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Question-20992




Question Number 20992 by NECx last updated on 09/Sep/17
Commented by NECx last updated on 09/Sep/17
please help guys
$$\mathrm{please}\:\mathrm{help}\:\mathrm{guys} \\ $$
Answered by $@ty@m last updated on 10/Sep/17
Let there be x girls.  ⇒no. of boys=((110)/(100))x  ⇒no. of children, i.e.          no. of boys+girls=x+((110)/(100))x         Again,  Let no. of men=y  ⇒no. of women=((115)/(100))y   ⇒no. of adults=((215)/(100))y  −(2)  ATQ  ((215)/(100))y×((120)/(100))=((210)/(100))x  ⇒x=((215)/(100))×((120)/(210))y  ⇒x=((43)/(35))y −−−(3)  ∴ no. of boys=((110)/(100))x=((110)/(100))×((43)/(35))y                               =((473)/(350))y −−(4)  Given,  population<6000  ((210)/(100))x+((215)/(100))y<6000  210x+215y<600000  210×((43)/(35))y+215y<600000  258y+215y<600000  473y<600000  y<1268.5 −−(5)    (2)⇒y∣20  (3)⇒y∣35  (4)⇒y∣350  LCM(20,35,350)=700  ∴ (5)⇒ y=700  ⇒no. of men=700  ⇒no. of women=((115)/(100))×700=805  no. of girls=((43)/(35))×700=860  no. of boys=((473)/(350))×700=946
$${Let}\:{there}\:{be}\:{x}\:{girls}. \\ $$$$\Rightarrow{no}.\:{of}\:{boys}=\frac{\mathrm{110}}{\mathrm{100}}{x} \\ $$$$\Rightarrow{no}.\:{of}\:{children},\:{i}.{e}. \\ $$$$\:\:\:\:\:\:\:\:{no}.\:{of}\:{boys}+{girls}={x}+\frac{\mathrm{110}}{\mathrm{100}}{x} \\ $$$$\:\:\:\:\: \\ $$$${Again}, \\ $$$${Let}\:{no}.\:{of}\:{men}={y} \\ $$$$\Rightarrow{no}.\:{of}\:{women}=\frac{\mathrm{115}}{\mathrm{100}}{y}\: \\ $$$$\Rightarrow{no}.\:{of}\:{adults}=\frac{\mathrm{215}}{\mathrm{100}}{y}\:\:−\left(\mathrm{2}\right) \\ $$$${ATQ} \\ $$$$\frac{\mathrm{215}}{\mathrm{100}}{y}×\frac{\mathrm{120}}{\mathrm{100}}=\frac{\mathrm{210}}{\mathrm{100}}{x} \\ $$$$\Rightarrow{x}=\frac{\mathrm{215}}{\mathrm{100}}×\frac{\mathrm{120}}{\mathrm{210}}{y} \\ $$$$\Rightarrow{x}=\frac{\mathrm{43}}{\mathrm{35}}{y}\:−−−\left(\mathrm{3}\right) \\ $$$$\therefore\:{no}.\:{of}\:{boys}=\frac{\mathrm{110}}{\mathrm{100}}{x}=\frac{\mathrm{110}}{\mathrm{100}}×\frac{\mathrm{43}}{\mathrm{35}}{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{473}}{\mathrm{350}}{y}\:−−\left(\mathrm{4}\right) \\ $$$${Given}, \\ $$$${population}<\mathrm{6000} \\ $$$$\frac{\mathrm{210}}{\mathrm{100}}{x}+\frac{\mathrm{215}}{\mathrm{100}}{y}<\mathrm{6000} \\ $$$$\mathrm{210}{x}+\mathrm{215}{y}<\mathrm{600000} \\ $$$$\mathrm{210}×\frac{\mathrm{43}}{\mathrm{35}}{y}+\mathrm{215}{y}<\mathrm{600000} \\ $$$$\mathrm{258}{y}+\mathrm{215}{y}<\mathrm{600000} \\ $$$$\mathrm{473}{y}<\mathrm{600000} \\ $$$${y}<\mathrm{1268}.\mathrm{5}\:−−\left(\mathrm{5}\right) \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\Rightarrow{y}\mid\mathrm{20} \\ $$$$\left(\mathrm{3}\right)\Rightarrow{y}\mid\mathrm{35} \\ $$$$\left(\mathrm{4}\right)\Rightarrow{y}\mid\mathrm{350} \\ $$$${LCM}\left(\mathrm{20},\mathrm{35},\mathrm{350}\right)=\mathrm{700} \\ $$$$\therefore\:\left(\mathrm{5}\right)\Rightarrow\:{y}=\mathrm{700} \\ $$$$\Rightarrow{no}.\:{of}\:{men}=\mathrm{700} \\ $$$$\Rightarrow{no}.\:{of}\:{women}=\frac{\mathrm{115}}{\mathrm{100}}×\mathrm{700}=\mathrm{805} \\ $$$${no}.\:{of}\:{girls}=\frac{\mathrm{43}}{\mathrm{35}}×\mathrm{700}=\mathrm{860} \\ $$$${no}.\:{of}\:{boys}=\frac{\mathrm{473}}{\mathrm{350}}×\mathrm{700}=\mathrm{946} \\ $$
