CAT
AreYou
PreparingFor
CAT2017?
Problem Solving Techniques Booklet
By Ashank Dubey (CAT 100%iler)
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CAT
QUANTTECHNIQUESBYTESTCRACKER
INDEX
(1)CONSTANTPRODUCTRULE
(2)POWERCYCLE
(3)“MINIMUMOFALL”REGIONSINVENNDIAGRAMS
(4)LAST2DIGITSTECHNIQUE
(5)SIMLIARTODIFFERENTGROUPING(P&C)
(6)APPLICATIONOFFACTORIALS
(7)GRAPHICALDIVISIONFORGEOMETRY
(8)ASSUMPTIONMETHOD
(1)CONSTANTPRODUCTRULE(1/x)&1/(x+1)
Thisrulecanbeappliedwhenwehavetwoparameterswhoseproductisconstant,orinotherwords,whentheyareinversely
proportionaltoeachother.
e.g.)TimexSpeed=Distance,Pricexconsumption=Expenditure,Lengthxbreadth=Area
A1/xincreaseinoneoftheparameterswillresultina1/(x+1)decreaseintheother,Parameteriftheparametersare
inverselyproportional.
Let’sseetheapplicationwiththefollowingexamples
1. Amantravelsfromhishometoofficeat4km/hrandreacheshisoffice20minlate.Ifthespeedhadbeen6km/hrhewould
havereached10minearly.Findthedistancefromhishometooffice?
Solution:Assumeoriginalspeed=4km/hr.Percentageincreaseinspeedfrom4to6=50%or½.1/2increaseinspeedwillresult
in1/3decreaseinoriginaltime=30minutes.(fromgivendata).Originaltime=90minutes=1.5hoursàAnswerisDistance=
4x1.5=6km
2. A20%increaseinpriceofsugar.Findthe%decreaseinconsumptionafamilyshouldadoptsothattheexpenditureremains
constant
Solution:Herepricexconsumption=expenditure(constant)
Usingthetechnique,20%(1/5)increaseresultsin16.66%(1/6)decreaseinconsumption.Answer=16.66%
(2)POWERCYCLE
b
Thelastdigitofanumberoftheforma fallsinaparticularsequenceororderdependingontheunitdigitofthenumber(a)and
thepowerthenumberisraisedto(b).Thepowercycleofanumberthusdependsonits’unitdigit.
Considerthepowercycleof2
1
5
2 =2, 2 =32 2
6
2 =4
2 =64 3
7
2 =8
2 =128 4
8
2 =16 2 =256 th
Asitcanbeobserved,theunitdigitgetsrepeatedafterevery4 powerof2.Hence,wecansaythat2hasapowercycleof
2,4,8,6withfrequency4.
Thismeansthat,anumberoftheform
4k+1
2 willhavethelastdigitas2
4k+2
2 willhavethelastdigitas4
4k+3
2 willhavethelastdigitas8
4k+4
2 willhavethelastdigitas6(wherek=0,1,2,3…)
Thisisapplicablenotonlyfor2,butforallnumbersendingin2.
Thereforetofindthelastdigitofanumberraisedtoanypower,wejustneedtoknowthepowercycleofdigitsfrom0to9,
whicharegivenbelow
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Unitdigit
0
1
2
3
4
5
6
7
8
9
Powercycle
0
1
2,4,8,6
3,9,7,1
4,6
5
6
7,9,3,1
8,4,2,6
9,1
Frequency
1
1
4
4
2
1
1
4
4
2
Forexample
75
1. Findtheremainderwhen3 isdividedby5.
(a)Expresspowerintheform,4k+xwherex=1,2,3,4.Inthiscase75=4k+3.
(b)Takethepowercycleof3whichis3,9,7,1.Sincetheformis4k+3,takethethirddigitinthecycle,whichis7
Anynumberdividedby5,theremainderwillbethatoftheunitdigitdividedby5.Hencetheremainderis2.
