CIRCULAR REASONING

　　CIRCULAR REASONING INVOLVES ASSUMING AS A PREMISE¹ THAT WHICH YOU ARE TRYING

　　TO PROVE. INTUITIVELY, IT MAY SEEM THAT NO ONE WOULD FALL FOR SUCH AN ARGUME

　　NT. HOWEVER, THE CONCLUSION MAY APPEAR TO STATE SOMETHING ADDITIONAL, OR THE

　　ARGUMENT MAY BE SO LONG THAT THE READER MAY FORGET THAT THE CONCLUSION WAS

　　STATED AS A PREMISE.

　　EXAMPLE:

　　THE DEATH PENALTY IS APPROPRIATE FOR TRAITORS² BECAUSE IT IS RIGHT TO EXECUTE

　　THOSE WHO BETRAY THEIR OWN COUNTRY AND THEREBY³ RISK THE LIVES OF MILLIONS.

　　THIS ARGUMENT IS CIRCULAR BECAUSE "RIGHT" MEANS ESSENTIALLY⁴ THE SAME THING A

　　S "APPROPRIATE." IN EFFECT, THE WRITER IS SAYING THAT THE DEATH PENALTY IS A

　　PPROPRIATE BECAUSE IT IS APPROPRIATE.

　　SHIFTING THE BURDEN OF PROOF

　　IT IS INCUMBENT⁵ ON THE WRITER TO PROVIDE EVIDENCE OR SUPPORT FOR HER POSITIO

　　N. TO IMPLY THAT A POSITION IS TRUE MERELY BECAUSE NO ONE HAS DISPROVED IT I

　　S TO SHIFT THE BURDEN OF PROOF TO OTHERS.

　　EXAMPLE:

　　SINCE NO ONE HAS BEEN ABLE TO PROVE GOD’S EXISTENCE, THERE MUST NOT BE A GOD

　　..

　　THERE ARE TWO MAJOR WEAKNESSES IN THIS ARGUMENT. FIRST, THE FACT THAT GOD’S

　　EXISTENCE HAS YET TO BE PROVEN DOES NOT PRECLUDE⁶ ANY FUTURE PROOF OF EXISTEN

　　CE. SECOND, IF THERE IS A GOD, ONE WOULD EXPECT THAT HIS EXISTENCE IS INDEPE

　　NDENT OF ANY PROOF BY MAN.

　　UNWARRANTED ASSUMPTIONS

　　THE FALLACY OF UNWARRANTED ASSUMPTION IS COMMITTED WHEN THE CONCLUSION OF AN

　　ARGUMENT IS BASED ON A PREMISE (IMPLICIT OR EXPLICIT⁷) THAT IS FALSE OR UNWA

　　RRANTED. AN ASSUMPTION IS UNWARRANTED WHEN IT IS FALSE--THESE PREMISES⁸ ARE U

　　SUALLY SUPPRESSED OR VAGUELY⁹ WRITTEN. AN ASSUMPTION IS ALSO UNWARRANTED WHEN

　　IT IS TRUE BUT DOES NOT APPLY IN THE GIVEN CONTEXT--THESE PREMISES ARE USUA

　　LLY EXPLICIT.

　　EXAMPLE: (FALSE DICHOTOMY)

　　EITHER RESTRICTIONS¹⁰ MUST BE PLACED ON FREEDOM OF SPEECH OR CERTAIN SUBVERSIV

　　E ELEMENTS IN SOCIETY WILL USE IT TO DESTROY THIS COUNTRY. SINCE TO ALLOW TH

　　E LATTER TO OCCUR IS UNCONSCIONABLE, WE MUST RESTRICT FREEDOM OF SPEECH.