Commented by NECx last updated on 10/Sep/17
wow..... this is great sir.  i′m most grateful sir.Words cant  express how happy I am. Thank  you
$$\mathrm{wow}…..\:\mathrm{this}\:\mathrm{is}\:\mathrm{great}\:\mathrm{sir}. \\ $$$$\mathrm{i}'\mathrm{m}\:\mathrm{most}\:\mathrm{grateful}\:\mathrm{sir}.\mathrm{Words}\:\mathrm{cant} \\ $$$$\mathrm{express}\:\mathrm{how}\:\mathrm{happy}\:\mathrm{I}\:\mathrm{am}.\:\mathrm{Thank} \\ $$$$\mathrm{you} \\ $$
Commented by NECx last updated on 10/Sep/17
what do you mean by  (2)⇒y∣20  (3)⇒y∣35  (4)⇒y∣350  l.c.m(20,35,350)=700
$$\mathrm{what}\:\mathrm{do}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{by} \\ $$$$\left(\mathrm{2}\right)\Rightarrow\mathrm{y}\mid\mathrm{20} \\ $$$$\left(\mathrm{3}\right)\Rightarrow\mathrm{y}\mid\mathrm{35} \\ $$$$\left(\mathrm{4}\right)\Rightarrow\mathrm{y}\mid\mathrm{350} \\ $$$$\mathrm{l}.\mathrm{c}.\mathrm{m}\left(\mathrm{20},\mathrm{35},\mathrm{350}\right)=\mathrm{700} \\ $$
Commented by $@ty@m last updated on 10/Sep/17
∵ y is no. of men  ∴y∈N   Similarly no. of adults=((215)/(100))y∈N  which is possible only if y is a multiple of 20.  that is what I meant by   (2)⇒y∣20   and likewise for other two expessions.  Further, to fulfill the three   conditions simltaneously  we′ll have to find the Least Common  Multiple (LCM) of 20, 35 and 350  which is 700.  ∴y∈{700, 1400, ......}  but in ineqation(5), we have  y<1268.5  ⇒y=700  Are all your doubts clear now?
$$\because\:{y}\:{is}\:{no}.\:{of}\:{men} \\ $$$$\therefore{y}\in\mathrm{N} \\ $$$$\:{Similarly}\:{no}.\:{of}\:{adults}=\frac{\mathrm{215}}{\mathrm{100}}{y}\in\mathrm{N} \\ $$$${which}\:{is}\:{possible}\:{only}\:{if}\:{y}\:{is}\:{a}\:{multiple}\:{of}\:\mathrm{20}. \\ $$$${that}\:{is}\:{what}\:{I}\:{meant}\:{by}\: \\ $$$$\left(\mathrm{2}\right)\Rightarrow\mathrm{y}\mid\mathrm{20}\: \\ $$$${and}\:{likewise}\:{for}\:{other}\:{two}\:{expessions}. \\ $$$${Further},\:{to}\:{fulfill}\:{the}\:{three}\: \\ $$$${conditions}\:{simltaneously} \\ $$$${we}'{ll}\:{have}\:{to}\:{find}\:{the}\:{Least}\:{Common} \\ $$$${Multiple}\:\left({LCM}\right)\:{of}\:\mathrm{20},\:\mathrm{35}\:{and}\:\mathrm{350} \\ $$$${which}\:{is}\:\mathrm{700}. \\ $$$$\therefore{y}\in\left\{\mathrm{700},\:\mathrm{1400},\:……\right\} \\ $$$${but}\:{in}\:{ineqation}\left(\mathrm{5}\right),\:{we}\:{have} \\ $$$${y}<\mathrm{1268}.\mathrm{5} \\ $$$$\Rightarrow{y}=\mathrm{700} \\ $$$${Are}\:{all}\:{your}\:{doubts}\:{clear}\:{now}? \\ $$
Commented by NECx last updated on 10/Sep/17
yeah.... i understand .  please which topic in maths does  this question come from?