Sometimes,youmaygetaquestioninthetermofvariables,whereyouneedtosubstitutevaluestogettheanswerinthefastest
waypossible.
Forexample,
3^4n
2. Findtheunitdigitof7
3^4
81.
Putn=1,theproblemreducesto7 ,whichis7 Since81=4k+1,takethefirstdigitinthepowercycleof7,whichis7.
1957
1982
3. Whatisthefirstnonzerointegerfromtherightin8330 +8370 ?
(a)3
(b)1
(c)9
(d)noneofthese
Solution:
1982
8370 willendwithmorenumberofzeroessoweneedtoconsideronlythefirstpart.Rightmostnon-zerointegerofthe
1957
1957
expressionwillbe=unitdigitof833 =unitdigitof3 Since,1957=4k+1,takethefirstdigitinthepowercycleof3,whichis3.
1!+2!+3!+…..+13!
1!+2!+3!....+28!
1!+2!+3!+…....+32!
1!+2!+3!+…....+67!
4. IfN=(13)
+(28)
+(32)
+(67)
thentheunitdigitofNis
(a)4
(b)8
(c)2
(d)noneofthese
Solution:
BasedonPowerCycle
After4!Everynumberisoftheform4k+4,hereinallfourtermspowerbecomes4k+1.Sotakingthefirstdigitfromthepower
cyclesof3,8,2,and7wewillgettheunitdigitasi.e3+8+2+7=4.Ans=0
3)Usefultechniquetofindthelast2digitsofanyexpressionoftheformab
Dependingonthelastdigitofthenumberinquestion,wecanfindthelasttwodigitsofthatnumber.
Wecanclassifythetechniquetobeappliedinto4categories.
TYPE
METHOD
EXAMPLES
67
1)Numbersendingin1
Thelastdigitisalways1.
1)21 =__41(2x7=__4)
nd
87
The2 lastdigit=productoftens
2)41 =__81(4x7=__8)
digitofbase*unitdigitofthe
167
power.
3)1261 =__21(6x7=__2)
67
In21 ;2isthetensdigitofbaseand 124
7istheunitdigitofpower
4)31 =__21(3x4=__2)
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2)Numberendingwith5
Tens power Unit
Second
digit
digit
last
of
digit
base
Odd
Odd
5
7
Odd
Even
5
2
Even Odd
5
2
Even Even
5
2
Changethepowersothatthebase
endswith1andthenusethesame
techniqueasforthosenumbers
endingwith1.
4
4
2
eg)3 ,7 &9 allwillendin1.
3)Numbersendingin3,7,9
4)Forevennumbers(2,4,6,8)
Usethepatternofthenumber1024
10
=2 i.e.
10
x2 raisedtoevenpowerendswith
76and
10
x2 raisedtooddpowerendswith
24.
34
e.g.)1555 =__25
288
4 72
4
e.g.)17 =(17 ) (takingpower4as7 willendin1.
2
2 72
(17 x17 ) 72
2
=(_89*_89) (aslast2digitsof17 =89)
72
=(_21) (aslast2digitsof89*89=21)
Answer=__41(as2*2=4)
788
10 78
8
e.g.)2 =(2 ) x2 =
__76x__56=__56.
Itisalsoimportanttonotethat,
F 76multipliedbythelast2digitsofanypowerof2willendinthesamelast2digits
E.g.76x04=04,76x08=08,76x16=16,76x32=32
2
2
2
2
2
2
2
2
2
F Thelasttwodigitsof,x ,(50-x) ,(50+x) ,(100-x) willalwaysbethesame.Forexamplelast2digitsof12 ,38 ,62 ,88 ,112 ….
willallbethesame.
2
2
2
2
2
2
2
2
Also,lasttwodigitsof11 =39 =61 =89 =111 =139 =161 =189 andsoon
3.Tofindthesquaresofnumbersfrom30-70wecanusethefollowingmethod
2
Tofind41 Step1:Differencefrom25willbefirst2digits=16
Step2:Squareofthedifferencefrom50willbelast2digits=81.Answer=1681.
