小升初数学难题The math problem in the early stages.docx
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小升初数学难题The math problem in the early stages.docx
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小升初数学难题Themathproblemintheearlystages
小升初数学难题(Themathproblemintheearlystages)
A,
Severallitersinbrine,thefirsttimetojoinsaltaccountedfor8%ofsaltwaterafteracertainamountofwater,asecondtimetheyjoinwiththesameamountofwaterforthefirsttime,thesaltaccountedfor5%ofsaltwater,ifthethirdtimeafterjoiningwiththesameamountofwaterforthefirsttime,thesaltaccountedforaboutafewpercentofthesaltwater?
(adecimalbeforethepercent)
(1)thefirstsolution:
Examinationsite:
concentrationproblem.Analysis:
firstafteraddingwatercontainingsaltbrineratedownto8%,thesecondtimeinasmanysaltratedroppedto5%afterwater,itnotchangetheamountofsugardoesnotchange,thequalityofthewater,wemayhavetojointheXoriginallysomebrineas1,thenaddwaterforthefirsttimeafterthesaltcontentisthesecondtimethe(1+X)X8%saltcontentafteraddingwateris(1+X+X)X5%,sothatwecanworkouthowmuchwaterisaddedXis,accordingtothemeaningofthesaltcontentandaddthesameamountofwaterforthethirdtimeaftersaltcontent.Answer:
solution:
setacupofwatertoX,
Afterthefirstadditionofxcupofwater,thesaltpercentageinbrinechangedto8%-thatis,sugar(1+x)*8%,
Thesecondtimeyouaddthesameamountofwater,thesaltinthesaltwaterisapercentageof5%,whichissugar(1+x+x)by5%,
Soyouget1plusxtimes8isequalto1plusxplusxtimes5%
X=1.5,
Forthethirdtime,addthesameamountofwater,thesaltofsaltwater(1+1.5)by8%(1+1.5x3)isapproximately0.036=3.6%,
Sotheamountofsaltinthesaltwaterafterthethirdadditionofthesameamountofwateris3.6%.
A:
thethirdtimeafterjoiningthesameamountofwater,thenthebrinesalinityis3.6%.Thereview:
thistopicmainlyexaminesweshouldseizethesaltisnotavariable,becauseoftheaddedwater,saltcontentwouldreduce,wemayhavetojointhewaterasx,theoriginalbrineas1,joinedbythefirstandsecondwaterwecanworkoutx,andx,wecandirectlyintothecalculationforthethirdtimeafteraddingwatersaltcontent.
(2)secondsolution:
Holdthesaltinthesaltwaterunchanged.8%ofthesaltwaterandsalt:
water=8:
(100-8)=2=they40:
4605%ofthesaltwaterandsalt:
water=5:
100-5)=5=60officersgoesoutwaterincreasedby760-760=460(a)thethird300addthesameamountofwater,waterwithatotalof760+300=1060(a)brine:
1060+40=1100(a)thethirdbrineconcentrationafteraddthesameamountofwateris:
40memberspresent1100=3.6%
Second,
Threevolumesofthesamebottlearefilledwithanalcoholsolution,andalcoholandwaterare2:
1,3:
1,4:
1.Whatistheratioofalcoholtowaterwhenmixedwiththreebottlesofalcohol?
(a)
A.133:
47b.131:
49
C.33:
12d.3:
1
Answer:
a,
Analytic:
setthebottlevolumeis1,becauseofalcoholandwateris2:
1,3:
1and4:
1,soafterthreebottlesofalcoholsolutionblending,ratioofalcoholandwater(2/3+3/4+4/5):
(1/3+1/4+1/5)=133:
47
Three,
Aconcentrationof40%and10%ofthesugarwaterinthetwocups,togetherwithaconcentrationof30%sugarwater;Ifyouaddanother300gramsofsugarwater,theconcentrationbecomes25%.
Sohowmanygramsofsugarwaterdoyouhave?
(1)firstsolution:
300*20%=60(g),setXcanbeatotalbrine,30%ofbrineisheavy(X-300),inaccordancewiththequestiontoX-60=25%to30%(X-300)solutiontoX=600,30%ofthesaltwater:
X-300=300-300=300X300=300(g)30%(g):
aconcentrationof40%salineXg,theconcentrationof10%saltwater(300-)Xg,accordingtothequestionto40%+10%(300-)XXX=200=90solutionanswer:
40%oftheoriginal200gramsofsaltwater.
Nhasa,bcups,awater,andafruitjuice.Firstpourthecupofwaterintothecupanddoubletheliquidinthecup.Pourthecupofjuiceintoacupanddoubletheliquidinthecup.......IfIpourfourtimes,whatistheconcentrationofthetwocupsofjuice?
Process:
Theconcentrationofpurewateris0%andtheconcentrationofpurejuiceis100%
Thefirstonewas0%
Bconcentrations(100%+0%)/2=50%
Thesecondnailconcentration(50%+0%)/2=25%
Bconcentration50%
Thethirddoseis25%
Bconcentrations(25%+50%)/2=37.5%
Thefourthconcentration(37.5%+25%)/2=31.25%
Bconcentration37.5%
Becauseeachtimeyoupouravolumeofsolutionfromeachother'scup,theconcentrationofthenewsolutionistheaveragevalueoftheconcentrationofthetwosolutions.So:
thefirsttime:
a:
0%b:
50%thesecondtime:
a:
25%b:
50%afterthethirdtime;A:
25%,b:
(50+25)/2%=37.5%andthefourthtime:
a:
(25+37.5)/2%=31.25%b:
37.5%;a:
31.25%,b:
(37.5+31.25)/2%=34.375%,andthepurealcoholinthefinalsecondglassis34.375%ofthealcoholsolution
Four:
Hasareservoircontaining9conduit,oneofthemastheinletpipe,theremainingeightroottothesameoutletpipe.Inletpipewithuniformspeedtothereservoirwaterflooding.Latersomeonewantstoopentheoutletpipe,thewaterofthepoolallrowsoflight(atthismomenthasinjectedsomewaterinthepool).Iftheeightoutletpipeisopenedentirely,need3hourstoarrangeallofthewaterinthepoollight;Ifyoucanonlyopen5outletpipes,youneed6hourstodrainallthewaterinthetank.