$$\mathrm{yeah}….\:\mathrm{i}\:\mathrm{understand}\:. \\ $$$$\mathrm{please}\:\mathrm{which}\:\mathrm{topic}\:\mathrm{in}\:\mathrm{maths}\:\mathrm{does} \\ $$$$\mathrm{this}\:\mathrm{question}\:\mathrm{come}\:\mathrm{from}? \\ $$
Commented by $@ty@m last updated on 10/Sep/17
HaaHaaHaa  the same question I was going  to ask you....  It seems to be “Applcation of ineqality”  but actually perception of “Percentage” and  “Number Theory” are also involved.  More than theory, it needs logic  to solve.
$${HaaHaaHaa} \\ $$$${the}\:{same}\:{question}\:{I}\:{was}\:{going} \\ $$$${to}\:{ask}\:{you}…. \\ $$$${It}\:{seems}\:{to}\:{be}\:“{Applcation}\:{of}\:{ineqality}'' \\ $$$${but}\:{actually}\:{perception}\:{of}\:“{Percentage}''\:{and} \\ $$$$“{Number}\:{Theory}''\:{are}\:{also}\:{involved}. \\ $$$${More}\:{than}\:{theory},\:{it}\:{needs}\:{logic} \\ $$$${to}\:{solve}. \\ $$
Commented by NECx last updated on 10/Sep/17
hmmmm.... i hear you sir.
$$\mathrm{hmmmm}….\:\mathrm{i}\:\mathrm{hear}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by mrW1 last updated on 10/Sep/17
C=children in %  A=adult in %  B=boys in %  G=girls in %    C+A=100%=1  C=1.2×A  ⇒C=(6/(11))  ⇒A=(5/(11))    B+G=C=(6/(11))  B=1.1×G  ⇒B=((22)/(77))  ⇒G=((20)/(77))    W+M=A=(5/(11))  W=1.15×M  ⇒W=((115)/(473))  ⇒M=((100)/(473))    let N=total population  ⇒Boys=((22)/(77))N=22×(N/(77))  ⇒Girls=((20)/(77))N=20×(N/(77))  ⇒Women=((115)/(473))N=115×(N/(473))  ⇒Men=((100)/(473))N=100×(N/(473))    LCM(77,473)=3311  ⇒N=3311k with k∈N  ⇒Boys=22×((3311k)/(77))=946k  ⇒Girls=20×((3311k)/(77))=860k  ⇒Women=115×((3311k)/(473))=805k  ⇒Men=100×((3311k)/(473))=700k    since N<6000, ⇒k=1  ⇒946 boys, 860 girls, 805 women, 700 men
$$\mathrm{C}=\mathrm{children}\:\mathrm{in}\:\% \\ $$$$\mathrm{A}=\mathrm{adult}\:\mathrm{in}\:\% \\ $$$$\mathrm{B}=\mathrm{boys}\:\mathrm{in}\:\% \\ $$$$\mathrm{G}=\mathrm{girls}\:\mathrm{in}\:\% \\ $$$$ \\ $$$$\mathrm{C}+\mathrm{A}=\mathrm{100\%}=\mathrm{1} \\ $$$$\mathrm{C}=\mathrm{1}.\mathrm{2}×\mathrm{A} \\ $$$$\Rightarrow\mathrm{C}=\frac{\mathrm{6}}{\mathrm{11}} \\ $$$$\Rightarrow\mathrm{A}=\frac{\mathrm{5}}{\mathrm{11}} \\ $$$$ \\ $$$$\mathrm{B}+\mathrm{G}=\mathrm{C}=\frac{\mathrm{6}}{\mathrm{11}} \\ $$$$\mathrm{B}=\mathrm{1}.