2
Tofind43 Step1:Differencefrom25willbefirst2digits=18
Step2:Squareofthedifferencefrom50willbelast2digits=49.Answer=1849
Combiningallthesetechniqueswecanfindthelast2digitsforanynumberbecauseeveryevennumbercanbewrittenas2*an
oddnumber
4)“MINIMUMOFALL”REGIONSINVENNDIAGRAMS
1. Inasurveyconductedamong100meninacompany,100menusebrandA,75usebrandB,80usebrandC,90usebrandD&
60usebrandEofthesameproduct.Whatistheminimumpossiblenumberofmenusingallthe5brands,ifallthe100men
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Sumofthedifferencefrom100=(100-100)+(100-75)+(100-80)+(100-90)+(100-60)=95
Againtakethedifferencefrom100=5(answer)
Thismethodwillbeexplainedindetailinthenextbooklet.
5)SIMILARTODIFFERENTGROUPINGINPERMUTATION&COMBINATION
AllquestionsinPermutationandCombinationfallinto4categories,andifyoumasterthese4categories,youcanunderstandall
conceptsinP&Ceasily.
(i)SimilartoDifferent
(ii)DifferenttoSimilar
(iii)SimilartoSimilar
(iv)DifferenttoDifferent
Inthisbooklet,wewilllookatthefirstcategory;i.e.SimilartoDifferent,whichentailsthenumberofwaysofdividing‘n’
identical(similar)thingsinto‘r’distinct(different)groups
(a)NOLIMITQUESTIONS
Letmeexplainthiswithanexample.SupposeIhave10identicalchocolatestobedividedamong3people.The10chocolates
needtobedistributedinto3partswhereapartcanhavezeroormorechocolates.
Soletusrepresentchocolatesbyblueballs.Thestraightredlinesareusedtodividethemintoparts.Soyoucanseethatfor
dividinginto3parts,youneedonlytwolines.
st
nd
rd
Supposeyouwanttogive1 person1chocolate,2 3chocolatesand3 6chocolates.Thenyoucanshowitas:
Supposeyouwanttogiveoneperson1chocolate,anotherperson6chocolatesandanotherone3,thenitcanberepresented
as:
Nowiffirstpersongets0,secondgets1andthirdgets9chocolatesthenitcanberepresentedas:
Nowsupposeyouwanttogivefirstperson0,secondalso0andthirdallof10thenyoucanshowitlike:
So,fordividing10identicalchocolatesamong3personsyoucanassumetohave12(10balls+2sticks)thingsamongwhichten
areidenticalandrest2aresameandofonekind.Sonumberofwaysinwhichyoucandistributetenchocolatesamong3
peopleisthesameinwhichyoucanarrange12thingsamongwhich10areidenticalandofonekindwhile2areidenticalandof
onekindwhichcanbedonein12!/(10!2!)
The above situation is same as finding the number of positive integral solutions of a + b + c = 10. a, b, c is the number of
chocolatesgiventodifferentpersons.
st
OrelseIcanalsosayhowmanyintegralpointslieonthelinea+b+c=10inthe1 quadrant.Inboththecasestheansweris
12
C2.
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(b)LOWERLIMITQUESTIONS
Nowsupposewehavearestrictionthatthegroupscannotbeemptyi.e.intheaboveexampleall3personsshouldgetatleast
1.
Youhavetodividetenchocolatesamong3personssothateachgetsatleastone.Sointhestartonlygivethemoneeach.This
you will do in just 1 way as all the chocolates are identical. Now, you are left with 7 chocolates and you have to divide them
among3peopleinsuchthatwaythateachgets0ormore.Youcandothiseasilyasexplainedaboveusingballsandsticks.
Numberofways=9!/(7!2!).