　　THE CONCLUSION ABOVE IS UNSOUND BECAUSE

　　(A) SUBVERSIVES¹¹ DO NOT IN FACT WANT TO DESTROY THE COUNTRY

　　(B) THE AUTHOR PLACES TOO MUCH IMPORTANCE ON THE FREEDOM OF SPEECH

　　(C) THE AUTHOR FAILS TO CONSIDER AN ACCOMMODATION BETWEEN THE TWO ALTERNATIV

　　ES

　　(D) THE MEANING OF "FREEDOM OF SPEECH" HAS NOT BEEN DEFINED

　　(E) SUBVERSIVES ARE A TRUE THREAT TO OUR WAY OF LIFE

　　THE ARGUER OFFERS TWO OPTIONS: EITHER RESTRICT FREEDOM OF SPEECH, OR LOSE TH

　　E COUNTRY. HE HOPES THE READER WILL ASSUME THAT THESE ARE THE ONLY OPTIONS A

　　VAILABLE. THIS IS UNWARRANTED. HE DOES NOT STATE HOW THE SO-CALLED "SUBVERSI

　　VE ELEMENTS" WOULD DESTROY THE COUNTRY, NOR FOR THAT MATTER, WHY THEY WOULD

　　WANT TO DESTROY IT. THERE MAY BE A THIRD OPTION THAT THE AUTHOR DID NOT MENT

　　ION; NAMELY, THAT SOCIETY MAY BE ABLE TO TOLERATE THE "SUBVERSIVES" AND IT M

　　AY EVEN BE IMPROVED BY THE DIVERSITY OF OPINION THEY OFFER. THE ANSWER IS (C

　　).

　　APPEAL TO AUTHORITY

　　TO APPEAL TO AUTHORITY IS TO CITE AN EXPERT’S OPINION AS SUPPORT FOR ONE’S O

　　WN OPINION. THIS METHOD OF THOUGHT IS NOT NECESSARILY FALLACIOUS. CLEARLY, T

　　HE REASONABLENESS OF THE ARGUMENT DEPENDS ON THE "EXPERTISE¹²" OF THE PERSON B

　　EING CITED AND WHETHER SHE IS AN EXPERT IN A FIELD RELEVANT TO THE ARGUMENT.

　　APPEALING TO A DOCTOR’S AUTHORITY ON A MEDICAL ISSUE, FOR EXAMPLE, WOULD BE

　　REASONABLE; BUT IF THE ISSUE IS ABOUT DERMATOLOGY AND THE DOCTOR IS AN ORTH

　　OPEDIST, THEN THE ARGUMENT WOULD BE QUESTIONABLE¹³.

　　PERSONAL ATTACK

　　IN A PERSONAL ATTACK (AD HOMINEM), A PERSON’S CHARACTER IS CHALLENGED INSTEA

　　D OF HER OPINIONS.

　　EXAMPLE:

　　POLITICIAN: HOW CAN WE TRUST MY OPPONENT TO BE TRUE TO THE VOTERS? HE ISN’T

　　TRUE TO HIS WIFE!

　　THIS ARGUMENT IS WEAK BECAUSE IT ATTACKS THE OPPONENT’S CHARACTER, NOT HIS P

　　OSITIONS. SOME PEOPLE MAY CONSIDER FIDELITY¹⁴ A PREREQUISITE¹⁵ FOR PUBLIC OFFICE

　　.. HISTORY, HOWEVER, SHOWS NO CORRELATION¹⁶ BETWEEN FIDELITY AND GREAT POLITICA

　　L LEADERSHIP.

　　--

　　I WOULD FLY YOU TO THE MOON AND BACK

　　IF YOU’LL BE IF YOU’LL BE MY BABY

　　GOT A TICKET FOR A WORLDSWHERESWE BELONG

　　SO WOULD YOU BE MY BABY

　　TESTPREP充分性精解转载SMTH 2001-10-14 10:51:58发信人: YKK (我不说话并不代表我不在乎),信区: ENGLISHTEST

　　标题: (GMAT)TESTPREP充分性精解

　　发信站: BBS水木清华站(FRI OCT 12 16:07:05 2001)

　　DATA SUFFICIENCY

　　----------------------------------------------------------------------------

　　----

　　INTRODUCTION DATA SUFFICIENCY

　　MOST PEOPLE HAVE MUCH MORE DIFFICULTY WITH THE DATA SUFFICIENCY PROBLEMS THA

　　N WITH THE STANDARD MATH PROBLEMS. HOWEVER, THE MATHEMATICAL KNOWLEDGE AND S

　　KILL REQUIRED TO SOLVE DATA SUFFICIENCY PROBLEMS IS NO GREATER THAN THAT REQ

　　UIRED TO SOLVE STANDARD MATH PROBLEMS. WHAT MAKES DATA SUFFICIENCY PROBLEMS

　　APPEAR HARDER AT FIRST IS THE COMPLICATED DIRECTIONS. BUT ONCE YOU BECOME FA

　　MILIAR WITH THE DIRECTIONS, YOU’LL FIND THESE PROBLEMS NO HARDER THAN STANDA

　　RD MATH PROBLEMS. IN FACT, PEOPLE USUALLY BECOME PROFICIENT¹⁷ MORE QUICKLY ON

　　DATA SUFFICIENCY PROBLEMS.