(1)firstsolution:
Analysis:
ifyouopenapipethatcandrain"1portion"perhour,then8waterpipeswilldrain8*3=24(portions)for3hours.5outofthewaterpipefor6hourstotaldischarge5*6=30(part);Intwocases,thewaterintakeinthreehoursis30-24=6(part).Theintakeofwaterperhouris6and3=2(part).In4.5hours,theoriginalwaterinthepoolandthewaterinthewaterinletare8*3+(4.5-3)*2=27(part).
Solution:
toopenawaterpipethatcandischarge"1portion"perhour,8outletpipeswillbedrainedfor3hours,8*3=24(part);5outofthewaterpipefor6hourstotaldischarge5x6=30(part).
30-24=6(part),thesesixarethewaterinthe"6-3=3"hours.
(30-24)tosign(6-3)=6to3=2(part),this"2"istheinletwaterperhour.
[8x3+(4.5-3)x2
24+1.5x=[2]present4.5
27memberspresent=4.5
=6(root)
Answer:
itisnecessarytoopen6waterpipesatthesametime.Thekeyistoopenapipeanddrain"1"everyhour.
Five:
Theexhibitionopenedat9o'clock,buttherewerelongqueuesforadmission.Fromthefirstaudience,thenumberofspectatorsperminuteisthesame.Iftherearethreeentranceopenings,therewillbenoqueueat9:
9.Iftherearefiveentranceopenings,therewillbenoqueueat9:
5.Sowhattimedoesthefirstaudiencearrive?
(1)firstsolution:
Solution:
thenumberofpeoplewhopassbyoneminuteisone.Thespeedofpeople'sarrival:
(3x9-5x5)isthenumberofpeoplewhoarewaitingfor:
3times9-0.5times9=22.5.Thesepeopleareaccumulatingattherateof0.5perminute,sothepreviousqueueis22.5=45minutes,whichmeansthatthefirstaudiencearrivesat8:
15.
Knowledgesummary
(1)thephalanxproblem
Numberofouteredges-2=thenumberofinnersides
(numberofouteredges-1)x4=outerperimeter
Numberofouteredges-2-hollowsidelength2=realarea
Thetrainoverpasses
Thelengthofthecar+thebridge=speed*time
Thelengthofthecarandthelengthofthecar
Thelengthofthelengthofthecaristhespeeddifference
Theencounterandproblemofthetrainwiththedriverontherideroronanothertrain
Thelengthofthecar=thespeedandthecross-time
Carlength=speeddifferentialxchaseandtime
1.Partyaandpartybtwopipeseparateopen,filledwithwater,ittakes20hoursrespectively,16hours.Cwaterpipes,aloneinapoolto10hours,ifthereisnowaterpool,atthesametimeopentwopipes,partyaandpartybafter5hours,andthenopenthedrainc,qpoolfilledwithorhowmanyhours?
Solution:
1/20plus1/16isequalto9/80,whichistheefficiencyofab
So9over80times5is45over80,whichis5hoursofwater
So1minus45over80isequalto35over80
The35/80sign(9/80-1/10)=35means35hourstofill
A:
itwilltake35hoursafter5hourstofillthepool.
Itwilltake20daysfortheteamtocompleteawatercanal.Theteamwilltake30daystocomplete.Ifthetwoteamsworktogether,theefficiencyoftheirworkwillbereducedduetotheimpactofeachother'sconstruction.Theefficiencyoftheteam'sworkisfouroutoffive.
Theteam'sefficiencyisninetenthsoftheoriginal.Itisnowplannedfor16daystocompletethecanal,andthenumberofdaysofcooperationbetweenthetwoteamsisasfewaspossible.
Theworkingefficiencyofabis1/20,andtheworkingefficiencyofabis1/20*4/5+1/30*9/10=7/100,whichcanbeseenfromtheeffectof>bontheworkof>.
Becauseoftherequirementthat"thenumberofdaysofcooperationbetweenthetwoteamsisasfewaspossible",sothefastoneshouldbedone,andin16daysitwillbetoolateforthecooperationtobecompleted.Onlyinthiswaycan"thenumberofdaysofcooperationbetweenthetwoteamsbeasfewaspossible".
Thecooperationtimeisxdays,andtheonlytimeis(16-x)days
1over20times16minusxplus7over100timesxis1
X=10
Answer:
abminimumcooperation10days
3.Awork,aandbshouldbecompletedin4hours.Bandcshouldbecompletedin5hours.Now,pleaseaskaandctodoitfor2hours.Theremainingbwillbecompletedin6hours.Howmanyhoursdoesittaketofinishthisworkalone?
Solution:
1/4indicatestheworkloadof1hourofcooperation.1/5representstheworkloadof1hourforthecooperationofb/c
(1/4+1/5)*2=9/10meansthatahasdone2hours,bh
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