Theabovesituationissameasfindingthenumberofpositivenaturalnumbersolutionsofa+b+c=10.a,b,cisthenumber
ofchocolatesgiventodifferentpersons.
st
OrelseIcanalsosayhowmanyintegralnaturalpointslieonthelinea+b+c=10inthe1 quadrant.
9
Inboththecasestheansweris C2.
NowsupposeIchangethequestionandsaythatnowyouhavetodivide10chocolatesamong3personsinsuchawaythatone
getsatleast1,secondatleast2andthirdatleast3.
It’sassimpleasthelastone.Firstfullfilltherequiredcondition.
st
Give1 person1,second2andthird3andthendividetheleft4(10–1–2-3)chocolatesamongthose3insuchawaythateach
getsatleast1.
6
Thisissameasarranging4ballsand2stickswhichcanbedonein C2ways.
Theabovesituationissameasfindingthenumberofpositiveintegralsolutionsof
a+b+c=10suchthata≥1,b≥2,c≥3.a,b,cisthenumberofchocolatesgiventodifferentpersons.
9
Inthiscasetheansweris C2.
2. Rajeshwenttothemarkettobuy18fruitsinall.Ifthereweremangoes,bananas,applesandorangesforsaletheninhow
manywayscouldRajeshbuyatleastonefruitofeachkind?
17
18
21
21
(a) C3 (b) C4 (c) C3 (d) C4
Solution:
ThisisaGroupingtype1SimilartoDistinctquestion,withalowerlimitcondition.
M+B+A+O=18
Removeonefromeachgroup,therefore4issubtractedfrombothsides.Theproblemchangesto
M+B+A+O=14
Usingourshortcut,Theansweristhearrangementof14ballsand3sticks
17
i.e. C3
3. Therearefourballstobeputinfiveboxeswhereeachboxcanaccommodateanynumberofballs.Inhowmanywayscan
onedothisif:
(a)Ballsaresimilarandboxesaredifferent
(1)275 (2)70 (3)120 (4)19
Solution:Whentheballsaresimilarandtheboxesaredifferent,it’sagroupingtype1question
8
A+B+C+D+E=4,whereA,B,C,D,Earethedifferentboxes.Thenumberofwaysofselectionanddistribution= C4=70
4. Thenumberofnon-negativeintegralsolutionsofA+B+C≤10
(a)84 (b)286 (c)220 (d)noneofthese
Bynon-negativeintegralsolutions,theconditionsimplythatwecanhave0andnaturalnumbervaluesforA,B,C,andD
Toremovethesign≤addanotherdummyvariablex4.Theproblemchangesto
A+B+C+D=10
Thisisanexampleofgroupingtype1(SimilartoDistinct)
Itisthearrangementof10ballsand3sticks.
13
Usingtheshortcutofballsandsticks,Thereforetheansweris C3=286
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6)APPLICATIONOFFACTORIALS
AthoroughunderstandingofFactorialsisimportantbecausetheyplayapivotalrolenotonlyinunderstandingconceptsin
NumbersbutalsootherimportanttopicslikePermutationandCombination
DefinitionofFactorialàN!=1x2x3x…x(n-1)xn
E.g.1)5!=1x2x3x4x5=120e.g.2)3!=1x2x3=6
LetusnowlookattheapplicationofFactorials
(I)Highestpowerinafactorialorinaproduct
QuestionsbasedonhighestpowerinafactorialareseenyearafteryearinCAT.Questionsbasedonthiscanbecategorized
basedonthenatureofthenumber(primeorcomposite)whosehighestpowerwearefindinginthefactorial,i.e
(ii)Highestpowerofaprimenumberinafactorial:
Tofindthehighestpowerofaprimenumber(x)inafactorial(N!),continuouslydivideNbyxandaddallthequotients.
1. Thehighestpowerof5in100!
Solution:
100/5=20;20/5=4;Addingthequotients,its20+4=24.Sohighestpowerof5in100!=24
ALTERNATIVEMETHOD
2
2
100/5+100/5 =20+4=24(Wetakeupto5 asitisthehighestpowerof5whichislessthan100)
(iii)Highestpowerofacompositenumberinfactorial
Factorizethenumberintoprimes.Findthehighestpowerofalltheprimenumbersinthatfactorialusingthepreviousmethod.