　　THE DIRECTIONS

　　THE DIRECTIONS FOR DATA SUFFICIENCY QUESTIONS ARE RATHER COMPLICATED. BEFORE

　　READING ANY FURTHER, TAKE SOME TIME TO LEARN THE DIRECTIONS COLD. SOME OF T

　　HE WORDING IN THE DIRECTIONS BELOW HAS BEEN CHANGED FROM THE GMAT TO MAKE IT

　　CLEARER. YOU SHOULD NEVER HAVE TO LOOK AT THE INSTRUCTIONS DURING THE TEST.

　　DIRECTIONS: EACH OF THE FOLLOWING DATA SUFFICIENCY PROBLEMS CONTAINS A QUEST

　　ION FOLLOWED BY TWO STATEMENTS, NUMBERED (1) AND (2). YOU NEED NOT SOLVE THE

　　PROBLEM; RATHER YOU MUST DECIDE WHETHER THE INFORMATION GIVEN IS SUFFICIENT

　　TO SOLVE THE PROBLEM.

　　THE CORRECT ANSWER TO A QUESTION IS

　　A IF STATEMENT (1) ALONE IS SUFFICIENT TO ANSWER THE QUESTION BUT STATEMENT

　　(2) ALONE IS NOT SUFFICIENT;

　　B IF STATEMENT (2) ALONE IS SUFFICIENT TO ANSWER THE QUESTION BUT STATEMENT

　　(1) ALONE IS NOT SUFFICIENT;

　　C IF THE TWO STATEMENTS TAKEN TOGETHER ARE SUFFICIENT TO ANSWER THE QUESTION

　　, BUT NEITHER STATEMENT ALONE IS SUFFICIENT;

　　D IF EACH STATEMENT ALONE IS SUFFICIENT TO ANSWER THE QUESTION;

　　E IF THE TWO STATEMENTS TAKEN TOGETHER ARE STILL NOT SUFFICIENT TO ANSWER TH

　　E QUESTION.

　　NUMBERS: ONLY REAL NUMBERS ARE USED. THAT IS, THERE ARE NO COMPLEX NUMBERS.

　　DRAWINGS: THE DRAWINGS ARE DRAWN¹⁸ TO SCALE ACCORDING TO THE INFORMATION GIVEN

　　IN THE QUESTION, BUT MAY CONFLICT WITH THE INFORMATION GIVEN IN STATEMENTS

　　(1) AND (2).

　　YOU CAN ASSUME THAT A LINE THAT APPEARS STRAIGHT IS STRAIGHT AND THAT ANGLE

　　MEASURES CANNOT BE ZERO.

　　YOU CAN ASSUME THAT THE RELATIVE POSITIONS OF POINTS, ANGLES, AND OBJECTS AR

　　E AS SHOWN.

　　ALL DRAWINGS LIE IN A PLANE UNLESS STATED OTHERWISE.

　　EXAMPLE:

　　IN TRIANGLE ABC TO THE RIGHT, WHAT IS THE VALUE OF Y?

　　(1) AB = AC

　　(2) X = 30

　　EXPLANATION: BY STATEMENT (1), TRIANGLE ABC IS ISOSCELES. HENCE, ITS BASE AN

　　GLES ARE EQUAL: Y = Z. SINCE THE ANGLE SUM OF A TRIANGLE IS 180 DEGREES, WE

　　GET X + Y + Z = 180. REPLACING Z WITH Y IN THIS EQUATION AND THEN SIMPLIFYIN

　　G YIELDS X + 2Y = 180. SINCE STATEMENT (1) DOES NOT GIVE A VALUE FOR X, WE C

　　ANNOT DETERMINE THE VALUE OF Y FROM STATEMENT (1) ALONE. BY STATEMENT (2), X

　　= 30. HENCE, X + Y + Z = 180 BECOMES 30 + Y + Z = 180, OR Y + Z = 150. SINC

　　E STATEMENT (2) DOES NOT GIVE A VALUE FOR Z, WE CANNOT DETERMINE THE VALUE O

　　F Y FROM STATEMENT (2) ALONE. HOWEVER, USING BOTH STATEMENTS IN COMBINATION,

　　WE CAN FIND BOTH X AND Z AND THEREFORE Y. HENCE, THE ANSWER IS C.

　　NOTICE IN THE ABOVE EXAMPLE THAT THE TRIANGLE APPEARS TO BE A RIGHT TRIANGLE

　　.. HOWEVER, THAT CANNOT BE ASSUMED: ANGLE A MAY BE 89 DEGREES OR 91 DEGREES,

　　WE CAN’T TELL FROM THE DRAWING. YOU MUST BE VERY CAREFUL NOT TO ASSUME ANY M

　　ORE THAN WHAT IS EXPLICITLY¹⁹ GIVEN IN A DATA SUFFICIENCY PROBLEM.

