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Question Number 20992 by NECx last updated on 09/Sep/17

Commented by NECx last updated on 09/Sep/17

$$\mathrm{please}\mathrm{help}\mathrm{guys}$$

Answered by $@ty@m last updated on 10/Sep/17

$$Lettherebexgirls.$$$$\Rightarrow no.ofboys=\frac{110}{100}x$$$$\Rightarrow no.ofchildren,i.e.$$$$no.ofboys+girls=x+\frac{110}{100}x$$$$$$$$Again,$$$$Letno.ofmen=y$$$$\Rightarrow no.ofwomen=\frac{115}{100}y$$$$\Rightarrow no.ofadults=\frac{215}{100}y-\left(2\right)$$$$ATQ$$$$\frac{215}{100}y\times \frac{120}{100}=\frac{210}{100}x$$$$\Rightarrow x=\frac{215}{100}\times \frac{120}{210}y$$$$\Rightarrow x=\frac{43}{35}y---\left(3\right)$$$$\therefore no.ofboys=\frac{110}{100}x=\frac{110}{100}\times \frac{43}{35}y$$$$=\frac{473}{350}y--\left(4\right)$$$$Given,$$$$population<6000$$$$\frac{210}{100}x+\frac{215}{100}y<6000$$$$210x+215y<600000$$$$210\times \frac{43}{35}y+215y<600000$$$$258y+215y<600000$$$$473y<600000$$$$y<1268.5--\left(5\right)$$$$$$$$\left(2\right)\Rightarrow y\mid 20$$$$\left(3\right)\Rightarrow y\mid 35$$$$\left(4\right)\Rightarrow y\mid 350$$$$LCM(20,35,350)=700$$$$\therefore \left(5\right)\Rightarrow y=700$$$$\Rightarrow no.ofmen=700$$$$\Rightarrow no.ofwomen=\frac{115}{100}\times 700=805$$$$no.ofgirls=\frac{43}{35}\times 700=860$$$$no.ofboys=\frac{473}{350}\times 700=946$$

Commented by NECx last updated on 10/Sep/17

$$\mathrm{wow}.....\mathrm{this}\mathrm{is}\mathrm{great}\mathrm{sir}.$$$${\mathrm{i}}^{\prime }\mathrm{m}\mathrm{most}\mathrm{grateful}\mathrm{sir}.\mathrm{Words}\mathrm{cant}$$$$\mathrm{express}\mathrm{how}\mathrm{happy}\mathrm{I}\mathrm{am}.\mathrm{Thank}$$$$\mathrm{you}$$

Commented by NECx last updated on 10/Sep/17

$$\mathrm{what}\mathrm{do}\mathrm{you}\mathrm{mean}\mathrm{by}$$$$\left(2\right)\Rightarrow \mathrm{y}\mid 20$$$$\left(3\right)\Rightarrow \mathrm{y}\mid 35$$$$\left(4\right)\Rightarrow \mathrm{y}\mid 350$$$$\mathrm{l}.\mathrm{c}.\mathrm{m}(20,35,350)=700$$

Commented by $@ty@m last updated on 10/Sep/17

$$\because yisno.ofmen$$$$\therefore y\in \mathrm{N}$$$$Similarlyno.ofadults=\frac{215}{100}y\in \mathrm{N}$$$$whichispossibleonlyifyisamultipleof20.$$$$thatiswhatImeantby$$$$\left(2\right)\Rightarrow \mathrm{y}\mid 20$$$$andlikewiseforothertwoexpessions.$$$$Further,tofulfillthethree$$$$conditionssimltaneously$$$${we}^{\prime }llhavetofindtheLeastCommon$$$$Multiple\left(LCM\right)of20,35and350$$$$whichis700.$$$$\therefore y\in \{700,1400,......\}$$$$butinineqation\left(5\right),wehave$$$$y<1268.5$$$$\Rightarrow y=700$$$$Areallyourdoubtsclearnow?$$

Commented by NECx last updated on 10/Sep/17

$$\mathrm{yeah}....\mathrm{i}\mathrm{understand}.$$$$\mathrm{please}\mathrm{which}\mathrm{topic}\mathrm{in}\mathrm{maths}\mathrm{does}$$$$\mathrm{this}\mathrm{question}\mathrm{come}\mathrm{from}?$$

Commented by $@ty@m last updated on 10/Sep/17

$$HaaHaaHaa$$$$thesamequestionIwasgoing$$$$toaskyou....$$$$Itseemstobe‘‘Applcationof{ineqality}^{″}$$$$butactuallyperceptionof‘‘{Percentage}^{″}and$$$$‘‘Number{Theory}^{″}arealsoinvolved.$$$$Morethantheory,itneedslogic$$$$tosolve.$$

