Home Aptitude Algebra Reasoning English PlacementPapers Technical GD's HRinterviewQuestions General BuyCRTBook

AlligationRuleandMixturesandReplacements

Alligationrulehelpsustofind,inwhatratiotwomixtureswithdifferentconcentrationsaretobemixedtogetatargetconcentration.

AssumethatinanengineeringcollegeECEbranchaverageaggregateis80%andthatofCSEis68%.Inwhatratiothestudentsintheseclassesaretobemixedtogetanaggregateof76%?

TakeECEclassaverageas ,CSEclassaverageas .Ifthereare studentsinECEand studentsinCSEthentheoverallaverage iscalculatedasfollows

Ifwerearrangethisequation

Toapplythisruleeasilywefollowasmalldiagram

*

*

2#

g g*

*

*

*

2#

g g g *

*

2#

*

*

g g *

2#

*

2#

*

*

2#

2#

search...

Submitaplacementquestion/Interviewexperience:Clickhere

Feedback/Suggestons:ClickHere

Follow 1k

Author:4.6kLike

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.

4

2

*

4

2

*

PracticeProblems

1.Apersoncovers800kmpartlyataspeedof50kmphandpartlyataspeedof150kmph,in10hoursoverall.Whatisthedistancecoveredatthespeedsof150kmph?

Hisaveragespeedfortheentirejourneyis

Nowweneedtofindinwhatratioheneedstotravel800km,partlyat50kmph,and150kmphtogetaveragespeedof80kmph

Soforevery70partsofthetimetravelledat50kmph,hehastotravel30partsare150kmph.

Asthetotaltimeis10hourshemusthavetravelled =10hoursat150kmph

2.ThecostofoilisRs.100perkilogram.Afteradulterationwithanotheroil,whichcostsRs.50perkilogram,Ram

2

2

2

2

g

sellsthemixtureatRs.96perkilogrammakingaprofitof20%.Inwhatratidoeshemixthetwokindsofoil?

Beforeapplyingthealligationruleweneedtomakesurethatalltheperametersareinthesameunits.Here2costspricesand1sellingpricewasgiven.Soweshouldconverthesellingpriceintocostprice.Given

Nowweapplythealligationrule.

SoThesetwomustbemixedintheratio3:2.

3.Twosolutionsofmilkandwaterarecombinedintheratio2:3byvolume.Theresultantsolutionisa40%milksolution.Findthemilkconcentrationinthefirstsolutioniftheconcentrationofmilkinthesecondis60%?Itwasgiventhat areintheratio2:3andsecondsolutionconcentrationis60andresultantsolutionconcentrationis40.

g

g

*

*

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?

4

4

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.

g

O

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.

'$ *$g

Y

7

O

g

,g

,g

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

' g $

g

g

'$

'$

'$

g

g

g

g g g

g

g

Home

520Like