　　ELIMINATION²⁰

　　DATA SUFFICIENCY QUESTIONS PROVIDE FERTILE GROUND FOR ELIMINATION. IN FACT,

　　IT IS RARE THAT YOU WON’T BE ABLE TO ELIMINATE SOME ANSWER-CHOICES. REMEMBER

　　, IF YOU CAN ELIMINATE AT LEAST ONE ANSWER CHOICE, THE ODDS²¹ OF GAINING POINT

　　S BY GUESSING ARE IN YOUR FAVOR.

　　THE FOLLOWING TABLE SUMMARIZES HOW ELIMINATION FUNCTIONS WITH DATA SUFFICIEN

　　CY PROBLEMS.

　　STATEMENT CHOICES ELIMINATED

　　(1) IS SUFFICIENT B, C, E

　　(1) IS NOT SUFFICIENT A, D

　　(2) IS SUFFICIENT A, C, E

　　(2) IS NOT SUFFICIENT B, D

　　(1) IS NOT SUFFICIENT AND (2) IS NOT SUFFICIENT A, B, D

　　EXAMPLE 1: WHAT IS THE 1ST TERM IN SEQUENCE S?

　　(1) THE 3RD TERM OF S IS 4.

　　(2) THE 2ND TERM OF S IS THREE TIMES THE 1ST, AND THE 3RD TERM IS FOUR TIMES

　　THE 2ND.

　　(1) IS NO HELP IN FINDING THE FIRST TERM OF S. FOR EXAMPLE, THE FOLLOWING SE

　　QUENCES EACH HAVE 4 AS THEIR THIRD TERM, YET THEY HAVE DIFFERENT FIRST TERMS

　　:

　　0, 2, 4

　　-4, 0, 4

　　THIS ELIMINATES CHOICES A AND D. NOW, EVEN IF WE ARE UNABLE TO SOLVE THIS PR

　　OBLEM, WE HAVE SIGNIFICANTLY INCREASED OUR CHANCES OF GUESSING CORRECTLY--FR

　　OM 1 IN 5 TO 1 IN 3.

　　TURNING TO (2), WE COMPLETELY IGNORE THE INFORMATION IN (1). ALTHOUGH (2) CO

　　NTAINS A LOT OF INFORMATION, IT ALSO IS NOT SUFFICIENT. FOR EXAMPLE, THE FOL

　　LOWING SEQUENCES EACH SATISFY (2), YET THEY HAVE DIFFERENT FIRST TERMS:

　　1, 3, 12

　　3, 9, 36

　　THIS ELIMINATES B, AND OUR CHANCES OF GUESSING CORRECTLY HAVE INCREASED TO 1

　　IN 2.

　　NEXT, WE CONSIDER (1) AND (2) TOGETHER. FROM (1), WE KNOW "THE 3RD TERM OF S

　　IS 4." FROM (2), WE KNOW "THE 3RD TERM IS FOUR TIMES THE 2ND." THIS IS EQUI

　　VALENT TO SAYING THE 2ND TERM IS 1/4 THE 3RD TERM: (1/4)4 = 1. FURTHER, FROM

　　(2), WE KNOW "THE 2ND TERM IS THREE TIMES THE 1ST." THIS IS EQUIVALENT TO S

　　AYING THE 1ST TERM IS 1/3 THE 2ND TERM: (1/3)1 = 1/3. HENCE, THE FIRST TERM

　　OF THE SEQUENCE IS FULLY²² DETERMINED²³: 1/3, 1, 4. THE ANSWER IS C.

　　EXAMPLE 2: IN THE FIGURE TO THE RIGHT, WHAT IS THE AREA OF THE TRIANGLE?

　　(1)

　　(2) X = 90

　　RECALL THAT A TRIANGLE IS A RIGHT TRIANGLE IF AND ONLY IF THE SQUARE OF THE

　　LONGEST SIDE IS EQUAL TO THE SUM OF THE SQUARES OF THE SHORTER SIDES (PYTHAG

　　OREAN THEOREM). HENCE, (1) IMPLIES THAT THE TRIANGLE IS A RIGHT TRIANGLE. SO

　　THE AREA OF THE TRIANGLE IS (6)(8)/2. NOTE, THERE IS NO NEED TO CALCULATE T

　　HE AREA--WE JUST NEED TO KNOW THAT THE AREA CAN BE CALCULATED. HENCE, THE AN

　　SWER IS EITHER A OR D.

