小升初数学 难题Word文件下载.docx
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小升初数学 难题Word文件下载.docx
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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:
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:
25%b:
50%afterthethirdtime;
A:
25%,b:
(50+25)/2%=37.5%andthefourthtime:
(25+37.5)/2%=31.25%b:
37.5%;
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.
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"
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)
itisnecessarytoopen6waterpipesatthesametime.Thekeyistoopenapipeanddrain"
1"
everyhour.
Five:
Theexhibitionopenedat9o'
clock,buttherewerelongqueuesforadmission.Fromthefirstaudience,thenumberofspectatorsperminuteisthesame.Iftherearethreeentranceopenings,therewillbenoqueueat9:
9.Iftherearefiveentranceopenings,therewillbenoqueueat9:
5.Sowhattimedoesthefirstaudiencearrive?
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?
1/20plus1/16isequalto9/80,whichistheefficiencyofab
So9over80times5is45over80,whichis5hoursofwater
So1minus45over80isequalto35over80
The35/80sign(9/80-1/10)=35means35hourstofill
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
abminimumcooperation10days
3.Awork,aandbshouldbecompletedin4hours.Bandcshouldbecompletedin5hours.Now,pleaseaskaandctodoitfor2hours.Theremainingbwillbecompletedin6hours.Howmanyhoursdoesittaketofinishthisworkalone?
1/4indicatestheworkloadof1hourofcooperation.1/5representstheworkloadof1hourforthecooperationofb/c
(1/4+1/5)*2=9/10meansthatahasdone2hours,bhasdone4hours,andchasdone2h
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