root of 300 and the square root of 200.Copyright © 2007 Examkrackers, Inc.
16. 16. 12 . VERBAL REASONING & MATHEMATICAL TECHNIQUESQuestionsSolve the following problems without a calculator. Try not touse a pencil. 2.3 X la x 5.2 x 10-5 1. A. 1.2 X 10-1 B. 2.8 C. 3.1 x 10 .D. 5.6 X 102 2. (2.5 X 10-7 x 3.7 X 10-<5) + 4.2 X 10 A. B. 1.3 5.1 X X 10- 11 10- 10 I ! C. 4.2 X 10 .D. 1.3 X 10 15 3. [(1.1 X 10-4) + (8.9 X 10-5)]1/2 A. 1.1 X 10-2 B. 1.4 X 10-2 c. 1.8 X 10- D. 2.0 X 10- 4. 112(3.4 X 102 ) (2.9 X 108) A. 1.5 X 10 18 B. 3.1x1018 C. 1.4 X 1019 D. 3.1 X 10 19 1.6 X 10-19 x 15 5. 36 2 A. 1.9 X 10-21 B. 2.3 X 10-17 C. 1.2 X 10-9 D. 3.2 X 10-9 ,-OT x LOUT S! JaMSUB pBXa al.J.L jJaJJOJ S! I "£ jUBOy)lril!SIf alB SJaqumu J"l[IO al[L paJJOJ S!;) r SJ&MSUV Copyright © 2007 Examkrackers, Inc.
17. 17. INTRODUCTION TO THE MCAT INCLUDING MCAT MATH . 13i.8 ProportionsOn the MeAT, proportional relationships between variables can often be used tocircumvent lengthy calculations or, in some cases, the MeAT question simply asksthe test taker to identify the relationship directly. When the MeAT asks for thechange in one variable due to the change in another, they are making the assump-tion that all other variables remain constant.In the equation F = rna, we see that if we double F while holding rn constant, a dou-bles. If we triple F, a triples. The same relationship holds for m and F. This type ofrelationship is called a direct proportion. x2 x2 I=ml F and a are directly proportional.Notice that if we change the equation to F = rna + b, the directly proportional rela-tionships are destroyed. Now if we double F while holding all variables besides aconstant, a increases, but does not double. In order for variables to be directly pro-portional to each other, they must both be in the numerator or denominator whenthey are on opposite sides of the equation, or one must be in the numerator whilethe Other is in the denominator when they are on the same side of the equation.In addition, all sums or differences in the equation must be contained in paren-theses and multiplied by the rest of the equation. No variables within the sumsor differences will be directly proportional to any other variable.If we examine the relationship between m and a, in F :::: rna, we see that when F isheld constant and rn is doubled, a is reduced by a factor of 2. This type of relation-ship is called an inverse proportion. Again the relationship is destroyed if we addb to one side of the equation. In order for variables to be inversely proportional toeach other, they must both be in the numerator or denominator when they are onthe same side of the equation, or one must be in the numerator while the other isin the denominator when they are on opposite sides of the equation. In addition,all sums or differences in the equation must be contained in parentheses andmultiplied by the rest of the equation. No variables within the sums or differ-ences will be directly proportional to any other variable. x2 F=/zl x2 m and a are inversely proportional.If we examine a more complicated equation, the same rules apply. However, wehave to take care when dealing with exponents. One method to solve an equationusing proportions is as follows. Suppose we are given the following equation: 4 APnr Q= 8TJLThis is Poiseuilles Law. The volume flow rate Q of a real fluid through a horizon-tal pipe is equal to the product of the change in pressure t.P,1t, and the radius of thepipe to the fourth power r" divided by 8 times the viscosity TJ and the length L ofthe pipe.Water (TJ = 1.80 X 10-3 Pa s) flows through a pipe with a 14.0 cm radius at 2.00 Lis.An engineer wishes to increase the length of the pipe from 10.0 m to 40.0 m with-Copyright © 2007 Examkrackers, Inc.