\mathrm{1}×\mathrm{G} \\ $$$$\Rightarrow\mathrm{B}=\frac{\mathrm{22}}{\mathrm{77}} \\ $$$$\Rightarrow\mathrm{G}=\frac{\mathrm{20}}{\mathrm{77}} \\ $$$$ \\ $$$$\mathrm{W}+\mathrm{M}=\mathrm{A}=\frac{\mathrm{5}}{\mathrm{11}} \\ $$$$\mathrm{W}=\mathrm{1}.\mathrm{15}×\mathrm{M} \\ $$$$\Rightarrow\mathrm{W}=\frac{\mathrm{115}}{\mathrm{473}} \\ $$$$\Rightarrow\mathrm{M}=\frac{\mathrm{100}}{\mathrm{473}} \\ $$$$ \\ $$$$\mathrm{let}\:\mathrm{N}=\mathrm{total}\:\mathrm{population} \\ $$$$\Rightarrow\mathrm{Boys}=\frac{\mathrm{22}}{\mathrm{77}}\mathrm{N}=\mathrm{22}×\frac{\mathrm{N}}{\mathrm{77}} \\ $$$$\Rightarrow\mathrm{Girls}=\frac{\mathrm{20}}{\mathrm{77}}\mathrm{N}=\mathrm{20}×\frac{\mathrm{N}}{\mathrm{77}} \\ $$$$\Rightarrow\mathrm{Women}=\frac{\mathrm{115}}{\mathrm{473}}\mathrm{N}=\mathrm{115}×\frac{\mathrm{N}}{\mathrm{473}} \\ $$$$\Rightarrow\mathrm{Men}=\frac{\mathrm{100}}{\mathrm{473}}\mathrm{N}=\mathrm{100}×\frac{\mathrm{N}}{\mathrm{473}} \\ $$$$ \\ $$$$\mathrm{LCM}\left(\mathrm{77},\mathrm{473}\right)=\mathrm{3311} \\ $$$$\Rightarrow\mathrm{N}=\mathrm{3311k}\:\mathrm{with}\:\mathrm{k}\in\mathbb{N} \\ $$$$\Rightarrow\mathrm{Boys}=\mathrm{22}×\frac{\mathrm{3311k}}{\mathrm{77}}=\mathrm{946k} \\ $$$$\Rightarrow\mathrm{Girls}=\mathrm{20}×\frac{\mathrm{3311k}}{\mathrm{77}}=\mathrm{860k} \\ $$$$\Rightarrow\mathrm{Women}=\mathrm{115}×\frac{\mathrm{3311k}}{\mathrm{473}}=\mathrm{805k} \\ $$$$\Rightarrow\mathrm{Men}=\mathrm{100}×\frac{\mathrm{3311k}}{\mathrm{473}}=\mathrm{700k} \\ $$$$ \\ $$$$\mathrm{since}\:\mathrm{N}<\mathrm{6000},\:\Rightarrow\mathrm{k}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{946}\:\mathrm{boys},\:\mathrm{860}\:\mathrm{girls},\:\mathrm{805}\:\mathrm{women},\:\mathrm{700}\:\mathrm{men} \\ $$
Commented by NECx last updated on 10/Sep/17
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by NECx last updated on 10/Sep/17
sorry but i really need to ask this;    why was it necessary to find the  LCM of the denominators?      Thanks for always helping.