　　TURNING TO (2), WE SEE IMMEDIATELY THAT WE HAVE A RIGHT TRIANGLE. HENCE, AGA

　　IN THE AREA CAN BE CALCULATED. THE ANSWER IS D.

　　EXAMPLE 3: IS P < Q ?

　　(1) P/3 < Q/3

　　(2) -P + X > -Q + X

　　MULTIPLYING BOTH SIDES OF P/3 < Q/3 BY 3 YIELDS P < Q.

　　HENCE, (1) IS SUFFICIENT. AS TO (2), SUBTRACT X FROM BOTH SIDES OF -P + X >

　　-Q + X, WHICH YIELDS -P > -Q.

　　MULTIPLYING BOTH SIDES OF THIS INEQUALITY BY -1, AND RECALLING THAT MULTIPLY

　　ING BOTH SIDES OF AN INEQUALITY BY A NEGATIVE NUMBER REVERSES THE INEQUALITY

　　, YIELDS P < Q.

　　HENCE, (2) IS ALSO SUFFICIENT. THE ANSWER IS D.

　　EXAMPLE 4: IF X IS BOTH THE CUBE OF AN INTEGER AND BETWEEN 2 AND 200, WHAT I

　　S THE VALUE OF X?

　　(1) X IS ODD.

　　(2) X IS THE SQUARE OF AN INTEGER.

　　SINCE X IS BOTH A CUBE AND BETWEEN 2 AND 200, WE ARE LOOKING AT THE INTEGERS

　　:

　　WHICH REDUCE TO

　　8, 27, 64, 125

　　SINCE THERE ARE TWO ODD INTEGERS IN THIS SET, (1) IS NOT SUFFICIENT TO UNIQU

　　ELY DETERMINE THE VALUE OF X. THIS ELIMINATES CHOICES A AND D.

　　NEXT, THERE IS ONLY ONE PERFECT SQUARE, 64, IN THE SET. HENCE, (2) IS SUFFIC

　　IENT TO DETERMINE THE VALUE OF X. THE ANSWER IS B.

　　EXAMPLE 5: IS CAB A CODE WORD IN LANGUAGE Q?

　　(1) ABC IS THE BASE WORD.

　　(2) IF C IMMEDIATELY FOLLOWS B, THEN C CAN BE MOVED TO THE FRONT OF THE CODE

　　WORD TO GENERATE ANOTHER WORD.

　　FROM (1), WE CANNOT DETERMINE WHETHER CAB IS A CODE WORD SINCE (1) GIVES NO

　　RULE FOR GENERATING ANOTHER WORD FROM THE BASE WORD. THIS ELIMINATES A AND D

　　..

　　TURNING TO (2), WE STILL CANNOT DETERMINE WHETHER CAB IS A CODE WORD SINCE N

　　OW WE HAVE NO WORD TO APPLY THIS RULE TO. THIS ELIMINATES B.

　　HOWEVER, IF WE CONSIDER (1) AND (2) TOGETHER, THEN WE CAN DETERMINE WHETHER

　　CAB IS A CODE WORD:

　　FROM (1), ABC IS A CODE WORD.

　　FROM (2), THE C IN THE CODE WORD ABC CAN BE MOVED TO THE FRONT OF THE WORD:

　　CAB.

　　HENCE, CAB IS A CODE WORD AND THE ANSWER IS C.

　　UNWARRANTED ASSUMPTIONS

　　BE EXTRA CAREFUL NOT TO READ ANY MORESINTOSA STATEMENT THAN WHAT IS GIVEN.

　　?THE MAIN PURPOSE OF SOME DIFFICULT PROBLEMS IS TO LURE²⁴ YOUSINTOSMAKING AN U

　　NWARRANTED ASSUMPTION.

　　IF YOU AVOID THE TEMPTATION, THESE PROBLEMS CAN BECOME ROUTINE.

　　EXAMPLE 6: DID INCUMBENT I GET OVER 50% OF THE VOTE?

　　(1) CHALLENGER C GOT 49% OF THE VOTE.

　　(2) INCUMBENT I GOT 25,000 OF THE 100,000 VOTES CAST.

　　IF YOU DID NOT MAKE ANY UNWARRANTED ASSUMPTIONS, YOU PROBABLY DID NOT FIND T

　　HIS TO BE A HARD PROBLEM. WHAT MAKES A PROBLEM DIFFICULT IS NOT NECESSARILY

　　ITS UNDERLYING²⁵ COMPLEXITY²⁶; RATHER A PROBLEM IS CLASSIFIED AS DIFFICULT IF MA

　　NY PEOPLE MISS IT. A PROBLEM MAY BE SIMPLE YET CONTAIN A PSYCHOLOGICAL TRAP

　　THAT CAUSES PEOPLE TO ANSWER IT INCORRECTLY.