19. 19. INTRODUCTION TO THE MCAT INCLUDING MCAT MATH • 15Questions: 4. The kinetic energy E of an object is given by E = Ih mv where m is the objects mass and v is the ve1ocity,of the object. If the velocity of an object decreases by a factor of 1. The coefficient of surface tension is given by the equation 2 what will happen its kinetic energy? y = (F - mg)l(2L), where F is the net force necessary to A. Kinetic energy will increase by a factor of 2. pull a submerged wire of weight mg aud length L through B. Kinetic energy will increase by a factor of 4. the surface of the fluid in question. The force required to C. Kinetic energy will decrease by a factor of 2. remove a submerged wire from water was measured and D. Kinetic energy will decrease by a factor of 4. recorded. If an equal force is required to remove a separate submerged wire with the same mass but twice the length from fluid x, what is the coefficient of surface 5. Elastic potential energy in a spring is directly propor- tension for fluid x. (Yw",,, = 0.073 mN/m) tional to the square ofthe displacement of one end of the A. 0.018 mN/m spring from its rest position while the other end remains B. 0.037 mN/m fixed. If the elastic potential energy in the spring is 100 J C. 0.073 mN/m when it is compressed tq half its rest length, what is its D. 0.146 mN/m energy when it is compressed to one fourth its rest length. A. 50J B. 150 J 2. A solid sphere rotating about a central axis has a moment C. 200 J of inertia D. 225 J I =~ MR2 3 where R is the radius of the sphere and M is its mass. Although Callisto, a moon of Jupiter, is approximately the same size as the planet Mercury, Mercury is 3 times as dense. How do their moments of inertia compare? A. The moment of inertia for Mercury is 9 times . greater thau for Callisto. B. The moment of inertia for Mercury is 3 times OOI x z: ue1fl,alBa12 S! A2,aua a1fl snljL greater than for Callisto. .z: UB1fl lalBa.6i 10 ,fl ue1fllalBa12 S! ,Sl ,Sl C. The moment of inertia for Mercury is equal to the JO 10pBJ B Aq paSBalJII! S! A2,aua a1fl sn1fl Sl moment of inertia for Callisto. JO lOpBJ B Aq paSBalJII! S! lUalUaJBlds!p aljL lUJ D. The moment of inertia for Callisto is 3 times greater S,c S1 lUalUaJBlds1p IBll!J aljl puB lUJ OS S! walU than for Mercury. -aJBlds1p IBH!II! a1fl ua1fl C-aJ!oljJ po02 B SABMIB S! OOllnq 1fl2ual2upds AUB asn um aM) lsallB 2uOllUJ OOI 21I!1ds B au¢lBllI! aM H palJO S! a s 3. The force of gravity on an any object due to earth is given by the equation F = G(moM/?) where G is the gravita- ,A Ol IBUO!podOld AIpal!p S! :J. -pallO:> S! a Ii tional constant, M is the mass of the earth, mf> is the mass of the object and r is the distance between the center of 6 Aq pap!A!p aq lsnlU mass of the earth and the center of mass of the object. If :I :I Ol IBUOBlOdOld AlaSlaAII! S! J JO alBnbs a rocket weighs 3.6 x 106 N at the surface of the earth a1fl aJII!S 1flIBa a1fl JO laluaJ a1fl lUOlJ !!PBl what is the force on the rocket due to gravity when . ljPBtI aalln S! H os 1fllBtI a1fl JO aJBpnS aljl lUOl] the rocket has reached an altitude of 1.2 x 104 km? !!PBl l[!lBtI OMl S! amlal"S al[.L aJBJlls s,l[!IBa (G = 6.67 X 10-11 Nm/kg, radius of the earth = 6370 km, al[! ol 1fllBtI al[! JO laluaJ aljl lUOIJ aJUBlS!P mass of the earth = 5.98 x 1024 kg) aljl S! J pald!ll S! J tBl[! aas Ol ASBa S! t1 UOBBl -ou J!J!luaps ljHM po02 alB nOA n pallo, S! g .€ I A. 1.2 x 10sN B. 4.3 x 105 N ,,!palI! JO lUdlUOlU Ol I"uoB C. 4.8 X 10 N -lodold ATl::lal~p S~ SSI?W aA~SSI?UI aIOUl saUlH-" D. 9.6 X 10 N : S! AllJJaw Jasuap salUH : S! AllJlaw pue az~s aUlI?S al[1- alI? sa~poq al{l a::lll!S ·p.U.IOJ S1 g ·z SJ8MSUYCopyright~) 2007 Examkrackers, Inc.