$$\mathrm{sorry}\:\mathrm{but}\:\mathrm{i}\:\mathrm{really}\:\mathrm{need}\:\mathrm{to}\:\mathrm{ask}\:\mathrm{this}; \\ $$$$ \\ $$$$\mathrm{why}\:\mathrm{was}\:\mathrm{it}\:\mathrm{necessary}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{LCM}\:\mathrm{of}\:\mathrm{the}\:\mathrm{denominators}? \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{always}\:\mathrm{helping}. \\ $$
Commented by mrW1 last updated on 11/Sep/17
(N/(77)) and (N/(473)) must be integer, i.e.  N must be a multiple of 77 and 473.  ⇒N=LCM(77,473)×k=3311k
$$\frac{\mathrm{N}}{\mathrm{77}}\:\mathrm{and}\:\frac{\mathrm{N}}{\mathrm{473}}\:\mathrm{must}\:\mathrm{be}\:\mathrm{integer},\:\mathrm{i}.\mathrm{e}. \\ $$$$\mathrm{N}\:\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{77}\:\mathrm{and}\:\mathrm{473}. \\ $$$$\Rightarrow\mathrm{N}=\mathrm{LCM}\left(\mathrm{77},\mathrm{473}\right)×\mathrm{k}=\mathrm{3311k} \\ $$
Answered by ajfour last updated on 10/Sep/17
((boy)/(girl))=((11x)/(10x)) ;  ((women)/(men))=((23y)/(20y))  ((children)/(adult))=(6/5).  ⇒     ((21x)/(43y))=(6/5)   ⇒      35x=86y   Also    21x+43y < 6000  or          42x+35x <12000                  x<((12000)/(77)) <155((65)/(77))  x is also a multiple of 86, so    x=86 , y=35  no. of boys = 11x=946  no. of girls = 10x=860  no. of women = 23y=805  no. of men = 20y=700  total population =3311 .
$$\frac{{boy}}{{girl}}=\frac{\mathrm{11}{x}}{\mathrm{10}{x}}\:;\:\:\frac{{women}}{{men}}=\frac{\mathrm{23}{y}}{\mathrm{20}{y}} \\ $$$$\frac{{children}}{{adult}}=\frac{\mathrm{6}}{\mathrm{5}}. \\ $$$$\Rightarrow\:\:\:\:\:\frac{\mathrm{21}{x}}{\mathrm{43}{y}}=\frac{\mathrm{6}}{\mathrm{5}}\: \\ $$$$\Rightarrow\:\:\:\:\:\:\mathrm{35}{x}=\mathrm{86}{y} \\ $$$$\:{Also}\:\:\:\:\mathrm{21}{x}+\mathrm{43}{y}\:<\:\mathrm{6000} \\ $$$${or}\:\:\:\:\:\:\:\:\:\:\mathrm{42}{x}+\mathrm{35}{x}\:<\mathrm{12000} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}<\frac{\mathrm{12000}}{\mathrm{77}}\:<\mathrm{155}\frac{\mathrm{65}}{\mathrm{77}} \\ $$$${x}\:{is}\:{also}\:{a}\:{multiple}\:{of}\:\mathrm{86},\:{so} \\ $$$$\:\:{x}=\mathrm{86}\:,\:{y}=\mathrm{35} \\ $$$${no}.\:{of}\:{boys}\:=\:\mathrm{11}{x}=\mathrm{946} \\ $$$${no}.\:{of}\:{girls}\:=\:\mathrm{10}{x}=\mathrm{860} \\ $$$${no}.\:{of}\:{women}\:=\:\mathrm{23}{y}=\mathrm{805} \\ $$$${no}.\:{of}\:{men}\:=\:\mathrm{20}{y}=\mathrm{700} \\ $$$${total}\:{population}\:=\mathrm{3311}\:. \\ $$
Commented by NECx last updated on 10/Sep/17
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by $@ty@m last updated on 10/Sep/17
BEST SOLUTION
$$\mathbb{BEST}\:\mathbb{SOLUTION} \\ $$
Commented by NECx last updated on 10/Sep/17
seems to be a really nice solution  for others but i′m really confused.  This is my question here guys ,  35x=86y        .....how?
$$\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a}\:\mathrm{really}\:\mathrm{nice}\:\mathrm{solution} \\ $$$$\mathrm{for}\:\mathrm{others}\:\mathrm{but}\:\mathrm{i}'\mathrm{m}\:\mathrm{really}\:\mathrm{confused}. \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{my}\:\mathrm{question}\:\mathrm{here}\:\mathrm{guys}\:, \\ $$$$\mathrm{35x}=\mathrm{86y}\:\:\:\:\:\:\:\:…..\mathrm{how}? \\ $$
Commented by ajfour last updated on 11/Sep/17
follows from line before ..
$${follows}\:{from}\:{line}\:{before}\:.. \\ $$
Commented by NECx last updated on 11/Sep/17
oh... its clear
$$\mathrm{oh}…\:\mathrm{its}\:\mathrm{clear} \\ $$

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