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AlligationRuleandMixturesandReplacements
Alligationrulehelpsustofind,inwhatratiotwomixtureswithdifferentconcentrationsaretobemixedtogetatargetconcentration.
AssumethatinanengineeringcollegeECEbranchaverageaggregateis80%andthatofCSEis68%.Inwhatratiothestudentsintheseclassesaretobemixedtogetanaggregateof76%?
TakeECEclassaverageas ,CSEclassaverageas .Ifthereare studentsinECEand studentsinCSEthentheoverallaverage iscalculatedasfollows
Ifwerearrangethisequation
Toapplythisruleeasilywefollowasmalldiagram
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Eventhoughalligationruleiswidelyappliedinmixtures,wecanseethisruleinmanyareasacrossarithmetic.
Competitiveexamiswhereyouneedtosolveproblemsinextremetimeconstraints.Soyouneedtodevelopclearunderstandingofconceptstosolveproblemsquickly.Letusdiscussaclassicproblemtounderstandhowsolidconceptsenableustoclearthetestseasily.
Acaskcontainsalcoholsolutionwithconcentration18%.If6Litersofthismixtureistakenoutandreplacedwithwater,thenconcentrationdropsto15%.Whatistheoriginalvolumeinthecask.Explanation:
Youcanobservefromthediagram,concentrationremainsthesameafter6literswereremovedfromthecask.Concentrationchangesinverselyproportionaltothevolumeofwateraddedtothemixture.Sochangeintheconcentrationhappenedonlyfromsecondstagetothirdchange.
Method1:Weknowthattotalalcoholcomponentinthemixtureisequaltoalcoholcomponentinthemixturethathasbeentakenoutplusremainingalcoholinthecask.soassumeinitialorfinalvolumeisVliters.18%(V)=18%(6)+15%(V)3%(V)=18%(6)V=36
Method2:Volumeandconcentrationareinverselyproportionaltoeachother.IV=initialvolumeIC=initialconcentrationFV=FinalvolumeFC=FinalconcentrationIVxIC=FVxFC(V6)x18%=Vx15%(V6)/V=6/5SoV=36
Method3:Wecanalsosolvethisproblembyusingalligationrule.Itstatesthatinwhatrationtwocomponentsaremixedtogetatargetedconcentration.Wecanapplythisruletothisproblemforthesecondstage.Weaddedwaterwhichisat0%concentrationtoamixtureof18%concentrationtogetasolutionwith15%concentration.
SoweunderstandfromtheabovediagramMixtureandwatershouldbemixedintheration15:3togetdesiredconcentration15%.Butweknowthat3unitsofwaterisequalto6litersso15unitsofmixtureisequalto30liters.Totalvolumeisequalto30+6=36.Pleasenotethatinthesecondstagethevolumeisequalto(V6)
Method4:InitialCondition18%=(A:W)=18:82FinalCondition15%=(A:W)=15:85
WeknowthatthereisnochangeinthePurealcoholcomponentfromsecondstagetothirdstage.sowecanequatealcoholcomponentintheabovetwoequationbymultiplyingwithappropriatenumbers.Nowweobserveachangeinthewatercomponentsfrom41to51.Thisisduetothewaterweaddedtothemixture.Weadded6litersofwaterwhichisequalto10unitschangeinthemixture.so10Units=6Liters60Units=36Liters
Method5:Wecanusethisformula
Heren=1becausewemadethissubstitutiononlyonce.
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PracticeProblems
1.Apersoncovers800kmpartlyataspeedof50kmphandpartlyataspeedof150kmph,in10hoursoverall.Whatisthedistancecoveredatthespeedsof150kmph?
Hisaveragespeedfortheentirejourneyis
Nowweneedtofindinwhatratioheneedstotravel800km,partlyat50kmph,and150kmphtogetaveragespeedof80kmph
Soforevery70partsofthetimetravelledat50kmph,hehastotravel30partsare150kmph.
Asthetotaltimeis10hourshemusthavetravelled =10hoursat150kmph
2.ThecostofoilisRs.100perkilogram.Afteradulterationwithanotheroil,whichcostsRs.50perkilogram,Ram
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sellsthemixtureatRs.96perkilogrammakingaprofitof20%.Inwhatratidoeshemixthetwokindsofoil?
Beforeapplyingthealligationruleweneedtomakesurethatalltheperametersareinthesameunits.Here2costspricesand1sellingpricewasgiven.Soweshouldconverthesellingpriceintocostprice.Given
Nowweapplythealligationrule.
SoThesetwomustbemixedintheratio3:2.
3.Twosolutionsofmilkandwaterarecombinedintheratio2:3byvolume.Theresultantsolutionisa40%milksolution.Findthemilkconcentrationinthefirstsolutioniftheconcentrationofmilkinthesecondis60%?Itwasgiventhat areintheratio2:3andsecondsolutionconcentrationis60andresultantsolutionconcentrationis40.