Taketheleastpower.
2. Tofindthehighestpowerof10in100!
Solution:Factorize10=5*2.
1.Highestpowerof5in100!=24 2.Highestpowerof2in100!=97
Therefore,theanswerwillbe24,becausetogeta10,youneedapairof2and5,andonly24suchpairsareavailable.Sotakethe
lesseri.e.24istheanswer.
3. Highestpowerof12in100!
2
2
Solution:12=2 *3.Findthehighestpowerof2 and3in100!
Firstfindoutthehighestpowerof2.
Listingoutthequotients:100/2=50;50/2=25;25/2=12;12/2=6;6/2=3;3/2=1
2
2
Highestpowerof2=50+25+12+6+3+1=97.Sohighestpowerof2 =48(outof972’sonly48canmake2 )
Nowforthehighestpowerof3.100/3=33;33/3=11;11/3=3;3/3=1;
Highestpowerof3=48,Highestpowerof12=48
(iv)Numberofzeroesintheendofafactorialoraproduct
Findingthenumberofzeroesformsthebaseconceptforanumberofapplicationquestions.Inbase10,numberofzerosinthe
enddependsonthenumberof10s;i.e.effectively,onthenumberof5s
F InbaseN,numberofzeroesintheendhighestpowerofNinthatproduct
4. Findthenumberofzeroesin13!Inbase10
Solution:Weneedtoeffectivelyfindthehighestpowerof10in13!=Highestpowerof5in13!Asthispowerwillbelesser.
13/5=2
5. Findthenumberofzeroesintheendof15!inbase12.
2
Solution:Highestpowerof12in15!=highestpowerof2 *3in15!=Highestpowerof3in15!=5
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(V)Numberoffactorsofanyfactorial
Thereisatechnique,whichcanbeusedtofindthenumberoffactorsinafactorial
6. Findthefactorsof12!
10
5
2
STEP1:Primefactorize12!i.e.findoutthehighestpowerofallprimefactorstill12(i.e.2,3,5,7,11).12!=2 x3 x5 x7x11
STEP2:Usetheformula
m
n
N=a *b (a,baretheprimefactors).Thennumberoffactors=(m+1)(n+1)
Thenumberoffactors=(10+1)(5+1)(2+1)(1+1)(1+1)=792.Answer=792
(VI)Rightmostnonzerointegerinafactorial
7. Whatisthelastnon–zerodigitof20!
Solution:Usingabovepropertywecanwrite:
1x2x3x4x5x6x7x8x9x10=4x
11x12x13x14x15x16x17x18x19x20=4x
Solastnon–zerodigitof20!= xlastnon–zerodigitof4!=lastnon–zerodigitof6x4=4
APPLICATIONQUESTIONBASEDONFACTORIAL
8. Howmanynaturalnumbersaretheresuchthattheirfactorialsareendingwith5zeroes?
10!is1x2x3x4x(5)x6x7x8x9x(2x5).Fromthiswecanseethathighestpowerof5till10!is2.Continuinglikethis,10!-14!,highest
powerof5willbe2.Thenext5willbeobtainedat15=(5*3).
Therefore,from15!To19!-Thehighestpowerof5willbe3.
20!-24!–HighestPower=4,In25,wearegettingoneextrafive,as25=5x5.Therefore,25!to29!,wewillgethighestpowerof
5as6.Theanswertothequestionistherefore,0.Therearenonaturalnumberswhosefactorialsendwith5zeroes.