　　THE ABOVE PROBLEM IS DIFFICULT BECAUSE MANY PEOPLE SUBCONSCIOUSLY²⁷ ASSUME THA

　　T THERE ARE ONLY TWO CANDIDATES. THEY THEN FIGURE THAT SINCE THE CHALLENGER

　　RECEIVED 49% OF THE VOTE THE INCUMBENT RECEIVED 51% OF THE VOTE. THIS WOULD

　　BE A VALID²⁸ DEDUCTION²⁹ IF C WERE THE ONLY CHALLENGER (YOU MIGHT ASK, "WHAT IF

　　SOME PEOPLE VOTED FOR NONE-OF-THE-ABOVE?" BUT DON’T GET CARRIED AWAY WITH FI

　　NDING EXCEPTIONS. THE WRITERS OF THE GMAT WOULD NOT SET A TRAP THAT SUBTLE).

　　BUT WE CANNOT ASSUME THAT. THERE MAY BE TWO OR MORE CHALLENGERS. HENCE, (1)

　　IS INSUFFICIENT³⁰.

　　NOW, CONSIDER (2) ALONE. SINCE INCUMBENT I RECEIVED 25,000 OF THE 100,000 VO

　　TES CAST, I NECESSARILY RECEIVED 25% OF THE VOTE. HENCE, THE ANSWER TO THE Q

　　UESTION IS "NO, THE INCUMBENT DID NOT RECEIVE OVER 50% OF THE VOTE." THEREFO

　　RE, (2) IS SUFFICIENT TO ANSWER THE QUESTION. THE ANSWER IS B.

　　NOTE, SOME PEOPLE HAVE TROUBLE WITH (2) BECAUSE THEY FEEL THAT THE QUESTION

　　ASKS FOR A "YES" ANSWER. BUT ON DATA SUFFICIENCY QUESTIONS, A "NO" ANSWER IS

　　JUST AS VALID AS A "YES" ANSWER. WHAT WE’RE LOOKING FOR IS A DEFINITE ANSWE

　　R.

　　CHECKING EXTREME CASES

　　?WHEN DRAWING A GEOMETRIC FIGURE OR CHECKING A GIVEN ONE, BE SURE TO INCLUDE

　　DRAWINGS OF EXTREME CASES AS WELL AS ORDINARY ONES.

　　EXAMPLE 1: IN THE FIGURE TO THE RIGHT, AC IS A CHORD AND B IS A POINT ON THE

　　CIRCLE. WHAT IS THE MEASURE OF ANGLE X?

　　ALTHOUGH IN THE DRAWING AC LOOKS TO BE A DIAMETER, THAT CANNOT BE ASSUMED. A

　　LL WE KNOW IS THAT AC IS A CHORD. HENCE, NUMEROUS CASES ARE POSSIBLE, THREE

　　OF WHICH ARE ILLUSTRATED³¹ BELOW:

　　IN CASE I, X IS GREATER THAN 45 DEGREES; IN CASE II, X EQUALS 45 DEGREES; IN

　　CASE III, X IS LESS THAN 45 DEGREES. HENCE, THE GIVEN INFORMATION IS NOT SU

　　FFICIENT TO ANSWER THE QUESTION.

　　EXAMPLE 2: THREE RAYS EMANATE³² FROM A COMMON POINT AND FORM THREE ANGLES WITH

　　MEASURES P, Q, AND R. WHAT IS THE MEASURE OF Q + R ?

　　IT IS NATURAL TO MAKE THE DRAWING SYMMETRIC AS FOLLOWS:

　　IN THIS CASE, P = Q = R = 120, SO Q + R = 240. HOWEVER, THERE ARE OTHER DRAW

　　INGS POSSIBLE. FOR EXAMPLE:

　　IN THIS CASE, Q + R = 180. HENCE, THE GIVEN INFORMATION IS NOT SUFFICIENT TO

　　ANSWER THE QUESTION.

　　PROBLEMS:

　　1. SUPPOSE 3P + 4Q = 11. THEN WHAT IS THE VALUE OF Q?

　　(1) P IS PRIME.

