GMT考试--Testprep数学精解(5)
CIRCULAR REASONING INVOLVES ASSUMING AS A PREMISE 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 TRAITORS BECAUSE IT IS RIGHT TO EXECUTE
THOSE WHO BETRAY THEIR OWN COUNTRY AND THEREBY RISK THE LIVES OF MILLIONS.
THIS ARGUMENT IS CIRCULAR BECAUSE "RIGHT" MEANS ESSENTIALLY 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 INCUMBENT 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 PRECLUDE 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 EXPLICIT) THAT IS FALSE OR UNWA
RRANTED. AN ASSUMPTION IS UNWARRANTED WHEN IT IS FALSE--THESE PREMISES ARE U
SUALLY SUPPRESSED OR VAGUELY 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 RESTRICTIONS 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) SUBVERSIVES 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 "EXPERTISE" 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 QUESTIONABLE.
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 FIDELITY A PREREQUISITE FOR PUBLIC OFFICE
.. HISTORY, HOWEVER, SHOWS NO CORRELATION 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
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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 PROFICIENT 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 DRAWN 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 EXPLICITLY GIVEN IN A DATA SUFFICIENCY PROBLEM.
ELIMINATION
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 ODDS 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 FULLY DETERMINED: 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 LURE 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 UNDERLYING COMPLEXITY; 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 SUBCONSCIOUSLY 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 VALID DEDUCTION 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 INSUFFICIENT.
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 ILLUSTRATED 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 EMANATE 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 PERIMETER 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 INITIALLY LISTED AT A PRICE THAT WOULD HAVE GIVEN THE STORE A
PROFIT OF 20 PERCENT OF THE WHOLESALE 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 DIGITS 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 TEMPTING. 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 STANDING 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