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Fromtheaboveweknowthat20,40xmustbeintheratio2:3
4.Ashopkeepersold45kg.ofgoods.Ifhesellssomequantityatalossof3%andrestat17%profit,making5%profitonthewhole,findthequantitysoldatprofit.
Note:175=12,and5(3)=5+3=8Therefore,Ratioofquantitiessoldatprofitandatloss=8:12=2:3
Therefore,Quantitysoldatprofit= x45=18kg.
5.Amixtureof70litresofwineandwatercontains10%water.Howmuchwatershouldbeaddedtomake25%waterintheresultingmixture?
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Therefore,Theratiois75:15=5:1.Therefore,Forevery5litresofmixture,1litreofwaterisadded.
Therefore,For70litres,watertobeadded= x70=14litres.
AlternativeMethod:Water=10%of70litres=7litresWine=707=63litresNow,inthenewmixture,wateris25%andwineis75%.Hence,innewmixturewineis3timesofwater.
Therefore,Waterinnewmixture= x63=21litres
Therefore,Watertobeadded=217=14litres
6.AmanpurchasedahorseandacowforRs.5000.Hesellsthehorseat20%profitandthecowat10%loss.Ifhegains2%onthewholetransaction,thecostofthehorseis:
Therefore,Ratiobetweencostofahorseandthatofacow=12:18=2:3.
Therefore,Costofthehorse= x5000=Rs.2000
7.Inanexamination,astudentgets3marksforeverycorrectanswerandloses1markforeverywronganswer.Ifhescores'0'marksinapaperof100questions,howmanyofhisanswerswerecorrect?
Therefore,Ratioofcorrectandwronganswers=1:3
Therefore,Correctanswers= x100=25
8.Thebattingaverageofacricketplayeris72runsperinning.Inthenext4inning,hecouldscoreonly80runsandtherebydecreaseshisbattingaverageby2runs.Whatistotalnumberofinningplayedbyhimtilllastmatch?
Averageoflast4inning= =20runs
Averageofinningplayedearlier=72runsNewaverage=722=70runs
Ratioofinningplayed(beforeandafter)=50:2=25:1Given,inningsplayed(after)=4Therefore,Inningplayed(before)=25x4=100Therefore,Totalinningplayed=100+4=104AlternativeMethod:
Averagerunsinthelast4inning= =20runs
Shortfrompreviousaverage=7220=52Short(total)=52x4But,shortruns(perinning)=2runs
Therefore,Totalinningplayed= =104inning
MixturesandReplacements
Theproblemsrelatedtomixturesbasedontwoimportantconcepts.AlligationruleandInverseproportionalityrulearethetwo.
Intheseproblemsweareaskedtofindtheresultantconcentrationaftermixingtwoorthreecomponentsorthefinalconcentrationwhenonecomponentofthemixtureisbeingreplacedbyanothercomponentwhichismostlyonethecomponentsofthemixture.
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Thegeneralformulaforreplacementsisasfollows:
HereFC=FinalconcentrationIC=Inititalconcentrationx=replacementquantityV=Finalvolumeafterreplacementn=numberofreplacements
Note:AlwaysrememberFCandICaretheconcentrationsofthesecondcomponentinthemixture."x"istheconcentrationofthefirstcomponent.
9.Fromasolutioncontainingmilkandwaterintheratio3:4,10Lisremovedandreplacedbywater.Iftheresultantsolutioncontainsmilkandwaterintheratio1:2thenwhatwastheamountoftheoriginalsolution?Herealsowearereplacingwithwater.SoFCandICmustbemilkconcentrations.Initialconcentrationofthemilk=3/7Finalconcentrationofthemilk=1/3Applyingformula
10.10%ofasolutionofmilkandwaterisremovedandthenreplacedwiththesameamountofwater.Iftheresultingratioofmilkandwateris2:3,findtheratioofmilkandwaterintheoriginalsolution.Applyingformula:
Here2/5isthemilkconcentration.
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11.Abeakerhad20Lofalcoholglycerolmixtureintheratio4:1byvolume.Inthefirstround,4Lofthemixtureisremovedandreplacedwithglycerol.Inthesecondround,5Loftheresultantsolutionisremovedandreplacedwithglycerol.Finally,10Loftheresultantmixtureisremovedandreplacedwithglycerol.WhatisthefinalquantityofglycerolinthemixtureHerewearereplacingthemixturewithglycerol.SowehavetotakeAlcoholconcentrationsforICandFC.Initialconcentrationofalcoholis4/5=80%
Applyingtheformulaforthefirstreplacement:
Here istheconcentrationafterfirstreplacement.
Secondreplacement:
ThirdReplacement:
Nowsubstitutingthe and in weget
12.Inamixtureof80L,milkandwaterareintheratio7:3.If24Lofthismixtureisreplacedby16Lifmilk,findthefinalratioofmilkandwater.Finalvolumeofthemixture=8024+16=72Replacementquantity=16Applyingformula,
SoMilkandWaterafterreplacement=23:7
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