　　(2) Q = -2P

　　(1) IS INSUFFICIENT. FOR EXAMPLE, IF P = 3 AND Q = 1/2, THEN 3P + 4Q = 3(3)

　　+ 4(1/2) = 11. HOWEVER, IF P = 5 AND Q = -1, THEN 3P + 4Q = 3(5) + 4(-1) = 1

　　1. SINCE THE VALUE OF Q IS NOT UNIQUE, (1) IS INSUFFICIENT.

　　TURNING TO (2), WE NOW HAVE A SYSTEM OF TWO EQUATIONS IN TWO UNKNOWNS. HENCE

　　, THE SYSTEM CAN BE SOLVED TO DETERMINE THE VALUE OF Q. THUS, (2) IS SUFFICI

　　ENT, AND THE ANSWER IS B.

　　2. WHAT IS THE PERIMETER³³ OF TRIANGLE ABC ABOVE?

　　(1) THE RATIO OF DE TO BF IS 1: 3.

　　(2) D AND E ARE MIDPOINTS OF SIDES AB AND CB, RESPECTIVELY.

　　SINCE WE DO NOT EVEN KNOW WHETHER BF IS AN ALTITUDE, NOTHING CAN BE DETERMIN

　　ED FROM (1). MORE IMPORTANTLY, THERE IS NO INFORMATION TELLING US THE ABSOLU

　　TE SIZE OF THE TRIANGLE.

　　AS TO (2), ALTHOUGH FROM GEOMETRY WE KNOW THAT DE = AC/2, THIS RELATIONSHIP

　　HOLDS FOR ANY SIZE TRIANGLE. HENCE, (2) IS ALSO INSUFFICIENT.

　　TOGETHER, (1) AND (2) ARE ALSO INSUFFICIENT SINCE WE STILL DON’T HAVE INFORM

　　ATION ABOUT THE SIZE OF THE TRIANGLE, SO WE CAN’T DETERMINE THE PERIMETER. T

　　HE ANSWER IS E.

　　3. A DRESS WAS INITIALLY³⁴ LISTED AT A PRICE THAT WOULD HAVE GIVEN THE STORE A

　　PROFIT OF 20 PERCENT OF THE WHOLESALE³⁵ COST. WHAT WAS THE WHOLESALE COST OF

　　THE DRESS?

　　(1) AFTER REDUCING THE ASKING PRICE BY 10 PERCENT, THE DRESS SOLD FOR A NET

　　PROFIT OF 10 DOLLARS.

　　(2) THE DRESS SOLD FOR 50 DOLLARS.

　　CONSIDER JUST THE QUESTION SETUP. SINCE THE STORE WOULD HAVE MADE A PROFIT O

　　F 20 PERCENT ON THE WHOLESALE COST, THE ORIGINAL PRICE P OF THE DRESS WAS 12

　　0 PERCENT OF THE COST: P = 1.2C. NOW, TRANSLATING (1)SINTOSAN EQUATION YIELD

　　S:

　　P - .1P = C + 10

　　SIMPLIFYING GIVES

　　..9P = C + 10

　　SOLVING FOR P YIELDS

　　P = (C + 10)/.9

　　PLUGGING THIS EXPRESSION FOR PSINTOSP = 1.2C GIVES

　　(C + 10)/.9 = 1.2C

　　SINCE WE NOW HAVE ONLY ONE EQUATION INVOLVING THE COST, WE CAN DETERMINE THE

　　COST BY SOLVING FOR C. HENCE, THE ANSWER IS A OR D.

　　(2) IS INSUFFICIENT SINCE IT DOES NOT RELATE THE SELLING PRICE TO ANY OTHER

　　INFORMATION. NOTE, THE PHRASE "INITIALLY LISTED" IMPLIES THAT THERE WAS MORE

　　THAN ONE ASKING PRICE. IF IT WASN’T FOR THAT PHRASE, (2) WOULD BE SUFFICIEN

　　T. THE ANSWER IS A.

　　4. WHAT IS THE VALUE OF THE TWO-DIGIT NUMBER X?

　　(1) THE SUM OF ITS DIGITS³⁶ IS 4.

　　(2) THE DIFFERENCE OF ITS DIGITS IS 4.

　　CONSIDERING (1) ONLY, X MUST BE 13, 22, 31, OR 40. HENCE, (1) IS NOT SUFFICI

　　ENT TO DETERMINE THE VALUE OF X.

　　CONSIDERING (2) ONLY, X MUST BE 40, 51, 15, 62, 26, 73, 37, 84, 48, 95, OR 5

　　9. HENCE, (2) IS NOT SUFFICIENT TO DETERMINE THE VALUE OF X.

　　CONSIDERING (1) AND (2) TOGETHER, WE SEE THAT 40 AND ONLY 40 IS COMMON TO TH

　　E TWO SETS OF CHOICES FOR X. HENCE, X MUST BE 40. THUS, TOGETHER (1) AND (2)

　　ARE SUFFICIENT TO UNIQUELY DETERMINE THE VALUE OF X. THE ANSWER IS C.