7)USEOFGRAPHICALDIVISIONINGEOMETRY
Let’slookatatechniquewhichwillhelpyousolveageometryquestioninnotime
1. ABCDisasquareandEandFarethemidpointsofABandBCrespectively.FindtheratioofArea(ABCD):Area(DEF)?
Let’sdividethefigureusingdottedlinesasshowninFigureB.AreaofABCD=100%.AreaAEDG=50%.ThenAreainshadedregion
1(AED)=25%.Similarly,DCFH=50%.Areainshadedregion2(DCF)=25%.NowAreaofEOFB=25%..Areaofshadedregion3(BEF)
=12.5%.Totalareaoutsidetriangle=62.5%.Areainsidetriangle=100-62.5=37.5%.Requiredratio=100/37.5=8:3
8)ASSUMPTIONmethod
Thereareatleast10questionsoutof25inCAT2007whereyoucanapplythistechnique.Thisinvolvesassumingsimplevalues
forthevariablesinthequestions,andsubstitutinginansweroptionsbasedonthosevalues.Assumptionhelpstotremendously
speeduptheprocessofevaluatingtheanswerasshownbelow.
2
2
2
1. k&2k arethetworootsoftheequationx –px+q.Findq+4q +6pq=
2
3
3
(a)q (b)p (c)0(d)2p
Solution:Assumeanequationwithroots1&2(k=1)
2
=>p(sumofroots)=3andq(productofroots)=2.Substituteinq+4q +6pq=54.Lookintheansweroptionsfor54on
3
3
substitutingvaluesofp=3andq=2.weget2p =54.=>Answer=2p .
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CAT
3
3
2. Let‘x’bethearithmeticmeanandy,zbethetwogeometricmeansbetweenanytwopositivenumbers.Thevalueof(y +z )/
(xyz)is
(a)2 (b)3 (c)1/2 (d)3/2
Solution:AssumeaGP1111thenx=1,y=1z=1.Answeronsubstitution=2,whichwillmakethecalculationevenfaster,
halfoftheproblemsinAlgebracanbesolvedusingassumption.Thisisnotdirectsubstitution.Inthenexte.g.:seehowyoucan
usethesametechniqueinanequationquestion.
vSOMEACTUALQUESTIONSFROMCATSOLVEDUSINGTHISTECHNIQUE
1. ConsiderthesetS={2,3,4……2n+1),wherenisapositiveintegerlargerthan2007.DefineXastheaverageofoddintegersinS
andYastheaverageoftheevenintegersinS.WhatisthevalueofX-Y?
(a)1
(b)n/2 (c)(n+1)/2n
(d)2008
(e)0
Solution:Thequestionisindependentofn,whichisshownbelow.
Taken=2.ThenS={2,3,4,5).X=4andY=3.X-Y=1,Taken=3.thenS={2,3,4,5,6,7}.X=5andY=4.X-Y=1
Henceyoucandirectlymarktheansweroption(a).Youcansolvethequestioninlessthan60seconds.
2. LetSbethesetofallpairs(i,j)where1≤i<j≤nandn≥4.AnytwodistinctmembersofSarecalledfriends,iftheyhaveone
constituentofthepairsincommonand“enemies”otherwise.Forexample,ifn=4,thenS={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}.
Here(1,2),(1,3)arefriendsbut(1,4),(2,3)areenemies.Forgeneraln,consideranytwomembersofSthatarefriends.How
manyothermembersofSwillbecommonfriendsofthesemembers?
2
2
(a)2n-6 (b)n/2(n-3)
(c)n-2 (d)(n -7n+16)/2 e)(n -5n+8)/2
Solution:Wewillstartwithn=5,thenyouwillgettwooptions(c)and(d).
Nowtaken=6,letstaketwomembers(1,2)and(1,3).Thentheircommonfriendswillbe(1,4),(1,5),(1,6)and(2,3).i.eFour
commonfriends.Sosubstituten=6amongansweroptionsandcheckwhereyougetanswer4.Onlyoption(c).
There were more questions which could be solved using similar strategies. The methods given above clearly show that for
someonewithgoodconceptualknowledgeandrightstrategiesthequantsectionisacakewalk.
………………………………………………………………………………………………………………………………………………….....................................
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