Commented by NECx last updated on 10/Sep/17

$$\mathrm{hmmmm}....\mathrm{i}\mathrm{hear}\mathrm{you}\mathrm{sir}.$$

Answered by mrW1 last updated on 10/Sep/17

$$\mathrm{C}=\mathrm{children}\mathrm{in}\mathrm{\%}$$$$\mathrm{A}=\mathrm{adult}\mathrm{in}\mathrm{\%}$$$$\mathrm{B}=\mathrm{boys}\mathrm{in}\mathrm{\%}$$$$\mathrm{G}=\mathrm{girls}\mathrm{in}\mathrm{\%}$$$$$$$$\mathrm{C}+\mathrm{A}=100\mathrm{\%}=1$$$$\mathrm{C}=1.2\times \mathrm{A}$$$$\Rightarrow \mathrm{C}=\frac{6}{11}$$$$\Rightarrow \mathrm{A}=\frac{5}{11}$$$$$$$$\mathrm{B}+\mathrm{G}=\mathrm{C}=\frac{6}{11}$$$$\mathrm{B}=1.1\times \mathrm{G}$$$$\Rightarrow \mathrm{B}=\frac{22}{77}$$$$\Rightarrow \mathrm{G}=\frac{20}{77}$$$$$$$$\mathrm{W}+\mathrm{M}=\mathrm{A}=\frac{5}{11}$$$$\mathrm{W}=1.15\times \mathrm{M}$$$$\Rightarrow \mathrm{W}=\frac{115}{473}$$$$\Rightarrow \mathrm{M}=\frac{100}{473}$$$$$$$$\mathrm{let}\mathrm{N}=\mathrm{total}\mathrm{population}$$$$\Rightarrow \mathrm{Boys}=\frac{22}{77}\mathrm{N}=22\times \frac{\mathrm{N}}{77}$$$$\Rightarrow \mathrm{Girls}=\frac{20}{77}\mathrm{N}=20\times \frac{\mathrm{N}}{77}$$$$\Rightarrow \mathrm{Women}=\frac{115}{473}\mathrm{N}=115\times \frac{\mathrm{N}}{473}$$$$\Rightarrow \mathrm{Men}=\frac{100}{473}\mathrm{N}=100\times \frac{\mathrm{N}}{473}$$$$$$$$\mathrm{LCM}(77,473)=3311$$$$\Rightarrow \mathrm{N}=3311\mathrm{k}\mathrm{with}\mathrm{k}\in \mathbb{N}$$$$\Rightarrow \mathrm{Boys}=22\times \frac{3311\mathrm{k}}{77}=946\mathrm{k}$$$$\Rightarrow \mathrm{Girls}=20\times \frac{3311\mathrm{k}}{77}=860\mathrm{k}$$$$\Rightarrow \mathrm{Women}=115\times \frac{3311\mathrm{k}}{473}=805\mathrm{k}$$$$\Rightarrow \mathrm{Men}=100\times \frac{3311\mathrm{k}}{473}=700\mathrm{k}$$$$$$$$\mathrm{since}\mathrm{N}<6000,\Rightarrow \mathrm{k}=1$$$$\Rightarrow 946\mathrm{boys},860\mathrm{girls},805\mathrm{women},700\mathrm{men}$$

Commented by NECx last updated on 10/Sep/17

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by NECx last updated on 10/Sep/17

$$\mathrm{sorry}\mathrm{but}\mathrm{i}\mathrm{really}\mathrm{need}\mathrm{to}\mathrm{ask}\mathrm{this};$$$$$$$$\mathrm{why}\mathrm{was}\mathrm{it}\mathrm{necessary}\mathrm{to}\mathrm{find}\mathrm{the}$$$$\mathrm{LCM}\mathrm{of}\mathrm{the}\mathrm{denominators}?$$$$$$$$$$$$\mathrm{Thanks}\mathrm{for}\mathrm{always}\mathrm{helping}.$$

Commented by mrW1 last updated on 11/Sep/17

$$\frac{\mathrm{N}}{77}\mathrm{and}\frac{\mathrm{N}}{473}\mathrm{must}\mathrm{be}\mathrm{integer},\mathrm{i}.\mathrm{e}.$$$$\mathrm{N}\mathrm{must}\mathrm{be}\mathrm{a}\mathrm{multiple}\mathrm{of}77\mathrm{and}473.$$$$\Rightarrow \mathrm{N}=\mathrm{LCM}(77,473)\times \mathrm{k}=3311\mathrm{k}$$

Answered by ajfour last updated on 10/Sep/17

$$\frac{boy}{girl}=\frac{11x}{10x};\frac{women}{men}=\frac{23y}{20y}$$$$\frac{children}{adult}=\frac{6}{5}.$$$$\Rightarrow \frac{21x}{43y}=\frac{6}{5}$$$$\Rightarrow 35x=86y$$$$Also21x+43y<6000$$$$or42x+35x<12000$$$$x<\frac{12000}{77}<155\frac{65}{77}$$$$xisalsoamultipleof86,so$$$$x=86,y=35$$$$no.ofboys=11x=946$$$$no.ofgirls=10x=860$$$$no.ofwomen=23y=805$$$$no.ofmen=20y=700$$$$totalpopulation=3311.$$

Commented by NECx last updated on 10/Sep/17

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by $@ty@m last updated on 10/Sep/17

$$\mathbb{BEST}\mathbb{SOLUTION}$$

Commented by NECx last updated on 10/Sep/17

$$\mathrm{seems}\mathrm{to}\mathrm{be}\mathrm{a}\mathrm{really}\mathrm{nice}\mathrm{solution}$$$$\mathrm{for}\mathrm{others}\mathrm{but}{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{really}\mathrm{confused}.$$$$\mathrm{This}\mathrm{is}\mathrm{my}\mathrm{question}\mathrm{here}\mathrm{guys},$$$$35\mathrm{x}=86\mathrm{y}.....\mathrm{how}?$$

Commented by ajfour last updated on 11/Sep/17

$$followsfromlinebefore..$$

Commented by NECx last updated on 11/Sep/17

$$\mathrm{oh}...\mathrm{its}\mathrm{clear}$$

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