　　5. IF X AND Y DO NOT EQUAL 0, IS X/Y AN INTEGER?

　　(1) X IS PRIME.

　　(2) Y IS EVEN.

　　(1) IS NOT SUFFICIENT SINCE WE DON’T KNOW THE VALUE OF Y. SIMILARLY, (2) IS

　　NOT SUFFICIENT. FURTHERMORE, (1) AND (2) TOGETHER ARE STILL INSUFFICIENT SIN

　　CE THERE IS AN EVEN PRIME NUMBER--2. FOR EXAMPLE, LET X BE THE PRIME NUMBER

　　2, AND LET Y BE THE EVEN NUMBER 2 (DON’T FORGET THAT DIFFERENT VARIABLES CAN

　　STAND FOR THE SAME NUMBER). THEN X/Y = 2/2 = 1, WHICH IS AN INTEGER. FOR AL

　　L OTHER VALUES OF X AND Y, X/Y IS NOT AN INTEGER. (PLUG IN A FEW VALUES TO V

　　ERIFY THIS.) THE ANSWER IS E.

　　6. IS 500 THE AVERAGE (ARITHMETIC MEAN) SCORE ON THE GMAT?

　　(1) HALF OF THE PEOPLE WHO TAKE THE GMAT SCORE ABOVE 500 AND HALF OF THE PEO

　　PLE SCORE BELOW 500.

　　(2) THE HIGHEST GMAT SCORE IS 800 AND THE LOWEST SCORE IS 200.

　　MANY STUDENTS MISTAKENLY THINK THAT (1) IMPLIES THE AVERAGE IS 500. SUPPOSE

　　JUST 2 PEOPLE TAKE THE TEST AND ONE SCORES 700 (ABOVE 500) AND THE OTHER SCO

　　RES 400 (BELOW 500). CLEARLY, THE AVERAGE SCORE FOR THE TWO TEST-TAKERS IS N

　　OT 500. (2) IS LESS TEMPTING³⁷. KNOWING THE HIGHEST AND LOWEST SCORES TELLS US

　　NOTHING ABOUT THE OTHER SCORES. FINALLY, (1) AND (2) TOGETHER DO NOT DETERM

　　INE THE AVERAGE SINCE TOGETHER THEY STILL DON’T TELL US THE DISTRIBUTION OF

　　MOST OF THE SCORES. THE ANSWER IS E.

　　7. THE SET S OF NUMBERS HAS THE FOLLOWING PROPERTIES:

　　I) IF X IS IN S, THEN 1/X IS IN S.

　　II) IF BOTH X AND Y ARE IN S, THEN SO IS X + Y.

　　IS 3 IN S?

　　(1) 1/3 IS IN S.

　　(2) 1 IS IN S.

　　CONSIDER (1) ALONE. SINCE 1/3 IS IN S, WE KNOW FROM PROPERTY I THAT 1/(1/3)

　　= 3 IS IN S. HENCE, (1) IS SUFFICIENT.

　　CONSIDER (2) ALONE. SINCE 1 IS IN S, WE KNOW FROM PROPERTY II THAT 1 + 1 = 2

　　(NOTE, NOTHING IN PROPERTY II PREVENTS X AND Y FROM STANDING³⁸ FOR THE SAME N

　　UMBER. IN THIS CASE BOTH STAND FOR 1.) IS IN S. APPLYING PROPERTY II AGAIN S

　　HOWS THAT 1 + 2 = 3 IS IN S. HENCE, (2) IS ALSO SUFFICIENT. THE ANSWER IS D.

　　8. WHAT IS THE AREA OF THE TRIANGLE ABOVE?

　　(1) A = X, B = 2X, AND C = 3X.

　　(2) THE SIDE OPPOSITE A IS 4 AND THE SIDE OPPOSITE B IS 3.

　　FROM (1) WE CAN DETERMINE THE MEASURES OF THE ANGLES: A + B + C = X + 2X + 3

　　X = 6X = 180

　　DIVIDING THE LAST EQUATION BY 6 GIVES: X = 30

　　HENCE, A = 30, B = 60, AND C = 90. HOWEVER, DIFFERENT SIZE TRIANGLES CAN HAV

　　E THESE ANGLE MEASURES, AS THE DIAGRAM BELOW ILLUS

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