20. 20. 16 VERBAL REASONING & MATHEMATICAL TECHN IQUES i.9 Graphs The MCAT requirE," that you recognize the graphical relationship between two vari- ables in certain types of equations. The three graphs below are the most commonly used. You should memorize them. The first is a directly proportional relationship; the second is an exponential relationship; and the third is an inversely proportional relationship. t y t y y x .. x .. x- y ~ nx y~ IIXNote: n is greater than zero for the graph Notice that, if we add apositive constant b to the right side, the graph is simplyof y ~ nx, and n is greater than one for raised vertically by an amount b. If we subtract a positive constant b from the rightthe othertwo gra phs. side, the graphs are shifted downwarQs. y t y t y b b-l-- - bAs long as the value of n is within the x .. x • x ..given parameters, the general shape of y ~ nx + b y ~ 1/,, + bthe graph will not change. When graphs are unitless, multiplying the right side of an equation by a positive constant w ill not change the shape of the graph. If one side of the equation is nega- y I must have been tive, or multiplied by a negative constant, the graph is reflected across the x axis. multiplied by a negative constant. Whenever the M CAT asks you to identify the graphical relationship between two variables you should assume that all other variables in the equation are constants unless told otherwise. Next, manipulate the equation into one of the above forms (with or without the added constant b, and choose the corresponding graph. If you are unsure of a graphical relationship, plug in 1 for all variables except the variables in the question and then plug in 0, 1, and 2 for x and solve for y. Plot your results and look for the general corresponding shape. Copyright @ 2007 Examkrackers, Inc.
21. 21. INTRODUCTION TO THE M CAT INCLUDING M CAT M ATH . 17Questions 3. Which of the following graphs shows the relationship between frequency and wavelength, of electromagnetic radiation through a vacuum? (c = VA) 1. The height of an object dropped from a building in the A. C. absence of air resistance is gi ven by the equation h = ho + Vof + 1 gr, where ho and v are the initial height /2 t / t Q and velocity respectively and g is - 10 m/s- If Y, is zero which graph best represents the relationship betwee n h ;> ;> and I? A. C. A____ ho h, A---- .., t t B. D. j~ t ..::: t ____ 1---- ;> B. D. A____ h" A- - - - !<> 4. Which of the following gra phs best describes the magnitude of the eleclrostatic fo rce F = k( qq)/r created by an object with negative charge on an object with a 1---- 1 ---- positive charge as the distance r between them changes? A. C. 2. Whi ch of the following graphs best describes the magnitude of th e force (F) on a spring obeying Hooke; A. = law (F - ktu) as it is compressed to dxOliiX? C. 1 1 r ____ r ____ t .... t .... B. D. B. tJ.x ~ Llxmax D. tJ.x ---- ~mat 1 1 .... r ____ r---- Ax ---- Ax mu tJ.x ---- LlxmuCopy righ t © 2007 Examk rackers. Inc.
22. 22. :)UI s)a:p e)~w ex 3 LOOl @ +116!lA o J d Answers 1. A is correct. Since v, is zero we have h = h, + l/2gf. Since g is in the opposite direction to h, and ho is a constant we can rew rite this equa- tion as h =_1/2 gl + h, where g = 10. This is the same form as y = X. The negative sign flips the graph vertically. In addition a constant has been added to the right side so the graph inter- cepts the y axis at h,. 2. A is correct. The question asks for magnitude. Thus the negative sign is ignored and the equa- tion has the form y = nx. 3. D is correct. Manipulation of thls formula pro- duces v =ciA.. Which is in the form of y = Ilx. 4. D is correct. The form of this equation is y = IIX. The negative can be ignored because the question asks for magnitude. 5. D is correct. The equation has the form y = nx wh ere n is l i t. _ _M .-M ~ ~ (1 ·s .-M _ _M . ~ :J V (11M = J ) i,~unpew e ,(q :>uop M IJOM pue J J:>Mod U:>OMloq dlqs - uo~.IOJ oIJ1 SOliu)SUOUJop sqdu£l j/U1MolloJ 041 JO 401% S S3nDI NH:J3.l lo:J I ~oV<3H ~oVoJ lI 9NIN OSo3 ~ lo8 ~ 3 i 8~